Chapter - 1 Solid State (1 - 70) - | Unique Book...

Solid State

"A true friend is one soul in two bodies." -Aristotle

Introduction :The physical state of matter is the interplay of intermolecular forces of attraction like dipole-dipole interactions, dipole-induced dipole interactions, London forces, hydrogen bonding etc.

Chapter

SYLLABUS1.1 CLASSIFICATION OF SOLIDS

(a) Characteristic properties of solids

(b) Classification of Solids

(c) Properties of crystalline solids

(d) Properties of amorphous solids

(e) Anisotropy

(f) Isotropy

(g) Isomorphous

(h) Polymorphous

(i) Glass

(j) Distinguish between Crystalline solids &

Amorphous Solids

(k) Classification of crystalline solids

(l) Solid ice is lighter than water

(m) Ionic solids are hard and brittle

(n) Allotropic forms of carbon

Questions and Answers

Theoretical MCQs

Advanced MCQs

1.2 UNIT CELL & BRAVAIS LATTICES

(a) Crystal lattice

(b) Lattice point

(c) Unit cell

(d) Distinguish between crystal lattice and unit

cell

(e) Types of unit cells

(f) Seven types of unit cells or Fourteen Bravais

lattice

(g) Number of lattice points in one unit cell of

the crystal structure

(h) Number of atoms in one unit cell of the

crystal structure

(i) Solved examples


Theoretical MCQs

Advanced MCQs

1.3 PACKING IN SOLIDS

(a) Coordination number

(b) Linear packing in one dimension OR Close

packing in one dimension (Stage - I)

(c) Close packing in two dimensions. (Stage -

II) OR AAAA type and ABAB type of two

dimensional arrangement. OR Square close

packing and hexagonal close packing in

two dimension

(d) AAAA type of three dimensional packing :

1Chapter

2

Unique Solutions® S.Y.J.C. Science - Chemistry - Part I

(stage III) or Explain simple cubic structure

(SCC)

(e) ABAB type of three dimensional packing

OR Hexagonal close packing in three

dimension

(f) ABCABC type of three dimensional packing

or cubic close packed structure (ccp) in

three dimension

(g) Distinguish between hexagonal close

packing and cubic close packing.

(h) Distinguish between between tetrahedral

void and octahedral void.

(i) Coordination numbers in the following :

(j) Define

(k) Calculation of packing efficiency in

crystals

(l) Obtain a relation for the density of the crystal


Theoretical MCQs

1.4 PACKING EFFICIENCY AND DENSITY OF

UNIT CELL

(a) Packing in voids of ionic solids

(b) Radius ratio for ionic compounds

(c) Packing of ions in cubic closed packed

(ccp) or hexagonal closed packed (hcp)

structures

(d) Structure of NaCl unit cell (rock salt

structure)


Theoretical MCQs

Advanced MCQs

1.5 IMPERFECTIONS OR DEFECTS IN

CRYSTAL STRUCTURE

(a) Defects in crystal structure

(b) Types of defects in the solids or crystal

structures

(c) Point defects

ADDITIONAL IMPORTANT INFORMATION


Theoretical MCQs

1.6 ELECTRICAL AND MAGNETIC

PROPERTIES OF CRYSTALLINE SOLID

(a) Electrical properties of solids

(b) Band theory or origin of electrical properites

in solids

(c) Semiconductors and their types

(d) Uses of semiconductors

(e) Origin of magnetic properties in solids

(f) Diamagnetism

(g) Paramagnetism

(h) Ferromagnetism

(i) Antiferromagnetism

(j) Ferrimagnetism

(k) Gouy’s method for determining magnetic

properties of the substances


Theoretical MCQs

Advanced MCQs

HOURS BEFORE EXAM

NUMERICALS WITH SOLUTIONS

NUMERICALS FOR PRACTICE

HIGHER ORDER THINKING SKILLS

Chapter - 1 Solid State

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1.1 CLASSIFICATION OF SOLIDS : Concept Explanation :

(a) Characteristic properties of solids :1. Solid of fixed composition has fixed mass, volume, shape and density.2. Solid state is generally denser than liquid or gaseous state of the same matter, except ice

(solid H2O) which is lighter than liquid water.3. Most of the solids are hard, incompressible and rigid. Some of the solids like K, Na, P are

soft. The intermolecular distances are shortest hence, intermolecular forces of attractionare stronger compared to liquid and gaseous state and hence, solids cannot be compressedand they are rigid.

4. All solids have characteristic melting points which depend on the extent of intermolecularforces present in solid state. Stronger the intermolecular forces of attraction, higher is themelting point.

5. The constituent particles of solids like molecules, atoms or ions have fixed stationary positionsin solid and can only oscillate about their equilibrium or mean positions. Hence, they havefixed shape and cannot be poured like liquids.

(b) Classification of solids :Depending on the presence or absence of orderly arrangement of the constituent particles,the solids are classified into two types :

1. Crystalline solids : A homogeneous solid in which the constituent particles like atoms,ions or molecules are arranged in a definite repeating pattern throughout the solid is calledcrystalline solid. For example, diamond, NaCl, K2SO4, etc.

2. Amorphous solids : A substance which appears like solid but does not have perfectlyordered crystalline structure of constituent particles is called amorphous solid.For example : tar, glass, plastics, rubber, butter, etc.

(c) Properties of crystalline solids :1. A homogeneous solid in which the constituent particles like atoms, ions or molecules are

arranged in a definite repeating pattern throughout the solid is called crystalline solid.2. A crystalline solid consists of a large number of small crystals called unit cells, each having

a definite characteristic geometrical shape and properties of the crystalline solid.3. The intermolecular forces of attraction are at maximum.4. The forces in the crystalline solid may involve ionic bonds, covalent bonds, hydrogen bonds

and van der Waals forces.5. All pure crystalline solids have sharp melting point.6. The physical properties like refractive index, electrical conductance etc. of crystalline solids

change with change in directions. The ability of crystalline solids to change values of physicalproperties when measured in different directions is called anisotropy.

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7. They have definite enthalpy of fusion and it depends upon arrangement of particles.8. When cut with a sharp edged tool, they split into two pieces and new surfaces are plain

and smooth.

(d) Properties of amorphous solids :1. A substance which appears like solid but does not have perfectly ordered crystalline structure

of constituent particles is called amorphous solid.2. An amorphous solid does not have regular arrangement of constituent particles.3. Amorphous solids do not have sharp melting points. With increase in temperature they

gradually soften, become less viscous and melt over a range of temperature.4. Amorphous solids are called supercooled liquids or pseudo solids.5. Physical properties do not change with change in directions hence, amorphous solids are

isotropic in nature.6. Amorphous solids behave like fluids and very slowly float under gravity.7. They do not have definite enthalpy of fusion.8. When cut with a sharp edged tool, they split into pieces with irregular surfaces.

(e) Anisotropy :Definition: The physical properties like refractiveindex, electrical conductance etc of crystalline solidschange with change in directions. The ability ofcrystalline solids to change values of physical propertieswhen measured in different directions is calledanisotropy.Explanation : This property is due to differentarrangement of constituents in different directions.Example: Refractive Index, electrical conductance,dielectric constant etc.(Different arrangements of constituent particles about different directions, AB, CD and EF.)

(f) Isotropy :Definition: The ability of amorphous solids to exhibit identical physical properties eventhough measured in different directions is called isotropy.

(g) Isomorphous :Definition : When two or more crystalline substances have the same crystal structure,they are said to be isomorphous and the phenomenon is called as isomorphism. For example,Cr2O3 and Fe2O3, NaF and MgO.


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(h) Polymorphous :Definition : A single substance which crystallizes in two or more forms under differentconditions is called polymorphous and this phenomenon is called as polymorphism. Forexample, carbon exists as diamond and graphite, or sulphur exists as rhombic and monoclinicforms.

(i) Glass1. Glass is an optically transparent material produced by fusing together silicon dioxide (SiO2),

sodium oxide (Na2O), boron oxide (B2O3) and a trace amount of transition metal oxidesto impart colour to the glass.Types of Glass :Quartz glass : Only silicon dioxide, SiO2.Pyrex glass : 60-80 % SiO2, 10-25 % B2O3 and remaining amount of Al2O3.Soda lime glass : 75 % SiO2, 15% Na2O and 10% CaO.Red glass : trace amount of gold and copper.Yellow glass : UO2.Blue glass : CoO or CuO.Green glass : Fe2O3 or CuO.

(j) Distinguish between Crystalline solids and Amorphous solids.

Crystalline Solids

1. These solids have definite characteristicshape.

2. They have sharp melting point.

3. When cut with a sharp edged tool, theysplit into two pieces and new surfaces areplain and smooth.

4. They have definite enthalpy of fusion.5. They are anisotropic.6. They are true solids.

7. They have regular arrangement of theconstituent particles. They are said toexhibit long range order.

8. Example : Copper, Silver, Iron, Commonsalt, Zinc sulphide etc.

Amorphous Solids

1. These solids have irregular shape.

2. They do not have definite melting point.They gradually softens over a range oftemperature.

3. When cut with a sharp edged tool, theycut into two pieces with irregular surfaces.

4. They do not have definite enthalpy of fusion.5. They are isotropic.6. They are pseudo-solids and super-cooled

liquids.7. They do not have regular arrangement of

constituent particles. They may have shortrange order.

8. Example : Glass, Rubber, Plastic etc.

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(k) Classification of crystalline solids :Crystalline solids are classified into four main types as follows :

1. Molecular Solids : These are crystalline substances in which the constituent particles aremolecules. The molecules are held together by dispersion forces or London forces, dipole-dipole forces or hydrogen bonds. These are further subdivided into :(a) Polar molecular solids : In such solids, molecules are formed by polar covalent

bonds. For example HCl, NH3(s), solid SO2 etc. In these, molecules are held byrelatively stronger dipole-dipole forces. These solids are soft non-conductor of electricity,have low melting and boiling points and exists in gaseous state under normal conditionsof temperature and pressure. In polar molecules, there is separation of positive andnegative charges hence, the molecule arrange themselves in such a way that oppositecharges of neighbouring molecules are brought together.

(b) Non-polar molecular solids : Crystalline solids in which the constituent particles areeither atoms like those of noble gases or non-polar molecules. e.g. solid helium, solidargon, solid CO2, solid hydrogen etc. In these, the constituent particles are held byweak-weak dispersion or London forces. These are also soft solids having very lowmelting point and are non-conductor of electricity. These are liquid or gaseous stateat room temperature and pressure and are volatile.

(c) Hydrogen bonded molecular solids : The molecule of such solids contain hydrogenbonds between them. For example, consider solid water (ice). The negative end of onemolecule (Oδ-) attracts the positive end of another molecule (Hδ+) forming hydrogenbond. These solids contain hydrogen bond between H and F, O or N. These exists asliquids or even gases at normal conditions of temperature and pressure and solidify oncooling. These solids are non-conductor of electricity and have higher melting andboiling points compared to non-polar and polar molecular solids.

2. Ionic Solids : These solids consist of positively and negatively charged ions arranged indefinite manner throughout the solid. The constituent ions are held together by strongelectrostatic force of attraction (coulombic forces). The charges on ions and their arrangementis such that they balance each other and molecule is electrically neutral.The actual arrangement of ions in ionic solids depends upon :(i) size of cation and anion, (ii) the charge on the ions, and (iii) the ease with which anionscan be polarized.The three dimensional arrangement of cations and anions is such that the opposite forcesare completely compensated or neutralized. Hence, these solids are hard and brittle andpossess higher melting point. As the +ve and –ve charges compensated, there is no freeelectrons in the solid hence, are non-conductor of electricity. However, when fused (molten)or dissolved in aqueous solution, the ions get separated and they conduct electricity. Thecommon examples of ionic solids are NaCl, KCl, KNO3, LiF, Na2SO4 etc.


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3. Metallic solids : In metallic solids, the constituent particles are metal ions i.e. positive ions(kernels) immersed in a sea of electrons. The solids are formed by the atoms of sameelement which contribute one or more electrons towards sea of mobile electrons.Since, the electrostatic force of attraction between the positively charged kernels and freemobile electrons is very strong, metallic bond is a strong bond. The free mobile electronsare delocalized and are responsible for electrical and thermal conductivity. In metallic solids,atoms are arranged one over the other in the form of spheres. Under the influence of stress,these layers slips one over the other and since, electrons are delocalized, they quickly adjustthe new environment without breaking the solid. i.e., they are malleable and ductile.

4. Covalent Solids : Covalent or network solids are those in which the constituents particlesare atoms which are held together by covalent bonds. Depending on the nature of covalentbond they are hard or brittle, giant molecules and have high melting points. Depending uponthe availability of mobile electrons, they may be either good conductor or insulators. e.g.,diamond, graphite and fullerene which are allotropic modification of carbon. The otherexample of covalent solids are carborundum (silicon carbide), quartz (SiO2), boron nitride(BN) etc.

(l) Solid ice is lighter than water1. Ice has hexagonal three dimensional crystal structure formed due to intermolecular hydrogen

bonding which leave almost half the space vacant.2. The structures of liquid water and solid ice are almost identical.3. On melting of ice, some hydrogen bonds are broken and vacant spaces are occupied by

water molecules, and they are accommodated more closely.4. In the solid ice, vacant space is more than in liquid water. Therefore, the density of ice is

less than water.5. Hence, ice is lighter than water and ice floats on the surface of water.

The structure of ice

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(m) Ionic solids are hard and brittle :1. In ionic crystalline solids, constituent particles are

positively charged cations and negatively chargedanions placed at alternate lattice points.

2. The ions are held by strong coulombic electrostaticforces of attraction compensating opposite forces.Hence, they are hard.

3. Since there are no free electrons, they are not malleable and on applying a shearing force,ionic crystals break into small units. Hence they are brittle.

(n) Allotropic forms of carbon :Carbon has three allotropic modifications :

1. Diamond2. Graphite3. Fullerene1. Diamond : In diamond, each carbon is sp3 hybridised and is bonded to four other carbon

atom by single covalent bond. Each carbon atom lies at the centre of regular tetrahedronand the other four carbon atoms are present at the corners of tetrahedron. Hence, thereis three dimensional network of strong covalent bonds.This makes diamond extremely hard crystal with very high melting point (3843K).As all the valence electrons of carbon are strongly held in carbon-carbon bonds, it is poorconductor of electricity.

2. Graphite : In graphite, each carbon atom is sp2 hybridised and is bonded to three othercarbon atoms by single covalent bond while the fourth electron form π-bond with othercarbon atom. Thus, it forms hexagonal rings in two dimension. These array of rings formlayers. The layers are separated by a distance of 340pm. The large distance between theselayers indicate that only weak Van der Waal’s forces exists which hold these layers together,is responsible for the soft nature of graphite. The electrons forming π-bonds are delocalizedand are relatively free to move under the influence of electric field. Hence, it is goodconductor of electricity.

3. Fullerene : In 1985, new allotrope of carbon wasdiscovered by passing high power laser on carbon and foundto contain sixty carbon atoms i.e., C60. It had soccer-ballshape i.e. hollow sphere. Like graphite, each carbon is sp2

hybridised and the delocalized molecular orbits are spreadover the complete structure of Fullerene.Fullerene react with transition metal to form a catalyst. Withpotassium, it reacts to form K35C60 which is super-conductor.

The structure of C60,Buckminster - fullerene


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MeltingPoint

Very low

Low

Low

High

Fairlyhigh

Veryhigh

Type of Solid

1. Molecular solids(a) Non polar(b) Polar(c) Hydrogen

bonded

2. Ionic solids

3. Metallic solids

4. Covalent ornetwork solids

ConstituentParticles

Molecules

Ions

Positiveions in asea ofdelocalisedelectrons

Atoms

Bonding/Attractive

Forces

Dispersion orLondon forcesDipole-dipoleinteractionsHydrogenbonding

Coulombic orelectrostatic

Metallicbonding

Covalentbonding

Examples

Ar, CCl4H2, I2, CO2

HCl, SO2

H2O (ice)

NaCl, MgO,ZnS, CaF2

Fe, Cu, Ag,Mg

SiO2

(quartz),SiC, C(diamond),AIN,C(graphite)

PhysicalNature

Soft

Soft

Hard

Hard butbrittle

Hard butmalleableandductile

Hard

Soft

Electricalconductivity

Insulator

Insulator

Insulator

Insulatorsin solidstate butconductorsin moltenstate andin aqueoussolutions

Conductorsin solidstate aswell as inmoltenstate

Insulators

Conductor(exception)

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Questions and Answers• Answer in short :

*1. Give characteristics of solid state?Ans. The characteristics of solid states are :

(a) They have definite three dimensional arrangement of constituent (atoms / ions / molecules).(b) They have definite shape, size and volume.(c) They have definite melting points and boiling points.(d) They have long range order.

*2. Give the classification of solids.Ans. On the basis of orderly arrangement of constituent particles (atoms, ions or molecules), solids

are classified into two type.(a) Crystalline solids (b) Amorphous solids

*3. Classify the following solids into different types :(a) Plastic (b) P4 molecule(c) S8 molecule (d) Iodine molecule(e) Tetra phosphorus decoxide (f) Ammonium phosphate(g) Brass (h) Rubidium(i) Graphite (j) Diamond(k) NaCl (l) Silicon

Ans.Types of solids

Non Polar moleculear solids Amorphous Metallic Covalent Ionic

I2 Plastic Brass Graphite AmmoniumDiamond phosphateSilicon NaClP4 moleculeS8 moleculeTetra

Rubidium PhosphorousDecoxide

*4. Explain crystalline solids and amorphous solids.Ans. Refer 1.1 (b)

5. Define : (a) Aniosotropy (b) Isotropy (c) Isotropy (d) PolymorphousAns. Refer 1.1 (e), (f), (g), (h)


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*6. What is glass? Distinguish between crystalline solids and amorphous solids? Give examples.Ans. Glass is an optically transparent material produced by fusing together silicon dioxide (SiO2),

sodium oxide (Na2O), boron oxide (B2O3) and a trace amount of transition metal oxides toimpart colour to the glass.Ans. Refer 1.1 (j)

*7. Explain : (a) Molecular solids (b) Hydrogen bonded molecular solids(c) Ionic solids (d) Metallic solids(e) Covalent solids (f) Coordination number

Ans. Refer 1.1 (k)(f) Coordination number : The number of nearest neighbours of any constituent particles

is called its coordination number.*8. Write a note on : (a) Diamond (b) Graphite (c) Fullerene.

Ans. (a) Diamond : In diamond, each carbon is sp3 hybridised and is bonded to four other carbonatom by single covalent bond. Each carbon atom lies at the centre of regular tetrahedronand the other four carbon atoms are present at the corners of tetrahedron. Hence, thereis three dimensional network of strong covalent bonds.This makes diamond extremely hard crystal with very high melting point (3843K). As allthe valence electrons of carbon are strongly held in carbon-carbon bonds, it is poor conductorof electricity.

(b) Graphite : In graphite, each carbon atom is sp2 hybridised and is bonded to three othercarbon atoms by single covalent bond while the fourth electron form ð-bond with othercarbon atom. Thus, it forms hexagonal rings in two dimension. These array of rings formlayers. The layers are separated by a distance of 340pm. The large distance between theselayers indicate that only weak Van der Waal’s forces exists which hold these layers together,is responsible for the soft nature of graphite. The electrons forming ð-bonds are delocalizedand are relatively free to move under the influence of electric field. Hence, it is goodconductor of electricity.

(c) Fullerene : In 1985, new allotrope of carbon was discovered by passing high power laseron carbon and found to contain sixty carbon atoms i.e., C60. It had soccer-ball shape i.e.hollow sphere. Like graphite, each carbon is sp2 hybridised and the delocalized molecularorbits are spread over the complete structure of Fullerene. Fullerene react with transitionmetal to form a catalyst. With potassium, it reacts to form K35C60 which is super-conductor.[Refer fig. 1.1 (n) - 3]

*9. Explain why : (a) Ionic solids are hard and brittle, (b) Solid ice is lighter than water.Ans. (a) [Refer 1.1 (l)]

(b) [Refer 1.1 (m)]

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10. Classify the following solids as ionic, covalent, metallic, molecular or amorphous solid.(a) Tetraphosphorous decoxide (b) Graphite(c) SiC (d) Rubidium(e) I2 (f) Ammonium phosphate(g) Brass (h) Si(i) P4 (j) Solid CO2

(k) LiBrAns. Ionic : LiBr, (NH4)2SO4 Metallic : Rubidium

Molecular : P4O10, I2, Solid CO2 Covalent or Network : Graphite, SiC, SiAmorphous : Nil

11. Write a feature which will distinguish a metalic solid from ionic solid.Ans. Metallic solids are good conductor of heat and electricity while ionic solids are insulator in solid

state but are conductor in molten and aqueous solution.

12. Write a distinguishing feature of metallic solid.Ans. Metallic solids are good conductor of heat and electricity.

13. What is difference between glass and quartz while both are made from SiO4 tetrahedral?Ans. Glass is amorphous while Quartz is crystalline solid.

14. Some of very old glass objects appear slightly milky instead of being transparent. Why?Ans. This is because of some crystallisation in that region.

15. Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremelyhigh temperature. What type of solid is it?

Ans. Network or Covalent solid.

16. Ionic solid conduct electricity in molten state but not in solid state. Explain.Ans. In Ionic solids, the electrostatic force of attraction between oppositely charged ions are strong.

Hence, ions are not available for conductance. However, in molten or in aqueous solution, ionsare available for conductance.

17. What type of solids are electrical conductors, malleable?Ans. Metallic.

19. Classify the following as amorphous or crystalline solid.(a) Naphthalene (b) Teflon (c) Benzoic acid (d) Potassium nitrate(e) Cellophane (f) Fibre glass (g) Copper (h) Zinc sulphide

Ans. Amorphous solids : (b) Teflon (e) Cellophane (f) Fibre glassCrystalline solids : (a) Naphthalene (c) Benzoic acid (d) Potassium nitrate (g) Copper

(h) Zinc sulphide.


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Multiple Choice Questions :• Theoretical MCQs :

1. The solids in which the magnitude of physical properties changes with the direction of measurementare .......a) amorphous b) crystalline c) colloidal d) metalloid

*2. In solids the constituents particles may be .......a) atoms b) ionsc) molecules d) any one of the above three

*3. A single substance that exists in two or more forms is called .......a) polymorphous b) amorphousc) isomorphous d) mono-morphous

4. The force that hold kernels together in the crystal is called .......a) ionic bond b) covalent bond c) metallic bond d) hydrogen bond

*5. Graphite is a .......a) molecular crystal b) covalent crystal c) ionic crystal d) metallic crystal

*6. Diamond is a .......a) molecular crystal b) covalent crystal c) ionic crystal d) metallic crystal

*7. The major binding force of diamond is .......a) ionic bond b) covalent bondc) dipole-dipole attraction d) induced dipole-dipole attraction

*8. The major binding force in silicon is .......a) ionic bond b) covalent bondc) dipole-dipole attraction d) induced dipole-dipole attraction

• Advanced MCQs :9. The solids which are brittle and hard are .......

a) ionic solid b) molecular solid c) covalent solid d) metallic solid10. Pyrex glass is obtained by fusing together .......

a) SiO2, MgO and CuO b) SiO2, Na2CO3 and CaCO3

c) SiO2, B2O3 and Al2O3 d) SiO2 and Na2SiO3

*11. The major binding force in graphite is .......a) ionic bond b) covalent bond c) hydrogen bond d) London force

12. In which of the following pairs both the solids belong to same type?a) solid CO2, ZnS b) CaF2, Ca c) graphite, ice d) SiC, AlN

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1.2 UNIT CELL AND BRAVAIS LATTICES :

Concept Explanation :(a) Crystal lattice:

Definition : A regular arrangement of the constituent particles (atoms, ions or molecules)of a crystalline solid having similar environment in three dimensional space is called crystallattice or space lattice.

(b) Lattice point : Definition : A position occupied by a crystal constituent particle like an atom, ion or amolecule in the crystal lattice is called lattice point or lattice site.

(c) Unit cell :1. Unit cell : It is the smallest repeating structural unit of a crystalline solid (or crystal lattice)

which when repeated in different directions produces the crystal line solid (lattice).2. The crystal is considered to consist of an infinite number of unit cells.3. The unit cell possesses all the characteristics of the crystalline solid.4. A unit cell is characterised by following parameters :

(a) Edges or edge lengths : The intersection of twofaces of crystal lattice is called as edge. The threeedges denoted by a, b and c represent the dimensions(lengths) of the unit cell along three axes. These edgesmay or may not be mutually perpendicular.


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(b) Angles between the edges (or planes) : There are three angles between the edgesof the unit cell represented as α, β and γ.

(i) The angle α is between edges b and c.(ii) The angle β is between edges a and c.(iii) The angle γ is between edges a and b.

The crystal is defined with the help of these parameters of its unit cell.

(d) Distinguish between crystal lattice and unit cell :

(e) Types of unit cells :Basically unit cells are of two types as follows :

1. Primitive unit cells : The unit cells in which the constituent particles like atoms, ions ormolecules are present only at the corners of the unit cell are called primitive unit cells orsimple unit cells.

2. Non-primitive or centered unit cells : The unit cells in which the constituent particlesare present at corners as well as at some other positions in the unit cell are called non-primitive or centred unit cells. The centred unit cells are of three types :(a) Body centred unit cell : A unit cell in which the constituent particles are present at

the corners as well as at its body centre is called body centred unit cell.(b) Face centred unit cell : A unit cell in which the constituent particles are present at

the corners as well as at the centre of each face is called face centred unit cell or cubicclose packed (CCP) unit cell.

(c) End centred unit cell : A unit cell in which the constituent particles are present at thecorners as well as at the centres of two opposite faces is called end centred unit cell.

Crystal lattice

1. It is a regular arrangement of constituentparticles of a crystalline solids in threedimensional space.

2. It is made up of a large number of unitcells.

3. Crystal lattice is defined in terms ofproperties of unit cell.

4. Crystal lattice can be prepared, handledand studied experimentally.

5. It is a macroscopic system.

Unit cell

1. It is the smallest repeating structural unitwhich when repeated in space in alldimensions generates a crystal lattice.

2. It is one fundamental unit of crystal latticewhich possesses all the properties of thecrystal.

3. Unit cell describes the fundamentalproperties of the crystal lattice.

4. Unit cell is a hypothetical object whichis the smallest possible imaginary part ofthe crystal lattice and cannot be handledand studied individually.

5. It is a microscopic molecular size system.

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(f) Seven types of unit cells Or Fourteen Bravais lattices

1. Simple or primitive : It has points at allthe corners of the unit cell i.e. it consistof one atom at all the corners of cubeand has all edges (sides) equal i.e. a = b= c and α = β = γ = 90º.

2. Body centred cube : In this type of crystal, inaddition to the lattice point present at the corners,one atom (lattice point) is present at the bodycentre of the cube. The edge lengths are a = b= c and α = β = γ = 90º; e.g. Na, Rb, Fe, Ti, W,U, Zr, etc.

3. Face centred cube : In this type of crystal, thereare points at all the corners as well as at the centreof each face. The edge lengths are a = b = c andα = β = γ = 90º; e.g. Cu, Al, Ni, etc.


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(a) Tetragonal lattice :It is of two types. In this type of unit cell, two edgesare same while the third edge is different(i) Simple or primitive tetragonal : The

lattice points are present at the corners oftetragon. The edge lengths are a = b ≠ cand α = β = γ = 90º; e.g. SnO2

(ii) Body centred tetragonal : In this typeof unit cell, eight points are present at thecorners of tetragon and one point is presentat the centre. The edge lengths are a = b≠ c and α = β = γ = 90º; e.g. TiO2, CaSO4

etc.

(b) Orthorhombic lattice : In orthorhombic unit cell, all edges are different i.e. a ≠ b≠ c and all angles are same i.e. α = β = γ = 90º. It is further classified as :(a) simple or primitive (b) body centred orthorhombic (c) face centred and (d)end-centred orthorhombic unit cell.

3. Monoclinic lattice : In monoclinicunit cell, all edges are different, a≠ b ≠ c and α = β = 90º but γ ≠ 90ºi.e. two of the angles are same butthe third angle is different. It is alsoof two types :

(a) primitive monoclinic and(b) end-centred monoclinic

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4. Triclinic lattice : In this, all edges as wellas angles are different i.e. a ≠ b ≠ c and α≠ β ≠ γ ≠ 90º

5. Hexagonal lattice : This type of unit cellhas two same edge length i.e.a = b but the third edge has different length.Also two angles are same i.e.α = β = 90º but γ = 120º.

6. Rhombohedral lattice : In this type of unitcell, all edges have same length i.e. a = b =c. Two axial angles are of 90º but the thirdangle is more than 90º i.e. α = β ≠ γ

CRYSTAL SYSTEMS AND BRAVAIS LATTICES

Crystal System Types of Lattices

Cubic Simple, Body centred, Face centredTetragonal Simple, Body cent redOrthorhombic Simple, Body centred, Face centred, End centredMonoclinic Simple, End centredTriclinic SimpleHexagonal SimpleRhombohedral Simple


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Types of unit cells their edge lengths, angles and examples.

(g) Number of lattice points in one unit cell of the crystal structures.

1. Simple cubic2. Face centred cubic3. Body centred cubic4. Face centred tetragonal.Lattice points in one unit cell represent the positions of atoms, ions or molecules in the unit cell.1. Simple cubic unit cell : In this primitive unit cell, the lattice points are at 8 comers of the

unit cell. Hence there are 8 lattice points.2. Face centred cubic unit cell : In this unit cell, the lattice points are at 8 corners and 6 face

centres.(In cubic close packing unit cell, the lattice points are also at edge centres and body centre.)

3. Body centred cubic unit cell : In this, the lattice points are at 8 corners and one at bodycentre.

4. Face centred tetragonal cubic unit cell : In this, the lattice points are at 8 corners and 6 facecentres.

No. Crystal Type Edge Angle Examplesystem length

1. Cubic Simple/primitive a = b = c α = β = γ = 90º Polonium

2. Cubic Body centred a = b = c α = β = γ = 90º Fe, Rb, Na, Ti, W, U, Zr

3. Cubic Face centred a = b = c α = β = γ = 90º Cu, Al, Ni, Au, Ag, Pt

4. Tetragonal Primitive a = b ≠ c α = β = γ = 90º SnO2

5. Tetragonal Body centred a = b ≠ c α = β = γ = 90º TiO2, CaSO4

6. Orthorhombic Primitive a ≠ b ≠ c α = β = γ = 90º Rhombic sulphur

7. Orthorhombic Body centred a ≠ b ≠ c α = β = γ = 90º KNO3

8. Orthorhombic Face centred a ≠ b ≠ c α = β = γ = 90º BaSO4

9. Orthorhombic End centred a ≠ b ≠ c α = β = γ = 90º MgSO4; 7H2O

10. Monoclinic Primitive a ≠ b ≠ c α = β = 90º, γ ≠ 90º Monoclinic sulphur

11. Monoclinic End centred a ≠ b ≠ c α = β = 90º, γ ≠ 90º Na2SO4; 10H2O

12. Triclinic Primitive a ≠ b ≠ c α ≠ β ≠ γ ≠ 90º K2Cr2O7, H3BO3

13. Hexagonal Primitive a = b ≠ c α = β =90º, γ = 120º ZnO, BeO, CoS, SnS

14. Rhombohedral Primitive a = b = c α = β = γ ≠ 90º Calcite, NaNO3, FeCO3

20


(h) Number of atoms in one unit cell of the crystal structures :1. Simple cubic2. Body centred cubic3. Face centred cubic4. Hexagonal unit cell

1. Number of atoms in simple cubic (scc) crystal :In simple or primitive cubic unit cell, there are 8 atoms at 8corners. Each corner contributes l/8th atom to the unit cell.

∴ Number of atoms present in the unit cell = 1

× 8 = 18

Hence, the volume of the unit cell is equal to the volume of oneatom.

2. Number of atoms in body centred cubic (bcc) crystal :In this unit cell, there are 8 atoms at 8 corners and one additionalatom at the body centre. Each corner contributes l/8th atom, tothe unit cell, hence, due to 8 corners.

Number of atoms = 1

8 = 18

× atom.

An atom at the body centre wholly belongs to the unit cell.∴ Total number of atoms present in bcc unit cell = 1 + 1= 2.

Hence the volume of unit cell is equal to the volume of two atoms.

3. Number of atoms in face centred cubic (fcc cubic crystal) :In this unit cell, there are 8 atoms at 8 corners and 6 atoms at6 face centres. Each corner contributes l/8th atom to the unitcell, hence, due to 8 corners,

Number of atoms = 1

× 8 = 18

Each face centre contributes half of the atom to the unit cell, hence, due to 6 face centres,

∴ Number of atoms = 1

× 6 = 32

Total number of atoms present in fee unit cell = 1+ 3 = 4.Hence, the volume of the unit cell is equal to the volume of four atoms.


21

4. Hexagonal unit cell : In this unit cell, 12 atoms are present at 12 corners and 2 atomsare present at the centres of two opposite hexagonal faces.Each corner, contributes l/6th atom to the unit cell since it is common for 6 unit cells, hence,due to 12 corners.

Number of atoms = 1

× 12 = 26

Each face centre contributes half atom to the unit cell, hence, due to 2 face centres,Number of atoms = 1/2 × 2 = 1.Total number of atoms in hexagonal unit cell = 2+1=3.

(i) Solved Examples :

1. A compound formed by element A and B, crystallises in the cubic arrangement in which A atomsare at the corners of a cube and B atoms are at the face centres. What is the formula ofcompound?

Sol. The number of A atoms in the unit cell of cube = 1

8 ×8

= 1

B atoms are at the centre of face and each face is shared by two cubes. The number of B atoms

are = 1

6 ×2

= 3.

∴∴∴∴∴ The formula of compound is AB3

2. If three elements P, Q and R crystallises in a cubic solid lattice with P atoms at the corners,Q atoms at the cube centres and R atoms at the centre of edges, then write the formula of thecompound.

Sol. As P atoms are present at the corners of cube

∴∴∴∴∴ no. of P atoms in unit cell = 1

8 ×8

= 1

Q atoms are present at the cube centres∴∴∴∴∴ no. of Q atoms in unit cell = 1

R atoms are present at the edges. There are 12 edges and atom at each edge is shared by fouratoms.

∴∴∴∴∴ no. of R atoms in unit cell = 1

12 ×4

= 3

∴∴∴∴∴ Formula of compound is PQR3

3. Sodium crystallises in bcc unit cell. Calculate the approximate number of unit cell in 9.2g ofsodium (atomic mass of Na = 23)

Sol. No. of atoms present in 9.2g of sodium = 239.2 × 6.022 × 10

23 = 2.4088 × 1023 atoms

22


A bcc unit cell contains 2 atoms

∴∴∴∴∴ Number of unit cells = 232.4088 × 10

2 = 1.2044 × 1023

4. Silver crystallises with fcc unit cell. Each side of the unit cell has a length of 400 pm. Calculatethe radius of silver atom.

Sol. If a is the edge length of unit cell, then fcc diagonal = 2 afcc diagonal is also equals to 4r

∴ 4r = 2 a

r = 2 a

4=

1.414 × 400

4 = 141.4 pm

5. Gold crystallises in fcc unit cell (atomic radius = 0.144 nm). What is the length of the side ofthe cell?

Sol. For fcc unit cell, radius of atom, r = a

2 2a = r 2 2⋅ = 0.144 × 2 × 1.414 = 0.407 nm


1. What is a primitive cell?Ans. A primitive cell has particles (atoms) at all the corners of the unit cell.

2. Name the crystal system in which all the three axes are of equal length which are inclined atthe same angle but the angle is not equal to 90º.

Ans. Rhombohedral.

3. What is meant by the term coordination number? What is the coordination number of atoms in(a) a cubic close packing and (b) body centred cubic close packing.

Ans. It is number of nearest neighbours with which a given sphere is in contact.(a) 12 (b) 8

4. How many atoms can be assigned to it’s unit cell, if an element forms :(a) body centred cubic cell and (b) face centred cubic cell?

Ans. (a) 2 (b) 4

5. If a is the edge length of a body centred cubic structure and r is the radius of the atom, thenhow are these two related?

Ans. For a body centred cubic structure, edge length and radius is related as : r = 3

a4

.


23

6. How are unit cell and space lattice related?Ans. Space lattice is obtained by repeating the unit cell in three dimensions. The spatial arrangement,

stoichiometry and density of unit cell and space lattice are equivalent.

7. What is the total number of atoms present per unit cell in a face centred cubic structure?Ans. The total number of atoms present is 4.

8. If three elements X, Y and Z crystallises in a cubic solid with X atoms at the corners, Y atomsat the cube center and Z atoms at the faces of cube, then write the formula of the compound.

Ans. Atom X per unit cell = 1

8 ×8

= 1

Atom Y per unit cell = 1

Atom Z per unit cell = 1

6 ×2

= 3

∴∴∴∴∴ Formula of the compound = XYZ3

9. In a solid AB ‘A’ atoms have cubic close packing arrangement and B atoms occupy all theoctanedral sites. If all the face centred atoms are along one of the axis are removed, then whatwill be the resultant stoichiometry of the compound?

Ans. In cubic close packing structure, there are 8A at the corners of the cube and 6A atoms on theface centres. If all the face centred atoms along one of the axis are removed, it means removalof 2A atoms. Therefore, only 4A atoms will be left on faces.

∴∴∴∴∴ No. of A atoms per unit cell = 1 1

8 × + 4 ×8 2

= 3

No. of B atoms in ccp structure are 12 at edge corner and one at body centre. Therefore, no.

of B atoms per unit cell = 1

12 × + 14

= 4

∴∴∴∴∴ stoichiometry of compound = A3B4

10. Sodium crystallizes in bcc unit cell. Calculate the approximate number of unit cells in 9.2g ofsodium (At mass = 23) (CBSE sample paper 2011)

Ans. No. of atoms in 9.2 g of sodium =236.022 × 10 × 9.2

23 = 2.4088 × 1023 atoms

A bcc unit cell contains 2 atoms,

∴ No. of unit cells present =232.4088 × 10

2 = 1.2044 × 1023

11. Explain : Coordination number.Ans. Coordination number can be 4, 6, 8 or 12. Higher coordination number indicate that sphere are

compactly arranged, space between them is less and are denser. For example in primitive orsimple cubic unit cell, every sphere is in contact with four sphere i.e. a single sphere is surrounded

24


by total number of six spheres. Thus, it’s coordination number is six. Similarly, the coordinationnumber of body centred cube is eight while coordination number of face centred cube is twelve.

*12. What is unit cell? Explain Bravais Lattice.Ans. Unit cell is smallest part (position) which when repeated in different direction generates the

entire lattice. Bravais has proved that crystals do not have simple lattice i.e. having lattice pointonly at the corners. They can be arranged in maximum of fourteen types. The arrangement iscalled Bravais Lattice.

*13. Explain with the help of a diagram.(a) Seven types of unit cell. (b) Three types of cubic cells.(c) Two types of tetragonal unit cells. (d) Four types of orthorhombic unit cells.(e) Two types of monoclinic unit cell. (f) Triclinic unit cell(g) Primitive hexagonal unit cell.

Ans. Refer 1.2 (f)

*14. Give the number of lattice points in one unit cell of the crystal structures.(a) Simple cubic.(b) Face centred cubic.(c) Body centred cubic(d) Face centred tetragonal.

Ans. (a) No. of lattice point in simple cube : In simple cube, the atoms are present at the cornersof cube. Atoms present at the corner contribute 1/8 to each cube because it is spared by8 cube.

∴∴∴∴∴ The no. of atoms present in each unit cell (lattice point)

= 8 corners atoms × 1

8 atoms per unit cell

= 1 atom(b) Face centred cube : It has points at all the corners as well as at the centre of each of six

faces which is shared by two unit cells. Therefore, contribution of each of atom at the faceper unit cell is 1/2.Thus, No. of atoms present at corners per unit cell



= 1 atomNo. of atoms present at faces per unit cell

= 6 atoms at the faces × 1


= 3 atoms∴∴∴∴∴ Total no. of atoms per unit cell = 1 + 3 = 4 atoms.


25

(c) Body centred cube : It has points at all the corners as well as body centre of the cube. Thepoints (atoms) present at the corner is shared by 8 other cube. Thus, the contribution ofeach corner atoms is only 1/8th.Thus, the no. of atoms present at the corner per unit cell



= 1 + 1= 2 atoms

15. Distinguish between Crystal Lattice and Unit Cell.

Ans. Ref. 1.2 (d)


1. The crystal system with unit cell dimension, a = b ≠ c and α = β = 90º and γ = 120º has .......structure.

a) rhombohedrals b) hexagonal c) cubic d) tetragonal2. In the face centred unit cell, the points are present at the .......

a) corners and face centres of the unit cell b) corners of the unit cellc) face centres of the unit cell d) corners and centres of the unit cell

3. The number of atoms in body centred cube is .......a) 1 b) 2 c) 4 d) 8

4. The number of atoms present in a face-centred cube is .......a) 1 b) 2 c) 3 d) 4

5. The most unsymmetrical and symmetrical systems are respectively .......a) tetragonal, cubic b) triclinic, cubicc) rhombohedral, hexagonal d) orthorhombic, cubic

• Advanced MCQs :6. The relation a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90º belong to the crystal system .......

a) triclinic b) monoclinic c) rhombic d) cubic7. If the edge length in a crystal is a = b ≠ c and the angle between them is α = β = γ = 90º, what

type of crystal lattice is?a) monoclinic b) cubic c) orthorhombic d) tetragonal

26


1.3 PACKING IN SOLIDS : Concept Explanation :

(a) Coordination number :The coordination number of constituent particle of the crystal lattice is the number ofparticles surrounding a single particle in the crystal lattice.More the coordination number more tightly the particles are packed in the crystal lattice.

(b) Linear packing in one dimension OR close packing in one dimension:(Stage - I)The constituent particles of thecrystalline solid arerepresented by spheres ofequal size.

In one dimensional arrangement, the spheres are arranged in a tow, touching each other.Each sphere is in contact with two of its neighbours. Thus the coordination number is 2.

(c) Close packing in two dimensions. (Stage - II) ORAAAA type and ABAB type of two dimensional arrangement. ORSquare close packing and hexagonal close packing in two dimensionTwo dimensional close packing crystal structure can be generated by placing one dimensionallinear crystal structure over another to form multiple layers. This staking of linear rowsmay be taking place in two different ways giving two different two dimensional structuresas follows :

1. AAAA type two dimensional close packing or square close packing :(a) Different one dimensional

identical rows are placed oneover the other.

(b) In this arrangement eachsphere in a row is placedover another sphere ofanother row.

(c) In this, spheres havehorizontal as well as verticalalignment. All the rows ofspheres are identical inplanar structure.


27

(d) All crests of spheres as well as all the depressions formed by the arrangement arealigned.

(e) If the first row is labelled as A type, then second and all subsequent rows are alsoidentical, hence are of A type. Therefore, this planar two dimensional close packingis called AAAA…… type packing.

(f) In this arrangement, each sphere is in contact with four other spheres around it. Hence,the coordination number of each sphere is four and the packing is called two dimensionalor planar square close packing.

(g) In this, packing efficiency is 52.4 %. In this comparatively empty space is more. Theshape of void is square.

2. ABAB type two dimensional packing or hexagonal close packing :(a) In this arrangement, the

crests of the spheres ofone row are placed intothe depressions ortroughs formedbetween adjacentspheres of next row.

(b) This gives closest typepacking of spheres andreduces the unoccupiedor void spaceconsiderably.

(c) In this arrangement, the positions of spheres of neighbouring rows do not match eachother since crests of spheres of one row are in contact with depressions or troughs

of next row.

(d) If one row of spheres is labelled as A then the next row will be B, third row will againbe A, fourth row B and so on. Hence this planar or two dimensional close packing is

called ABAB... type packing.

(e) In this arrangement, each sphere is in contact with or touching six other spheres aroundit shown by a hexagon obtained by joining the centres of six spheres.

(f) Hence, the coordination number of each sphere is six and the packing is called two

dimensional or planar hexagonal close packing.(g) In this, the packing efficiency is 60.4% which is more than linear close packing.

(h) The size of hole is very small and has triangular shape.

28


(d) AAAA type of three dimensional packing: (stage - III) OR Explain simplecubic structure (SCC) :All the real structures of crystalline solids are three dimensional structures obtained bystaking one above the other two dimensional AAAA type square close packed crystallinestructures or two dimensional ABAB type hexagonal close packed structures.

1. Three dimensional close packing from two dimensional AAAA type square closepacked layers :(a) In this arrangement, AAAA

type of two dimensional crystallayers are placed over theother such that all the spheresof the successive layers areexactly above the spheres ofthe lower layers.

(b) All spheres of different layersare perfectly alignedhorizontally and verticallyforming unit cells havingprimitive or simple cubicstructure.

(c) Since all the layers are identical and if each layeris labelled as layer A, then whole three dimensionalcrystal lattice will be of AAAA... type.

(d) Each sphere is in contact with six surroundedspheres, four being in the same plane, one in theabove layer and one in the below layer. Hence thecoordination number of each sphere is six.

(e) This arrangement is called AAAA type or Simplecubic structure.

(f) In this arrangement the packing efficiency is 52.4% and empty or void space is 47.6%.

(e) ABAB type of three dimensional packing OR Hexagonal close packingin three dimension :

1. Placing second layer over the first layer :(a) Consider a two dimensional hexagonal close packed layer 'A' with triangular voids 'a'

and 'b'.


29

(b) Place a similar layer above it such that the spheres of the second layer are placed inthe depressions of the first layer.

(c) Since the spheres are aligned differently let us call the second layer as 'B'.(d) It is observed that not all the triangular voids of the first layer are covered. In this

arrangement, if triangular voids of the type 'b' are occupied then 'a' type are unoccupied,i.e., only half of the voids of first layer are occupied.

(e) When a triangular void of the first layer is covered by a sphere of the second layer(or vice versa), tetrahedral void is formed.

(f) The remaining half of the triangular voids of the first layer are covered by the triangularvoids of the second layer to form a octahedral void.

30


2. Placing third layer over the second layer :(a) Tetrahedral voids of the second layer may

be covered by the spheres of the third layer.(b) In this case, the spheres of the third layer

are exactly aligned with those of the first layer.(c) Thus the pattern of spheres is repeated in

alternate layers.(d) This pattern is called as ABAB and this

structure is called hexagonal close packed(hcp).

(e) In this, packing efficiency is 74%. Thecoordination number of each sphere is 12.(each sphere has 6 neighboring spheres inits own layer, 3 spheres in the layer aboveand 3 spheres in the layer below it.).

(f) ABCABC type of three dimensional packing OR Cubic close packedstructure (ccp) in three dimension :

1. Placing second layer over the first layer:(a) Consider a two dimensional hexagonal close packed layer 'A' with triangular voids 'a'

and 'b'.(b) Place a similar layer above it such that the spheres of the second layer are placed in

the depressions of the first layer.(c) Since the spheres are aligned differently let us call the second layer as 'B'.(d) It is observed that not all the triangular voids of the first layer are covered. In this

arrangement, if triangular voids of the type 'b' are occupied then 'a' type are unoccupied,i.e., only half of the voids of first layer are occupied.

(e) When a triangular void of the first layer is covered by a sphere of the second layer(or vice versa), a tetrahedral void is formed.


31

(f) The remaining half of the triangular voids of the first layer are covered by the triangularvoids of the second layer to form a octahedral void.

2. Placing third layer over the second layer :(a) The third layer may be placed

in such a way that its spherescover the octahedral voids.

(b) In this case, the spheres of thethird layer are not aligned withthose of either the first or thesecond layer.

(c) Let us call the third layer asC.

(d) This pattern is called asABCABC and this structure iscalled Cubic close packed(ccp).

(e) This arrangement is also calledface-centered cubic structure.

(f) In this, packing efficiency is74%. The coordination numberof each sphere is 12.

(g) Distinguish between hexagonal close packing and cubic close packing.

Hexagonal close packing

1. In this three dimensional building of unit cell,

spheres of third layer are placed on

tetrahedral voids of the second layer.

2. In this, the spheres of third layer lie directly

above the spheres of first layer.

3. In this first and third layers are identical.

4. In this first and fourth layers are different.

5. The arrangement of packing is ABAB type.

Cubic close packing

1. In this three dimensional building of unit cell,

the third layer is placed over octahedral voids

created by the first two layers.

2. In this, the spheres of third layer do not lie

above the spheres of first layer.

3. In this first and third layers are different.

4. In this first and fourth layers are identical.

5. The arrangement of packing is ABCABC

type.

Close cubic packing (ccp)

32


(h) Distinguish between tetrahedral void and octahedral void.

(i) Coordination numbers in the following :1. At the corner of scc2. Body centred cubic (bcc) structure3. Face centred cubic (fcc) structure

1. At the corner of scc : If we consider an atom at thecomer of simple cubic structure (scc) structure then itis in contact with six identical equidistant spheres,namely three from three corners of the unit cells (1, 2,3) and three of neighbouring unit cells, (4, 5, 6). Hencecoordination number of the sphere is 6.

2. Body centred cubic (bcc) structure : In this unitcell, 8 eight atoms are present at eight corners whileone additional atom is at the body centre. The atom atthe body centre is in contact with eight identical andequidistant atoms. Hence its coordination numberis 8.

Tetrahedral void

1. This vacant space or void is surroundedby four atomic spheres.

2. The atom in tetrahedral void is in contactwith four atoms placed at four corners ofa tetrahedron.

3. This void is formed when a triangularvoid made by three coplanar spheres, isin contact with fourth sphere above orbelow it.

4. The coordination number of a tetrahedralvoid is 4.

5. There are two tetrahedral voids per sphereor atom.

6. The tetrahedral voids are present on thebody diagonals of the unit cell.

Octahedral void

1. This vacant space or void is surroundedby six atomic spheres.

2. The atom in octahedral void is in contactwith six atoms placed at six corners of anoctahedron.

3. Octahedral void is formed by two sets ofequilateral triangles pointing in oppositedirections with six spheres.

4. The coordination number of an octahedralvoid is 6.

5. There is one octahedral void per sphereor atom.

6. The octahedral voids are present on theedges and 1 at the body centre of theunit cell.


33

3. Face centred cubic (fcc) structure : In this unit cell, eight atoms are present at eightcorners and six atoms are present at six face centres. Each sphere in the structure is incontact with surrounding twelve equidistant neighbouring spheres, six in the same layer,three in the layer above and three in the layer below. Hence the coordination number ofan atom or sphere is 12.

(j) Definition :1. Packing fraction : The fraction of the total space in unit cell occupied by the constituent

particles is called packing fraction.Volume occupied by all constituent particles

Packing fraction =Volume of the unit cell

2. Packing efficiency : It is the percentage of total space in the unit cell occupied by theconstituent particles like atoms, ions or molecules of the crystal.

Packing efficient = Volume occupied by all constituent particles

× 100Volume of the unit cell

= Packing fraction × 100

3. Void space in the crystal : A vacant space or a space not occupied by the constituentparticles in the unit cell is called a void space. It is given by,Percentage of void space = 100 packing efficiency

ORFraction of void space = 1 Packing fraction.

(k) Calculation of packing efficiency in following crystals :1. Simple cubic (scc) structure2. Body centred cubic (bcc) structure3. Face centred cubic (fcc) structure OR Cubic close packing (ccp) structure.

1. Packing efficiency in simple cubic (scc) crystal : In this primitiveunit cell, 8 atoms of element are present at 8 corners. Two atomsare in contact along the edge AB of length a of the unit cell.If r is the radius of the atom, then AB = a = 2r

∴ r = a

2Since each corner atom contributes l/8th of the atom.

34


Total number of atoms in unit cell = 1

× 88

= 1 atom

Volume of unit cell = a3 = (2r)3 = 8r3

Volume of 1 atom = 34r

3π

Packing fraction = Volume occupied by 1 atom

Volume of unit cell

= 3

3

4/3 r

8r

π

= 6

π

= 0.524∴ Packing efficiency = 0.524 × 100 = 52.4 %

Percentage of void space = 100 – 52.4 = 47.6 %

2. Body centred cubic (bcc) structure : In this unit cell, there are 8 atoms at 8 cornersand one atom at the body centre.

∴ Total number of atoms = 1

8 × + 1 = 1 + 1 = 28

The atoms are in contact along the body diagonal BF. Let a be the edge length and r theradius of an atom.Consider a triangle BCE.BE2 = BC2 + CE2 = a2 + a2 = 2a2

Consider triangle BEF.BF2 = BE2 + EF2

= 2a2 + a2

= 3a2

BF = 3a

From figure, BF = 4r

∴ 4r = 3a

∴ R =3

a4

Volume of unit cell = a3 =

3

4× r

3

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

= 364

r3 3


3π

bcc structure

r

r r

r

A

B C

D

E

a

a

F


35

Packing fraction =Volume occupied by 2 atom

Volume of unit cell

=

3

3

42 × r

364

r3 3

π

=3

8

π

=3 × 3.142

8= 0.68

∴ Packing efficiency = 0.68 × 100 = 68 %∴ % of void space = 100 – 68 = 32 %

3. Face centred cubic (fcc) structure : In this unit cell, there are 8 atoms at 8 corners and6 atoms at 6 face centres.

Number of atoms due to 8 corners = 1

8 ×8

= 1

Number of atoms due to 6 face centres = 1

× 62

= 3

∴ Total number of atoms in the unit cell = 1+ 3 = 4The atoms are in contact along the face diagonal BD. Let a be the edge length and r, the radius of an atom.Consider a triangle BCD.

BD2 = BC2 + CD2

= a2 + a2 = 2a2

∴ BD = 2aFrom figure, BD = 4r

∴ 4r = 2a

∴ r =2

a4

= a

2 2

r =a

2 2

Hence, a = 2 2

Volume of unit cell = a3 = ( )3

2 2r = 316 2r

36



3π

∴ Packing fraction =Volume occupied by 4 atoms

Volume of unit cell

=

3

3

4r

3416 2r

π⎛ ⎞⎜ ⎟⎝ ⎠

= 3

3

16/3 r

16 2r

π

= 3 2

π

= 3.142

3 × 1.414= 0.74

∴ Packing efficiency = 0.74 ×100 = 74 %∴ % of void space = 100 – 74 = 26 %

(l) Obtain a relation for the density of the crystal :Consider a cubic unit cell of edge length a. Then the volume of the unit cell is = V = a ×a × a = a3. Let M be the atomic mass of the constituting element or metal. If NA is the

Avogadro number, then the mass of one atom will be A

M

N .

If z is the number of atoms present in the unit cell, then the mass of unit cell will be,

Mass of unit cell = Mass of z atoms = z = A

M

NThen the density of the unit cell is,

Density of unit cell =Mass of unit cell

Volume of the unit cell

d = A3

Mz ×

N

a

∴ d = 3A

z × M

a × N


37


1. What is coordination number ?Ans : Refer 1.3 (a)

2. Explain linear packing in one dimension. OR Close packing in one dimension.Ans : Refer 1.3 (b)

3. Explain the following :Planar packing arrangement of spheres. ORClose packing in two dimensions. ORAAAA type and ABAB type of two dimensional arrangement. ORSquare close packing and hexagonal close packing in two dimension

Ans : Refer 1.3 (c)

4. Explain AAAA type of three dimensional packing. Or Explain simple cubic structure (SCC).Ans : Refer 1.3 (d)

5. Explain ABAB type of three dimensional packing or Explain hexagonal close packing structure.Ans : Refer 1.3 (e)

6. Explain ABCABC type of three dimensional packing or Explain Cubic close packed structure(ccp).

Ans : Refer 1.3 (f)

*7. Distinguish between hexagonal close packing and cubic close packing.Ans : Refer 1.3 (g)

*8. Distinguish between tetrahedral void and octahedral void.Ans : Refer 1.3 (h)

9. Find the coordination numbers in the following :(a) At the corner of scc(b) Body centred cubic (bcc) structure(c) Face centred cubic (fcc) structure.

Ans : Refer 1.3 (i)

10. What is packing fraction ?Ans : Refer 1.3 (j) (1)

11. What is packing efficiency ?Ans: Refer 1.3 (j) (2)

12. What is a void space in the crystal ?Ans : Refer 1.3 (j) (3)

38


*13. Calculate packing efficiency in the following crystals :(a) Simple cubic (scc) structure

(b) Body centred cubic (bcc) structure

(c) Face centred cubic (fcc) structure OR Cubic close packing (ccp) structure.Ans : Refer 1.3 (k)

14. Obtain a relation for the density of the crystal.Ans : Refer 1.3 (l)

15. Write the packing fraction, packing efficiency and void space for the following crystal structures:

(1) SCC (2) BCC (3) FCC (4) HCP

Ans. Crystal structure Packing fraction Packing efficiency Void space

(1) SCC 0.524 52.4% 47.6 %(2) BCC 0.68 68% 32%(3) FCC 0.74 74% 26%(4) HCP 0.74 74% 26%

*16. Explain with the help of neat diagrams AAAA and ABAB and ABCABC type of three dimensionalpackings OR Explain the packing and voids in ionic solids.

Ans. Refer 1.3


1. Which type of crystal structure is obtained by ABAB arrangement of spheres?

a) tetragonal b) hexagonal c) octahedral d) cubic2. ABC ABC ....... closest packing of spheres in solid gives

a) cubic close packing b) hexagonal close packingc) tetragonal close packing d) octahedral close packing

• Advanced MCQs :

*3. The ratio of close packed atoms to tetrahedral holes in cubic packing is .......

a) 1 : 1 b) 1 : 2 c) 2 : 1 d) 1 : 3

*4. The ratio of close packed atoms to octahedral holes in hexagonal close packing is .......

a) 1 : 1 b) 1 : 2 c) 2 : 1 d) 1 : 3


39

1.4 PACKING EFFICIENCY AND DENSITY OF UNIT CELL :

Concept Explanation :

(a) Packing in voids of ionic solids :

1. Ionic solids are formed from cations and anions.

2. The number of oppositely charged ions are appropriately

adjusted such that the charges are balanced and the

compound formed is electrically neutral.

3. Cations and anions are always at alternate positions for

more stability.

4. Anions occupy the lattice points while the relatively smaller cations occupy either tetrahedral

or octahedral voids.

5. Generally, if cation is relatively smaller than anions, then the cations occupy tetrahedral

voids while bigger cations occupy octahedral voids.

6. In case, the size of cation is very large then, the packing of anions is adjusted by separating

them and larger cations are accommodated in larger cubic holes.

(b) Radius ratio rule for ionic compounds :1. The ratio of radius of cation (r+) to the radius of anion (r-) is known as radius ratio of the

ionic solid and represented as

Radius ratio = +

–

r

r2. The radius ratio rule is useful in predicting the crystalline structures of ionic solids.

3. The structure of an ionic compound depends upon stoichiometry of it and the sizes of ions.

4. If r+ and r− are the radii of cation and anion respectively, then r+/r− represents the radius ratio.

5. Greater the radius ratio, greater is the coordination number of cation.

Radius ratio in ionic crystals

Radius ratio Coordination number Crystalline structure Example

0.155 – 0.225 3 Plane triangular B2O3

0.225 – 0.414 4 Tetrahedral ZnS

0.414 – 0.732 6 Octahedral NaCl

0.732 – 1 8 Cubic CsCl

40


(c) Packing of ions in cubic closed packed (ccp) or hexagonal closedpacked (hcp) structures :

1. In case of cubic closed packed (ccp) or hexagonal closed packed (hcp) structures, there

are two tetrahedral holes per anion and one octahedral hole per anion in the array.

2. In ionic compounds with formula AB, if the cation has a smaller size it occupies tetrahedral

voids. However, since the number of tetrahedral voids are double the number of anions,

only half of the tetrahedral voids are occupied and the other half are vacant. This type of

structure is known as Zinc blende structure. Eg: ZnS, CuCl etc.

3. In ionic compounds with formula AB, if the cation has a bigger size it occupies octahedral

voids. However, since the number of octahedral voids are equal to the number of anions,

all the octahedral voids are occupied. This type of structure is known as rock salt structure.

Eg: NaCl, AgBr etc.

4. Consider the ionic crystals (A2B) of Li2O, Li2S, Na2O, Na2S, etc. in which size of cation

is less than the size of anion (r+ < r−). There are total four anions, while eight cations are

packed in eight tetrahedral voids. This type of structure is called Antifluorite structure

5. In ionic compounds with formula AB2, the cation has a bigger size it forms ccp structure.

Since the number of anions are twice the number of cations they occupy all the tetrahedral

voids. This type of structure is known as Fluorite structure. Eg: BaCl2, SrF2 etc

(d) Structure of NaCl unit cell (rock salt structure) :1. NaCl unit cell has face centred cubic (fcc) structure.

2. Na+ ions exist in octahedral voids at 12 edge centres and at body centre.Since edge centre atom contributes 1/4 th atom, the total number of Na+ ions will be,1

× 12 + 14

(body centre) = 4


41

3. Cl− ions exist at 8 corners and 6 face centres. Hence, the total number of Cl− ions will

be, 1 1

× 8 + × 68 2

= 4

4. Thus, unit cell contains 4Na+ and 4Cl− ions, or 4NaCl molecules.5. The coordination number of Na+ and Cl% is 6.6. The edge length of NaCl unit cell : a = 2(rc + ra)

(KCl crystal has the same structure like NaCl.)


*1. Explain the packing in voids of ionic solids.Ans. Refer 1.4 (a)

2. What is radius ratio ?Ans. Refer 1.4 (b)

*3. Explain radius ratio rule for ionic compounds.Ans. Refer 1.4 (b)

4. Explain the packing of ions in cubic closed packed (ccp) or hexagonal closed packed (hcp)structures.

Ans. Refer 1.4 (c)

5. Explain the structure of NaCl unit cell. (rock salt structure)Ans. Refer 1.4 (d)


1. The number of nearest neighbours around each atom is a face centred cube is .......a) 4 b) 6 c) 8 d) 12

2. In a body centred cube, the number of nearest neighbouring atoms are .......a) 2 b) 4 c) 6 d) 8

3. In ionic compound has radius ration (r+/r–) in the range of 0.225 to 0.414, what is the structureand coordination number of compound?a) cubic and 8 b) octahedral and 6c) tetrahedral and 4 d) trigonal and 3

4. The inter ionic distance for cesium chloride crystal is .......

a)2

a3

b)3

a2

c) 3a d)2a

3

42


*5. The packing efficiency for a body centred cubic structure is .......a) 0.42 b) 0.53 c) 0.68 d) 0.82

*6. An ionic crystal lattice has r+/r– radius ratio of 0.524. its coordination number is .......a) 2 b) 4 c) 6 d) 8

7. In hcp mode of staking, a sphere has coordination number .......a) 4 b) 6 d) 8 d) 12

8. If edge of bcc crystal of an element is ‘a’ cm. M is the atomic mass and NA, is the Avogadro’snumber, then the density of crystal is .......

a) 3

4M

a NAb)

A

3

2N

Ma c)3 A

Ma N d)

3

A

Ma

1N

9. In which of the following structures, the anion has maximum coordination number?a) NaCl b) ZnS c) CaF2 d) Na2O

• Advanced MCQs :10. In the closest packing of atoms .......

a) the size of tetrahedral void is greater than that of octahedral voids.

b) the size of tetrahedral void is smaller than that of octahedral voids.c) the size of tetrahedral and octahedral voids are same.d) the size of tetrahedral void is greater than that of octahedral voids.

11. Compound AxBy crystallises in hexagonal cubic close packing lattice. The element A occupy

2/3 of tetrahedral voids. Calculate x and y .......a) x = 1, y = 2 b) x = 2, y = 3 c) x = 3, y = 4 d) x = 4, y = 3

*12. The number of octahedral sites per sphere in fcc structure is .......

a) 1 b) 2 c) 3 d) 4*13. The number of tetrahedral sites per sphere in ccp structure is, .......

a) 1 b) 2 c) 3 d) 4

*14. An ionic compound AxBy occurs in fcc type crystal structure with B ion at the centre of eachface and A ion occupying corners of the cube. Give the formula AxBy.a) AB3 b) AB4 c) A3B d) A4B

15. The packing fraction for a body centred cube is .......a) 0.42 b) 0.54 c) 0.68 d) 0.74


43

1.5 IMPERFECTIONS OR DEFECTS IN CRYSTAL STRUCTURE : Concept Explanation :

(a) Defects in crystal structure :1. Any deviation from perfectly ordered arrangement of atoms, ions or molecules in the

crystal lattice is called a defect.2. Crystalline solids are formed by regular repetition of large number of unit cells in all directions.3. There is a short range as well as a long range order in the arrangement of constituents

particles in solids.4. The real crystals are never perfect. They always contain some imperfection in the crystal

formation.5. The crystalline solids are formed by cooling a substance in the liquid state.6. If the crystallisation takes place faster, then the orderly arrangement of constituent particles

cannot take place which produces defect in solids.7. The defects can be minimized by carrying out crystallisation at the slowest rate, so that

orderly arrangement of particles can take place.8. The crystal defects change the original physical and chemical properties of crystalline solids.

(b) Types of defects in the solids or crystal structures :The defects in crystalline solids are of two types viz.,1. Point defect 2. Line defect.

1. Point defects are further classified as :(a) Vacancy defect or Schottky defect(b) Interstitial defect or Frenkel defect(c) Impurity defect :This is further classified as :(i) Substitution impurity defect (ii) Interstitial impurity defect.

2. Line defects are further classified as :(a) Edge dislocation (b) Screw dislocation.

(c) Point defects :1. Point defect : The defect is due to a fault in the arrangement of a point i.e. a constituent

particle, e.g. an atom or an ion or a molecule in the crystalline solid.2. The point defects are classified as follows :

(a) Vacancy defect or Schottky defect :(i) Sometimes during crystallisation, some of the places of the constituent particles

remain unoccupied and the defect generated is called Vacancy defect.

44


(ii) In case of ionic solids, the vacanciesare produced due to absence of cations

and anions in stoichiometric proportions

from their positions. This defect iscalled Schottky Defect.

(iii) In these ionic solids, electrical neutrality

is maintained but the actual density isless than the expected value.

(iv) These defects are observed in the

solids with cations and anions of almostequal size. E.g. NaCl, KCl, CsCl, etc.

(b) Interstitial defect or Frenkel defect :(i) When cation or anion from ionic solid

leaves a regular lattice site andoccupies a place between the regularlattice points or interstitial points. Thisdefect is called Frenkel or interstitialdefect.

(ii) Frenkel defect arises in case of ionicsolids with relatively smaller cationicsize which can fit into interstitial spaceand difference in ionic radii is large.

(iii) This defect is observed in AgCl solidby Ag+ ions, in ZnS solid by Zn2+ ions,etc.

(iv) Frenkel defect does not alter the density of the crystalline solid.(c) Impurity defect :

(i) This defect arises when a cation from its regular site in ionic crystal lattice isreplaced by different cations.

(ii) If the impurity cation is substituted in the place of regular cation, then it is calledsubstitution impurity defect.


45

(iii) If the impurity of cation is present in the interstitial positions then it is calledinterstitial impurity defect.

(iv) The interstitial impurity defect changes the properties of the original crystallinesolid.

(v) Alloys are formed by substitution defects. E.g. Brass, in which Cu metal issubstituted by zinc metal. Stainless steel involves an interstitial impurity defect bycarbon atoms.

ADDITIONAL IMPORTANT INFORMATION :Non-Stoichiometric defects:1. The defects discussed so far do not disturb the stoichiometry of the crystalline substance.2. However, in some solids the constituent elements are in non-stoichiometric ratio due to

defects.3. These defects are of two types :

(a) Metal excess defect(b) Metal deficiency defect.

(a) Metal excess defect :1. Metal excess defect due to anionic vacancies:

(i) Alkali metal halides like NaCl and KCl showthis type of defect.

(ii) When NaCl crystals are heated in anatmosphere of Sodium vapour, the Na atomsare deposited on the surface of the crystal.

(iii) The Cl– ions diffuse to the surface of thecrystal and combine with Na atoms to giveNaCl.

(iv) This happens by loss of electron by Na atomsto form Na+ ions.

(v) The released electrons diffuse into the crystal and occupy anionic sites.(vi) As a result the crystal has excess of sodium.(vii) The anionic sites occupied by unpaired electrons are called F-centres.(viii) They impart yellow colour to the crystals of NaCl.(ix) The colour results by excitation of these electrons when they absorb energy from

the visible light.2. Metal excess defect due to presence of extra cations at interstitial sites :

(i) ZnO is white in colour at room temperature.(ii) On heating it loses oxygen and turns yellow.

(iii) 2+ –2

1ZnO Zn + O + 2e

2⎯⎯→

46


(iv) Now there is excess of Zn2+ in the crystal which moves to interstitial sites andelectrons to the neighbouring interstitial sites.

(b) Metal deficiency defect :(i) There are many solids which are difficult to prepare in the stoichiometric

composition and contain less amount of metal.(ii) Eg : in FeO the actual composition varies from Fe0.93O to Fe0.96O. In these

crystals some Fe+2 ions are missing and the loss of positive charge is made upby the presence of Fe+3.


*1. Explain interstitial defects and impurity defects?Ans. Refer 1.5 (c) 2(b, c)

*2. Explain the terms : (a) Schottky defect, (b) Frenkel defectAns. (a) Refer 1.5 (c) (ii) (b) Refer 1.5 (b) (i)

*3. What are point defects?Ans. The defects which is due to a fault produced in the arrangement of a point i.e. a constituent

particle like atom, ion or molecule in a crystalline solid.

4. What is meant by defects in crystal structure or solid?Ans. Refer 1.5 (a)

5. Why do defects in the crystalline solids arise ?Ans. Refer 1.5 (a)

6. Mention the types of defects in the solids or crystal structures.Ans. Refer 1.5 (b)

7. Explain point defects.Ans. Refer 1.5 (c)

8. Write the type of defect in the following : (a) KC1 (b) AgCl (c) Brass (d) Stainless steel(e) ZnS (f) Bronze.

Ans. Substance Type of defect

(a) KCl Schottky defect or vacancy defect(b) AgCl Frenkel defect or interstitial defect(c) Brass Substitution impurity defect(d) Stainless steel Interstitial impurity defect(e) ZnS Schottky defect or vacancy defect(f) Bronze Substitution impurity defect


47

Multiple Choice Questions :

• Theoretical MCQs :*1. In crystalline solid few of the cations moved from their positions into the interstitial .......

position. The defect is called as, .......a) interstitial defect b) Frenkel defect c) Schottky defect d) line defect

*2. Due to Frenkel defect the density of ionic solid .......a) increases b) decreases c) remains same d) fluctuates

1.6 Electrical and Magnetic properties of crystalline solid : Concept Explanation :

Solids exhibit an interesting range of variation electrical conductivity over 27 orders of magnitude

ranging from 10–20 to 107 Ohm–1m–1. Based on their conductivity, solids are classified into :

(i) Conductors (ii) Insulators and (iii) Semi-conductors

(a) Electrical propertiesof solids :1. Conductors : The solid which allow the current to flow through it is called conductor.

They have conductivity in the range of 104 to 107 Ohm–1m-1. e.g. metals.2. Insulators : The solids which do not allow passage of electric current through them are

called insulators. They have very low conductivity ranging between 10–20 to 10–10

Ohm–1 m–1. e.g. wood, rubber, sulphur, phosphorus etc.3. Semi-conductors : The solids whose conductivities are in between that of conductors and

insulators are called semiconductors. Their conductivities are in the range of 10–6 to 104

Ohm–1 m–1 and the conductivity is mainly due to presence of an impurity or some defects.In most of the solids, the conductivity is through the movement of electron under the influenceof electric field while in ionic solid, it is by ions. In solids (metals), thus, conductivity stronglydepends on the number of valence electron available for conductance. The electricalconductivity of conductors, insulators and semi conductors can be explained on the basisof band theory.

(b) Band theory OR Origin of electrical properties in solids :1. Metals are good conductors of heat and electricity. In order to explain the conducting

properties of metals a band theory was developed.2. The theory assumes that the atomic orbitals of the metal atoms overlap to form molecular

orbitals which are spread all over the crystal structure.3. With increase in number of atoms participating in crystal formation, the number of molecular

orbitals containing electrons is greatly increased.

48


4. As the number of molecular orbitals increases, the energy difference between adjacentmolecular orbitals decreases and when it becomes very less, the orbitals merge into oneanother forming continuous bands which extent over the entire crystal.

5. The interaction between these large number of orbitals leads to formation of bonding andanti bonding molecular orbitals.

6. All the molecular orbitals are very close to each other and cannot be distinguished fromone another and therefore, these orbitals are collectively called a band.

7. There are two types of bands of molecular orbitals as(a) Valence band : The atomic orbitals with filled electrons from the inner shells form

valence bands, where there are no free mobile electrons since they are involved inbonding.

(b) Conduction band : Atomic orbitals which are partially filled or empty on overlappingform closely placed molecular orbitals giving conduction bands where electrons aredelocalized and can conduct, heat and electricity.

8. In metallic crystals, the valence bands and conduction bands overlap or are very close andenergy difference between them is very less, so that electrons from valence bands caneasily be excited to conduction bands.A metal crystal is considered as an electron sea model and viewed as a three dimensionalarray of metal cations immersed in a sea of mobile electrons. Consider magnesium metal,

12Mg having electronic configuration 1s2, 2s2, 2p6, 3s2, 3pº, 3dº. In metallic crystal it existsas Mg2+. The atomic orbitals 1s2, 2s2, 2p6, after overlapping form valence bands whileempty orbitals, 3sº, 3pº, 3dº form conduction bands.

9 In insulators, energy difference between valence band and conduction band is very high,hence high energy is required to excite the electrons from valence band to conduction band.


49

10. If the energy differencebetween valence bandand conduction band ismoderate, then thesubstance in ordinarycondition is non-conductor,but if heated it becomesconductor due to transitionof electrons intoconduction band. Suchconductors aresemiconductors.

(c) Semiconductors and their types :1. Semiconductors : The substances like silicon, germanium which have poor electrical

conductance at low temperature but the conductance increases with the increase intemperature are called semiconductors.

2. Their conductivity lies between metallic conductors and insulators.3. The energy difference between valence band and conduction band

is relatively small, hence, the electrons from valence band can beexcited to conduction band by heating.

4. Types of semiconductors :There are two types of semiconductors as follows :(i) Intrinsic semiconductors(ii) Extrinsic semiconductors obtained by doping Si.The extrinsic semiconductors are of three types.They are (a) n-type semiconductors, (b) p-type semiconductorsand (c) n-p type semiconductors.

5. Intrinsic semiconductor : Silicon, 14Si, is a semiconductorhaving electronic configuration 1s2, 2s2, 2p6, 3s2, 3p2,3d0, 4s0.All the 4 electrons in Si are involved in the covalent bondingin the tetrahedrally situated neighbouring Si atoms.On heating some of valence band electrons go to conductionband and silicon becomes electric conductor.Silicon and germanium are the examples of intrinsicsemiconductors.

50


6. n-type semiconductors :(a) n-type semiconductor is an extrinsic

semiconductor obtained by dopingsilicon with external electron richimpurity like fifth group element. Forexample, arsenic.

(b) Arsenic has five valence electrons, outof which four are involved in covalentbonding with neighbouring Si atomswhile one electrons remains free anddelocalised.

(c) These free electrons increase the electrical conductivity of the semiconductor.(d) The semiconductors with extra non-bonding free

electrons are called n-type semiconductors.7. p-type semiconductors.

(a) p-type semiconductor is an extrinsicsemiconductor obtained by doping Siliconwith external electron deficient impurity likethird group element. For example, boron.

(b) Boron (5B) has only 3 valence electronswhich form covalent bonds with theneighbouring Si atoms, while one bond hasshortage of one electron.

(c) This creates a vacancy or a hole, hence the electron from neighbouring Si atom jumpsinto this hole creating a vacancy in itself. This process continues, i.e., positive holesmove in one direction while electrons moves in opposite direction.

(d) Due to electron deficient positions, this semiconductor is called p-type semiconductor.(e) When p-type semiconductor is connected to the external source of electricity, electrons

from neighbouring silicon atoms jump into the holes so that electrons move towardspositive electrode and holes migrate towards negative electrode.

(f) Hence, electrical conduction in p-type semiconductor is due to electrons and holes.

(d) Uses of semiconductors :(a) They are used in transistors, digital computers and cameras.(b) They are used in solar cells and television sets.(c) By combining n-type and p-type semiconductors, n-p junctions are formed which are

effectively used in rectifiers or to convert light energy into electrical energy.


51

(e) Origin of magnetic properties in solids :. (a) While electrons are revolving round the

nucleus in various orbits, they are alsospinning about their own axis.

(b) A spinning charge generates a magneticfield. Hence, spinning electrons act like tinymagnets.

(c) If an orbital contains only one electron thenit may spin either clockwise or anticlockwise.This generates magnetic field.

(d) However, if the orbital contains two electrons then one electron spins in clockwisedirection and other spins in anticlockwise direction, then their spins are balanced andmagnetic moments and magnetic properties are cancelled and the substance is said tobe diamagnetic.

(e) When a substance contains one or more unpairedelectrons spinning in same direction, then theirmagnetic moments and magnetic properties addand the substance is said to be paramagnetic.

(f) Diamagnetism :1. The magnetic properties of a substance arise due to

presence of the electrons.2. An electron while revolving around the nucleus, also

spins around its own axis and generates a magneticmoment and a magnetic property.

3. If an atom or a molecule of the substance contains all electrons paired, spinning clockwiseand anticlockwise, their magnetic moments and magnetic properties get cancelled. Hencethey oppose and repel the applied magnetic field. This phenomenon is called diamagnetismand the substance is said to be diamagnetic.For example : Zn, Cd, H2O, NaCl, etc.

(g) Paramagnetism :1. The magnetic properties of a substance arise due to

the presence of electrons.2. An electron while revolving around the nucleus, also spins around its own axis and generates

a magnetic moment and magnetic properties.

52


3. If an atom or a molecule contains one or more unpaired electrons spinning in same direction,clockwise or anticlockwise, then the substance is associated with net magnetic momentand magnetic properties. They experience a net force of attraction when placed in themagnetic field. This phenomenon is called paramagnetism and the substance is said to beparamagnetic.For example, O2, Cu2+, Fe3+, Cr3+, NO, etc.

(h) Ferromagnetism :1. The magnetic properties of a substance arise

due to the presence of electrons.2. An electron while revolving around the nucleus

also spins around its own axis and generatesa magnetic moment and magnetic properties.

3. The substances which possess unpaired electrons and high paramagnetic character andwhen placed in a magnetic field are strongly attracted and show permanent magnetic momenteven when the external magnetic field is removed are said to be ferromagnetic. They canbe permanently magnetised.

4. In the solid state, the metal ions of ferromagnetic substance are grouped together into smallregions called domains, where each domain acts as a tiny magnet.For example : Fe, Co, Gd, CrO2, etc.

(i) Antiferromagnetism :Substances like MnO showing anti-ferromagnetism have domain structure similar toferromagnetic substance, but their domains are oppositely oriented and cancel out eachother’s magnetic moment.

(j) FerrimagnetismFerrimagnetism is observed when the magnetic moments of the domains in the substanceare aligned in parallel and antiparallel directions in unequal number. They are weakly attractedby magnetic field as compared to ferromagnetic substances. Eg: Fe3O4 etc.


53

(k) Gouy’s method for determining magnetic properties of the substances :Principle of Gouy’s method :

1. An electron in an atom while revolvingaround nucleus also spins around its ownaxis generating a magnetic moment andmagnetic field.

2. The resultant magnetic properties dependupon, the number of paired and unpairedelectrons.(a) If all the electrons are paired, their

magnetic moments and magneticproperties are cancelled and thesubstance is said to be diamagnetic.Such substance repels the appliedmagnetic field.

(b) If a substance contains one or more unpaired electrons, then the substance is associatedwith a net magnetic moment and magnetic properties. Such a substance is paramagneticin nature, which is attracted when placed in the applied external magnetic field.

Procedure :1. The substance is weighed in the absence and presence of the magnetic field,2. A diamagnetic substance is pushed upward when placed in the magnetic field between two

magnets. Hence, the weight of the substance appears less in the presence of magnetic fieldthan in the absence of magnetic field,

3. A paramagnetic substance is pulled down due/to attraction when placed in the magneticfield, hence the weight of the substance appears more in the presence of magnetic field.

4. If the substance is ferromagnetic, it is pulled down more weighing still higher in the presenceof magnetic field.Hence, from the changes in weights* in and out of magnetic field, magnetic properties ofthe substances can be decided.


*1. What are semiconductors? Describe the two main types of semiconductors?Ans. Refer 1.6 (c)

*2. Explain Band theory?Ans. Refer 1.6 (b)

*3. Explain the origin of electrical properties in solids.Ans. Refer 1.5(e)

54


Conductor1. A substance which conducts

heat and electricity to a greaterextent is called conductor.

2. In this, conduction bands andvalence bands overlap or arevery closely spaced.

3. There is no energy differenceor very less energy differencebetween valence bands andconduction bands.

4. There are free electrons in theconduction bands.

Insulator1. A substance which cannot

conduct heat and electricityunder any conditions is calledinsulator.

2. In this, conduction bands andvalence bands are far apart.

3. The energy differencebetween conduction bandsand valence bands is verylarge.

4. There are no free electronsin the conduction bands andelectrons can't be excitedfrom valence bands toconduction bands due tolarge energy difference.

Semiconductor1. A substance which has poor

electrical conductance at lowtemperature but higherconductance at highertemperature is calledsemiconductor.

2. In this, conduction bandsand valence bands are spacedclosely.

3. The energy differencebetween conduction bandsand valence bands is small.

4. The electrons can be easilyexcited from valence bands toconduction bands by heating.

4. Explain n-type semiconductors.Ans. Refer 1.6(c)6

5. Explain p-type semiconductors.Ans. Refer 1.6(c)7

*6. Classify the following semiconductors into n or p-type.(a) B doped with Si (b) As doped with Si (c) P doped with Si (d) Ge doped with In.

Ans. Semiconductor Type

(a) B doped with Si p-type(b) As doped with Si n-type(c) P doped with Si n-type(d) Ge doped with In p-type

7. What are the uses of semiconductors ?Ans. Refer 1.6 (d)

*8. Distinguish between conductor, insulator and semiconductor.Ans.


55

5. The conductance decreaseswith the increase intemperature.

6. E.g., Metals, alloys.7. The conducting properties

can't be improved by addingthird substance.

8.

5. No effect of temperature onconducting properties.

6. Wood, rubber, plastics.7. No effect of addition of any

substance.

8.

5. Conductance increaseswith the increase intemperature.

6. E.g. Si, Ge7. By doping, conducting

properties improve.E.g. n-type, semiconductors.

8.

9. Identify the following substances as paramagnetic, diamagnetic or ferromagnetic :(a) Sodium 11Na (b) Magnesium 12Mg (c) 20Ca2+ cation(d) 17Cl– anion (e) Iron 26Fe (f) 27Co atom(g) 28Ni atom

Ans. Paramagnetic (a) Sodium 11NaDiamagnetic (b) Magnesium 12Mg

(c) 20Ca2+ cation(d) 17C1– anion

Ferromagnetic (e) Iron 26Fe(f) 27Co atom(h) 28Ni atom

*10. Explain the origin of magnetic properties in solids.Ans. Refer 1.5 (e)

*11. Explain diamagnetism.Ans. Refer 1.5 (f)

*12. Explain paramagnetism.Ans. Refer 1.5 (g)

13. Explain ferromagnetism.Ans. Refer 1.5 (h)

56


14. Explain Antiferromagnetism.Ans. Refer 1.5 (i)

15. Explain FerrimagnetismAns. Refer 1.5 (j)

16. Explain Gouy's method for determining magnetic properties of the substances.Ans. Refer 1.5 (k)


1. To get n–type doped semi-conductor, the impurity to be added to silicon should havefollowing number of valance electron?a) 1 b) 2 c) 3 d) 5

2. A p-type material is electrically .......a) positive b) negativec) neutral d) depends on the concentration of p-impurities

*3. Semiconductors are manufactured by addition of impurities of .......a) p-block elements b) s-block elements c) lanthanoids d) actinoids

*4. p-type semi conductor is formed when trace amount of impurity is added to silicon. The numberof valence electrons in the impurity atom must be .......a) 3 b) 5 c) 1 d) 2

*5. n-type semiconductor is formed when trace amount of impurity is added to silicon. The numberof electrons in the impurity atom must be .......a) 3 b) 5 c) 1 d) 2

• Advanced MCQs :6. Which group element is added to germanium in order to get n-type semiconductor?

a) 12 b) 13 c) 14 d) 157. The addition of traces of Aluminium in the crystal lattice of Germanium gives rise to which type

of semiconductor?a) none of the above b) p-type c) n-p junction d) n-type

HOURS BEFORE EXAM Isomorphous : Solids which have same crystal structure. Their constituent particles have

same ratios. e.g. NaF and MgO, NaNO3 and CaCO3 etc. Anisotropy : The substances which have physical properties different in different direction.

e.g. refractive index, electric conductance, dielectric constant, thermal conductivity etc. Isotropy : The solids in which physical properties do not change with the direction of

measurement. eg. amorphous solids.


57

Crystal Lattice : Regular three dimensional arrangement of points in space are calledspace point or crystal lattice.

Unit cell : Smallest part of crystal lattice which when repeated in different direction generatesthe entire lattice.

Cubic Tetragonal Orthorhombic Monoclinic Triclinica ≠ b ≠ c, α ≠ β ≠ γ ≠ 90º

Primitivea ≠ b ≠ c, αα = β = 90º, = γ = 90º

End-centereda ≠ b ≠ c,α = β = 90º, γ = 90º

Hexagonala = b ≠ cα = β = 90ºγ = 120º

Rhombohedrala = b = c,α = β = γ ≠ 90º

1 2 3 4 5

6 7

Primitivea = b ≠ c,α = β = γ = 90º

Body centereda = b ≠ c,α = β = γ = 90º

Simple orprimitivea = b = c

α = β = γ = 90º

Body centreda = b = c

α = β = γ = 90º

Face centereda = b = c

α = β = γ = 90º

Body centreda ≠ b ≠ cα = β = γ = 90º

Primitivea ≠ b ≠ cα = β = γ = 90º

End-centreda ≠ b ≠ cα = β = γ = 90º

Face-centreda ≠ b ≠ cα = β = γ = 90º

Classification of solids

Crystallinee.g. NaCl, ZnS, Copper, Gold etc.

Molecular solids Ionic solidse.g. NaCl

Covalent orNetwork solids

e.g. diamond, graphite

Metallic solidse.g. metals

Polar molecularsolids e.g. HCl,NH3, solid SO2

Non-polarmolecular solide.g. solid CO2,

solid Ar

Hydrogen bondedmolecular solid

e.g. solid water (ice)

Amorphouse.g. Glass, rubber, plastic

58


Coordination number : Number of particles/atoms/spheres surrounding a single particle.

Number of atoms per unit cell of cubic lattice

Defects in solids : Some irregularity arises during crystallisation of solids, known asimperfections or defects. These are.

Point defects

arises due to faculty arrangementof constituent particles.

Vacancy defect Cations and anions

remain unoccupied instoichiometric proportion.

Electrical neutrality ismaintained.

Result in decrease indensity of solid.

Interstitial defect

Frankel defect Some constituent particle

occupy interstitial position. Shown by ionic solids.

Large difference in ionic sizei.e. between cation & anion.e.g. AgCl, ZnS, AgBr, AgI.

Schottky defect Shown in by ionic solid only. Cation and anion have almost

same size. Cation and anion remain absent in

stoichiometric ratios. Electrical neutrality is maintained. Decreases the density of solid.

Impurity defect♦ Regular cation in crystal

is substituted by othercation.

arises due to deviation from the ideal arrangementof entire row of lattice point

Line defects

Simple or primitive

1) 8 corner atoms × 1

8

= 1 atom/unit cell

2) r = a

2

3) Packing efficiency = 52.4%4) Density of unit cell (ρ) =

3A

mass of unit cell n × M=

volume of unit cell a N⋅

∴∴∴∴∴ density (P) = 3

n × M

a NA⋅

Body centred

1) 8 corner atoms × 1

8 = 1

2) No. of atoms at bodycenter position = 1

3) Total No. of atoms/unitcell = 2

4) r = a

2 2

5) Packing efficiency = 68%

Face centred1) No. of corner atoms

= 8 × 1

8 = 1

2) No. of atoms at face

centre = 6 × 1

2 = 3

3) Total No. of atoms/unit cell= 3 + 1 = 4

4) r = 3a

4

5) Packing efficiency = 74%


59

Electrical Conductivity :Electrical conductivity in solid is due to free mobile electrons which can be explained on thebasis of band theory. Based on electrical conductivity solids are classified as

Magnetic properties of solids :Magnetic property in solids arises due to presence of unpaired electrons as each electron actsas tiny bar magnet. Based on magnetic behaviour solids are classified as

Conductors The energy difference

between the valence band andconduction band i.e. band gapis negligibly small.

Conducts at room temperatureor even at slightly lower temp.

Semiconductors Band gap is slightly greater than

conductors but lower than that ofinsulators.

Conductivity increases with increasein temperature.

Basically. they are insulator at anabsolute zero. e.g. Si, Ge.

Insulators Band gap is large.

They do not conduct atany temperature.

n-type Prepared by doping silicon or Germanium with

group 15 element like As, P etc. Conductivity increases due to presence of

excess of electrons.

p-type Prepared by doping Si or Ge with group 13

elements like Ga or In. Conductivity increases due to (creation of

holes (+ve) in the crystal structure.

Diamagnetic All electrons are paired i.e. their

spin is paired ( ). Their magnetic moment is zero.

They are weakly repelled byexternal magnetic field.e.g. water, benzene, NaCl etc.

Paramagnetic All electrons are not spin paired.

They have net magnetic moment.

They are weakly attractedtowards external magnetic field.

Loose their magnetism inabsence of external magneticfield. e.g. O2–,Cu2+, Fe+3 etc.

Ferromagnetic All spins are aligned in one

direction only. Very strongly attracted by

applied magnetic field. They can be permanently

magnetised. They produce strong

magnetic field. Magnetic moment is high.

Anti-ferromagnetic Alignment of spin is in opposite direction.

Very strongly attracted towards appliedmagnetic field.

Net magnetic moment is zero due tocompensation of domains.

Ferrimagnetism Alignment of spin is parallel and anti parallel

direction i.e. two electrons spin in parallel while3rd electron has spins in opposite direction.

They have net magnetic moment. Weakly attracted by applied magnetic field. Becomes paramagnetic on heating. e.g. Fe2O4,

CuFe2O4, MgFe2O4 etc.

60


FORMULAS

Type of atom Volume Unit Cell No. of atomscontribution per unit cell

At the corner of unit cell 1/8 Simple cubic 1At the body centre 1 Body centered cubic 1 + 1 = 2At the face centre ½ Face Centered cubic 1 + 3 = 4

Hexagonal 1 + 2 = 3

Property hcp ccp bccType of packing AB AB A... ABC ABC A... AB AB A...Available space occupied 74% 74% 68%Coordination number 12 12 8

Relationship between radius of constituent particle (r) and edge length (a) :a. Simple cubic unit cell : a = 2r

b. Face centred unit cell : a = 2 2r

c. Body centred unit cell : a = 4r

3

Crystal structure Packing fraction Packing efficiency Void space1. SCC 0.524 52.4% 47.6 %2. BCC 0.68 68% 32%3. FCC 0.74 74% 26%4. HCP 0.74 74% 26%

DENSITY OF THE UNIT CELL ( )ρ : It is defined as the ratio of mass per unit cell

to the total volume of unit cell. d = 3A

z × M

a × NWhere Z = Number of particles per unit cellM = Atomic mass or molecular massNA = Avogadro number (6.022 × 1023 mol–1)α = Edge length of the unit cell = α pm = α × 10–10 cma3 = volume of the unit cellThe density of the substance is same as the density of the unit cell.Radius ratio in ionic crystals

Radius ratio Coordination number Crystalline structure Example0.155 − 0.225 3 Plane triangular B2O3

0.225 − 0.414 4 Tetrahedral ZnS0.414 − 0.732 6 Octahedral NaCl

0.732 − 1 8 cubic CsCl


61

NUMERICALS WITH SOLUTIONS :*1. Niobium is found to crystallize with bcc

structure and found to have density of8.55g cm–3. Determine the atomic radiusof niobium if its atomic mass is 9.

Given :Crystal structure of Niobium = bccdensity (d) = 8.55 g cm–3

Atomic mass of Niobium = 93To find :

the atomic radius of niobium = ?Solution :

Formula = d = A

n M

V N

⋅

⋅

No. of atoms per unit cell (n) in bcc = 2Total volume of unit cell = a3

∴∴∴∴∴ d = 3 23

2 × 93

a × 6.023 ×10

a3= 36.12 × 10–24

a = –243 36.12 ×10

= –83 × 3.304 × 10

4

= 1.431 × 10–8 cmor 14.31 × 10 cm

a = 14.31 nm

*2. Copper crystallises into a fcc structure andthe unit cell has length of edge3.61 × 10–8 cm. Calculate the density ofcopper if the molar mass of Cu is 63.5 gmol.

Given :Crystal structure = fccedge length (a) = 3.61 × 10–8 cm

To find :density (d) = ?

Solution :

Formula = d = A

n M

V . N

⋅

molar mass = 63.5 g molNo. of atoms per unit cell (n) in fcc = 4Total volume of unit cell = a3 = (3.61 ×10–8 cm)3 = 47.045 × 10–24 cm3

∴∴∴∴∴ d = –24 23

4 × 63.5

47.045 × 10 × 6.022 × 10

d = 8.93 g cm-3.

*3. Silver crystallises in fcc structure with edgelength of unit cell, 4.07 × 10–8 cm and ifdensity of metallic silver is 10.5 g cm–3.Calculate the molecular mass of silver.

Given :Type of crystal structure = fccedge length (a) = 4.07 × 10–8 cmdensity (2δ) = 10.5 g cm3–

To find :molar mass of silver = ?

Solution :

Formula = d = A

n M

V N

⋅

⋅

M = Ad v N

n

⋅ ⋅

No. of atoms per unit cell (n) in fcc = 4Total volume of unit cell

= (v) = a3

= (4.07 × 10–8 cm)3

= 47.045 × 10–24 cm3

62


∴∴∴∴∴ M =2310.5 × 67.419 × 6.022 × 10

4

M = 106.5

*4. Determine the density of cesium chloridewhich crystallises in a bcc structure withthe edge length of unit cell, 4.07 × 10–8 cmand if density of metallic silver is 10.5 gcm–3. Calculate the molecular mass ofsilver.

Given :edge length (a) = 412.1 pm= 4.12 × 10–8 cmType of crystal structure = bccAtomic mass of cesium = 133Atomic mass of chlorine = 35.5

∴∴∴∴∴ Molar mass of CsCl = 168.5To find :

Density of cesium chloride (CsCl) = ?Solution :

Formula = d = A

n M

V N

⋅

⋅

No. of atoms per unit cell of bcc = 2

∴∴∴∴∴ d = –8 3 23

2 × 168.5

(4.12 × 10 ) (6.022 × 10 )

= 8.001 g cm3.

*5. Unit cell of iron crystal has edge length of288 pm and density of 7.86 g cm–3.Determine the type of crystal lattice (Fe =56) (Hint : Determine number of atoms ofunit cell. It comes to 2, hence, bcc type)

Given :edge length (a) = 288 pm = 2.88 × 10–8 cmdensity (d) = 7.869 cm–3

Atomic mass of iron = 56Type of crystal lattice = ?

Solution :

volume of one unit cell =M

d

=56

7.86 = 7.125g cm3–

Total volume of unit cell= a3

= (2.88 5 10–8)3

= 23.888 × 10–24 cm3

No. of unit cell in 56g metal

=Total volume

Vol. of one unit cell

=–2423.888 × 10

7.125

= 2.982 × 1023 unit cells.No. of atoms per unit cell

=Avogadro's number

No. of unit cells

=23

–24

6.022 × 10

2.982 × 10

= 2.019Since the number of atoms per unit cellis 2, it is bcc type.

*6. An atom crystallises in fcc crystal latticeand has a density of 10 g cm–3 with unitcell edge length of 100 pm. Calculatenumber of atoms present in 1 g of crystal.

Given :Density (d) = 10 g cm3

edge length (a) = 100 pm = 1 × 10–8 cm


63

To find :No. of atoms in 1 g crystal = ?

Solution :

volume of 1 g of metal =mass

density

=1

10

= 0.1 g cm3

Total volume of unit cell = a3

= (1 × 10–8)3

= 1 × 10–24 cm3

No. of atoms in limits volume

= –24

0.1

1× 10

= 1 × 1023 unit cellsFace entered cube contains 4 atoms perunit cell.

∴∴∴∴∴ Number of atoms in 1g = 4 × 1 × 1023

Number of atoms in 1g = 4 × 1023 atoms.

*7. An element A and B constitute bcc typecrystalline structure. Element A occupiesbody centre position and B is at the cornersof cube. What is the formula of thecompound? What are the coordinationnumbers of A and B? (Formula AB,coordination numbers of A = 8, and B = 8).

Solution :No. of atoms occupying body centeredposition = 1No. of atoms at the corners per unit cell

=1

8 ×8

= 1

∴∴∴∴∴ Formula of compound is AB

Since, in body centred cube, each atom issurrounded by eight nearest neighbour, itscoordination number is 8.

∴∴∴∴∴ Coordination No. of A = 8coordination No. of B = 8

*8. Atoms C and D form fcc crystallinestructure. Atom C is present at the cornersof the cube and D is the faces of the cube.What is the formula of the compound.

Solution:As, atom C, occupies corners of the cube,no. of atoms present at the corners per unit

cells = 1

8 corners ×8

= 1

No. of atoms present at the faces per unit

cell = 1

6 ×2

= 3

∴∴∴∴∴ Formula of compound is CD3.

9. A cubic unit cell contains atoms A at thecorners, atoms B at face centres and atomC at the body centre. What is the formulaof the crystalline compound ?

Solution :Atoms A are at 8 corners, atoms B at the 6face centres and one atom C at body centre.

Total number of atoms of 1

A ×8 18

= =

Total number of atoms of 1

B ×6 32

= = .

One atom of C at the body centre.Therefore the unit cell contains one atomof A, three atoms of B and one atom of C.Hence, the formula of the compound isAB3C.

∴∴∴∴∴ The formula of the crystallinecompound is AB3C.

64


10. In a cubic crystalline structure of zincblende (ZnS), sulphide ions are at cornersand face centres while zinc ions occupyhalf of tetrahedral voids. Find in the unitcell :

(a) Number of Zn2 + ions(b) Number of S2− ions(c) Number of ZnS molecules(d) Molecular formula of zinc blende.

Solution :S2− ions are at 8 corners and 6 facecentres. Zn2+ ions occupy half oftetrahedral voids.Number of S2− ions in unit cell

=

1 1× 8 + × 6

8 2(corners) (face centres)

= 1+ 3=4Cubic unit cell has 8 tetrahedral voids. Sincehalf of them are occupied by Zn2+ ions,there are 4Zn2+ ions in the unit cell.Hence, number of zinc sulphide (ZnS)molecules is 4. Molecular formula of zincblende is ZnS.(a) Number of S2−−−−− ions = 4(b) Number of Zn2+ ions = 4(c) Number of ZnS molecules = 4(d) Molecular formula of zinc blende

= ZnS.

11. In a crystalline compound, atoms A occupyccp lattices while atoms B occupy 2/3 rdtetrahedral voids. What is the formula ofthe compound ?

Solution :In ccp unit, lattice points are 8 corners and

6 face centres where atoms A are present.∴ Number of A atoms

= 1 1× 8 + × 6

8 2(corners) (face centres)

= 1 + 3 = 4In cubic unit cell, there are 8 tetrahedralvoids. Hence,

Number of B atoms = 2 16

× 8 =3 3

Hence, the formula should be A4B16/3 orA12B16 or A3B4.

∴∴∴∴∴ The formula of the compound is A3B4.

12. Atoms of elements A and B form a moleculeAxBy and crystallises in hcp lattice. Theelement A occupies 2/3 of tetrahedralvoids. What is the formula of AxBy ?

Solution :Atoms A are present in 2/3 tetrahedralvoids while the atoms B are present at 12corners and 2 face centres of hcp crystallattice.Hence, total number of B atoms is,

1 1× 12 + × 2 = 2 + 1 = 3

6 2

For each B atom, there will be 2 tetrahedralvoids. Hence, total tetrahedral voids willbe 3 × 2 = 6.The number of A atoms in voids will be,

2× 6 = 4

3.

Thus, hcp unit cell contains 4 atoms of Aand 3 atoms of B, hence the formula ofAxBy is, A4B3.


65

[Alternative Method :In hcp crystal lattice, the number oftetrahedral voids is twice the number of

anions of element B. Hence for 2

3

occupancy of tetrahedral voids by Aindicates half the occupancy of lattice pointsby anion B. Hence, the ratio of cation of

A and anion B will be 2 1

:3 2

or 4 : 3.

Therefore, the molecular formula willbe A4B3.

13. A compound AmBn with A in the form ofcation and B in the form of anion form ccptype crystal lattice. The cations of A occupyall tetrahedral voids. Predict the formulaof compound.

Solution :(a) In the ccp type crystalline lattice, cations of

A occupy all the tetrahedral voids while anionsof B occupy 8 corners and 6 face centres.

(b) Total number of anions of B

= 1 1

× 8 + × 68 2

= 1 + 3 =4

(c) Number of octahedral voids = Number ofanions = 4

∴ Number of cations of A = 4.(d) Hence, ccp unit cell will contain 4 cations

of A and 4 anions of B.Hence, AmBn = A4B4.Therefore, the formula of thecompound is AB.

14. In the ionic solid AB, the radius of cation

A+ is 1.3oA while that of anion B– is 1.76

0A . Predict the structure of AB andcoordination number of A+.

Solution :Given : Radius of cation,

A+ = r+ = 1.285oA

Radius of anion, B− = r_ = 1.76oA

Structure of AB = ?Coordination number of A+ = ?Radius ratio is,

+

–

r 1.285= 0.730

r 1.76

Since the radius ratio is 0.730 ≅ 0.732, thecrystalline structure is cubic andcoordination number of A+ is 8.

∴∴∴∴∴ Crystalline structure is cubic.Coordination number of A+ = 8.

15. If the coordination number of a cation inan ionic solid is 4. What is the type of voidoccupied by cation ?

Solution :Coordination number of cation is 4. Hence,the void must be tetrahedral.

16. An organic compound crystallises in anorthorhombic system with two moleculesper unit cell. The unit cell dimensions are12.05 ×10–8 and 15.05 × 10–8 cm and 2.69× 10–8 cm. If density of crystal is 1.419 gcm3. Calculate the molar mass ofcompound.

66


Given :Cell dimension = 12.05 × 10–8

and 15.05 × 10–8 cmand 2.69 × 10–8 cmd = 1.149 g cm3

To find :M = ?

Solution :Molar mass (M)

= Avogadro's number

density × Total ×No. of molecules per unit cell

M = Ad × V × N

n

= –24 23(1.419)(12.05×15.05×2.69×10 )(6.022×10 )

2

= 209 g mo1–1

17. Lithium borohydride, LiBH4 in anorthorhombic system with four moleculesper unit cell. The unit cell dimensions are,a = 6.81ºA, b = 4.43ºA and c = 7.14ºA. If themolar mass of LiBH4 is 21.76 g mol–1.Calculate the density of the crystal.

Given :Molar mass = 21.76 g mole–1.unit cell dimensions are, a = 6.81ºA,b = 4.43ºA and c = 7.14ºA

To find :d = ?

Solution :1–A = 10–8 cm

=N M

V NA

⋅

⋅

= –24 23

4 × 21.76

(6.81× 4.43 × 7.17 ×10 ) × (3.022 × 10 )

d = 0.668 g cm3–.

18. A metallic element exists as a cubic lattice.Each edge of the unit cell is 2.88 × 10–8 cm.The density of metal is 7.20 g cm3– Howmany unit cell there will be in 100g of metal?

Given :d = 7.20 g cm–3

Total volume of unit cell = 2.88 × 10–8 cmTo find :

no. of unit cell = ?Solution :

Total volume of unit cell= a3 = (2.88 × 10–8) = 23.9 × 10–24 cm3.

Volume of 100 g of metal

=mass

density = 100

7.20 = 13.9 cm3.

No. of unit cells in this volume

=Volume of 100g of metal

Volume of unit cell

= –24

13.9

23.9 × 10

= 5.82 × 1023

∴∴∴∴∴ Number of unit cells in 100g of metalis 5.82 × 1023.

19. Sodium crystallises in a bcc unit cell.Calculate the approximate number of unitcells in 9.2g of sodium. (at mass of Na =23g) [CBSE sample paper 2011]

Given :unit cell = 9.2 g of sodium

To find :number of unit cells = ?

Solution :No. of atoms in 9.2 g of sodium


67

=wt. of sodium × Avogadro's number

Atomic mass

=239.20× 6.022 × 10

23= 2.4088 × 1023

A bcc unit cell contains 2 atoms

∴∴∴∴∴ No. of unit cells =232.4088 × 10

2

No. of unit cells = 1.2044 × 1023

20. Sodium metal crystallises in bcc lattice withcell edge, 4.29ºA. What is radius of sodiumatom?

Given :a = 4.29ºA

To find :r = ?

Solution :For a bcc, the radius of atoms is

r =3

a4

=1.732 × 4.29

4

r = 1.86ºA

21. A compound AB has rock salt typestructure. The formula wt. of AB is 6.023y amu and the closest A–B distance is

y 13 nm. Where y is an arbitrary number.

Find the density of. (I.I.T. 2004)Given :

formula wt. of AB = 6.023 y amu

distance = 13 nm

To find :density = ?

Solution :

= 3A

n × formula mass of substance

a × N

AB has rock salt structure i.e. f.c.c. type.Number of atoms per unit cell (n) is 4.formula mass = 6.023 yN0

3– kga = 2yr3nm

= 2yr3 × 10–9m

d =

–3

123 –9 33

4 × 6.023y × 10

6.022 × 10 × (2y × 10 )d = 5.0 kg m3–.

22. Silver has cubic unit cell with a cell edgeof 408 pm. It’s density is 10.6 g cm–3. Howmany atoms of silver are there in unit cell?What is the structure of silver?

Given :edge length = 408 pm = 408 × 10–8 cmdensity = 10.6 g cm3.Atomic mass of silver = 108

To find :density = ?

Solution :

d = 3A

nM

a N

∴∴∴∴∴ n =3

Ad × a × N

M

= –8 3 2310.6×(408×10 ) ×108×6.022×10

108

= 4.02No. of atoms per unit cell = 4

∴∴∴∴∴ No. of atoms per unit cell of silver is4, it has face centred cubic crystal.

68


23. Copper crystallises in fcc entered cubiclattice and has density of 8.930 g cm–3 at293K. Calculate the radius of copper atom.(at mass of Ca = 63.55)

Given :density = 8.930 g cm3.Atomic mass of Ca = 63.55

To find :radius = ?

Solution :

d = 3A

nM

a N

Number of atoms (n) in fcc unit cell = 4

∴∴∴∴∴ 8.93 = 3 23

4 × 63.55

a × 6.022 × 10

a3 = 23

4 × 63.55

8.93 × 6.022 × 10

= 4.73 × 10–23cm3

a = (4.73 × 10–23)r3

= 3.616 × 10–8cmfor fcc

r =2a

4 =

–82 × 3.616 × 10

4

=–81.141× 3.616 × 10

4

= 127.8 × 10–8cm or 127.8 pmradius = 127.8 × 10–8cm or 127.8 pm


1. Potassium iodide has unit cell with cell edge of 7.5 pm. The density of KI is3.12 g cm3–. How many potassium and iodide ions are contained in it?

(Ans. : 4K+8 4I–)

2. Rubidium atomic mass 85.51 crystallises in a body centred cubic lattice with density of 1.51 gcm3–. If the radius of rubidium atom is 248 pm, calculate Avagadro’s number.

(Ans. : 6.023 × 1023)

3. Lithium metal has body centred cubic close packing. It’s density is 0.53 g cm3– and its molarmass is 6.94 g mol. Calculate the volume of the unit cell of lithium.

(Ans. : 4.35 × 1023 cm3)

4. The density of copper metal is 8.95 of cm3–. If the radius of atoms is 127.8 pm. is the copperunit cell simple cubic body centred or face centred cubic. (At mass of Ca = 63.54)

5. A metallic element ‘x’ exists as a cubic lattice. Each edge of the unit cell is 2.90ºA and densityof metals is 7.20 g cm3. How many unit cell will be present in 100 g of the metal?

(Ans. : 5.7 × 1023)


69

6. A cubic solid is made of two elements X and Y. Atoms Y are present at the corners of cubeand atoms X at the body centre. What is the formula of the compound? What are the coordinationsof X and Y? (CBSE sample paper 2008) (Ans. : XY, 8 each)

7. When atoms are placed at the corners of all 12 edges of a cube. How many atoms are presentper unit cell? (Ans. : 1)

8. A unit cell consists of a cube in which there are A atoms at the corners and B atoms at thecentres and A atoms are missing from two corners in each unit cell. What is the simplest formulaof the compound? (PB S.B.E. 2011) (Ans. : AB4)

HIGHER ORDER THINKING SKILLS (HOTS)1. Atoms of element B form hcp lattice and those of the element A occupy 2/3rd of tetrahedral voids.

What is the formula of compound formed by the element A and B (NCERT).Ans. The number of tetrahedral voids formed is twice the number of element B and only 2/3rd of

it is occupied by atoms of element A.

∴ The ratio of atoms of element and B = 2

2 × :13

= 4 : 3

∴ The formula of the compound is A4B3

2. A solid made of two elements P and Q. Atoms Q are in ccp arrangement, while atoms P occupyall the tetrahedral sites. What is the formula of the compound? (PB.S.B.E 2002).

Ans. The ccp lattice is formed by the element Q. The number of octahedral voids generated wouldbe equal to the number of atoms Q. The number of tetrahedral voids is double the number ofoctahedral voids.

∴∴∴∴∴ The ratio of P : Q = 2 : 1∴∴∴∴∴ The formula of compound = P2Q

3. How many unit cells are present in a cube-shaped ideal crystal of NaCl of mass 1.00 g?(a) 2.57 × 1021 (b) 5.14 × 1021 (c) 1.28 × 1021 (d) 1.71 × 1021

Ans. The correct answer is (a)

Amount of NaC =1.00g

58.5g/mol

No. of unit cells in 1.0 g of NaCl =23 –16.02 × 10 mol 1.00g

×4 58.5g/mol

= 2.57 × 1021

70


4. How many Cs+ ions occupy second nearest neighbour locations of a Cs+ ion in the structureof cesium chloride crystals?

Ans. The cesium chloride structure, each Cs+ is surrounded by eight Cl– as the nearest neighbour

at a distance of 3

r2

, where r = – –Cl – Cld

The structure shows that each Cs+ finds six Cs+ as the next nearest neighbours at a distance

of r, where r = – –Cl – Cld

5. The compound Y Ba2Cu3O7, which shows superconductivity, has copper in oxidation state .......Assume that the rare earth element yttrium is in its usual + 3 oxidation state.

Ans. The oxidation state of Y = +3Barium exists as Ba2+ and oxygen as O2– (as in oxides) if x is the charge on copper atom, then(+3) + 2 × (+2) + 3 × x + 7 × (–2) = 0

This gives, x = 7

3

Thus, the oxidation state of copper in the given compound is 7

3

2Chapter

SYLLABUS2.1 SOLUTIONS

(a) Various terms involved in solutions(b) Types of solutions(c) Ways of expressing concentrations of

solutions(d) Solubility of solute in solvent(e) Henry’s law(f) Solid solutions

Questions and AnswersTheoretical MCQsNumerical MCQsAdvanced MCQs

2.2 COLLIGATIVE PROPERTIES ANDLOWERING OF VAPOUR PRESSURE

(a) Colligative properties(b) Vapour pressure of liquids(c) Factors which affect vapour pressure(d) Lowering of vapour pressure(e) Relative lowering of vapour pressure(f) Raoult’s law

(g) Raoult’s law for a solution of non-volatilesolute in a volatile solvent

(h) Expression for relative lowering of vapour

pressure / Expression for lowering of vapourpressure

(i) Expression for molar mass of a solute andrelative lowering of vapour pressureQuestions and AnswersTheoretical MCQsNumerical MCQsAdvanced MCQs

2.3 ELEVATION OF BOILING POINT ANDDEPRESSION OF FREEZING POINT

(a) (1) Boiling point(2) Elevation in boiling point

(b) Relation between elevation in boiling pointand lowering of vapour pressure

(c) Molal elevation constant or Ebullioscopicconstant

(d) Relation between elevation of boiling pointand molar mass of the solute

(e) (1) Freezing point(2) Depression in freezing point(3) Cause for depression of freezing point

(f) Depression of freezing point and loweringof vapour pressure graphically

Solutions and

Colligative Properties"Imagination is more important than knowledge."-Einstein

Introduction :The Study of Chemistry is incomplete without the study of solutions. Solutions are mixturesof two or more components. Depending on sizes of the components, the mixtures are classifiedinto three types :1. A coarse mixture 2. A colloidal dispersion 3. A true solution

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(g) Molal depression constant or Cryoscopicconstant

(h) Relation between molecular mass of soluteand depression in freezing point

(i) Appplications of depression of freezing pointQuestions and AnswersTheoretical MCQsNumerical MCQsAdvanced MCQs

2.4 OSMOTIC PRESSURE, ABNORMALMOLECULAR MASS AND VAN’T HOFFFACTOR

(a) Semi permeable membrane(b) Osmosis(c) Abbe Nollet experiment(d) Types of osmosis and osmotic pressure

(1) Osmotic pressure(2) Isotonic solutions(3) Hypertonic solutions(4) Hypotonic solutions

(e) Osmosis in day to day life

(f) Laws of osmotic pressure(1) van’t Hoff - Boyle law(2) van’t Hoff - Charles law(3) van’t Hoff - Avogadro’s law(4) van’t Hoff’s equation for dilute solutions

(g) Determination of molecular mass fromosmotic pressure

(h) Reverse Osmosis(i) Abnormal molecular mass(j) van’t Hoff factor

(k) van’t Hoff factor (i) and degree ofdissociation (α)Questions and AnswersTheoretical MCQsNumerical MCQsAdvanced MCQs

HOURS BEFORE EXAM

NUMERICALS WITH SOLUTION


HIGHER ORDER THINKING SKILLS

1. A coarse mixture : It is formed when the sizes of the constituent components are relatively

bigger, e.g. mixture of salt and sugar.2. A colloidal dispersion : It is formed when the sizes of the particles dispersed in solvent

are in the range of 10-7 cm to 10-4 cm. Colloidal particles carry positive or negative charge

which stabilizes colloidal dispersion e.g. ferric hydroxide sol, arsenic sulphide sol, etc. Colloidalsolutions are heterogeneous and can be easily separated.

3. A true solution : A true solution is formed when soluble substances are dissolved in the solvent.

The sizes of the particles dissolved are very small of the order of 10-8 cm. True solutions arehomogeneous and cannot be separated into components by simple mechanical methods.

The extent to which solute dissolves in solvent to form homogeneous solution depends on nature

of solute and solvent, because the solute particles are very small. The components of truesolutions cannot be separated by simple physical methods like centrifugation, filtration etc.

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Chapter - 2 Solutions and Colligative Properties

2.1 SOLUTIONS : Concept Explanation :

(a) Various terms involved in solutions :1. Solution : A true solution is defined as a homogeneous mixture of two or more substances,

the composition of which is not fixed and may be varied within certain limits.2. Solute : The component that constitutes smaller part of solution is called solute.3. Solvent : The component that constitutes larger part of the solution is called solvent.4. Solvation : The process of solvation is the interaction of solvent molecules with solute

particles to form aggregates.5. Hydration or Aquation : When water is a solvent then the process of solvation is called

hydration or Aquation.6. Binary solution : Solution with two components in it are called binary solutions.

Similarly for ternary and quaternary solutions.7. Aqueous solutions : Solutions in which water is used as a solvent are called aqueous solutions.8. Non – Aqueous solutions : Solutions in which solvent used is other than water are called

non – aqueous solutions.9. Concentration of Solution : It is defined as the amount of solute dissolved in a specific

amount of solvent.10. Dilute solutions : Solutions containing relatively less amount of solute are called dilute solutions.11. Concentrated solutions : Solutions containing relatively more amount of solute are called

concentrated solutions.

(b) Types of solutions :Solute Solvent Common Examples

Gaseous SolutionsGas Gas Mixture of non reacting gases, air.Liquid Gas Chloroform vapours mixed with nitrogen gas, water

vapour in air (humidity).Solid Gas I2 in air, dust or smoke particles in air.Liquid SolutionsGas Liquid Oxygen dissolved in water, CO2 dissolved in water.Liquid Liquid Ethanol dissolved in water.Solid Liquid Sucrose or salt in water, benzoic acid in benzene.Solid SolutionsGas Solid Solution of hydrogen in palladium, pumice stone

(phenomenon of adsorption of gases over metals).Liquid Solid Mercury with sodium (amalgam) and crystalline

salts (CuSO4;5H2O).Solid Solid Copper dissolved in gold, alloys like, brass , bronze.

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(c) Ways of expressing concentrations of solutions :1. Percentage by mass (W/W) :

The mass of solute in gram dissolved in solvent to form 100 gram of solution is calledpercentage by mass.

Mass percentage (% by weight) = mass of solute

× 100mass of solution

2. Percentage by volume (V/V) :It is defined as the ratio of number of parts by volume of the solute to one hundred partsby volume of the solution.

Percentage by Volume of solute = volume of solute

× 100volume of solution

Note : Volume is temperature dependent quantity, hence, percentage by volumechanges with change of temperature.

3. Mole fraction (x) :The mole fraction of any component of a solution is defined as the ratio of number ofmoles of that component present in the solution to the total number of moles of all componentsof the solution.Let us suppose that a solution contains n2 moles of solute and n1 moles of the solvent.Then,

Mole fraction of solute (x2) = 2

1 2

n

n + n

Mole fraction of solvent (x1) = 1

1 2

n

n + n

x1 + x2 = 1 (For binary solution containing 1 solute and 1 solvent)

Note :(1) It may be noted that the mole fraction is independent of temperature.(2) It does not have any units.(3) The sum of mole fractions of all the constituents of a solution is unity.

4. Molarity (M) :It is defined as the number of moles of solute present in 1 dm3 volume of the solution.

Molarity (M) = 3

Number of moles of solute

Volume of solution in litre or dm

No. of moles of a substance (n) = mass of the substance

Molar mass of the substance

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Note :(1) Molarity has one disadvantage. It changes with temperature because of

expansion or contraction of the liquid with temperature.(2) Units of Molarity : mol dm–3 OR Mol lit–1

(3) Molar = 1M, Decimolar = 0.1M , CentiMolar = 0.01M, Millimolar = 0.001M

5. Molality (m) :(a) It is defined as the number of moles of solute dissolved in 1 kg of solvent.

Molality = No. of moles of solute in kg

Mass of solvent in kg

(b) Units of Molality : Mol / Kg.(c) Molality is considered better as compared to molarity because the molarity changes

with temperature because of expansion or contraction of the liquid with temperature.However, molality does not change with temperature because mass of the solventdoes not change with change in temperature.

6. Parts per million :It is the mass or volume of solute in gram or cm3 per 106 g or 106cm3 of the solution.

Parts per Million (ppm) = 6Massor volumeof Solute

× 10TotalMassor volumeof solution

(d) Solubility of solute in solvent :1. Solubility : The maximum amount of solute that dissolves completely in a given amount

of solvent at a constant temperature is called solubility.It is expressed in terms of mol dm-3. Solubility changes with temperature.

2. Saturated solution : A solution in which no more solute can be dissolved at the giventemperature and pressure is called as saturated solution.In a saturated solution an equilibrium exists between dissolution and crystallization.

3. Unsaturated solution : It contains less amount of solute than required for forming saturatedsolution.

4. Super saturated solution : It contains excess of solute than required for formation ofsaturated solution.

5. Factors affecting the solubility of a solid in a liquid :(a) Nature of solute and the solvent : “Like dissolves like” is the general rule followed.

Polar solutes are soluble in polar solvents. E.g.: NaCl in water.Non polar solutes are more soluble in non-polar solvents. Eg: Iodine in CCl4.

(b) Temperature : The solubility of a solid in liquid is affected by temperature.

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Generally solubility increases with increase in temperature if the dissolution processis endothermic. If the dissolution process is exothermic then solubility increases withdecrease in temperature. However, this is not always true.

(c) Pressure : Solids are incompressible hence, change in pressure has no effects onsolubility of solids in liquids.

6. Factors affecting solubility of a gases in liquids :(a) Nature of solute and solvent : The principle of “like dissolves like ” is also applicable

here. Polar gases like HCl and NH3 are highly soluble in polar solvent like water.While O2, H2 and N2 being non polar is less soluble in water.

(b) Effect of temperature : According to Charles law, volume of a given mass of agas increases with increase in temperature.Therefore, the volume of a given mass of dissolved gas in a solution also increaseswith increase in temperature. So that it becomes impossible for the solvent toaccommodate gaseous solute in it and gas bubbles out.Hence, solubility of gas in liquid decreases with increase in temperature.

(c) Effect of pressure : Gases are highly compressible in nature.Hence, solubility of gases is greatly influenced by change of external pressure of thegas. Increase of external pressure increases the solubility of gas.In case ‘b’ when pressure of the gas is increased, its solubility also increases.

(d) Effect of addition of soluble salt : Solubility of dissolved gas is suppressed whena soluble salt is added to the solution of gas. Eg : When table salt is dissolved in carbonateddrink, the solubility of CO2 in soft-drink decreases and CO2 escapes out producingeffervescence.

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(e) Henry’s law :1. The law states that “the solubility of a gas in a liquid at constant temperature is proportional

to the pressure of the gas above the solution”.If ‘S’ is the solubility of the gas in mol dm-3 and ‘P’ is the pressure of the gas in atmosphere,then according to Henry’s lawS ∝ PS = KH × P(KH = proportionality constant called as Henry’s constant and has unit mol dm–3 atm–1)

2. Henry’s Constant : It is defined as solubility of gas in mol dm–3 at 1 atmosphere pressureat reference temperature. Henry’s constant depends upon the nature of the gas, the natureof the solvent and the temperature.

3. Marine life like fish prefer to stay at lower level in the sea.(a) In summer during a hot day, temperature at the surface of water is relatively high

and therefore, the solubility of oxygen in the upper layer is minimum.(b) At the same time the temperature of water at lower level is much less.(c) Water at lower level therefore, has more dissolved oxygen.(d) Hence, marine life prefer to stay at lower level in the sea.

(f) Solid solutions :1. A solid solution of two or more metals or of a metal or metals with one or more non

metals is called an alloy or solid solutions.2. Duralumin is an alloy of mainly aluminium, copper, magnesium and manganese. It is light;

strong as steel and is used in aircraft construction.3. Lead hardened by 10–20% antimony is used in bearings, bullets etc. It is resistant to acids

and is therefore, used for manufacturing of lead storage battery plates.4. Alloy of antimony with tin and copper is called Babbitt metal and it is antifriction alloy

used in machine bearings.5. Chromium is used in steel alloys which are very hard and strong. Stainless steel contains

chromium which makes it resistant to corrosion.6. Alloy spiegeleisen containing 5 to 20% manganese in iron and ferromanganeous containing

70 to 80% manganese are used for making very hard steels for rails, safes and heavy machinery.7. Alloy manganin containing 84% Cu, 12%Mn and 4% Ni has almost zero temperature coefficient

of electrical resistance and used in instruments used for making electrical measurements.8. Alloys of mercury with other metals are called amalgams. This property of mercury to

form amalgams is used for extracting metals from ores.

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Question and Answers• Answer in short :

*1. Define the terms : (a) Solution, (b) Solute, (c) Solvent, (d) Concentration.Ans. (a) Solution : A true solution is defined as a homogeneous mixture of two or more substances,

the composition of which is not fixed and may be varied within certain limits.(b) Solute : The component that constitutes smaller part of solution is called solute.(c) Solvent : The component that constitutes larger part of the solution is called solvent.(d) Concentration : It is defined as the amount of solute dissolved in a specific amount of solvent.

*2. What are the different types of solutions?

Ans. Solute Solvent Common ExamplesGaseous SolutionsGas Gas Mixture of non reacting gases, air.Liquid Gas Chloroform vapours mixed with nitrogen gas, water

vapour in air (humidity).Solid Gas I2 in air, dust or smoke particle in air.Liquid SolutionsGas Liquid Oxygen dissolved in water, CO2 dissolved in water.Liquid Liquid Ethanol dissolved in water.Solid Liquid Sucrose or salt in water, benzoic acid in benzene.Solid SolutionsGas Solid Solution of hydrogen in palladium, pumic stone

(phenomenon of adsorption of gases over metals).Liquid Solid Mercury with sodium (amalgams) and crystalline

salts (CuSO4. 5H2O).Solid Solid Copper dissolved in gold (alloys), brass, bronze.

*3. Explain the different ways of expressing concentration of a solution.Ans. Refer 2.1(c) 1, 2, 3, 4, 5, 6.

*4. Define : (a) Percentage by weight, (b) Mole fraction, (c) Molality, (d) Molarity.Ans. (a) Percentage by mass (W/W) : The mass of solute in gram dissolved in solvent to form

100 gram of solution is called percentage by mass.

Mass percentage (% by weight) = mass of solute


(b) Mole fraction : The mole fraction of any component of a solution is defined as the ratioof number of moles of that component present in the solution to the total number of molesof all components of the solution.

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Let us suppose that a solution contains n2 moles of solute and n1 moles of the solvent.Then,

Mole fraction of solute (x2) = 2

1 2

n

n + n

Mole fraction of solvent (x1) = 1

1 2

n

n + nx1 + x2 = 1 (For binary solution containing 1 solute and 1 solvent)

(c) Molality (m) :(1) It is defined as the number of moles of solute dissolved in 1 kg of solvent.

Molality = Number of moles of the solute in kg


(2) Units of Molality : Mol / Kg .(3) Molality is considered better as compared to molarity because the molarity changes

with temperature because of expansion or contraction of the liquid with temperature.However, molality does not change with temperature because mass of the solventdoes not change with change in temperature.

(d) Molarity (M) : It is defined as the number of moles of solute present in 1 dm3 volumeof the solution.

Molarity = 3

Number of moles of solute

Volume of solution in litre or dm

No. of moles of a substance (n) = mass of the substance

Molar mass of the substance

*5. Explain why concentration in terms of molality is preferred in comparison to molarity.

Ans. Molality is considered better as compared to molarity because the molarity changes with

temperature because of expansion or contraction of the liquid with temperature. However,

molality does not change with temperature because mass of the solvent does not change with

change in temperature.

6. Define solubility and explain the factors affecting solubility of a solid in a liquid.

Ans. (a) Solubility : The maximum amount of solute that dissolves completely in a given amount

of solvent at a constant temperature is called solubility.

Solubility is expressed in grams of solute per 100cm3 of solvent.

It is also expressed in terms of mol dm-3. Solubility changes with temperature.(b) Factors affecting the solubility of a solid in a liquid :

(1) Nature of solute and the solvent : “Like dissolves like” is the general rule followed.Polar solutes are soluble in polar solvents. E.g.: NaCl in water.

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Non polar solutes are more soluble in non-polar solvents. Eg: Iodine in CCl4.(2) Temperature : The solubility of a solid in liquid is affected by temperature.

Generally solubility increases with increase in temperature if the dissolution processis endothermic. If the dissolution process is exothermic then solubility increases withdecrease in temperature. However, this is not always true.

(3) Pressure : Solids are incompressible hence, change in pressure has no effects onsolubility of solids in liquids.

7. Define solubility and explain the factors affecting solubility of a gas in a liquid.Ans. (a) Solubility : The maximum amount of solute that dissolves completely in a given amount

of solvent at a constant temperature is called solubility.Solubility is expressed in grams of solute per 100 cm3 of solvent.It is also expressed in terms of mol dm-3. Solubility changes with temperature.

(b) Factors affecting solubility of a gases in liquids :(1) Nature of solute and solvent : The principle of “like dissolves like ” is also applicable

here. Polar gases like HCl and NH3 are highly soluble in polar solvent like water.While O2, H2 and N2, being non polar is less soluble in water.

(2) Effect of temperature : According to Charles law, volume of a given mass of agas increases with increase in temperature.Therefore, the volume of a given mass of dissolved gas also increases with increasein temperature. So that it becomes impossible for the solvent to accommodate gaseoussolute in it and gas bubbles out.Hence, solubility of gas in liquid decreases with increase in temperature.

(3) Effect of pressure : Gases are highly compressible in nature.Hence, solubility of gases is greatly influenced by change of external pressure of thegas. Increase of external pressure increases the solubility of gas.In case ‘b’ when pressure of the gas is increased, its solubility also increases.

(4) Effect of addition of soluble salt : Solubility of dissolved gas is suppressed whena soluble salt is added to the solution of gas. Eg : When table salt is dissolved in carbonateddrink, the solubility of CO2 in soft-drink decreases and CO2 escapes out producingeffervescence. [Refer fig. 2.1 d (6)]

*8. State and explain Henry’s law.Ans. The law states that “the solubility of a gas in a liquid at constant temperature is proportional

to the pressure of the gas above the solution”.If ‘S’ is the solubility of the gas in mol dm-3 and ‘P’ is the pressure of the gas in atmosphere,then according to Henry’s lawS ∝ PS = KH × P(KH = proportionality constant called as Henry’s constant and has unit mol dm–3 atm–1)

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9. Define Henry’s constant.Ans. It is defined as solubility of gas in mol dm–3 at 1 atmosphere pressure at reference temperature.

Henry’s constant depends upon the nature of the gas, the nature of the solvent and the temperature.10. Give reason : Marine life like fish prefer to stay at lower level in the sea.

Ans. Marine life like fish prefer to stay at lower level in the sea, because :(a) In summer during a hot day, temperature at the surface of water is relatively high and

therefore, the solubility of oxygen in the upper layer in minimum.(b) At the same time the temperature of water at lower level is much less.(c) Water at lower level therefore, has more dissolved oxygen.

*11. Explain solid solution.Ans. Refer 2.1 (f)


*1. A molal solution is one that contains one mole of a solute in .......a) 1 L of solvent b) 1000g of solventc) 1 L of solution d) 22.4 litres of solution

*2. Which of the following is independent of temperature? (MHT-CET 2008)a) Normality b) Molarity c) Molality d) Formality

*3. Molarity of solution depends on .......a) Temperature b) Nature of solute dissolvedc) Mass of solvent d) Pressure

• Numerical MCQs

*4. 20g of NaOH (Molar mass = 40gmol–1)is dissolved in 500cm3 of water. Molalityof resulting solution is .......a) 0.1m b) 0.5mc) 1.5m d) 1.0m

Ans. Molality =moles of NaOH


=20

40 × 0.5kg

= 1m

5. A solution is 0.25% by weight. The weightof solvent containing 1.25g of solute wouldbe .......a) 500g b) 498.75gc) 500.25g d) 501.25g

Ans. % by weight =mass of solute


0.25 =1.25


mass of solution = 500g∴∴∴∴∴ mass of solvent = 500 – 1.25 = 498.75g

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6. Calculate the mole fraction of glucose insolution (molecular weight = 180g) when3.6g of glucose is added to 90ml water at25ºC to produce glucose solution .......a) 4.0 b) 0.4c) 0.04 d) 0.004

Ans. mole fraction of glucose

=moles of glucose

moles of glucose + moles of water

=3.6 /180

3.6 /180 + 90/18

=3.6

903.6 = 0.004

7. A centi-molar solution is diluted 10 times.Its molarity would become .......a) 0.01 b) 0.1c) 0.001 d) 0.005

Ans. Resulting molarity =original molarity

dilution

=0.01

10 = 0.001M

8. An aqueous solution of non-electrolyte ‘A’with molecular mass 60 contains 6g in500mL and has a density equal to 1.05gmL–1. The molality of solution is .......a) 1.25 b) 0.19c) 0.25 d) 0.30

Ans. Mass of solution = density × volume= 1.05 × 500= 525g

∴ Mass of solvent = 525 – 6 = 519g= 0.519kg

Molality (m) =moles of solute

mass of solvent in kg

=6

60 × 0.519 = 0.19m

9. 100ml of 0.1M solution A is mixed with20ml of 0.2M solution B. The final molarityof the solution is .......a) 0.12M b) 0.15Mc) 0.18M d) 0.21M

Ans. M =1 1 2 2

1 2

M V × M V

V + V

=0.1 × 100 + 0.2 × 20

100 + 20

= 0.12M

10. The amount of solute required to prepare10L of a decimolar solution is .......a) 0.01mol b) 0.2molc) 0.1mol d) 1mol

Ans. M =n

v , 0.1 =

n

10∴ n = 1 mol

11. What is molality of the solution preparedby dissolving 18g of glucose.(mol. mass – 180) in 500g of water?a) 1.2m b) 0.4mc) 0.1m d) 0.2m

Ans. m =1

n

W (kg)

= –3

18

180 × 500 × 10 (kg)= 0.2m

12. Ten grams of potassium chloride aredissolved in 103 kg of solution, its strengthmay be expressed as .......a) 1 ppm b) 10 ppmc) 100 ppm d) 1000 ppm

Ans. ppm =6mass of solute

×10mass of solution

83


=–3

63

10 × 10 (kg)× 10

10

= 10 ppm

13. A sugar syrup of weight 214.2 contains34.2g of sugar (molar mass – 342). Themolality of the solution is .......a) 0.0099 b) 0.56c) 0.28 d) 0.34

Ans. m =1

n

W (kg)

= –3

34.2

342 × (214.2 – 34.2) × 10 kg

= –3

1

10 ×180 × 10

=1

1.8 = 0.56m

14. The mole fraction of the solute in one molalaqueous solution is .......a) 0.027 b) 0.06c) 0.018 d) 0.009

Ans. Let the mass of H2O be 1kg

m =2

1

n

W (kg)

1 = 2n

1 , n2 = 1 mole

=1000

18 = 55.55 mol

x2 =2

1 2

n

n + n

x2 = 1

1 + 55.55

=1

0.01856.55

≈

• Advanced MCQs15. The molecular weight of KOH is 56. What

is the molarity of solution prepared bydissolving 84.0 gram of KOH in 500ml ofsolution? (Sept. 2009)a) 3 b) 5c) 2 d) 2.5

Ans. M = 2n

V

= –3

84

56 × 500 × 10 = 3M

16. If 100ml of 1M NaOH solution is dilutedto 1 dm3, the molarity of the resultingsolution is ........ (March 2010)a) 1M b) 0.1Mc) 10M d) 0.05M

Ans. M1V1 = M2V2

1 × 100 = M2 × 1000

∴∴∴∴∴ M2 = 1

10 = 0.1M

17. Number of moles present in 6dm3 ofsolution to form centimolar solution is .......

(MHT-CET 2008)a) 6 × 10–2 b) 6 × 10–3

c) 6 × 10–1 d) 6 × 10–4

Ans. M = 2n

V

0.01 = 2n

6∴∴∴∴∴ n2 = 0.06 mole

= 6 × 10–2

18. If molality of NaOH (molar mass = 40)solution is 1.25m, then calculate W/W%.

(MHT-CET 2009)a) 2.35% b) 1.18%c) 4.76% d) 9.52%

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Ans. m = 2

1

n

W (kg)Let the mass of solvent be 1 kg

1.25 =2W

40 × 1∴∴∴∴∴ W2 = 40 × 1.25 × 1 = 50g

% W/W =massof solute×100

mass of solute + massof solution

=50

× 10050 + 1000

=50

× 1001050

= 4.76%

19. 0.1dm3 of 2M acidic solution is dilutedsuch that the concentration becomes 0.1M,then the final volume of the solution.

(MHT-CET 2009)a) 1 dm3 b) 10 dm3

c) 2 dm3 d) 20 dm3

Ans. M1V1 = M2V2

2 × 0.1 = 0.1 × V2

∴∴∴∴∴ V2 = 2 dm3

20. 45g of glucose (Molar mass = 180) isdissolved in 100g of water. What is molalityof the solution? (MHT-CET 2010)a) 0.25m b) 0.025mc) 2.5m d) 0.0025m

Ans. m =2

1

n

W (kg)

= –3

45

180 × 100 × 10

= 2.5m

21. 25cm3 of 0.2M solution of H2SO4 is dilutedto 0.5 dm3, the resulting solution is .......

(MHT-CET2010)a) 0.01M b) 0.001Mc) 0.2M d) 0.002M

Ans. M1V1 = M2V2

0.2 × 25 × 10–3 (dm3)= M2 × 0.5 (dm3)

∴ M2 =–30.2 × 25 × 10

0.5 = 0.01 M

2.2 COLLIGATIVE PROPERTIES AND LOWERING OF VAPOURPRESSURE :


(a) Colligative properties :1. “Colligative properties are defined as the properties of solutions that depend only on the

number of solute particles in solution and not on the nature of solute particles.”2. Examples :

(a) Lowering of vapour pressure(b) Elevation of boiling point of solvent in solution(c) Depression of freezing point of solvent in solution(d) Osmotic pressure.

3. Explanation :(a) All colligative properties are connected together because these properties are based

on the same principle.

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(b) The properties depend upon number of solute particles, the particles may be atoms,molecules, ions or aggregates of molecules.

(c) The actual number of solute particles present in solution decides the value of colligativeproperties.

(d) It is convenient to restrict the discussion to solutions of non electrolytes in liquid solvent.(e) The non electrolytes that neither dissociate nor associate in solution.(f) Colligative properties can be used to determine molar masses of non electrolyte solutes.(g) Colligative properties are applicable only to dilute solutions with concentrations less

than or equal to 0.2M.

(b) Vapour pressure of liquids :1. The vapour pressure of a

substance is defined as thepressure exerted by the gaseousstate of that substance when itis in equilibrium with the solidof liquid phase.

2. At room temperature, themolecules of liquid are inconstant random motion.

3. The molecules at the surface of the liquid escape from the surface by absorbing energyfrom the surrounding.This process is called evaporation. If the vessel containing the liquid is closed with a lid,then the molecules in vapour phase are trapped in the empty space and are in continuousrandom motion.

4. During this motion, molecules collide with each other and also with the walls of the container,thereby losing their energy and returning back to liquid state. This process is calledcondensation.

5. Eventually the rate of evaporation and condensation becomes equal i.e. equilibrium is attained.At this point the number of molecules in gaseous phase remains constant. The pressureexerted by these molecules on the surface of liquid is called as vapour pressure of the liquid.

6. The vapour pressure of the liquid depends on the nature of the liquid and the temperature.With increase of intermolecular forces of attraction, vapour pressure of liquid decreasesand with rise of temperature vapour pressure of liquid increases.

(c) Factors which affect vapour pressure :1. Nature of solvent :

(a) All liquids exhibit tendency for evaporation.(b) If the intermolecular forces of attraction are weak the liquids evaporate readily and

86


are called volatile liquids. Eg: Ethyl acetate is the most volatile liquid. Other Egs. Ethylalcohol, Acetone etc.

(c) If intermolecular forces are strong, the liquids are less volatile and hence, exert lowervapour pressure. Eg: Oil, mercury etc.

2. Nature of solute :(a) Non-volatile solute : The vapour pressure of a solvent is lowered when a non–volatile

solute is dissolved to form a solution. This is due to the fact that when a non–volatilesolute is added, the surface area available for evaporation of volatile solvent is decreased.Eg: Sugar in water.

(b) Volatile solute : In case of volatile solute, the vapour pressure of the resulting solutionis generally increased. Eg: Ethyl alcohol mixed with water.

3. Temperature :(a) As temperature increases, vapour pressure of a

liquid increases.(b) The temperature at which the vapour pressure of

the liquid becomes equal to the atmospheric pressure,the liquid starts to boil. It is boiling point of the liquid( 0

bT ).

(d) Lowering of vapour pressure :The difference between the vapourpressure of pure solvent ( 0

1p ) and thevapour pressure of solvent fromsolution (p) containing a non-volatilesolute is called the lowering of vapour

pressure. Thus, 01p – p = Δp =

Lowering of vapour pressure.It is found that, the vapour pressureof the solution (p) containing a nonvolatile solute is always less than that

of the pure solvent 01p , i.e. 0

1p < p .

(e) Relative lowering of vapour pressure :The relative lowering of vapour pressure of a solution is the ratio of the lowering of vapourpressure of solvent from solution to the vapour pressure of the pure solvent.

Thus, Relative lowering of vapour pressure = 01

01

p – p

p

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(f) Raoult’s law :The law states that, the partial vapour pressure of any volatile component of a solutionis the product of vapour pressure of that pure component and the mole fraction of thecomponent in the solution. Consider a solution containing two volatile components A1 andA2 with mole fractions x1 and x2 respectively Let 0

1p and 02p be the vapour pressures

of the pure components A1 and A2 respectively.According to Raoult’s law, the partial vapour pressures p1 and p2 of two components aregiven by,

p1 = 01 1p × x and p2 = 0

2 2p × x

pT = p1 + p2

pT = 0 01 1 2 2p × + p ×x x

Note : The solution which obeys Raoult’s law over the entire range of concentration

is called ideal solution. If the solution does not obey Raoult’s law then the

solution is non–ideal.

(g) Raoult’s law for a solution of non-volatile solute in a volatile solventRaoult’s law :For a solution containing a non-volatile solute in the volatile solvent, the law states that,the vapour pressure of a solution is equal to the product of vapour pressure of the puresolvent ( 0

1p ) and mole fraction of the solvent (x1).Explanation : Consider a binary solution of a non-volatile solute in a volatile solvent.Let, 0

1p = Vapour pressure of pure solvent.p = vapour pressure of solutionx1 = Mole fraction of the solvent,

∴∴∴∴∴ p ∝ x1 OR p = 01 1p × x

The above graph illustrates that as mole fraction ofsolvent (x1) in solution increases the vapour pressureof the solvent in solution (p) also increases.

(h) Expression for relative lowering of vapour pressure ORExpression for lowering of vapour pressure :According to Raoult’s law, vapour pressure of the solution is equal to the product ofvapour pressure of the pure solvent and its mole fraction.

1. Since p = 01 1p × x Where, 0

1p = vapour pressure of pure solvent.

p = vapour pressure of solutionx1 = Mole fraction of the solvent,

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2. Lowering of vapour pressure = Δp = 01p – p

3. ∴∴∴∴∴ Δp = 0 01 1 1p – p × x

4. ∴∴∴∴∴ Δp = 01 1p (1 – )x

5. ∴∴∴∴∴ Δp = 01 2p × x (since, x1 + x2 = 1, therefore, 1 – x1 = x2 )

Hence, lowering of vapour pressure is the product of vapour pressure of pure solvent

( 01p ) and mole fraction of solute (x2).

6. From (5) we get, 01

p

p = x2

7. ∴∴∴∴∴01

01

p – p

p = x2

Hence, the relative lowering of vapour pressure is equal to the mole fraction of solute.which proves that it is a colligative property.

(i) Expression for molar mass of a solute and relative lowering of vapourpressure :

1. Relative lowering of vapour pressure is given by the formula,

01

01

p – p

p = x2 .........(i)

2. Let W2 g of solute of molar mass M2 be dissolved in W1 g of solvent of molar mass M1.Hence, number of moles of solvent, n1 and number of moles of solute, n2 in solution are

given as, n1 = 1

1

W

M and n2 = 2

2

W

M

3. The mole fraction of solute x2 is given by

x2 = 2

1 2

n

n + n = 2 2

1 1 2 2

W / M

W / M + W / M ........(ii)

4. Combining (i) and (ii). We get;01

01

p – p

p = x2 = 2 2

1 1 2 2

W / M

W / M + W / M .........(iii)

5. The relation (iii) is applicable only for dilute solutions. Hence, n1 >> n2 . Hence, n2 maybe neglected in comparison with n1 in the above equation. Thus, eqn. (iii) can be writtenas follows :

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01

01

p – p

p =

2 2

1 1

W / M

W / M

6. Therefore, 01

01

p – p

p =

2 1

1 2

W M

W M

Questions and AnswersAnswer in short :

*1. Define colligative properties. Give examples.Ans. Colligative properties are defined as the properties of solutions that depend only on the number

of solute particles in solution and not on the nature of solute particles.Examples : a) Lowering of vapour pressure;

b) Elevation of boiling point of solvent in solution; c) Depression of freezing point of solvent in solution;

d) Osmotic pressure.

*2. What is a vapour pressure of liquid?Ans. (a) The vapour pressure of a substance is defined as the pressure exerted by the gaseous

state of that substance when it is in equilibrium with the solid of liquid phase.(b) At room temperature, the molecules of liquid are in constant random motion.(c) The molecules at the surface of the liquid escape from the surface by absorbing energy

from the surrounding.This process is called evaporation. If the vessel containing the liquid is closed with a lid,then the molecules in vapour phase are tapped in the empty space and are in continuousrandom motion.

(d) During this motion, molecules collide with each other and also with the walls of the container,thereby losing their energy and returning back to liquid state. This process is calledcondensation.

(e) Eventually the rate of evaporation and condensation becomes equal i.e. equilibrium is attained.At this point the number of molecules in gaseous phase remains constant. The pressureexerted by these molecules on the surface of liquid is called as vapour pressure of theliquid.

(f) The vapour pressure of the liquid depends on the nature of the liquid and the temperature.With increase of intermolecular forces of attraction, vapour pressure of liquid decreasesand with rise of temperature vapour pressure of liquid increases. [Refer Fig. 2.2 (b)]

*3. Explain the effect of temperature on the vapour pressure of a liquid.Ans. The effect of temperature on the vapour pressure of a liquid.

(a) As temperature increases, vapour pressure of a liquid increases.

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(b) The temperature at which the vapour pressure of the liquid becomes equal to the atmosphericpressure, the liquid starts to boil. It is boiling point of the liquid ( 0

bT ). [Refer Fig. 2.2 (c) (3)]

*4. What are the factors which affect vapour pressure?Ans. (a) Nature of solvent :

(1) All liquids exhibit tendency for evaporation.(2) If the intermolecular forces of attraction are weak the liquids evaporate readily and

are called volatile liquids. Eg: Ethyl acetate is the most volatile liquid. Other Examples.Ethyl alcohol, Acetone etc.

(3) If intermolecular forces are strong, the liquids are less volatile and hence exert lowervapour pressure. Eg: Oil, mercury etc.

(b) Nature of solute :(1) Non-volatile solute : The vapour pressure of a solvent is lowered when a non–volatile

solute is dissolved to form a solution. This is due to the fact that when a non–volatilesolute is added, the surface area available for evaporation of volatile solvent is decreased.Eg: Sugar in water.

(2) Volatile solute : In case of volatile solute, the vapour pressure of the resulting solutionis generally increased. Eg: Ethyl alcohol mixed with water.

(c) Temperature :(1) As temperature increases, vapour pressure of a liquid increases.(2) The temperature at which the vapour pressure of the liquid becomes equal to the

atmospheric pressure, the liquid starts to boil. It is boiling point of the liquid ( 0bT ).

[Refer Fig. 2.2 (c) (3)]

*5. What is lowering of vapour pressure of a solution?Ans. The difference between the vapour pressure of pure solvent ( 0

1p ) and the vapour pressureof solvent from solution (p) containing a non-volatile solute is called the lowering of vapourpressure. Thus, 0

1p – p = ∆p = Lowering of vapour pressure.It is found that, the vapour pressure of the solution (p) containing a non-volatile solute is alwaysless than that of the pure solvent 0

1p , i.e. 01p < p . [Refer Fig.. 2.2 (d)]

*6. What is relative lowering of vapour pressure?Ans. The relative lowering of vapour pressure of a solution is the ratio of the lowering of vapour

pressure of solvent from solution to the vapour pressure of the pure solvent. Thus,

Relative lowering of vapour pressure = 01

01

p – p

p

7. State and explain Raoult’s law of vapour pressure.Ans. The law states that, the partial vapour pressure of any volatile component of a solution is the

product of vapour pressure of that pure component and the mole fraction of the componentin the solution. Consider a solution containing two volatile components A1 and A2 with mole

91


fractions x1 and x2 resp. Let 01p and 0

2p be the vapour pressures of the pure componentsA1 and A2 resp.According to Raoult’s law, the partial vapour pressures p1 and p2 of two components are givenby, p1 = 0

1 1p × x and p2 = 02 2p × x

pT = p1 + p2 = 0 01 1 2 2p × + p ×x x

*8. State Raoult’s law and obtain expression for lowering of vapour pressure when non -volatilesolute is dissolved.

Ans. According to Raoult’s law, vapour pressure of the solution is equal to the product of vapourpressure of the pure solvent and its mole fraction

∴∴∴∴∴ p = 01 1p × x Where, 0

1p = vapour pressure of pure solvent.p = vapour pressure of solutionx1 = Mole fraction of the solvent,

Lowering of vapour pressure = Δp = 01p – p

∴∴∴∴∴ Δp = 0 01 1 1p – p × x

∴∴∴∴∴ Δp = 01 1p (1 – )x

∴∴∴∴∴ Δp = 01 2p × x (since, x1 + x2 = 1, therefore, 1 – x1 = x2 ) ...(i)



From (i) we get, 01

p

p

Δ = x2

∴∴∴∴∴01

01

p – p

p = x2

Hence, the relative lowering of vapour pressure is equal to the mole fraction of solute; whichproves that it is a colligative property.

9. Derive an expression for relative lowering of vapour pressure.ORShow that the relative lowering in vapour pressure of an ideal solution is equal to mole fractionof the solute.

Ans. According to Raoult’s law, vapour pressure of the solution is equal to the product of vapourpressure of the pure solvent and its mole fraction

∴∴∴∴∴ p = 01 1p × x Where, 0

1p = vapour pressure of pure solvent.

p = vapour pressure of solutionx1 = Mole fraction of the solvent,

Lowering of vapour pressure = Δp = 01p – p

∴∴∴∴∴ Δp = 0 01 1 1p – p × x

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∴∴∴∴∴ Δp = 01 1p (1 – )x

∴∴∴∴∴ Δp = 01 2p × x (since, x1 + x2 = 1, therefore, 1 – x1 = x2 )



From (v) we get, 01

p

p

Δ = x2 ∴∴∴∴∴

01

01

p – p

p = x2

Hence, the relative lowering of vapour pressure is equal to the mole fraction of solute thisproves that it is a colligative property.

*10. Derive the relationship between relative lowering of vapour pressure and molar mass of solute.Ans. Relative lowering of vapour pressure is given by the formula,

01

01

p – p

p = x2 ...(i)

Let W2 g of solute of molar mass M2 be dissolved in W1 g of solvent of molar mass M1.Hence, number of moles of solvent, n1 and number of moles of solute, n2 in solution are given

as, n1 = 1

1

W

M and n2 = 2

2

W

MThe mole fraction of solute, x2 is given by

x2 = 2

1 2

n

n + n = 2 2

1 1 2 2

W / M

W / M + W / M ...(ii)

Combining (i) and (ii) We get;

01

01

p – p

p = x2 =

2 2

1 1 2 2

W / M

W / M + W / M ...(iii)

The relation (iii) is applicable only for dilute solutions. Hence, n1 >> n2 . Hence, n2 may beneglected in comparison with n1 in the above equation. Thus, eqn. (iii) can be written as follows

:01

01

p – p

p =

2 2

1 1

W / M

W / M

Therefore, 01

01

p – p

p =

2 1

1 2

W M

W M


• Theoretical MCQs :*1. According to the Raoult’s law, the relative lowering of vapour pressure is equal to the .......

a) Mole fraction of solvent b) Mole fraction of solutec) Independent of mole fraction of solute d) Molality of solution

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*2. Partial pressure of solvent in solution of non-volatile solute is given by equation, .......a) p = x2p

0 a) p0 = xp c) p = x1p0 d) p0 = x1p

*3. When partial pressure of solvent in solution of non-volatile solute is plotted against its molefraction, nature of graph is .......a) A straight line passing through originb) A straight line parallel to mole fraction of solventc) A straight line parallel to vapour pressure of solventd) A straight line intersecting vapour pressure axis

Ans. ∴∴∴∴∴ The graph of p v/s x1 will be a straight linepassing through origin.

*4. Lowering of vapour pressure of solution .......a) is a property of soluteb) is a property of solute as well as solventc) is a property of solventd) is a colligative property

*5. Vapour pressure of solution of a non-volatile solute is always .......a) Equal to the vapour pressure of pure solventb) Higher than vapour pressure of pure solventc) Lower than vapour pressure of pure solventd) Constant

*6. Relative vapour pressure lowering depends only on .......a) Mole fraction of solute b) Nature of solventc) Nature of solute d) Nature of solute and solvent

• Numerical MCQs :

7. In a very dilute solution the no. of molesof solvent are 10 times more than that ofsolute. The vapour pressure of the solutionwould be .......(Given vapour pressure of pure solvent =80 mm)a) 800 mm b) 88 mmc) 72 mm d) 720 mm

Ans. p = 01 1p × n

=1

1 2

n80 ×

n + n

=2

2 2

10n80 ×

10n + n

=2

2

10n80 ×

11n ≈ 72 mm

8. The vapour pressure of a pure liquid ‘A’is 70 torr at 27º C. It forms an ideal solutionwith another liquid ‘B’. The mole fractionof B is 0.2 at 27ºC. Find the vapourpressure of pure liquid B if vapour pressureof solution is 84 torr.a) 14 b) 56c) 140 d) 70

Ans. p = 0 0A A B Bp × + p ×x x

84 = 0B70 × 0.8 + p × 0.2

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11. The addition of non volatile solute into the pure solvent (Sept. 2008)a) Increases the vapour pressure of solventb) Decreases the boiling point of solventc) Decreases the freezing point of solventd) Increases the freezing point of solvent

2.3 ELEVATION OF BOILING POINT AND DEPRESSION OFFREEZING POINT :

Concept Explanation :(a) Boiling Point Elevation :1. Boiling Point : The boiling point of a liquid is the temperature at which its vapour pressure

becomes equal to the atmosphere pressure.2. Elevation in Boiling Point (ΔTb) : The difference in the boiling points of the solution

(T) and pure solvent (Tº) is called the elevation in boiling point (ΔTb).∴∴∴∴∴ ΔTb = T – Tº

Note : The vapour pressure of the solution containing non-volatile solute is lessthan that of the solvent. Therefore, the solution has to be heated to a highertemperature so that its vapour pressure becomes equal to the atmosphericpressure. Thus, the boiling point of the solution is always higher than thatof the pure solvent. (T > Tº)

84 – 56 = 0Bp × 0.2

28

0.2= 0

Bp

∴∴∴∴∴ 0Bp = 140

9. The vapour pressure of water at roomtemperature is 23.8 mm Hg. The vapourpressure of an aqueous solution of sucrosewith mole fraction 0.1 is equal to .......a) 2.39 mm Hgb) 2.42 mm Hgc) 21.42 mm Hgd) 21.44 mm Hg

Ans. p = 01 1p × n

= 23.8 × 0.9[n1 = 1 – n2 = 1 – 0.1 = 0.9]

= 21.42 mm Hg

• Advanced MCQs :10. Which is not a colligative property?

(MHT-CET 2009)a) Freezing pointb) Lowering of Vapour pressurec) Osmotic pressured) Elevation in boiling point

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(b) Relation between Elevation in boiling point and lowering of vapourpressure :

1. The boiling point of a liquid is the temperatureat which its vapour pressure becomes equalto the atmospheric pressure.

2. The vapour pressure of the solution containinga non volatile solute is less than that of thepure solvent.

3. Therefore, the solution has to be heated to ahigher temperature so that its vapour pressurebecomes equal to the atmospheric pressure.

4. Thus, the boiling point of the solution is alwayshigher than that of pure solvent.(T > To)

5. The curve AB gives the vapour pressures for the pure solvent and the curve CD givesthe vapour for the solution at different temperatures.

6. At point P the vapour pressure of solvent becomes equal to atmospheric pressure. At pointP the temperature is To. Hence by definition, To is the boiling point of pure solvent.

7. At point Q the vapour pressure of solution becomes equal to atmospheric pressure. Atpoint Q the temperature is T. Hence by definition, T is the boiling point of Solution.

8. Mathematically, elevation in boiling point, ΔTb may be expressed as:ΔTb = T – To

9. From the graph it is evident that ΔTb ∝ Δp = 01p – p (Lowering of vapour pressure.)

10. It has been experimentally found that the elevation in the boiling point ΔTb of a solutionis proportional to the molal concentration of the solution, i.e.ΔTb ∝ m

or ΔTb = Kb × molality(m)

Note : ΔTb ∝ m, Thus, elevation in boiling point is directly proportional to themolal concentration of the solute (i.e., number of molecules) and therefore,it is a colligative property.

(c) Molal elevation constant or Ebullioscopic constant :Molal boiling point elevation constant, Kb is defined as the elevation in boiling point producedby dissolving one mole of solute in 1kg of solvent (i.e. 1 molal solution)

. Units of Kb (S.I) = K kg mol–1 or K/m.e.g. : Water (Boiling point = 373K), Kb = 0.52 K kg mol–1

The value of Kb depends only on the nature of solvent and not on the nature of solute.

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(d) Relation between elevation of boiling point and molar mass of thesolute :

1. The elevation in boiling point (ΔTb) is useful in determining the molecular mass of the solute.2. Let W2 kg of a non-volatile solute is dissolved in W1 kg of the solvent and M2 is molar

mass of the solute in kg mol-1.

3. molality (m) =moles of solute

Weight of solvent in Kg = 2 2

1

W / M

W

4. m =2 2

1

W / M

W =

2

2 1

W

M × W

5. ΔTb = Kb × molality(m)

6. ΔTb = b 2

2 1

K × W

M × W[From (4) and (5)]

7. M2 =b 2

b 1

K × W

T × WΔFrom the relation, molecular mass of the solute can be calculated when the values of otherquantities are known.

Note : The experimental method to determine molecular mass of a non volatile soluteby determining boiling point of pure solvent and solution of known concentrationis called ebullioscopy.

(e) Freezing Point Depression :1. Freezing point : The freezing point of a liquid may be defined as the temperature at

which, the vapour pressure of solid is equal to the vapour pressure of liquid.

2. Depression in freezing point : The depression in freezing point may be defined as thedecrease in freezing point of the solution (difference in freezing points of the pure solventand the solution). It may be denoted by ΔTf. Mathematically, ΔTf = T – T°, where,T° and T represent the freezing points of pure solvent and the solution respectively.

3. Cause for depression of freezing point :(i) The freezing point of a liquid is the temperature at which the liquid and the solid have

the same vapour pressure.(ii) Addition of a non-volatile solute to a liquid decreases the freezing point, i.e., the freezing

point of the solution is less than that of the pure solvent.(iii) This happens due to lowering of vapour pressure of the solvent by the addition of

the non volatile solute.

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(f) Depression of freezing point and lowering of vapour pressuregraphically :

1. The freezing point of a liquid is the temperature at which the liquid and the solid havethe same vapour pressure.

2. Addition of a non-volatile solute to a liquid decreases the freezing point, i.e., the freezingpoint of the solution is less than that of the pure solvent.

3. This happens due to lowering of vapour pressure of the solvent by the addition of thenon volatile solute.

4. The curves BC, DE and AB represent the vapour pressures of the pure solvent, solutionand the solid respectively at different temperatures.

5. From the curves, it is clear that at point B the solid and liquid states of pure solvent havesame vapour pressure.

6. At point B the temperature is Tº which will be freezing point of pure solvent, Tº.7. When solute is dissolved in the solvent, the vapour pressure of solvent is lowered and

resulting solution can no longer freeze at temperature Tº8. A new equilibrium is established at point D, where

vapour pressure of solvent of the solution and solidsolvent becomes equal.

9. At point D the temperature is T which will befreezing point of solution.

10. The vapour pressure curve DE of the solution,always lies below the vapour pressure curve ofpure solvent. Hence, intersection of vapourpressure curve of solution and solid solvent canoccur only at a point lower than Tº.

11. Therefore, any solution must have freezing pointlower than that of pure solvent.

12. Thus, the depression in freezing point, ΔTf = Tº - T13. It is evident from the graph that ΔTf α pº – ps.14. But, ΔTf ∝ m (experimentally)

∴ ΔTf = Kf x m (molality)15. Kf is molal freezing point depression constant or cyroscopic constant.

(g) Molal depression constant or cyroscopic constant :Molal depression constant or cyroscopic constant is defined as depression in freezing pointproduced by dissolving 1 mole of solute in 1 kg of solvent (i.e., 1 molal solution).

1. Units of Kb (S.I) = K kg mol-1 or K/m.

98


2. e.g : Water (Freezing point 0oC.) Kf = 1.86 K kg mol-1

3. The value of Kf depends only on the nature of solvent.

(h) Relation between molecular mass of solute and depression in freezingpoint :

1. The depression in freezing point (ΔTf) is useful in determining the molecular mass ofthe solute.

2. Let W2 kg of a non-volatile solute is dissolved in W1 kg of the solvent and M2 is molarmass of the solute in kg mol-1.

3. molality (m) =moles of solute

Weight of solvent in kg = 2 2

1

W / M

M

4. m =2

2 1

W

M × W

5. ΔTf = Kf × molality(m)

6. ΔTf =f 2

2 1

K × W

M × W [From (4) and (5)]

7. M2 =f 2

f 1

K × W

T × W

From the relation, molecular mass of the solute can be calculated.

Note : The experimental method to determine molecular mass of a non volatile soluteby determining freezing point of pure solvent and solution of knownconcentration is called cryoscopy.

(i) Applications of depression of freezing point :1. In cold countries ice frozen on roads and footpaths, is melted by spraying salts like sodium

chloride or calcium chloride, ice melts into water as the salt depresses the freezing pointof water.

2. Also in cold countries de-icing of aeroplanes is done by depressing the freezing point ofwater.

3. Ethylene glycol is a common automobile antifreezer. It is water soluble and non volatile.It is also used as coolant in automobile radiators.

WhereW2 = Weight of solute in kg,W1= Weight of solvent in kg,M2 = Molecular wt of solute in kg mol–1

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*1. What is boiling point of liquid?Ans. Boiling Point : The boiling point of a liquid is the temperature at which its vapour pressure

becomes equal to the atmosphere pressure.

*2. What is elevation of boiling point?Ans. Elevation in Boiling Point (ΔTb) : The difference in the boiling points of the solution (T) and

pure solvent (Tº) is called the elevation in boiling point (ΔTb).∴∴∴∴∴ ΔTb = T – Tº

*3. What is molal elevation constant? Does it depend on nature of the solute?Ans. Molal boiling point elevation constant, Kb is defined as the elevation in boiling point produced

by dissolving one mole of solute in 1kg of solvent (i.e. 1 molal solution)Units of Kb (S.I) = K kg mol-1 or K/m.E.g. : Water (Boiling point = 373K), Kb = 0.52 K kg mol-1

The value of Kb depends only on the nature of solvent and not on the nature of solute.

4. Explain graphically Relation between Elevation in boiling point and lowering of vapour pressure.Ans. [Refer Fig. 2.3 b]

(a) The boiling point of a liquid is the temperature at which its vapour pressure becomes equalto the atmospheric pressure.

(b) The vapour pressure of the solution containing a non volatile solute is less the that of thepure solvent.

(c) Therefore, the solution has to be heated to a higher temperature so that its vapour pressurebecomes equal to the atmospheric pressure.

(d) Thus, the boiling point of the solution is always higher than that of pure solvent.(T > To)

(e) The curve AB gives the vapour pressures for the pure solvent and the curve CD givesthe vapour for the solution at different temperatures.

(f) At point P the vapour pressure of solvent becomes equal to atmospheric pressure. At pointP the temperature is To. Hence, by definition, To is the boiling point of pure solvent.

(g) At point Q the vapour pressure of solution becomes equal to atmospheric pressure.At point Q the temperature is T. Hence by definition, T is the boiling point of Solution.

(h) Mathematically, elevation in boiling point, ΔTb may be expressed as : ΔTb = T – To

(i) From the graph it is evident that ΔTb ∝ Δp = 01p – p (Lowering of vapour pressure.)

(j) It has been experimentally found that the elevation in the boiling point ΔTb of a solutionis proportional to the molal concentration of the solution, i.e.ΔTb ∝ m or ΔTb = Kb × molality(m)

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*5. Derive the relation between elevation of boiling point and molar mass of the solute.Ans. The elevation in boiling point (ΔTb) is useful in determining the molecular mass of the solute.

Let W2 kg of a non-volatile solute is dissolved in W1 kg of the solvent and M2 is molar massof the solute in kg mol-1.

molality (m) =moles of solute

Weight of solvent in Kg

=2 2

1

W / M

W

m =2 2

1

W / M

W = 2

2 1

W

M × W ...(i)

ΔTb = Kb × molality(m) ...(ii)

ΔTb = b 2

2 1

K × W

M × W[From (i) and (ii)]

M2 =b 2

b 1

K × W

T × WΔFrom the relation, molecular mass of the solute can be calculated when the values of otherquantities are known.

*6. Define freezing point of liquid.Ans. The freezing point of a liquid may be defined as the temperature at which, the vapour

pressure of solid is equal to the vapour pressure of liquid.

7. Define freezing point depression.Ans. Depression in freezing point : The depression in freezing point may be defined as the decrease

in freezing point of the solution (difference in freezing points of the pure solvent and the solution).It may be denoted by ΔTf.Mathematically, ΔTf = T0 – T, where,T0 and T represent the freezing points of pure solvent and the solution respectively.

*8 What causes depression in freezing point?Ans. Cause for depression of freezing point :

(a) The freezing point of a liquid is the temperature at which the liquid and the solid havethe same vapour pressure.

(b) Addition of a non volatile solute to a liquid decreases the freezing point, i.e., the freezingpoint of the solution is less than that of the pure solvent.

(c) This happens due to lowering of vapour pressure of the solvent by the addition of thenon volatile solute.

WhereW2 = Weight of solute in kg,W1= Weight of solvent in kg,M2 = Molecular wt of solute in kg mol–1

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9. Explain relation between Depression of freezing point and lowering of vapour pressure graphically.Ans. [Refer Fig. 2.3 (f)]

(a) The freezing point of a liquid is the temperature at which the liquid and the solid havethe same vapour pressure.

(b) Addition of a non volatile solute to a liquid decreases the freezing point, i.e., the freezingpoint of the solution is less than that of the pure solvent.

(c) This happens due to lowering of vapour pressure of the solvent by the addition of thenon volatile solute.

(d) The curves BC, DE and AB represent the vapour pressures of the pure solvent, solutionand the solid respectively at different temperatures.

(e) From the curves, it is clear that at point B the solid and liquid states of pure solvent havesame vapour pressure

(f) At point B the temperature is To which will be freezing point of pure solvent, To.(g) When solute is dissolved in the solvent, the vapour pressure of solvent is lowered and

resulting solution can no longer freeze at temperature To

(h) A new equilibrium is established at point D, where vapour pressure of solvent of the solutionand solid solvent becomes equal

(i) At point D the temperature is T which will be freezing point of solution.(j) The vapour pressure curve DE of the solution, always lies below the vapour pressure

curve of pure solvent. Hence, intersection of vapour pressure curve of solution and solidsolvent can occur only at a point lower than To.

(k) Therefore, any solution must have freezing point lower than that of pure solvent.(l) Thus, the depression in freezing point, ΔTf = To - T(m) It is evident from the graph that ΔTf α po – ps.(n) But, ΔTf ∝ m (experimentally)

∴ ΔTf = Kf x m (molality)(o) Kf is molal freezing point depression constant or cyroscopic constant.

*10. Define molal depression constant or cryoscopic constant.Ans. Molal depression constant or cyroscopic constant is defined as depression in freezing point

produced by dissolving 1 mole of solute in 1 kg of solvent (i.e., 1 molal solution).(a) Units of Kb (S.I) = K kg mol-1 or K/m.(b) Eg : Water (Freezing point 0oC.) Kf = 1.86 K kg mol-1

(c) The value of Kf depends only on the nature of solvent.

*11. How is molar mass of a non volatile solute related to the depression of freezing point? Derivean equation.

Ans. The depression in freezing point (ΔTf) is useful in determining the molecular mass of the solute.Let W2 kg of a non-volatile solute is dissolved in W1 kg of the solvent and M2 is molar massof the solute in kg mol-1.

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molality (m) = moles of solute

Weight of solvent in kg = 2 2

1

W / M

M

m =2

2 1

W

M × W ...(i)

ΔTf = Kf × molality(m) ...(ii)

ΔTf =f 2

2 1

K × W

M × W [From (i) and (ii)]

M2 =f 2

f 1

K × W

T × WΔFrom the relation, molecular mass of the solute can be calculated.

12. Give the applications of freezing point depression.Ans. (a) In cold countries ice frozen on roads and footpaths, is melted by spraying salts like sodium

chloride or calcium chloride, ice melts into water as the salt depresses the freezing pointof water.

(b) Also in cold countries de-icing of aeroplanes is done by depressing the freezing point ofwater.

(c) Ethylene glycol is a common automobile antifreezer. It is water soluble and non volatile.It is also used as coolant in automobile radiators.


• Theoretical MCQs :*1. When NaCl is added to water .......

a) Freezing point is raised b) Boiling point is depressedc) Freezing point does not change d) Boiling point is raised

*2. Molal elevation constant is elevation in boiling point produced by .......a) 1g of solute in 100g of solvent b) 100g of solute in 1000g of solventc) 1 mole of solute in one litre of solvent d) 1 mole of solute in one kg of solvent

*3. The determination of molar mass from elevation in boiling point is called asa) Cryoscopy b) Osmometry c) Ebullioscopy d) Spectroscopy

*4. If the mass is expressed in grams then Kb is given by .......

a)2 b 2

1

M × T × W

1000 × w

Δb)

2

b 1 2

W × 1000

T × W × MΔ

c)2 b 1

2

M × T × W

1000 × W

Δd)

1

b 2 2

W × 1000

T × W × MΔ

103


5. Which of the following aqueous solutions will have the lowest value of freezing point in the

centigrade scale?

a) 1% Urea b) 1% Glucose c) 1% Sucrose d) 1% Starch

Ans. nurea = 1

60nglucose =

1

180nsucrose =

1

342nstarch =

1

very large∴∴∴∴∴ nurea is highest.

∴∴∴∴∴ It will have lowest value of freezing point.

6. Kf is equal to depression in freezing point produced by ................. .

a) 1 molar solute b) 1 normal solute c) 1 molal solute d) all are correct


7. Which one of the following aqueoussolutions will have highest boiling point?a) 0.1 M ureab) 30g of glucose per dm3

c) 3.42g of sucrose in 100mld) 0.2 M glucose

Ans. Molarity in option B =30

180 × 1 = 1

6

= 0.16 M

Molarity in option C = 3.42

342 × 0.1

= 0.1 M∴∴∴∴∴ Molarity in option in highest i.e. 0.2 M

glucose.

8. The boiling point of a decimolal aqueoussolution of glucose would be....... (Given Kb = 0.52K kg mol–1)a) 100.52ºC b) 99.48ºCc) 99.94ºC d) 100.052º C

Ans. ΔTb = Kb × m= 0.52 × 0.1 = 0.052

T = Tº + ΔTb

= 100 + 0.052

= 100.052ºC

9. The boiling point of a solution containing2.62g of a substance A in 100g of wateris higher by 0.0512ºC than the boiling pointof pure water. The molar mass of thesubstance(Kb = 5.12 Km–1) is .......a) 131 g mol–1 b) 262 g mol–1

c) 26.2 g mol–1 d) 2620g mol–1

Ans. M2 =b 2

b 1

K W×

T WΔ

= –3

–3

5.12 2.62 × 10×

0.0512 100 × 10

= 2.62 kg mol–1 = 2620 g mol–1

10. The rise in the boiling point of a solutionscontaining 1.8g of glucose in 100g of asolvent is 0.1ºC. The molal elevationconstant of the liquid is .......a) 0.01 k/m b) 0.1 k/mc) 1 k/m d) 10 k/m

Ans. ΔTb = Kb × m

104


0.1 = b –3

1.8K ×

180 × 100 × 10

Kb = 1 k/m

11. The molal freezing point constant for wateris 1.86. If 342 g of cane sugar (molar mass= 342 g mol–1) is dissolved in 1000g ofwater, the solution will freeze at ...........a) 1.86ºC b) –1.86ºCc) –3.92ºC d) 3.92ºC

Ans. ΔTf = Kf × m

= –3

3421.86 ×

342 × 1000 × 10

= 1.86T = Tº – ΔTf

= 0 – 1.86= –1.86ºC

12. The amount of urea (molar mass = 60gmol–1) to be dissolved in 500g of water toproduce a depression of 0.186º in thefreezing point is ....... .(Kf for water = 1.86 Km–1)a) 0.6g b) 60gc) 3g d) 6g

Ans. ΔTf = Kf × m

0.186 = –3

mass of urea1.86 ×

60 × 500 × 10

∴∴∴∴∴ Mass of urea = 3g

• Advanced MCQs :13. 15g of non-volatile solute is dissolved in

250g of solvent, shows elevation in boilingpoint 0.01K. If gram molecular weight ofsolute is 60, the molal elevation constantfor solvent is ..... (MHT-CET 2010)a) 0.1 Km–1

b) 1 Km–1

c) 10 Km–1

d) 0.01 Km–1

Ans. ΔTb = Kb × m

0.01 = b –3

15K ×

60 × 250 × 10

Kb = 0.01 k/m

14. What will be freezing point of solutionwhen 0.08kg of ethylene glycol (molarmass = 62) is added to 400g of water. (Kf

of water = 1.86K kg/mol) (MHT-CET 2008)

a) 269 K b) 267 Kc) 279 K d) 297 K

Ans. ΔTf = Kf × m

=3

–3

0.08 × 101.86 ×

62 × 400 × 10

=186 × 8

62 × 4

= 6KΔTf = Tº – T6 = 273– T

∴∴∴∴∴ T = 273 – 6 = 267 K

105


2.4 OSMOTIC PRESSURE, ABNORMAL MOLECULAR MASS ANDVANT HOFF FACTOR :

Concept Explanation :(a) Semi permeable membrane :

A membrane which allows the solventmolecules to pass through it but not thesolute particles is called a semi-permeablemembrane.Eg : Natural : Animal bladder, Animal andPlant cell wall, Potato membrane,parchment paper, etc.Artificial : Copper ferrocyanide membraneCu2 [Fe(CN)6], Cellulose, Cellulosenitrate, etc.

(b) Osmosis :1. Osmosis was first studied by Abbe Nollet.2. Osmosis : When a solution is separated from a pure solvent by a semi-permeable membrane

(such as an animal bladder or copper ferrocyanide) then it is observedthat the solvent flows from the region of pure solvent into solution.

3. Similarly, if two solutions of different concentrations are separated by a semi-permeablemembrane then the solvent flows from solution of lower concentration into solution of higherconcentration. This process is called osmosis.

4. Definition : The spontaneous and unidirectional flow of the solvent molecules frompure solvent into the solution or from a solution of lower concentration to the solutionof higher concentration through a semipermeable membrane is called osmosis.

(c) Abbe Nollet experiment :1. The apparatus consists of a long stem thistle funnel.2. The mouth of the thistle funnel is closed by a semipermeable

membrane like pig’s bladder.3. Sucrose solution of some concentration is filled in the thistle

funnel.4. The thistle funnel is then placed in beaker containing water

in inverted position.5. The solvent and solution are separated from each other

by a semipermeable membrane.

106


6. There occurs a net flow of solvent molecules into the solution.7. As a result the original volume of solution increases and liquid level of solution increases.8. Hydrostatic pressure is developed due to liquid column in thistle funnel.9. The liquid level in thistle funnel rises until the excess pressure so produced is sufficient

to stop the flow of solvent molecules.10. This equilibrium hydrostatic pressure which stops the net flow of solvent into the solution

is equal to Osmotic pressure.

(d) Types of Osmosis and osmotic pressure :1. Osmotic pressure : The osmotic pressure may be defined as the excess mechanical

pressure which is applied on the solution to stop the flow of solvent molecules into thesolution through a semi-permeable membrane.It is generally, denoted by π. It is a colligative property.

2. Isotonic solutions : Solutions which have the same osmotic pressure at the sametemperature are called Isotonic solutions.

3. Hypertonic solutions : A solution having more osmotic pressure than some other solutionsis called hypertonic solutions.

4. Hypotonic solutions :A solution having less osmotic pressure than the other solutions is called hypotonic solutions.

Note : It is interesting to note that a 0.91% (w/v) solution of sodium chloride (knownas saline water) is isotonic with human blood corpuscles. In this solution, thecorpuscles neither swell nor shrink. Therefore, the medicines are mixed withsaline water before being injected into the veins. However, a 5% NaCl solutionis hypertonic solution and when red blood cells are placed in this solution,water comes out of the cells and they shrink. On the other hand, when redblood cells are placed in distilled water (hypotonic solution) water flows intothe cells and they swell or burst.

(e) Osmosis in day to day life :1. A raw mango kept in a concentrated salt solution loses water due to osmosis and shrivel

(shrinks and wrinkles) into pickle.2. A limp carrot can be placed in water making it firm once again. Water moves in carrot

due to osmosis.3. Blood cells placed in water containing less than 0.91% salt collapse due to loss of water

by osmosis.4. People eating lot of salt experience edema i.e. swelling of tissue cells due to water retention.

107


5. The fruits can be preserved by adding sugar. Since bacterium cell loses water due to osmosis,shrivels and dies.

6. In plants the leaves loose water by transpiration. The solute concentration in leaftherefore, increases. Due to osmosis water is pulled from soil and transported in upwarddirection.

(f) Laws of osmotic pressure :1. Boyle Van’t Hoff law : It states that “Temperature remaining constant, the osmotic pressure

of a dilute solution is directly proportional to the molar concentration of the solution orinversely proportional to the volume of the solution.”If π represents the osmotic pressure (in atmospheres) and C is the molar concentrationin moles/litre, we haveπ ∝ C at constant temperatureIf n moles of the substance are dissolved in V litres of the solution.

C = n

V i.e.

1C

V∝

The above results may, therefore, be written as

1

V∝π (at constant temperature)

or πV = constant (at constant temperature)2. van’t Hoff Charle’s law : Similar to Charles law for gases, it state that ‘for a given

concentration of a solution, the osmotic pressure is directly proportional to the absolutetemperature’.Mathematically, π ∝ T (for a given concentration)

T

π = constant (for a given concentration)

3. van’t Hoff Avogadro’s law :Statement : This law states that, Equal volumes of isotonic solutions at the same temperaturecontain equal number of moles of solute.Van’t Hoff equation for two different dilute solutions can be written as follows:π1V1 = n1RT1 and π2V2 = n2RT2

If π1 = π2 (Isotonic Solutions) and T1 = T2,

Then , 1

1

n

V = 2

2

n

V

Hence, When osmotic pressure and temperature are same, equal volumes of solutionswould contain equal number of moles of solute.

108


4. Vant-Hoff’s equation for dilute solutions : According to Van’t Hoff-Boyle’s law1

Vπ ∝ (at constant temperature)

According to van’t Hoff-Charle’s law,π ∝ T (at given concentration)Combining the above results, we get

T

Vπ ∝ or πV = kT … (i)

where k is called general solution constant.vant Hoff further proved that the value of k is same as R,Hence, πV = RTFurther if the solution contains n moles of the solute dissolved in V litres of solution, then

C = n

VSubstituting the value of C in equation (1), we get;

π = n

R TV

πV = n RT ... (ii)

(g) Determination of molecular mass from osmotic pressure :According to van’t Hoff equation,π = CRT

But C = n

Vwhere n is the number of moles of solute dissolved in V litre of the solution.

∴∴∴∴∴ π = n

RTV

The number of moles of solute n may be given as 2

2

W

M . Here W2 is the weight of thesolute and M2 is its molar massSubstituting the value of n in the above expression.

π = 2

2

W RT

VM or M2 = 2W RT

VπThus, the molar mass of the solute, M2, can be calculated.

(h) Reverse osmosis :1. Definition : If, a solution is separated from the pure solvent by a semi-permeable

membrane and the pressure applied on the solution (more than the osmotic pressure)the solvent starts flowing from the solution towards the pure solvent. This phenomenonis known as reverse osmosis.

109


2. Application of reverse osmosis :Desalination :(a) Sea water contains 3.5%w/w of dissolved salts.(b) Drinking water is produced from sea water by

a process called as desalination by reverseosmosis.

(c) In reverse osmosis high pressure greater thanosmotic pressure of sea water i.e., 30 atm. isapplied to force water from concentratedaqueous solution like sea water to pure waterside through a semi permeable membrane.

(d) The method can be used provided a suitable semipermeable membrane is developedwhich withstands the high pressure condition over a prolonged period.

(i) Abnormal molecular mass :1. Colligative properties of solutions depend on actual number of solute particles present in

the solution.2. In Dilute solution of non- electrolytes like urea, glucose etc. the solute remains in normal

molecular condition and does not undergo dissociation or association.3. In case of solutions of electrolytes, it has been observed that, observed colligative properties

and thus, molar mass determined by these methods do not agree with the expected ortheoretical values.

4. This is because of association or dissociation of the solute molecules in the solution.(a) Dissociation :

(i) Molecules of certain substances (acids, bases and salts) dissociate or ionize ina solvent to give two or more particles. For example, AB dissociates to givedouble number of particles as :

+ –AB A + B

(ii) Consequently, the total number of particles increases in solution and, therefore,the colligative property of such solutions will be large for solutes undergoingdissociation.

(iii) Since the colligative property is inversely proportional to the molar mass of thesolute, the molar mass of solute in such cases will be less than normal. The valueof colligative property depends on the degree of dissociation of the solute in thesolution.

(iv) For example :+ –NaCl Na + Cl⎯⎯→ : 2 particles+ 2–

2 4 4K SO 2K + SO⎯⎯→ : 3 particles3+ –

3AlCl Al + 3Cl⎯⎯→ : 4 particles

110


(b) Association :(i) In some non-polar solvents, the solute molecules undergo association, i.e. two,

three or even more molecules exist in combination with each other to form biggermolecules.

(ii) Therefore, the total number of molecules in solution become less than thenumber of molecules of the substance added. This leads to a decrease inthe value of colligative properties because colligative properties is proportionalto no. of solute particles.

(iii) Since the colligative property is inversely proportional to the molar mass of thesolute, the molar mass of solute in such cases will be greater than normal.

(iv) For example, in benzene solvent solutes like acetic acid, benzoic acid etc. existas dimers. This is due to hydrogen bonding between these molecules.

3 3 22CH COOH (CH COOH)

6 5 6 5 22C H COOH (C H COOH)

(j) vant Hoff factor :1. This factor is used to express the extent of dissociation or association of the solutes in

their solutions.2. It defines as the ratio of observed colligative property produced by a given concentration

of electrolyte solution to the property observed for the same concentration of non electrolytesolution.

3. Therefore,

i = observed value of colligative property

theoretical value of colligative property

Since colligative property α number of solute particles present in the solution. Hence,

i = number of solute particles present in solution

theoretical number of solute particles

Similarly, colligative property 1

molecular massα

i = theoretical molecular mass

observed molecular mass

4. If i = 1, solute molecules are in normal state and no association or dissociation has takenplace.

5. If i > 1 , the solute molecules are dissociated in the solution.6. If i < 1, the solute molecules are associated in the solution.

111


(k) vant Hoff factor (i) and Degree of dissociation (ααααα) :Consider ‘m’ moles of an electrolyte AxBy dissolved in 1kg of solvent. Let α be thedegree of dissociation of the electrolyte.

y+ x–x yA B xA + yB

Initial moles m 0 0moles at equilibrium m – mα x(mα) y(mα)Total no. of moles at equilibrium = mt = (m – mα) + x(mα) + y(mα)

= m(1 – α) + x(mα) + y(mα)= m [1 – α + xα + yα]= m[1 + xα + yα – α]= m[1 + α (x + y – 1)]

Let the total no. of ions produced by dissociation by one mole of eletrolyte AxBy be x + y = n′∴∴∴∴∴ mt = m (1 + α (n1 – 1))

Now vant Hoff factor, i =observed number of moles

theoretical number of moles

∴∴∴∴∴ i = tm

m =

m (1 + (n –1) )

m

′ α

∴∴∴∴∴ i = 1 + (n′ – 1) α

Thus, degree of dissociation α = i – 1

n – 1′

i =M (Theoretical)

M (observed) = 1 + (n ′ – 1) α

M (Theoretical)– 1

M (observed) = (n ′ – 1) α

α =M (Theoretical) – M (observed)

M (observed) (n – 1)′


1. Define : *a) Semi permeable membrane b) Osmosis.Ans. (a) Semi permeable membrane : A membrane which allows the solvent molecules to pass

through it but not the solute particles is called a semi-permeable membrane.e. g : Natural : Animal bladder, Animal and Plant cell wall, Potato membrane, parchmentpaper, etc.Artificial : copper ferrocyanide membrane Cu2 [Fe(CN)6], Cellulose, cellulose nitrate, etc.

112


Definition : The spontaneous and unidirectional flow of the solvent molecules from pure solventinto the solution or from a solution of lower concentration to the solution of higher concentrationthrough a semipermeable membrane is called osmosis.

*2. Explain the process of osmosis. or What is Osmosis?Ans. (a) Osmosis was first studied by Abbe Nollet.

(b) Osmosis : When a solution is separated from a pure solvent by a semi-permeable membrane(such as an animal bladder or copper ferrocyanide) then it is observed that the solventflows from the region of pure solvent into solution. [Ref. Fig. 2.4 (b)]

(c) Similarly, if two solutions of different concentrations are separated by a semi-permeablemembrane then the solvent flows from solution of lower concentration into solution of higherconcentration. This process is called osmosis.

(d) Definition : The spontaneous and unidirectional flow of the solvent molecules from puresolvent into the solution or from a solution of lower concentration to the solution of higherconcentration through a semipermeable membrane is called osmosis.

3. Explain Abbe Nollet experiment.Ans. (a) The apparatus consists of a long stem thistle funnel.

(b) The mouth of the thistle funnel is closed by a semipermeable membrane like pig’s bladder(c) Sucrose solution of some concentration is filled in the thistle funnel.(d) The thistle funnel is then placed in beaker containing water in inverted position.(e) The solvent and solution are separated from each other by a semipermeable membrane.(f) There occurs a net flow of solvent molecules into the solution.(g) As a result the original volume of solution increases and liquid level of solution increases.(h) Hydrostatic pressure is developed due to liquid column in thistle funnel.(i) The liquid level in thistle funnel rises until the excess pressure so produced is sufficient

to stop the flow of solvent molecules.(j) This equilibrium hydrostatic pressure which stops the net flow of solvent into the solution

is equal to Osmotic pressure.[Refer Fig. 2.4 (c)]

*4. Define : a) Osmotic pressure b) Isotonic solution c) Hypertonic solution d) hypotonic solution.Ans. (a) Osmotic pressure : The osmotic pressure may be defined as the excess mechanical

pressure which is applied on the solution to stop the flow of solvent molecules into thesolution through a semi-permeable membrane. It is generally, denoted by π. It is a colligativeproperty.

(b) Isotonic solutions : Solutions which have the same osmotic pressure at the sametemperature are called Isotonic solutions.

(c) Hypertonic solutions : A solution having more osmotic pressure than some other solutionsis called hypertonic solutions.

(d) Hypotonic solutions : A solution having less osmotic pressure than the other solutionsis called hypotonic solutions.

113


5. Write any four applications of osmosis in day to day life.Ans. (a) A raw mango kept in a concentrated salt solution loses water due to osmosis and shrivel

(shrinks and wrinkles) into pickle.(b) A limp carrot can be placed in water making it firm once again. Water moves in carrot

due to osmosis.(c) Blood cells placed in water containing less than 0.91% salt collapse due to loss of water

by osmosis.(d) People eating lot of salt experience edema i.e. swelling of tissue cells due to water retention.(e) The fruits can be preserved by adding sugar. Since bacterium cell loses water due to osmosis,

shrivels and dies.(f) In plants the leaves loose water by transpiration. The solute concentration in leaf therefore,

increases. Due to osmosis water is pulled from soil and transported in upward direction.

*6. State van’t Hoff’s-Boyle’s law.Ans. Boyle Van’t Hoff law : It states that “Temperature remaining constant, the osmotic pressure

of a dilute solution is directly proportional to the molar concentration of the solution or inverselyproportional to the volume of the solution.”If π represents the osmotic pressure (in atmospheres) and C is the molar concentration inmoles/litre, we have

π ∝ C at constant temperatureIf n moles of the substance are dissolved in V litres of the solution.

C = n

V i.e.

1C

V∝

The above results may, therefore, be written as

1

V∝π (at constant temperature)

or πV = constant (at constant temperature)

*7. State van’t Hoff-Charles law.Ans. van’t Hoff Charle’s law : Similar to Charles law for gases, it state that ‘for a given concentration

of a solution, the osmotic pressure is directly proportional to the absolute temperature’.Mathematically, π ∝ T (for a given concentration)

T

π

= constant (for a given concentration)

*8. State van’t Hoff Avogadro’s law.Ans. Statement : This law states that, Equal volumes of isotonic solutions at the same temperature

contain equal number of moles of solute.Van’t Hoff equation for two different dilute solutions can be written as follows:

114


π1V1 = n1RT1 π2V2 = n2RT2

If π1 = π2 (Isotonic Solutions)and T1 = T2,

Then , 1

1

n

V = 2

2

n

V

Hence, when osmotic pressure and temperature are same, equal volumes of solutions wouldcontain equal number of moles of solute.

*9. Derive van’t Hoff equation for osmotic pressure of a solution.Ans. Vant-Hoff’s equation for dilute solutions :

According to Van’t Hoff-Boyle’s law1

Vπ ∝ (at constant temperature)

According to Van’t Hoff-Charle’s law,π ∝ T (at given concentration)

Combining the above results, we getT

Vπ ∝

or πV = kT …(i)where k is called general solution constant.Vant Hoff further proved that the value of k is same as R,Hence, πV = RTFurther if the solution contains n moles of the solute dissolved in V litres of solution, then

C =n

VSubstituting the value of C in equation (1), we get;

π =n

R TV

πV = n RT ...(ii)

*10. How is molar mass of a solute determined from osmotic pressure measurements.Ans. According to van’t Hoff equation,

π = CRT

But C = n

Vwhere n is the number of moles of solute dissolved in V litre of the solution.

∴∴∴∴∴ π = n

RTV

115


The number of moles of solute n may be given as 2

2

W

M . Here W2 is the weight of the solute

and M2 is its molar mass

Substituting the value of n in the above expression.

π = 2

2

W RT

VM or M2 = 2W RT

VπThus, the molar mass of the solute, M2, can be calculated.

11. Explain reverse osmosis.Ans. (a) Definition : If, a solution is separated from the pure solvent by a semi-permeable

membrane and the pressure applied on the solution (more than the osmotic pressure)the solvent starts flowing from the solution towards the pure solvent. This phenomenonis known as reverse osmosis.

(b) Application of reverse osmosis :Desalination :(1) Sea water contains 3.5% w/w of dissolved salts.(2) Drinking water is produced from sea water by a process called as desalination by

reverse osmosis.(3) In reverse osmosis high pressure greater than osmotic pressure of sea water i.e.,

30 atm. is applied to force water from concentrated aqueous solution like sea waterto pure water side through a semi permeable membrane.

(4) The method can be used provided a suitable semipermeable membrane is developedwhich withstands the high pressure condition over a prolonged period.[Refer Fig. 2.4 (h)]

*12. Explain Abnormal molecular masses.Ans. (a) Colligative properties of solutions depend on actual number of solute particles present in

the solution.(b) In Dilute solution of non-electrolytes like urea, glucose etc. the solute remains in normal

molecular condition and does not undergo dissociation or association.(c) In case of solutions of electrolytes, it has been observed that, observed colligative properties

and thus, molar mass determined by these methods do not agree with the expected ortheoretical values.

(d) This is because of association or dissociation of the solute molecules in the solution.(1) Dissociation :

(i) Molecules of certain substances (acids, bases and salts) dissociate or ionize ina solvent to give two or more particles. For example, AB dissociates to givedouble number of particles as :

+ –AB A + B

116


(ii) Consequently, the total number of particles increases in solution and, therefore,the colligative property of such solutions will be large for solutes undergoingdissociation.

(iii) Since the colligative property is inversely proportional to the molar mass of thesolute, the molar mass of solute in such cases will be less than normal. The valueof colligative property depends on the degree of dissociation of the solute in thesolution.

(iv) For example :+ –NaCl Na + Cl⎯⎯→ : 2 particles+ 2–

2 4 4K SO 2K + SO⎯⎯→ : 3 particles3+ –

3AlCl Al + 3Cl⎯⎯→ : 4 particles

(2) Association :(i) In some non-polar solvents, the solute molecules undergo association, i.e. two, three

or even more molecules exist in combination with each other to form bigger molecules.(ii) Therefore, the total number of molecules in solution become less than the number

of molecules of the substance added. This leads to a decrease in the value of colligativeproperties because colligative properties is proportional no. of solute particles.

(iii) Since the colligative property is inversely proportional to the molar mass of thesolute, the molar mass of solute in such cases will be greater than normal.

(iv) For example, in benzene solvent solutes like acetic acid, benzoic acid etc. existas dimers. This is due to hydrogen bonding between these molecules.

3 3 22CH COOH (CH COOH)

6 5 6 5 22C H COOH (C H COOH)

*13. Write a note on van’t Hoff factor or Explain van’t Hoff factor.Ans (a) This factor is used to express the extent of dissociation or association of the solutes in

their solutions.(b) It defines as the ratio of observed colligative property produced by a given concentration

of electrolyte solution to the property observed for the same concentration of non electrolytesolution.

(c) Therefore, i = observed value of colligative property

theoretical value of colligative propertySince colligative property α number of solute particles present in the solution. Hence,

i = number of solute particles present in solution

theoretical number of solute particles

Similarly, colligative property 1

molecular massα

117


i = theoretical molecular mass

observed molecular mass

(d) If i = 1, solute molecules are in normal state and no association or dissociation has takenplace.

(e) If i > 1 , the solute molecules are dissociated in the solution.(f) If i < 1, the solute molecules are associated in the solution.

*14. Derive and expression for vant Hoff factor and degree of dissociation. OR

Derive the equations : (a) α = i – 1

n – 1′ (b) α = M(theoretical) – M (observed)


Ans. Consider ‘m’ moles of an electrolyte AxBy dissolved in 1kg of solvent. Let α be the degreeof dissociation of the electrolyte.

y+ –yA B A + yB x

x x⎯⎯→←⎯⎯

Initial moles m 0 0moles at equilibrium m – mα x(mα) y(mα)Total no. of moles at equilibrium = mt = (m – mα) + x(mα) + y(mα)

= m(1 – α) + x(mα) + y(mα)= m[1 – α + xα + yα]= m[1 + xα + yα – α]= m[1 + α (x + y – 1)]

Let the total no. of ions produced by dissociation by one mole of eletrolyte AxBy be x + y = n′

∴∴∴∴∴ mt = m (1 + α (n′ – 1))

Now vant Hoff factor, i =observed number of moles

theoretical number of moles

∴∴∴∴∴ i = tm

m =

m (1 + (n –1) )

m

′ α

∴∴∴∴∴ i = 1 + (n′ – 1) α

Thus, degree of dissociation α = i – 1

n – 1′

i =M (Theoretical)

M (observed) = 1 + (n′ – 1) α

M (Theoretical)– 1

M (observed) = (n′ – 1) α

α =M (Theoretical) – M (observed)


118


*15. Explain Abnormal osmotic pressure.Ans. Osmotic pressure is a colligative property, thus, depends upon the number of particles present

in the solute. Therefore, the calculated osmotic pressure is different from its experimental valiein some cases.This is because of association or dissociation of the solute particles in the solution.a) Dissociation:

i) Molecules of certain substances (acids, bases and salts) dissociate or ionize in a solventto give two or more particles. For example, AB dissociates to give double number

of particles as : + –AB A + B

ii) Consequently, the total number of particles increases in solution and, therefore, theosmotic presure of such solutions will be large for solutes undergoing dissociation.

iii) For example: NaCl+ –aNaCl N + Cl⎯⎯→ : 2 particles

2 –+2 4 4K SO 2K + SO⎯⎯→ : 3 particles

3+ –3AlCl Al + 3Cl⎯⎯→ : 4 particles

b) Associaton :i) In some non-polar solvents, the solute molecules undergo association, i.e. two, three

or even more molecules exist in combination with each other to form bigger molecules.ii) Therefore, the total number of molecules in solution become less than the number

of molecules of the substance added. This leads to a decrease in the value of osmoticpressure because osmotic pressure ∝ no. of solute particles.

iii) For example, in benzene solvent solutes like acetic acid, benzoic acid etc exist as

dimmers as. This is due to hydrogen bonding between these molecules.

3 3 22CH COOH (CH COOH) 6 5 6 5 22C H COOH (C H COOH)


• Theoretical MCQs :*1. In osmosis .......

a) Solvent molecules pass from high concentration of solute to low concentration.b) Solvent molecules pass from a solution of low concentration of solute to a solution

of high concentration of solute.c) Solute molecules pass from low concentration to high concentration.d) Solute molecules pass from high concentration to low concentration.

119


*2. The two solutions with same osmotic pressure are called .......a) Isotonic b) Isomeric c) Hypotonic d) Hypertonic

*3. The values of gas constant and solution constant .......a) Are different b) Almost identicalc) Gas constant is greater than solution constantd) Gas constant is smaller than solution constant

*4. Which of the following 0.1M aqueous solutions will exert highest osmotic pressure?a) NaCl b) BaCl2 c) MgSO4 d) Al2(SO4)3

*5. Abnormal molar mass is produced by .......a) Association of soluteb) Dissociation of solutec) Both association and dissociation of soluted) Separation by semipermeable membrane

*6. Which of the following aqueous solutions will have minimum elevation in boiling point?a) 0.1M KCl b) 0.05M NaCl c) 1M AlPO4 d) 0.1M MgSO4

*7. Which of the following will have maximum depression in freezing point? (MHT-CET 2009)a) 0.5M Li2SO4 b) 1M KCl c) 0.5M Al2(SO4)3 d) 0.5M BaCl2


8. A solution has an osmotic pressure of 0.821atm at 300K. Its concentration would be.......a) 0.66M b) 0.32Mc) 0.066M d) 0.033M

Ans. π = C R T0.821 = C × 0.0821 × 300

C =1

30 = 0.033M

9. 1 molar solution of a non-electrolytecompound will produce an osmotic pressure.......a) 1 atm b) 44.8 atmc) 10.0 atm d) 22.4 atm

Ans. π = C R T= 1 × 0.0821 × 273= 22.4 atm

10. The weight of glucose to be dissolved in250cm3 of water, to get a solution which isisotonic with 0.2M solution of urea, is .......a) 36g b) 9gc) 18g d) 4.5g

Ans. πglucose = πurea Cglucose = Curea

glucosen

V= 0.2

–3

mass of glucose

180 × 250 × 10 = 0.2

∴ Mass of glucose = 0.2 × 180 × 250 × 10–3

= 9g11. The Vant Hoff factor for non-electrolytes

like, urea, glucose etc. will be .......a) 2 b) 0.5c) 1.0 d) –1

120


12. The freezing point of 1 molal NaCl solution assuming NaCl to 100% dissociated in water is.........a) –1.86ºC b) –3.72ºC c) +1.86ºC d) +3.72ºC

Ans. ΔTf = i × Kf × m= 2 × 1.86 × 1 = 3.72ºC

ΔTf = Tº – T3.72 = 0 – T

∴ T = 0 – 3.72= –3.72ºC

• Advanced MCQs :13. Among the following equimolar aqueous solutions identify the one having highest boiling point.

(March 2008)a) Sodium chloride b) Sucrose c) Sodium sulphate d) Urea

Ans. urea i = 1, sucrose i = 1NaCl i = 2, Na2SO4 i = 3

HOURS BEFORE EXAM

Per cent by Weight (% w/w) = Mass of solute

× 100Mass of solution

Per cent weight by Volume (% w/v) = Mass of solute

× 100Volume of solution

Per cent by Volume (% v/v) = Volume of solute

× 100Volume of solution

Parts per million (ppm) = 6Mass of solute

× 10Total mass of solution

No. of moles (n) = Weight in gms

Molecular weight

No. of moles = 3Vol of gas in dm at S.T.P

22.4

.

No. of moles = 23

No.of particles (Atoms,moleculesor ions)

6.023×10

121


Mole fraction of solute = x2 =2

2 1

n

n + nn1 = No. of moles of solvent, n2 = No. of moles of solute.

x1 + x2 = 1 [For Binary solution]

Molarity = 3

No. of moles of solute

Volume of solution in litre or dmUnit of molarity = mol/lit or mol/dm3

It depends upon temperature.Molarity of pure water = 55.55

Resulting molarity of mixed solution = 1 1 2 2

1 2

[M V + M V ]

V + V

For Dilution: M1V1 = M2V2

[If solution is diluted x times, then its molarity decrease by x times]

For Neutralization : Ma × Va = Mb × Vb (For acids or bases containing same no. ofH and OH)

Normality (N) = 3

No. of gm equi. of solute

Volume of solution in dm OR = gm / lit. of solute

eq. wt. of solute

No. of gm eq.wt. of solute = weight in gms of solute

eq. wt. of solute

Eq. wt. = Molecular weight

No. of H or OH atoms = Atomic weight

Valency = Molecular weight

Total charge on cation or Anion

Units = gm Equivalent per litre.Normality changes with temperature.If solution is diluted x time, its normality decreases by x times.

Resulting Normality of mixed solution : N = 1 1 2 2

1 2

N V + N V

V + VFor dilution : N1V1 = N2V2

For Neutralization: Na × Va = Nb × Vb (Irrespective of No. of H and OH present in acidand base)1 N = Normal solution N/10 = Decinormal solutionN/100 = Centinormal solution N/1000 = Millinormal solution

Relationship between Normality and Molarity :Normality = Basicity × Molarity (for acid) Normality = Acidity × Molarity (for base)

122


Basicity for acid = No. of H+ present [eg. HCl – 1, H2SO4 – 2, CH3COOH = 1]Acidity for base = No. of OH present [eg. NaOH = 1, Ca(OH)2 = 2]

Molality = No. of moles of solute

Weight of solvent in kgMolality is independent of temperature.Units = moles/kg. Molality of pure water = 55.55

Lowering of vapour pressure = p10– p (where, p1

0 = vapour pressure of pure solvent)

(p = vapour pressure of solution containing a non volatile solute.)

Relative lowering of vapour pressure = 01

01

p – p

p

Raoults law (for 2 miscible volatile liquids)

p1 = 01 1p × x p1 = Partial vapour pressure of Solvent

p2 = 02 2p × x 0

1p = vapour pressure of pure solvent

p = p1 + p2 x1 = mole fraction of solvent

p = Total vapour pressure of solution Similarly for solute

Raoults law (for non-volatile solute in volatile solvent)

p = po × x1 (x1 = mole fraction of solvent) and 0p – p

p = x2 (x2 = mole fraction of solute)

Molecular weight of solute using Raoults law M2 = 0

2 1 10

1 1

W × M p×

W p – p(for dilute solutions)M2 = Molecular weight of solute, M1 = Molecular weight of solvent.W2 = Weight of solute in gms, W1 = Weight of solvent in gms

Boiling Point Elevation :Elevation in B.P. = Increase in B.P. of solvent is non-volatile solute is added to it.ΔTb = T – T0 Where Tb = B.P. of solution,T0 = B.P. of pure solvent.ΔTb ∝ molality (m) (Raoults law)ΔTb = Kb × m (Kb = Molal elevation B.P. constant = Ebullioscopic constant).Units of Kb = Kelvin kg/mole.Kb for water = 0.52 Kelvin kg/mol.Molecular weight of solute.

123


M2 =b 2

b 1

K × W

T × WΔ M2 = Molecular weight of solute in kg mol-1

Kb = Ebullioscopic constant W2 = Weight of solute in kgΔTb = T – Tº W1 = Weight of solvent in kg.

Freezing Point Depression :Depression in F.P. : When a non-volatile solute is added to volatile solvent, the F.P. ofsolvent decreases.ΔT = Tº – TT0 = F.P. of pure solvent.T = F.P. of solutionΔTf ∝ m (molality)ΔTf = Kf x m (Raoults law)Kf = Molal F.P. depression constant [Cryoscopic constant]Kf = Kelvin kg/mole.g. Kf of water = 1.86 Kelvin kg/mol.

M2 =f 2

f 1

K × W

T × WΔ

W2 = Weight of solute in kgW1 = Weight of solvent in kgM2 = Molecular weight of solute in kg mol-1

Important formulae related to Osmotic Pressure

1

1C

π

=2

2C

π

;1

1T

π

=2

2T

π

;

1

1n

π

= 2

2n

π

; πV = nRT ;

πV =w

RTM

π = C RT

πππππ V RNm–2 or Pa m3 8.314

kPa dm3 8.314Atm dm3 0.082

124



Type-I-Concentration of solutions*1. 34.2 g of glucose is dissolved in 400 g of

water. Calculate percentage by mass ofglucose solution.

Given :mass of glucose = 34.2gmass of water = 400 g

To find :% by mass of glucose

Solution :

% by mass of glucose

=2

mass of glucose× 100

masss of glucose × mass of H O

=34.2

× 100(34.2 + 400)

=34.2

× 100434.2

= 7.87% w/w

% by mass of glucose is 7.87 w/w

*2. A solution is prepared by dissolving certainamount of solute in 500g of water. Thepercentage by mass of a solute in solutionis 2.38. Calculate mass of solute. (12.19g)

Given :mass of water = 500g% by mass of solute = 2.38%

To find :mass of solute

Sol. :Let the mass of solute be xg % by massof solute

=mass of solute

× 100mass of solute × mass of water

2.38 = × 100+ 500

x

x

2.38(x + 500) = 100x 2.38x + 500 × 2.38 = 100x

500 × 2.38 = 100x – 2.38x500 × 2.38 = 97.62x

mass of solute = x = 500 × 2.38

97.62

mass of solute = 12.19 g.

*3. 4.6 cm3 of methyl alcohol is dissolved in25.2 g of water. Calculate (i) % by massof methyl alcohol (ii) mole fraction ofmethyl alcohol and water. (Given densityof methyl alcohol = 0.7952 gcm3 andC = 12, H = 1, O = 16)

Given :Volume of methyl alcohol = 4.6 cm3

mass of water = 25.2gdensity of methyl alcohol = 0.7952 g/cm3

To find :i) % by mass of methyl alcoholii) mole fraction of methyl alcohol and water

Solution :Density of methyl alcohol

=mass of methyl alcohol

volume of methyl alcohol

0.7952 = mass of methyl alcohol

4.6∴ mass of methyl alcohol = 0.7952 × 4.6

mass of methyl alcohol = 3.6579 gMolar mass of methyl alcohol (CH3OH)= 12 + 3 + 16 + 1 = 32moles of methyl alcohol

=mass of methyl alcohol

molar mass of methyl alcohol

=3.658

32 = 0.1143 moles

125


mole of water =mass of water

molar mass of water

=25.2

18 = 1.4 moles

mole fraction of methyl alcohol (x2)

=moles of methyl alcohol

moles of methyl alcohol × moles of water

=0.1143

0.1143 + 1.4 = 0.1143

1.5143 = 0.0754

mole fraction of methyl alcohol = 0.0754mole fraction of water (x1)

= 1 – x2

= 1 – 0.0754 = 0.9245mole fraction of water = 0.9245% by mass of methyl alcohol

=massof methylalcohol

×100mass of water + mass of methyl alcohol

=3.658

× 100(25.2 + 3.658)

=3.658

× 10028.858

= 12.68% w/w

% by mass of methyl alcohol =12.68% w/w

*4. 12.8 cm3 of benzene is dissolved in 16.8cm3 of xylene. Calculate % by volume ofbenzene.

Given :volume of benzene = 12.8 cm3

volume of xylene = 16.8 cm3

To find :% by volume of benzene

Solution:% by volume of benzene

=volumeof benzene

×100volume of benzene + volumeof xylene

=12.8

× 10012.8 + 16.8 =

12.8× 100

29.6= 43.24 % w/w

% by volume of benzene = 43.24% w/w

*5. Calculate mole fraction of HCl in solutionof HCl containing 24.8% of HCl by mass.(H = 1, Cl = 35.5)

Given :% by mass of HCl = 24.8%

To find :mole fraction of HCl = ?

Solution:HCl solution is 24.8% by mass.Hence, 100g HCl solution contain 24.8g .HCl and (100 – 24.8) = 75.2gmole of HCl =

n2 =mass of HCl

molar mass of HCl

=24.8

36.5 = 0.6794 mol

mole of water =

n1 =mass of water

molar mass of water

=75.2

18 = 4.178 moles

mole fraction of HCl =

x2 =moles of HCl

moles of HCl + moles of water

=0.6794

0.6794 + 4.178

=0.6794

4.8574= 0.1398mole fraction of HCl = 0.1398

126


Solution :HNO3 solution is 12.2% by massHence 100g of HNO3 solution containing12.2g HNO3 and (100 – 12.2)

= 87.8g of H2OMoles of HNO3

n2 =3

3

mass of HNO

molar mass of HNO

n2 =12.2

63 = 0.1936 mol

moles of H2O

n1 =2

2

mass of H O

molar mass of H O

= 87.8

18 = 4.878 mol

mole fraction of HNO3 (x2)

=3

3 2

moles of HNO

moles of HNO + moles of H O

=0.1936

0.1936 + 4.878 = 0.1936

5.0713= 0.0382

mole fraction of HNO3 (x2)= 0.0382Mole fraction of H2O (x1)

= 1 – mole fraction of HNO3 (x2)= 1 – 0.0382 = 0.9618

Mole fraction of H2O (x1) = 0.9618

molality (m) =3

2

moles of HNO

mass of H O in kg

= –3

0.1936

87.8 × 10 kg = 2.205m

molality (m) = 2.205mDensity of HNO3 solution

=3

3

mass of HNO solution

volume of HNO solution

*6. Calculate mole fraction of solute in its 2molal aqueous solution.

Given :molality = 2m

To find :mole fraction of solute

Solution:Let the mass of water be 1 kg = 1000g

molality =no. of moles of solute

mass of water in kg

2 =moles of solute

1∴∴∴∴∴ moles of solute (n2) = 2 mol∴∴∴∴∴ moles of water =

n1 =mass of water

molar mass of water

=1000

18 = 55.55 mole

mole fraction of solute (x2)

=moles of solute

moles of solute + moles of water

=2

2 + 55.55 = 0.0347

mole fraction of solute = 0.0347

*7. Calculate the mole fractions, molality andmolarity of HNO3 in a solution containing12.2% HNO3. Given density of HNO3 =1.038g cm–3, H = 1, N = 14, O = 16.

Given :(i) % by mass of HNO3 = 12.2%(ii) Density of HNO3 = 1.038 g cm–3

To find :(i) mole fraction of HNO3

(ii) mole fraction of H2O(iii) molality = ?(iv) molarity = ?

127


moles of water (n1)

=2

mass of water

molar mass of H O

=4.2

18 = 0.2333 moles

mole fraction of H2SO4 (x2)

=2 4

2 4 2

moles of H SO

moles of H SO + moles of H O

=0.9775

0.9775 + 0.2333 = 0.9775

1.211= 0.8071

mole fraction of H2SO4 (x2) = 0.8071Density of H2SO4 solution

=2 4

2 4

mass of H SO solution

volume of H SO solution

1.91 =2 4

100


∴∴∴∴∴ vol. of H2SO4 soln. =100

1.91= 52.36cm3

= 52.36 × 10–3 dm3

Molarity =2 4 2

2 4

moles of H SO (n )


–3

0.9775

52.36 × 10 = 18.67 M

Molarity = 18.67 M

*9. Aqueous solution of NaOH is marked 10%(w/w). The density of the solution is 1.070gcm–3. Calculate (i) molarity (ii) molality and(iii) mole fraction of NaOH and water. Na= 23, H = 10, O = 16.

Given :(i) % by mass of NaOH = 10% w/w(ii) Density = 1.07 g/cm3

To find :(i) Molarity = ?

1.038 =3

100

volume of HNO solution

∴ vol. of HNO3 soln. =100

1.038= 96.34 cm3

= 96.34 × 10–3 dm3

Molarity =3

3

Moles of HNO

Volume of HNO solution

= –3

0.1936

96.34 × 10

= 2.01 MMolarity = 2.01 M

*8. Sulphuric acid is 95.8% by mass. Calculatemole fraction and molarity of H2SO4 ofdensity 1.91 cm3.(H = 1, S = 32, O = 16.)

Given :(i) % by mass of sulphuric acid = 95.8%(ii) Density of H2SO4 solution = 1.91 g/cm3

To find :(i) mole fraction of H2SO4

(ii) molarity = ?Solution :

H2SO4 solution is 95.8% by massHence, 100g of H2SO4 solution contains95.8% of H2SO4 and 100 – 95.8 = 4.2gof water.moles of H2SO4 (n2)

=2 4

2 4

mass of H SO

molar mass of H SO

=95.8

98= 0.9775 molesmoles of H2SO4 (n2) = 0.9775 moles

128


(ii) Molality = ?(iii) Mole fraction of NaOH (x2) = ?(iv) Mole fraction of water (x1) = ?

Solution :NaOH solution is 10% by massHence, 100g of NaOH solution contains10 g of NaOH in 100 – 10 = 90 g of water

moles of NaOH(n2) = mass of NaOH

molar massof NaOH

= 10

40 = 0.25 mole

moles of H2O (n1)

=mass of water

molar mass of water

=90

18 = 5 moles

mole fraction of NaOH (x2)

=2

moles of NaOH

moles of NaOH + moles of H O

=0.25

0.25 + 5.0 = 0.25

5.25 = 0.0476

mole fraction of NaOH (x2) = 0.0476mole fraction of H2O (x1)

= 1 – mole fraction of NaOH (x2)

molality =2

moles of NaOH

mass of H O in kg= 1 – 0.0476= 0.9524

= –3

0.25

90 × 10= 2.777 m

molality = 2.777 mDensity of NaOH solution

=mass of NaOH solution

volume of NaOH solution

1.07 = 100

volume of NaOH solution

∴∴∴∴∴ volume of NaOH solution

=100

1.07= 93.46 cm3

= 93.46 × 10–3 dm3

molarity = 3

molesof NaOH

valumeof NaOHsolutionindm

= –3

0.25

93.46 × 10 = 2.675 M

*10. Battery acid is 4.22 M aqueous H2SO4

solution, and has density of 1.21 g cm–3.What is the molality of H2SO4?(H = 1, S = 32, O = 16)

Given :molarity of H2SO4 aqueous = 4.22 M,density = 1.21 g cm–3

To find :molality = ?

Solution:let the volume of H2SO4 solution1 dm3 = 1000 cm3

molarity

=2 4 2

32 4

moles of H SO (n )

volume of H SO solution in dm

4.22 = 2 4moles of H SO

1∴∴∴∴∴ moles of H2SO4 (n2) = 4.22 moles

moles of H2SO4 (n2)

=2 4

2 4

mass of H SO

molar mass of H SO

4.22 = 2 4mass of H SO

98

∴∴∴∴∴ mass of H2SO4 = 98 × 4.22= 413.5 g

Density =2 4

2 4

mass of H SO solution


129


1.21 = 2 4mass of H SO solution

1000

∴∴∴∴∴ mass of H2SO4 solution = 1.21 × 1000= 1210 g

mass of solvent (H2O)= mass of H2SO4 solution – mass of H2SO4

= 1210 – 413.5 = 796.5g= 796.5 × 10–3 kg

molality (m)

=2 4

2

moles of H SO

mass of solvent (H O)in kg

= –3

4.22

796.5 × 10 = 5.298 m

molality = 5.298 m

*11. 35% (W/W) solution of ethylene glycol inwater, an anti-freezer used in automobilesin radiators as a coolant. It lowers freezingpoint of water to –17.6ºC. Calculate themole fraction of the components.

Given :% w/w of ethylene glycol solution = 35%

To find :mole fraction of ethylene glycol and water

Solution :Ethylene glycol solution is 35% by mass.Hence, 100g of ethylene glycolsolution contain 35g of ethylene glycolin 100 – 35 = 65g of water.molar mass of ethylene glycol[HO – CH2 – CH2 – OH] = 62g mol–1

moles of Ethylene glycol(n2)

=mass of ethylene glycol

molar mass of ethylene glycol

=35

62 = 0.5645 moles

moles of water (n1)

=mass of water

molar mass of water

=65

18 = 3.611 moles

mole fraction of ethylene glycol (x2)

=molesof ethyleneglycol

molesof ethyleneglycol+molesof water

=0.5645

0.5645 + 3.611 = 0.5645

4.1755 = 0.1352

mole fraction of ethylene glycol (x2)= 0.1352mole fraction of water (x1)

= 1 – mole fraction of ethylene glycol (x2)= 1 – 0.1352 = 0.8648

mole fraction of water = 0.8648

12. Find ppm (parts per million) of solute if 0.5gof solute is present in 107 g of solution.

Given :mass of solute = 0.5g,mass of solution = 107 g

To find :ppm

Solution:parts per million (ppm)

=6mass of solute


=6

7

0.5× 10

10 = 0.05 ppm

parts per million of solute = 0.05 ppm

*13. The tolerable level of contamination ofchloroform in a given sample of drinkingwater is found to be 15 ppm by mass.

(i) Express it in percentage by mass(ii) Determine molality of chloroform in the

130


given sample of water containing 15 ppmof chloroform.

Given :(i) ppm by mass of chloroform in water

= 15 ppmTo find :

(i) % by mass(ii) molality

Solution :ppm by mass of chloroform in water is 15ppm. Hence, 106 g of solution contains 15gof chloroform.mass of water = 106 – 15 ≈ ≈ ≈ ≈ ≈ 106 g% by mass of chloroform

=mass of chloroform


= 6

15× 100

10 = 1.5 × 10–3 %

% by mass of chloroform = 1.5 × 10–3 %molar mass of chloroform (CHCl3)

= 12 + 1 + 3 (35.5)= 13 + 106.5= 119.5 g mol–1

moles of chloroform

=mass of chloroform

molar mass of chloroform

=15

119.5 = 0.1255 mole

molality =2

moles of chloroform

mass of H O in kg

= 6 –3

0.1255

10 × 10 (kg)= 1.255 × 10–4 m

molality = 1.255 × 10–4 m

*14. What is the concentration of dissolvedoxygen at 25ºC at 1 atmospheric pressureif partial pressure of oxygen is 0.22 atm?The Henry’s law constant for oxygen is1.3 × 10–3 mol dm–3 atm–1.

Given :(i) partial pressure of oxygen = Po2

= 0.22 atm(ii) Henry Law constant for oxygen = KH

= 1.3 × 10–3, mol dm–3 atm–1

To find :Concentration of dissolved oxygen (s)

Solution:

According to Henry’s law, S = KH × O2p

K = 1.3 × 10–3 mol dm–3 atm–1

p = 0.22 atmHence,S = 1.3 × 10–3 mol dm–3 atm–1 × 0.22 atmS = 2.86 × 10–4 mol dm–3

*15. The solubility of nitrogen gas at 1 atmpressure at 25ºC is 6.8 × 10–4 mol dm–3.Calculate the solubility of N2 gas fromatmosphere at 25ºC if atmospheric pressureis 1 atmosphere and partial pressure of N2

gas at this temperature and pressure is 0.78atm.

Given :(i) solubility of N2 gas at 1 atm

at 25ºC = 6.8 × 10–4 mol dm–3

To find :solubility of N2 gas at 0.78 atm at 25ºC= ?

Solution :According to Henry’s law

S = N 2Kp , S = 6.8 × 10–4 mol dm–3,

/////////////////////////

//

131


N 2p = 1 atm

Hence,S = 6.8 × 10–4 mol dm–3 = K.1 atmHence,K = 6.8 × 10–4 mol dm–3

= 6.8 × 10–4 mol dm–3 atm–1

p = 0.78 atm, S = ?Therefore,S = 6.8 × 10–4 mol dm–3 atm–1 × 0.78 atm

= 5.304 × 10–4 mol dm–3

Solubility of N2 is reduced to = 5.304× 10–4 mol dm–3

TYPE - II - PROBLEM BASED ON COLLIGATIVE

PROPERTIES, LOWERING OF VAPOUR PRESSURE,RAOULTS LAW, MOLECULAR WEIGHT BY

LOWERING IN VAPOUR PRESSURE :*16. The vapour pressure of 2.1% solution of

a non-electrolyte in water at 100ºC is 755mm Hg. Calculate the molar mass of thesolute.

Given :(i) % by mass of non-electrolyte in water = 2.1%(ii) Temp = 100ºC(iii) Vapour pressure of solution = p = 755 mm kg

To find :molar mass of the solute = ?

Solution :100ºC pure water will boil.

∴∴∴∴∴ v.p of pure water ( 01p ) = Atmospheric

pressure = 760 mm of kgSolution is 2.1% by massHence 100 g of solution contains 2.1 g ofsolute (w2) in 100 – 2.1 = 97.9 g of water(w1)

01

01

p – p

p=

2 1

2 1

W M×

M W

760 – 755

760=

2

2.1 18×

M 97.9

M2 = 2.1× 18 × 760

5 × 97.9 = 58.69 g mol–1

molar mass of the solute (M2) = 58.6gmol–1

*17. The vapour pressure of water at 20ºC is17 mm Hg. Calculate the vapour pressureof a solution containing 2.8g of urea(NH2CONH2) in 50g of water N = 14, C= 12, O = 16, H = 1. (16.71 mm Hg)

Given :

(i) vapour pressure of water = 01p = 17 mm

Hg(ii) mass of urea = W2 = 2.8 g(iii) mass of water = W1 = 50 g

To find :vapour pressure of solution (p) = ?

Solution :molar mass of urea (NH2CONH2)M2 = 60 g mol–1

01

01

p – p

p=

2 1

2 1

W M×

M W

17 – p

17=

2.8 18×

60 50

17 – p =2.8 × 18 × 17

60 × 5017 – p = 0.2856

∴∴∴∴∴ p = 17 – 0.2856 = 16.71 mm HgVapour pressure of solution is 16.71mm Hg

132


*18. In an experiment, 18.04g of mannitol weredissolved in 100g of water. The vapourpressure of water was lowered by 0.309mm Hg. from 17.535 mm Hg. Calculate themolar mass of mannitol. (184.3 g mol–1)

Given :(i) mass of mannitol = W2 = 18.04 g(ii) mass of water = W1 = 100 g(iii) lowering of vapour pressure

= 01p – p

= 0.309 mm Hg(iv) vapour pressure of pure water

= 01p = 17.535 mm Hg

To find :molar mass of mannitol (M2) = ?

Solution:01

01

p – p

p=

2 1

2 1

W M×

M W

0.309

17.535=

2

18.04 18×

M 100

∴ M2 = 18.04×18×17.535

0.309×100 = 184.3 g mol–1

molar mass of mannitol (M2) =184.3 gmol–1

*19. Calculate the mass of a nonvolatile solute(molar mass 40 × 10–3 kg/mol), dissolvedin 114 × 10–3 kg of octane to reduce itsvapour pressures to 80%.

Given :(i) vapour pressure reduces to 80%(ii) mass of octane = 114 × 10–3 kg (W1)(iii) molar mass of non-volatile solute M2

= 40 × 10–3 kg/molTo find :

mass of non-volatile solute = W2 = ?

Solution :vapour pressure reduces to 80%

If vapour pressure of pure octane ( 01p ) = 40

then vapour pressure of solution (p) = 80molar mass of octane (C8H18)

= M2 = 8 (12) + 18 (1)= 96 + 18 = 114 g mol–1

moles of octane (n1)

=mass of octane

molar mass of octane

=–3

–3

114 × 10

114 × 10n1 = 1 mole

01

01

p – p

p= x2

100 – 80

100=

2

2

n

1 + n

∴∴∴∴∴20

100=

2

2

n

1 + n

1

5=

2

2

n

1 + n

1 + n2 = 5n2

1 = 5n2 – n2

4n2 = 1

∴∴∴∴∴ n2 =1

4 = 0.25 mol

moles of solute = mass of solute

molar mass of solute

0.25 = –3

mass of solute

40 × 10∴ mass of solute = 0.25 × 40 × 10–3

= 10 × 10–3 kgmass of solute (W2) = 10 × 10–3 kg

133


TYPE III - PROBLEM BASED ON ELEVATION

OF BOILING POINT :*20. Calculate the mass in grams of an impurity

of molar mass 100g mol–1 which would berequired to raise the boiling point of 50g ofChloroform by 0.30 K. (Kb for chloroform =3.63 K kg mol–1). (0.4132 g)

Given :(i) molar mass of Impurity= M2 = 100 g mol–1

= 100 × 10–3 kg mol–1

(ii) mass of Chloroform= W1 = 50 g = 50 × 10–3 kg

(iii) Elevation of boiling point= ΔTb = 0.3 K

(iv) Kb of chloroform = 3.63 K kg mol–1

To find :mass of Impurity = W2 = ?

Solution :

M2 =b 2

b 1

K W×

T WΔ

100 × 10–3 =2

–3

W3.63×

0.3 50 × 10

∴∴∴∴∴ W2 =–3 –3100 × 10 × 50 × 10 × 0.3

3.63= 0.4132 × 10–3 kg = 0.4132g

mass of Impurity = 0.4132g

*21. Boiling point of a solvent is 80.2ºC. When0.419 g of the solute of molar mass 252.4g mol–1 was dissolved in 75g of the solvent,the boiling point of the solution was foundto be 80.256ºC. Find the molal elevationconstant.

Given :(i) Boiling point of solvent

= Tº = 80.2ºC

(ii) Boiling point of solution= T = 80.256ºC

(iii) Mass of solute = W2= 0.419 g = 0.419 × 10–3 kg

(iv) Mass of solvent= W1 = 75g = 75 × 10–3 kg

(v) molar mass of solute= M2 = 252.4 g mol–1

= 252.4 × 10–3 kg mol –1

To find :Molar elevation constant = Kb = ?

Solution :ΔTb = T – Tº

= 80.256 – 80.2 = 0.056ºC

M2 =b 2

b 1

K W×

T WΔ

252.4 × 10–3 =–3

b–3

K 0.419 × 10×

0.056 75 × 10

∴∴∴∴∴ Kb=–3 –3

–3

252.4 × 10 × 0.056 × 75 × 10

0.419 × 10= 2.53 K kg mol–1

Molar elevation constant = Kb = 2.53K kg mol–1

*22. A solution containing 0.5126g ofnaphthalene (molar mass = 128.17 g mol–1)in 50.0g of CCl4 gives a boiling pointelevation of 0.402K. While a solution of0.6216g of unknown solute in the samemass of the solvent gives a boiling pointelevation of 0.647K. Find the molar massof the unknown solute.(Kb for CCl4 = 5.03K kg mol–1 of solvent)

Given :For naphthalene solution :mass of Naphthalene = W2

= 0.5126 g = 0.5126 × 10–3 kg

134


(ii) molar mass M2

= 128.17 g mol–1

= 12.8.17 × 10–3 kg mol–1

(iii) mass of CCl4 = W1

= 50 g = 50 × 10–3 kg(iv) ΔTb = 0.402 K

For unknown solute :(i) mass of unknown solute

= W2 = 0.6216 g= 0.6216 × 10–3 kg

(ii) mass of CCl4 = W2 = 50g = 50 × 10–3 kg(iii) ΔTb = 0.647 K(iv) Kb for CCl4 = 5.03 K kg mol–1

To find :molar mass of unknown solute = M2 = ?

Solution :For the solution of Unknown solute

M2 =b 2

b 1

K W×

T WΔ

=–3

–3

5.03 × 0.6216 × 10

0.647 × 50 ×10= 0.09665 kg mol–1

M2 = 96.65 g mol–1

*23. Calculate molality of a solution of asubstance which on dissolving in benzeneboils at 353.71 K. Kb for benzene is 2.53K kg mol–1 and boiling point of purebenzene is 353.35K.

Given :(i) Boiling point of pure benzene

= Tº = 353.35 K(ii) Boiling point of solution= 353.71 K(iii) Kb for benzene = 2.53 K kg mol–1

To find :molality = ?

Solution :ΔTb = T – Tº

= 353.71 – 353.35 = 0.36 KΔTb = Kb × m0.36 = 2.53 × m

m =b

b

T

K

Δ =

0.36

2.53 = 0.1423 mol kg–1

molality = 0.1423 mol kg–1

*24. 3.795g of sulphur is dissolved in 100g ofCS2. This solution boils at 319.81 K. Whatis the molecular formula of sulphur insolution? The boiling point of CS2 is 319.45K. Given that Kb for CS2 = 2.42 K kgmol–1 and atomic mass of S = 32.

Given :(i) mass of sulphur

= W2 = 3.795 g = 3.795 × 10–3 kg(ii) mass of CS2

= W1 = 100 g = 100 × 10–3 kg(iii) boiling of CS2

= Tº = 319.45 K(iv) boiling point solution

= T = 319.81 K(v) Kb for CS2 = 2.42 K kg mol–1

(vi) Atomic mass of S = 32To find :

molecular formula of sulphur = ?Solution :

ΔTb= T – T°= 319.81 – 319.45 = 0.36 K

M2 =b 2

b 1

K × W

T × WΔ

=–1 –3

–3

2.42K kg mol ×3.795×10 kg

0.36K×100×10 kg

135


= 0.2551 kg mol–1

= 255.1 g mol–1

Atomic mass of S = 32∴ Number of atoms in a molecule of sulphur

=molar mass of S

atomic mass of S

=255.1

32 = 7.97 ≈ 8

Hence, molecular formula of S is S8 inCS2

*25. Boiling point of water at 750 mm of Hgis 99.63ºC. How much sucrose must beadded to 500 g of water so that it boils at100ºC?

Given :(i) Boiling point of water = Tº = 99.63ºC(ii) Boiling point of solution = T = 100ºC(iii) Mass of water = W1 = 500 g = 0.5 kg(iv) Kb of water = 0.52 K kg mol–1

To find :mass of sucrose = W2 = ?

Solution :molar mass of sucrose (C12H22O11)

M2 =12(12) + 22(1) + 11(16)=342 g mol–1

=342 × 10–3 kg mol–1

ΔTb =T – Tº=100 – 99.63= 0.37ºC

M2 =b 2

b 1

K W×

T WΔ

342 × 10–3 =2

–3

W0.52×

0.37 500 × 10

∴∴∴∴∴ W2 =–3 –3342×10 ×0.37×500×10

0.52

=121.7 × 10–3 kg = 121.7 gmass of sucrose = 121.7 g

TYPE IV - PROBLEM BASED ON FREEZING

POINT DEPRESSION :*26. An aqueous solution containing 12.5 × 10–3

kg of non-volatile compound in 0.1 kg ofwater freezes at 272.49 K. Determinemolar mass of the compound.(Kf for water = 1.86 K kg mol–1, B.P. ofwater = 273.15 K)

Given :(i) mass of solute = W2 = 12.5 × 10–3 kg(ii) mass of water = W1 = 0.1 kg(iii) freezing point of solution = T = 272.49K(iv) freezing point of water = Tº = 273.15 K(v) Kf of water = 1.86 K kg mol–1

To find :molar mass of solute = M2 = ?

Solution:ΔTf = Tº – T

= 273.15 – 272.49 = 0.66K

M2 =f 2

f 1

K W×

T WΔ

=–31.86 × 12.5 × 10

0.66 × 0.1= 352.2 × 10–3 kg mol–1

= 352.2 g mol–1

molar mass of solute = 352.2g mol–1

27. Calculate the freezing point of solutionprepared by dissolving 4.5g of glucose(Molar mass = 180g mol–1) in 250g ofbromoform. Given, freezing point of

136


Given :(i) For urea

mass of urea= W2 = 1.02 g = 1.02 × 10–3 kgmass of solvent= W1 = 98.5 g = 98.5 × 10–3 kgΔTf = 0.211 K

(ii) For unknown compoundmass of unknown compound= W2 = 1.6 g = 1.6 × 10–3 kgmass of solvent= W1 = 86g = 86 × 10–3 kgΔTF = 0.34 K

To find :molar mass of unknown compound= M2 = ?

Solution :molality of urea solution,

m =massof urea in kg

molar mass urea×massof solvent in kg

=–3

–3 –1 –3

1.02 × 10 kg

60 × 10 kg mol × 98.5 × 10 kg

= 0.17 mol kg–1

Kf = fT

m

Δ

= –1

0.211K

0.17 molkg = 1.241 K kg mol–1

M2 =f 2

f 1

K × W

T × WΔ

=–1 –3

–3

1.241K kg mol × 1.60 × 10 kg

0.34K × 86.0 × 10 kg= 67.90 × 10–3 kg mol–1

= 67.90 g mol –1

molar mass of unknown compound= M2 = 67.90 g mol –1

bromoform = 7.8ºC and Kf for bromoform= 14.4 K kg mol–1.

Given :(i) mass of glucose

= M2 = 4.5g = 4.5 × 10–3 kg(ii) molar mass of glucose

= 180 g mol–1

= 180 × 10–3 kg mol–1

(iii) mass of bromoform= W1 = 250g = 250 × 10–3 kg

(iv) freezing point of bromoform = Tº = 7.8ºC(v) Kf of bromoform = 14.4 K. kg mol–1

To find : freezing point of solution = T = ?

Solution :

ΔTf =f 2

2 1

K × W

M × W

ΔTf =–1 –3

–3 –1 –3

14.4 K kg mol × 4.5×10 kg

180×10 kg mol ×250×10 kg

∴∴∴∴∴ = 1.44 K = 1.44ºCΔTf = freezing point of solvent – freezingpoint of solutionHence, freezing point of solution

= freezing point of solvent – ΔTf

= 7.80ºC – 1.44ºC = 6.36ºCfreezing point of solution = T = 6.36ºC

*28. 1.02 g of urea when dissolved in 98.5 gof certain solvent decreases its freezingpoint by 0.211 K. 1.60 g of unknowncompound when dissolved in 86.0 g of thesame solvent depresses the freezing pointby 0.34 K. Calculate the molar mass ofthe unknown compound.(Urea NH2CONH2, N = 14, C = 12, O= 16, H = 1)

137


TYPE V - PROBLEMS BASED ON OSMOTIC

PRESSURE :29. Calculate the Osmotic pressure of 3.6 g

of glucose (molar mass = 180 g mol–1)dissolved in 100 ml of water at 298 K.(R = 0.0821 L atm mol–1 K–1)

Given :(i) mass of glucose = 3.6 g(ii) molar mass of glucose = 180 g mol–1

(iii) volume of water = 100 ml = 100 × 10–3 dm3

(iv) T = 298 K(v) K = 0.0821 L atm mol–1 K–1

To find :Osmotic pressure π = ?

Solution:π V = nRT

π × 100 × 10–3 =3.6

× 0.0821 × 298180

∴∴∴∴∴ π = –3

3.6 × 0.0821× 298

180 × 100 × 10 = 4.893 atm

Osmotic pressure πππππ = 4.893 atm

*30. 30 g of glucose dissolved in one litre ofwater has an osmotic pressure 4.91 atmat 303 K. If the osmotic pressure of theglucose solution is 1.5 atm at the sametemperature, what would be itsconcentration? (Molar mass of glucose is180 g mol–1)

Given :Case - I

(i) mass of glucose = 30 g(ii) volume of water = 1 L(iii) osmotic pressure = π1 = 4.91 atm(iv) T = 303 K

Case - IIπ2 = 1.5 atm

To find :concentration in Case II = C2 = ?

Solution :π1 = C1 RT and π2 = C2 RT

Hence, 1

2

π

π =

1

2

C

C

π1 = 4.91 atm, π2 = 1.5 atmC1 = mol L–1

(molar mass of glucose = 180 g mol–1)

Hence, C2 = 2

1

π

π C1 =

1.5atm 30× M

4.91atm 180

1

moles of glucoseC =

volume of solutionmass of glucose

=molar massof glucose+ volumeof solution

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

= 0.0509 MThe concentration in 2nd Case =0.0509 M.

*31. What is the mass of sucrose in its 1 Lsolution (molar mass = 342 g mol–1) whichis isotonic with 6.6 × 10–3 kg L–1 of urea(NH2CONH2)? (Given molar masses, H =1, C = 12, N = 14, O = 16 in g mol–1).

Given :Sucrose solution :

(i) Volume = 1 L(ii) Molar mass of sucrose = 342 g mol–1

Urea solution(i) Mass of urea = 6.6 × 10–3 kg(ii) Volume of solution = 1 L(iii) Molar mass of urea (NH2CON2) = 60 g

mol–1 solution are isotonicTo find :

mass of sucrose = ?

138


Solution:π = CRTFor isotonic solution π1 = π2

Hence, C1 = C2; C1 and C2 areconcentrations in mol L–1

Molarity of urea solution =mass of urea

molar massMolar mass of urea

= (14 + 2 +12 + 16 + 14 + 2) g mol–1

= 60 g mol–1 = 60 × 10–3 kg mol–1

Concentration of urea

=–3

–3

6.6 × 10 kg

60 × 10 kg/mol = 0.11M = C2

Since urea solution is isotonic with sucrosesolution, C1 = C2

Hence, C1 = –3

mass of sucrose

342 × 10 = 0.11

Mass of sucrose = 0.11 × 342 × 10–3 kgHence,mass of sucrose = 37.62 × 10–3 kg

*32. Osmotic pressure of a solution containing6.8 × 10–3 kg of protein per 1 × 10–4 m3

of solution is 3.02 × 103 Pa at 37ºC.Calculate the molar mass of protein. (R= 8.314 J K–1 mol–1).

Given :(i) mass of proteins

= W2 = 6.8 × 10–3 kg(ii) volume of solution

= 1 × 10–4 m3

(iii) osmotic pressure = π = 3.02 × 103 Pa(iv) R2 = 8.314J K–1 mol–1

(v) T = 37ºC + 273.15 = 310.15 KTo find :

molar mass of protein

Solution :πV = n RT

πV =2

2

W

M RT

3.02 × 103 × 1 × 10–4 = –3

2

6.8×10×8.314×310.15

M

M2 =–3

3 –4

6.8×10 ×8.314×310.15

3.02×10 ×1×10

= 58.06 kg mol–1

= 58060 g mol–1

molar mass of proteins = 58060 g mol–1

*33. At 298 K, 1000 cm3 of a solution containing4.34g of solute shows osmotic pressure of2.55 atm. What is the molar mass ofsolute? (R = 0.0821 L atm K–1 mol–1)

Given :(i) T = 298 K(ii) volume of solution = 1000 cm3 = 1 L(iii) mass of solute = W2 = 4.34 g(iv) osmotic pressure = π = 2.55 atm(v) R = 0.0821 L atm K–1 mol–1

To find :molar mass of solute = M2 = ?

Solution:πV = n RT

πV =2

2

WRT

M

2.55 × 1 = 2

4.34× 0.021 × 298

M

M2 =4.34 × 0.0821 × 298

2.55

= 41.64 g mol–1

molar mass of solute = 41.64 g mol–1

139


TYPE VI - PROBLEM BASED ON VANT

HOFF FACTOR :*34. 0.2m aqueous solution of KCl freezes at –

0.680ºC. Calculate vant Hoff factor andobserved osmotic pressure of solution at 0ºC.

Given :(i) molality of Aqueous KCl solution = 0.2m(ii) freezing point of solution = T = –0.680ºC(iii) Kf = 1.86 K kg mol–1

To find :(i) vant Hoff factor(ii) observed osmotic pressure at 0ºC

Solution :ΔTf (observed) = Tº – T

= 0 – (– 0.680)= 0.680ºC

ΔTf (Theoretical) = Kf × m= 1.86 × 0.2= 0.372º C

i =f

f

T (observed)

T (theoretical)

ΔΔ

=0.680

0.372 = 1.83

i = 1.83Let the volume of solution = 1 L = 1 kg(for dilute Aqueous solution)

molality (m) =moles of solute

mass of solvent in kg

0.2 = 2n

1∴∴∴∴∴ n solute = 0.2 moles

π (Theoretical) = CRT

=n

RTV

=0.2

× 0.082 × 2731

= 4.477 atm

i =(observed)

(Theoretical)

π

π

1.83 =(observed)

4.477

π

∴∴∴∴∴ πππππ (observed) = 1.83 × 4.477 = 8.19 atm

*35. 0.01 molal aqueous solution of K3Fe(CN)6

freezes at –0.062ºC. Calculate thepercentage dissociation of solute, Kf forwater is 1.86 K kg–1 mol–1.

Given :(i) molality of K3Fe(CN)6 = 0.01 M(ii) freezing point of solution = T = –0.062ºC(iii) Kf = 1.86 K kg–1 mol–1

To find :% dissociation of solute

Solution :+ 3–

3 6 6K Fe(CN) 3K + Fe(CN)⎯⎯→

∴∴∴∴∴ n′ = 4ΔTf (Theoretical) = Kf × m

= 1.86 × 0.01= 0.0186 K

ΔTf (observed) = Tº – T= 0 – (–0.062)= 0.062ºC

i =f

f

T (observed)

T (theoretical)

ΔΔ

=0.062

0.0186= 3.333

α =i – 1

n – 1′

=3.333 – 1

4 – 1 = 0.78

% dissociation = α × 100= 0.78 × 100 = 78%

percentage dissociation (% ααααα) = 78%

140


∴∴∴∴∴ = 0.538 mol kg–1

i =m(observed)

m(theoretical)

=0.538

0.5 = 1.076

vant Hoff factor = 1.076

– +2 2–CH FCOOH CH FCOO + H⎯⎯→

Q n′ = 2Degree of dissociation (α)

=i – 1

n – 1′

= 1.076 – 1

2 – 1 = 0.076

Degree of dissociation (ααααα) = 0.076

– +2 2CH FCOOH CH FCOO + H

Initialmoles m o oEqulb moles m – m m mα α α

⎯⎯→

Ka =– +

2–

2

(CH FCOO ) (H )

(CH FCOO )

=m × m

m – m

α α

α

a =2m

1 –

α

α

=20.5 (0.076)

1 – 0.076

=20.05 × (0.0760)

0.924Ka = 3.126 × 10–3

Dissociation constant Ka = 3.126 × 10–3

*37. 10g of monochlorobutyric acid wasdissolved in 250g of water. If dissociationconstant of the acid is 1.45 × 10–3 and Kf

= 1.86 K kg mol–1, calculate depressionof the freezing point.(C = 12, Cl = 35, H = 1, O = 16)

*36. 19.5g of monofluoro acetic acid wasdissolved in 0.5 kg of water. The freezingpoint of solution was observed to be –1.0ºC.Calculate the vant Hoff factor, degree ofdissociation and dissociation constant of acid.(Atomic masses C = 12, H = 1, F = 19,O = 16 and Kf =1.86 K kg–1 mol–1)

Given :(i) mass of monofluoro acetic acid

(CH2FCOOH) = 19.5 g(ii) mass of water = 0.5 kg(iii) freezing point of solution = T = –100ºC(iv) Kf = 1.86 K kg–1 mol–1

To find :(i) vant Hoff factor(ii) Degree of dissociation(ii) Dissociation constant of acid

Solution:ΔTf = Tº – T

= 0 – (–1) = 1ºC= 1 K

molar mass of CH2FCOOH = 78 g mol–1

moles of CH2FCOOH

=2

2

mass of CH FCOOH

molar mass of CH FCOOH

=19.5

78 = 0.25 moles in 0.5 kg water.

molality =2moles of CH FCOOH

mass of water in kg

=0.25

0.5 = 0.5 mol kg–1

ΔTf (observed) = Kf × m= 1.86 × m(observed)

1

1.86= m(observed)

141


Given :(i) mass of monochlorobutyric acid = 10g(ii) mass of water = 250g(iii) dissociation constant = Ka = 1.45 × 10–3

(iv) Kf = 1.86 K kg mol–1

To find :Depression in freezing point = ?

Solution :molar mass of monochlorobutyric acid= 122 g mol–1

[CH3CH2CHCICOOH]moles of monochlorobutyric acid

=mass of acid

molor mass of acid

=10

122 = 0.082 moles

molality =2

mass of acid

mass of H O in kg

= –3

0.082

250 × 10= 0.328 mol kg–1

– +3 2 3 2CH CH CHClCOOH CH CH CHClCOO + H

Initialmoles m 0 0molesat eqlb m – m m mα α α

→

Ka = – +

3 2

3 2

(CH CH CHClCOO ) (H )

(CH CH CHClCOOH)

1.45 × 10–3 =m × m

m (1 – )

α α

α

1.45 × 10–3 =2m

1 –

αα

[Q α is very small 1 – α ≈ 1]∴ 1.45 × 10–3 = mα2

1.45 × 10–3 = 0.328 × α2

∴ α2 =–31.45 × 10

0.328α2 = 4.42 × 10–3

∴ α = –444.2 × 10 = 6.648 × 10–2

α =i – 1

n – 1′ (n′ = 2)

6.648 × 10–2 =i – 1

2 – 16.648 × 10–2 + 1 = i

i = 1.06648ΔTf (theoretical) = Kf × m

= 1.86 × 0.328= 0.61 K

i = f

f

T (observed)

T (theoretical)

ΔΔ

∴ ΔTf (observed) = i × ΔTf (theoretical)= 1.06648 × 0.61= 0.65 K

Depression of freezing point = 0.65 K

*38. 8g of benzoic acid when dissolved in 100gof benzene lowers its freezing point by1.62K. Calculate degree of association ofbenzoic acid if it, forms dimers in benzeneKf for benzene is 4.9K kg mol–1

(C = 12, H = 1, O = 16)Given :

(i) mass of benzoic acid = W2 = 8g(ii) mass of benzene = W1 = 100g(iii) ΔTf = 1.62 K(iv) Kf = 4.9 K kg mol–1

To find :Degree of association of benzoic acid = ?

Solution :

. M2 =f 2

f 1

K W×

T W

=–3

–3

4.9 × 8 × 10

1.62 × 100 × 10= 241.9 × 10–3 kg mol–1

142


∴∴∴∴∴ M2 (observed) = 241.9 g mol–1

M2 (theoretical) of benzoic acid(C6H5COOH) is 122.Observed Molecular mass of benzoic acidis almost double the theoretical molecularmass.

∴∴∴∴∴ Benzoic acid must associate to form adimer.

6 5 6 52 C H COOH (C H COOH)

Initialmoles m om

molesat equlb m – m2

αα

Total moles in solution

=m

m – m +2

α

α

= m 1 –2

α⎛ ⎞⎝ ⎠

i =observed moles in solution

theoretical moles in solution

=m 1 –

2m

α⎛ ⎞⎝ ⎠

= 1 –2

α

i =Theoretical molecular mass

Observed molecular mass

=122

241.98 = 0.5041

1 –2

α

= i

0.5041 = 1 –2

α

2

α

= 1 – 0.5041 = 0.4958

∴∴∴∴∴ α = 2 × 0.4958∴∴∴∴∴ α = 0.9918

% association of benzoic acid = 99.18%

*39. 0.6 ml of glacial acetic acid with density 1.06gm L–1 is dissolved in 1 kg water and thesolution froze at –0.0205ºC. Calculate VantHoff factor and Kf for water is 1.86 K kgmol–1 (i = 1.041, Ka = 1.86 × 10–5)

Given :(i) volume of acetic acid = 0.6 ml(ii) density = 1.06 g mL–1

(iii) mass of water = 1 kg(iv) freezing point of solution = T = –0.0205ºC(v) Kf = 1.86 K kg mol–1

To find :(i) vant Hoff factor (i) = ?(ii) dissociation constant (Ka) = ?

Solution :

Density =mass of acetic acid

volume of acetic acid

1.06 =mass of acetic acid

0.6∴∴∴∴∴ mass of acetic acid = 1.06 × 0.6g

moles of acetic acid

=mass of acetic acid

molar mass of acetic acid

=1.06 × 0.6

60 = 0.0106 moles

molality (theoretical)

=2

mass of acetic acid

mass of H O in kg

=0.0106

1 = 0.0106 mol kg–1

ΔTf (observed) = Tº – T= 0 – (–0.0205)= 0.0205ºC

ΔTf (theoretical) = Kf × M(Theoretical)= 1.86 × 0.0106= 0.0197ºC

143


i =f

f

T (observed)

T (theoretical)

ΔΔ

= 0.0205

0.0197= 1.041

– +3 3CH COOH CH COO + H

∴∴∴∴∴ n′ = 2∴∴∴∴∴ Degree of dissociation (α)

= i – 1

n – 1′

= 1.041 – 1

2 –1 = 0.041

Degree of dissociation (α) = 0.041

– +3 3CH COOH CH COO + H

Initialmoles m 0 0molesat equlb m – m m mα α α

∴ Ka =– +

3

3

(CH COO ) (H )

(CH COOH)

=m × m

m – m

α α

α =

2m

1 –

α

α

=20.0106 × (0.041)

1 – 0.041 = 1.86 × 10–

dissociation constant (Ka) = 1.86 × 10–5

*40. Henry’s law constant for the molality ofmethane in benzene at 298 K is 4.27 × 105

mm Hg. Calculate the solubility of methanein benzene at 298 K under 760 mm Hg.

Given :Henry’s law constant (K) = 4.27 × 105 mmHg Pressure (P) = 760 mm Hg

To find :Solubility (s) = ?

Solution :p = KHm

m =H

p

k = 5

760mm Hg

4.27 × 10 mm Hg

= 1.78 × 10–3

Solubility (s) = 1.78 × 10–3

NUMERICALS FOR PRACTICETYPE - I - PROBLEM BASED ON CONCENTRATION OF SOLUTION :

*1. 6 g of urea was dissoved in 500g of water. Calculate percentage by mass of urea in solution.[Ans : 1.186% by mass]

*2. 58 cm3 of ethyl alcohol was dissloved in 400 cm3 of water to form 454 cm3 of solution ofethyl alcohol. Calculate percentage by volume of ethyl alcohol in water.

[Ans : 12.78 % by volume]

*3. 23g of ethyl alcohol (Molar mass 46g mol–1) is dissolved in 54 g of water (molar mass 18gmol–1). Calculate the mole fraction of ethyl alcohol and water in solution. [Ans : 1]

*4. A solution of NaOH (Molar mass 40g mol–1) was prepared by dissolving 1.6g of NaOH in500 cm3 of water. Calculate molarity of NaOH solution. [Ans : 0.08 mol dm–3]

*5. 11.11 g of urea was dissolved in 100 g of water. Calculate the molality of solution. (N = 14,H = 1, C = 12, O = 16) [Ans : 1.852 mol kg–1]

144


*6. 34.2 g of sugar was dissolved in water to produce 214.2 g of sugar syrup. Calculate molalityand mole fraction of sugar in the syrup (C = 12, H = 1, O = 16)

[Ans : 0.556 mol kg–1, 0.0099]

*7. Calculate molarity and molality of sulphuric acid solution of density 1.198g cm–3 containing27% by mass of sulphuric acid (molar mass of H2SO4 = 98g mol–1)

[Ans : 3.77 mol kg–1, 3.301 mol dm–3]

*8. Commercially available concentrated hydrochloric acid is an aqueous solution containing 38%HCl gas by mass. If its density is 1.1 g cm–3, calculate the molarity of HCl solution and alsocalculate mole fractions of HCl and H2O. [Ans : 11.45 mol dm–3, 0.232, 0.768]

TYPE - II - PROBLEMS BASED ON HENRY’S LAW :9. What is the solubility of oxygen if partial pressure of oxygen is 0.012 atm. The Henry’s law

constant for oxygen is 1.3 × 10–3 mol dm–3 atm–1. [Hint : S = KH × PO2][Ans : 1.56 × 10–5 mol dm–3]

TYPE - III - PROBLEMS BASED ON LOWERING OF VAPOUR PRESSURE :*10. The vapour pressure of a solution containing 13 × 10–3 kg of solute in 0.1 kg of water at

298 K is 27.371 mm Hg. Calculate the molar mass of the solute. Given that the vapour pressureof water at 298 K is 28.065 mm Hg. [Ans : 94.63 g mol–1]

*11. The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non volatilesolute of mass 2.175 × 10–3 kg is added to 39.0 × 10–3 kg of benzene. The vapour pressureof the solution is 600 mm Hg. What is the molar mass of the solute? (Given : atomic massesC = 12, H = 1) [Ans : 69.6 g mol–1]

*12. A solution is prepared from 26.2 × 10–3 kg of an unknown substance and 112.0 × 10–3 kgacetone at 313 K. The vapour pressure of pure acetone at this temperature is 0.526 atm. Calculatethe vapour pressure of solution if the molar mass of substance is 273.52 × 10–3 kg mol–1.(Given the atomic masses C = 12, H = 1, O = 16) [Ans : 0.500 atm]

TYPE IV - PROBLEMS BASED ON BOILING POINT ELEVATION :*13. 0.15 molal solution of a substance boils at 373.23 K. Calculate molal elevation constant of

water. (Given boiling point of water = 373.15 K) [Ans : 0.53 K kg mol–1]

*14. A solution is prepared by dissolving 1.9 × 10–2 kg ammonia in 400 g of water. Calculate elevationin boiling point if Kb for water is 0.52 K kg mol–1 (N = 14, H = 1, O = 16)[Ans : 1.45 K]

*15. A solution containing 1.21g of camphor (molar mass 152g mol–1) in 26.68g of acetone boilsat 329.95 K. The boiling point of pure acetone is 329.45 K. Calculate molal elevation constantfor acetone. [Ans : 1.676 K kg mol–1]

145


*16. A solution was prepared by dissloving certain amount of compound in 31.8g of CCl4 has aboiling point of 0.392 K higher than that of pure CCl4. If the molar mass of compound is 128gmol–1, calculate the mass of the solute dissolved. (Given Kb for CCl4 = 5.02 K kg mol–1)

[Ans : 0.317 g]TYPE V - PROBLEMS BASED ON FREEZING POINT DEPRESSION :

*17. 5.08g of a substance dissolved in 50g of water lowered the freezing point by 1.2 K. Calculatethe molar mass of the substance. The molar depression constant for water is1.86 K kg mol–1. [Ans : 157.48g mol–1]

*18. 0.440 × 10–3 kg of a substance (molar mass = 178.9 × 10–3 kg mol–1) dissolved in 22.2 ×10–3 kg of benzene lowered the freezing point of benzene by 0.567 K. Calculate the molaldepression constant for benzene. [Ans : 5.12 K kg mol–1]

*19. The observed depression in the freezing point of water for a particular solution is 0.087 K.Calculate the molality of the solution if molal depression constant for water is1.86 K kg mol–1. [Ans : 0.0467 mol kg–1]

*20. A solution of glucose (C6H12O6) was prepared by dissolving certain amount in 100g of water.The depression in freezing point was 0.0410 K. If molal depression constant for water is 1.86K kg mol–1, calculate the mass of glucose dissolved. (C = 12, H = 1, O = 16)

[Ans : 0.396 g of glucose]TYPE VI - PROBLEMS BASED ON OSMOTIC PRESSURE :

*21. Calculate the osmotic pressure of 4.5g of glucose (molar mass = 180g mol–1) dissolved in100 ml of water at 298 K (R = 0.0821 L atm mol–1K–1) [Ans : 6.116 atmospheres]

*22. A solution of cane sugar (molar mass 342g mol–1) containing 17.8g L–1 has an osmotic pressure1.2 atm. Calculate the temperature of the solution. (R = 0.082 L atm mol–1K–1)

[Ans : 281.4 K]

*23. A solution has an osmotic pressure of 3.90 × 105Nm–2 at 300 K. Calculate its volume containing1 mole of solute if solution of same solute has osmotic pressure of 2.82 × 105Nm–2 and contains1 mole of solute in 10.5 m3. [Ans : 7.59 m3]

*24. Calculate the volume of a solution containing 34.2g of cane sugar (Molar mass = 342gmol–1) which has an osmotic pressure 2.42 atm at 20ºC. [Ans : 0.9945 L]

*25. A solution of particular amount of organic substance of molar mass 196g mol–1 dissolved in2 litres of water gave an osmotic pressure of 0.54 atm at 12ºC. Calculate the mass of solutedissolved. (R = 0.0821 L atm K–1 mol–1) [Ans : 9.04g]

TYPE VII - PROBLEMS BASED ON VANT HOFF FACTOR :*26. Calculate the Vant Hoff factor of CdSO4 (molecular mass 208.4 g mol–1) if the dissolution

146


of 5.21g of CdSO4 in 500 ml of water gives a depression in freezing point of 0.168ºC(Kf of water = 1.86 K kg mol–1)Hint :

M2 =f 2

f (observed) 1

K × W

T × WΔ∴ ΔTf = 0.093ºC

(Theoretical)

i =f

f

T (observed)

T (Theoretical)

ΔΔ

i =0.168

0.093 = 1.8

Vant Hoff Factor (i = 1.8)

HIGHER ORDER THINKING SKILLS (HOTS)1. Calculate the molarity of water if its density is 1000kg/m3.

Sol. : Since density of water is 1000kg/m3

∴ 1000 kg of water = 1 m3

∴ 1 kg of water = 10–3m3

∴ 1000 g of water = 10–3 × 103dm3

= 1 dm3

No. of moles of water = Mass of water / molar mass of water= 1000/18= 55.55moles

Molarity = moles of water 55.55

Molarity = = 55.55MVolume 1

2. Among the 0.1M solutions of urea, NaCl, BaCl2, Na3PO4 and Al2(SO4)3, find solutions havinghighest and lowest vapour pressure, boiling point, freezing point. Find the one having highestand lowest value of colligative properties.

Sol. Vapour pressure and freezing point will be lowest while boiling point highest fo Al2(SO4)3.Vapour pressure and freezing point will be highest while boiling point lowest for NaCl.NaCl will have lowest value of all colligative properties.Al2(SO4)3 will have highest value of all colligative properties.

3Introduction :The energy of any body is defined as its capacity to do work. eg. a gas at higher temperaturehas more energy and hence, has greater capacity to do work than the same gas at lowertemperature.

ChemicalThermodynamics AndEnergetics

Chapter

SYLLABUS3.1 LIMITATIONS OF THERMODYNAMICS

3.2 BASIC CONCEPTS INTHERMODYNAMICS

(a) Types of system(b) Properties of system(c) State and state functions(d) Properties of state functions(e) Thermodynamic equilibrium(f) Types of processes(g) Adiabatic process(h) Reversible process(i) Irreversible process( j ) Distinction between Isothermal and

Adiabatic process

3.3 NATURE OF HEAT AND WORK(a) Work(b) Work during expansion(c) Work during contraction(d) Pressure-volume work(e) Expression for pressure-volume work(f) Free expansion(g) Concept of maximum work(h) Conditions for maximum work(i) Expression for maximum work( j ) Path dependence nature of work

(k) Heat (q)( l ) Units of energy and work

(m) Sign conventions of W and qQuestions and AnswersTheoretical MCQsNumerical MCQsAdvanced MCQs

3.4 INTERNAL ENERGY

3.5 FIRST LAW OF THERMODYNAMICS(a) Statements of first law of thermodynamics(b) Mathematical equation of first law of

thermodynamics(c) Modified first law of thermodynamics(d) First law of thermodynamics for various

processesQuestions and AnswersTheoretical MCQsNumerical MCQsAdvanced MCQs

3.6 ENTHALPY(a) Relationship between ΔH, ΔU & PΔ V(b) Relationship between ΔH and ΔU for

chemical reactions(c) Work done in chemical reaction(d) Conditions under which ΔH = ΔU

"A person starts to live when he can live outside himself." -Einstein


148

3.7 ENTHALPIES OF PHYSICAL CHANGES(a) Enthalpy of phase transition(b) Enthalpy of atomic or molecular changes(c) Enthalpy of solution(d) Enthalpy of dilution


3.8 THERMOCHEMISTRY(a) Enthalpy of chemical reactions(b) Standard enthalpy of formation(c) Standard enthalpy of combustion(d) Bond enthalpy(e) Reaction enthalpy from Bond enthalpy(f) Hess’s law of constant heat summation(g) Applications of Hess’s law


3.9 SECOND LAW OF THERMODYNAMICS

3.10 SPONTANEOUS PROCESS(IRREVERSIBLE PROCESS)

(a) Energy and spontaneity(b) Entropy

(c) Quantitative definition of entropy(d) Entropy and spontaneity (2nd law of

thermodynamics)Questions and AnswersTheoretical MCQsNumerical MCQsAdvanced MCQs

3.11 GIBBS ENERGY(a) Gibbs energy and spontaneity(b) Predicting spontaneity of a process in

terms of ΔG(c) Temperature of equilibrium between

spontaneous and nonspontaneousProcess

(d) ΔG and equilibrium constant

3.12 THIRD LAW OF THERMODYNAMICS(a) Standard molar entropy (Sº)(b) Usefulness standard molar entropy (Sº)


HOURS BEFORE EXAM NUMERICALS WITH SOLUTION NUMERICALS FOR PRACTICE HIGHER ORDER THINKING SKILLS

→ Energy has different forms such as kinetic energy, potential energy, heat energy, radiant energy,electrical energy, chemical energy.

→ All types of energy can be converted from one form to other. Eg. when water stored in adam falls down, its potential energy is converted into kinetic energy. In galvanic cells, the chemicalenergy is converted into electrical energy.

→ Energy can neither be created nor destroyed. When one form of energy disappears, the otherform of energy appear in equal magnitude. Thus, different forms of energy are relatedquantitatively to each other.

→ Definition : Thermodynamics is the branch of science that deals with the different formsof energy, the quantitative relationships between them and the energy changes that occurin physical and chemical processes.

Chapter - 3 Chemical Thermodynamics And Energetics

149

3.1 LIMITATIONS OF THERMODYNAMICS :

Concept Explanation :(a) It gives no information regarding the rates at which physical and chemical processes occur.(b) It deals only with initial and final states of the system.(c) It does not tell anything about the mechanism of the process.(d) It deals with the properties of macroscopic systems (i.e. the systems containing large number

of atoms, ions or molecules) such as temperature, pressure, volume etc.(e) Thermodynamics does not deals with the microscopic properties of the system such as

electronic structure of atom, constitution of molecule from atoms etc.

3.2 BASIC CONCEPTS IN THERMODYNAMICS :

Concept Explanation :Universe is divided into two parts :

(1) System : A portion of the universe chosen to study the thermodynamic properties is calleda system.

(2) Surroundings : The portion of the universe other than system is called surroundings. Areal or imagenery line which separates system from surroundings is called boundary. Anexchange of matter and energy between the system and its surroundings can take placethrough this boundary.

Example :A reaction mixture in a stoppered flask in a system. The surroundings may be a constanttemperature both in which the flask is immersed. The walls of the flask act as boundary.

(a) Types of system :The systems are classified into three types, on the basis of exchange of matter and energybetween the system and its surroundings.1. Open system 2. Closed system 3. Isolated system

1. Open system : A system that can exchange both matterand energy with its surrounding is called an open system.Example :(a) A sample of liquid water kept in an open beaker(b) A chemical reaction carried out in a test tube(c) Hot tea in an open cup.

2. Closed system : A system that can exchange onlyenergy but not matter with the surroundings is called aclosed system.


150

Example :(a) A sample of liquid water kept in a closed beaker(b) A chemical reaction carried out in a closed vessel.(c) Hot tea in a cup covered with saucer.

3. Isolated system : A system that neither exchange matternor energy with its surroundings is called an isolatedsystem.Example :(a) A sample of liquid water kept in a closed container

insulated from the surroundings.(b) A chemical reaction carried out in a thermos flask(c) Hot tea in a thermos flask.

(b) Properties of system :A macroscopic system has certain thermodynamic properties like temperature, pressure,volume, energy, density.These macroscopic properties are classified into two types according to their dependenceon the amount of the substance present in the system.

1. Extensive property : The property of a system whose magnitude depends on the amountof matter present in the system is called extensive property.Examples : Mass, volume, internal energy, heat capacity, work done, enthalpy, entropy,internal energy, heat of reaction etc.

Note : The volume of one mole of a gas is 22.4 dm3 at STP, but two moles of thegas occupy 44.8 dm3 at STP.

2. Intensive property : The property of a system whose magnitude is independent of theamount of matter present in the system is called its intensive property.Examples :Temperature, density, surface tension, viscosity, refractive index, melting point, boiling point,specific heat.


151

Note : The boiling point of water is 100ºC at 1 atm. pressure, no matter whether thevolume is 1L, 2L or 100 Litres.

The ratio of two extensive properties is an intensive property.For example the ratio of two extensive properties, mass and volume (m/v) is density, whichis an intensive property.

(c) State and State Functions :1. State of system :

(a) Every system has certain measurable thermodynamic properties such as temperature,pressure and volume.These properties are called state variables.

(b) A known set of state variables describes the state (condition) of the system at anytime.

2. Change of state : When there is a change in one or more state variables, then it is saidthat system is changed from one state to another state.

3. State function :(a) Definition : Any property of a system whose value depends on the current state

of the system and is independent of the path followed to reach that state is calledthe state function.

(b) For a particular state of the system, the state function has fixed value in the presentcondition of the system. It does not depend on previous history of the system.

(c) For example, 1g of water is placed in a container and maintained at 25ºC and 1 atm.pressure. The volume of water will be approximately 1cm3.All the quantities, 1g, 25ºC, 1 atm. and 1 cm3 describe the current state of waterand are independent of its previous history.

(d) Hence, the properties mass, temperature, pressure and volume are state functions.

Other examples–enthalpy (H), internal energy (U), number of moles, entropy (S) etc.


152

(d) Properties of state functions :The state functions have two very important properties.

1. To describe the state of a system, it is not necessary to specify the values of all the statefunctions. When the values of a certain minimum number of state functions are assigned,then the values of other state functions are automatically fixed.Example : Consider a gaseous system, in which equation of state is, PV = n RT. Thisequation shows relationship between different state functions.Here only three out of four state functions. (n, P, V and T) are independent and one is dependent.For a gaseous system of constant mass i.e. for fixed ‘n’, if we specify only two statefunctions, the third is automatically fixed.

2. When the state of a system is changed, the change in any state function depends onlyon the initial and final states of the system and not on the path.Consider a system which in changed from initial state A (P1,V1) to final state B (P2V2)by three different paths as shown in figure.The change in pressure ΔP = P2 – P1 and the change in volume, ΔV = V2 – V1 is thesame irrespective of the path.

Note : Suppose the height of the terrace ofa building is 100 feet. We can reach theterrace by different ways such as throughstaircase or lift.The distance travelled depends on the pathchosen. The distance travelled is therefore,a path function. But the attitude (100 feet)of the terrace does not depend on the pathchosen. Hence, altitude is a state function

(e) Thermodynamic equilibrium :1. A system is said to be in thermodynamic equilibrium when the state functions

of the system do not change with time.2. When a gas is enclosed in a cylinder fitted with a piston, it has a definite temperature,

pressure and volume. These properties remain constant as long as piston is motionless.This in an equilibrium state.

3. During motion of the piston temperature, pressure and volume keeps on changing. Thisstate in which the piston in moving is a non-equilibrium state.

4. When the motion of piston is stopped; temperature, pressure and volume again becomesconstant. Now the system will be in a new-equilibrium state.

(f) Types of processes :The transition of a system from one equilibrium state to another is called a process.


153

Some common types of processes are :-1. Isothermal process :

(a) Definition : A process in which the temperature of the system remains constantthroughout the transformation is called an isothermal process.

(b) In isothermal process heat can flow from the system to the surroundings or vice versain order to keep the temperature constant.

(c) In isothermal process since the temperature is constant therefore, the internal energy(U) of the system remains constant i.e. ΔU = 0.

(d) Example - Boiling of water at 100ºC.2. Isobaric process :

(a) Definition : A process in which the pressure of the system remains constant iscalled isobaric process.

(b) In laboratory most of the chemical reactions are carried out in test tube at constantatmospheric pressure i.e. ΔP = 0

3. Isochoric process :(a) Definition : A process in which volume of the system remains constant is called

isochoric process.(b) A chemical reaction carried out in closed vessel is an example of isochoric process

i.e. ΔV = 0

(g) Adiabatic process :1. Definition : A process in which there is no exchange of heat between the system and

its surroundings is called an adiabatic process (q = 0).2. In this process, the system is insulated from the surroundings.3. When the process is exothermic, the temperature of the system rises.4. When the process is endothermic, the temperature of the system falls.5. In adiabatic process, since the temperature of the system rises or falls, and hence, its

internal energy also increases or decreases. i.e. ΔU ≠ 0.6. Example any chemical reaction carried out in closed and insulated vessel.

Note : All the above mentioned processes can be conducted reversibly or irreversibly.

(h) Reversible process :Definition : Any process conducted in such a manner that at every stage the drivingforce is only infinitesimally greater than the opposing force and the process can bereversed by a slight increase in the opposing force is called a reversible process.

Features of reversible process :1. The driving and opposing forces are only infinitesimally different from each other.2. The process can be reversed at any point during the process by making infinitesimal

change in conditions.


154

3. The process takes place in an infinite number of steps.4. At the end of every step of the process the system attains mechanical equilibrium.5. Maximum work is obtained during a reversible process.6. The process takes place so slowly that the system is always in temperature-pressure

equilibrium with its surroundings.7. Reversible process is not a real process but it is an imaginary process.8. Example : Expansion of an ideal gas by very slightly lowering of external pressure.

(i) Irreversible process :1. In this process driving force is much more greater than opposing force.2. It is real process & practical process.3. It is spontaneous process.4. The direction of the process cannot be reversed by applying small change in conditions.5. It takes place in only one step.6. It is fast process and takes place in finite time.7. The system is in equilibrium only at the end of the process.8. Work obtained is not maximum.9. Examples : (a) All natural process are irreversible

(b) Diffusion of gases.(c) Flow of water from higher level to lower level.

(j) Distinction between Isothermal and Adiabatic processes

3.3 NATURE OF HEAT AND WORK : Concept Explanation :

(a) There are only two ways of changing the energy of a closed system :(i) by transfer of energy as work (ii) by transfer of energy as heat.

(b) Heat and work are considered equivalent and are interconvertible into each other. Theyhave same units.

Isothermal process

1. The temperature of the system remainsconstant (ΔT = 0).

2. There is an exchange of heat between thesystem and its surroundings. (q ≠ 0)

3. The internal energy of the system remainsconstant (ΔU = 0).

4. The system is not thermally isolated from itssurroundings. Example : Boiling of water at100ºC.

Adiabatic process

1. The temperature of the system increases ordecreases. (ΔT ≠ 0)

2. There is no exchange of heat between thesystem and its surroundings. (q = 0)

3. The internal energy of the system increasesor decreases. (ΔU ≠ 0)

4. The system is thermally isolated from itssurroundings. Example : Any chemical reactioncarried out in closed and insulated vessel.


155

(a) Work :1. Work is one of the means (method) by which a system can exchange energy with the

surroundings.2. In mechanics, the mechanical work is defined as the transfer of energy by which a body

is moved through a distance by the application of force (i.e. W = f × d)3. In chemical thermodynamics, the type of work involved is pressure-volume work. Hence,

work is also defined as the transfer of energy that can be used to change the height ofmass in the surroundings.

(b) Work during expansion :1. Consider a chemical reaction, 2 2 2 22H O ( ) 2H O( ) + O (g)l l⎯⎯→ ,

that takes place in a cylinder. Suppose some mass is kept on the piston.2. The gas (O2) pushes the piston upwards and hence, mass raised

in the surroundings.3. In the process of lifting the mass, the system loses energy to the

surroundings.4. If no heat transfer occurs, the loss of energy by the system is

equal to the work done by the system is equal to the workdone by the system on the surroundings.

(c) Work during contraction :1. Consider a reaction, 3 4NH (g) + HCl(g) NH Cl(s)⎯⎯→ that takes place

in a cylinder.2. In this case volume of gas decreases.3. The piston moves downwards and hence, decreases the height of the mass4. In this process, surroundings loses energy to the system.5. If no heat transfer occurs, the gain in energy by the system is equal

to the work done by the surroundings on the system.


156

(d) Pressure-volume work :1. The work (W), which is done due to expansion or compression of a gas against an external

opposing pressure (P) is called pressure-volume work.2. The product of PV = W, can be explained as -

PV = × VA

f =

32

× dd

f = f.d = W (Where, d = distance)

(e) Expression for pressure-volume work :1. Suppose a gas at pressure P, volume V1, and temperature T, is

enclosed in a cylinder fitted with a frictionless piston of area A.2. As the gas expands, it pushes the piston upward through a distance

d against an external opposing force f.Derivation :Step - 1 : The work done by the gas during expansion is given byW = opposing force × distance

= – f d⋅(negative sign shows that internal energy of the system duringexpansion decreases).Step - 2 : But, f = pex.A

∴∴∴∴∴ W = –pex × A × dStep - 3 : But A × d = Change in volume = ΔV = V2 – V1

∴∴∴∴∴ W = –pex.ΔV = –pex (V2 – V1)During expansion : work is done by the system on surroundings,

V2 > V1 ∴∴∴∴∴ W = –veDuring compression : work is done on the system by surroundings,

V2 < V1 ∴∴∴∴∴ W = +ve

(f) Free expansion :Expansion of a gas against zero opposing force (as in vacuum) is called free expansion.When pex = 0, then W = 0

(g) Concept of maximum work :1. When opposing force is zero, no work is done.2. When opposing force increases from zero, more and more work will be done by the system.3. However, if opposing force becomes greater than the driving force the process gets reversed.4. Therefore, opposing force must be infinitesimally smaller than driving force to obtain maximum

work.5. If P – pex = ΔP, Or pex = P – ΔP, then


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W = –(p – ΔP).ΔVThus, W tends to maximum as ΔP tends to zero.

(h) Conditions for maximum work :1. The process must be thermodynamically reversible.2. The process takes place in an infinite number of steps.3. The driving force is infinitesimally greater than the opposing force.4. The system is in mechanical equilibrium with its surroundings. (Since two pressures are

almost the same).

(i) Expression for maximum work (Here pressure is variable) :1. Suppose ‘n’ mole of an ideal gas is enclosed in a cylinder fitted with frictionless piston.2. The gas is expanded isothermally and reversibly in a number of infinitesimally small steps

from V1 to V2 at temperature T.3. During each step external pressure (pex) is made infinitesimally smaller than the pressure

of the gas (P), by removing masses gradually from the piston.

4. The volume of the gas is increased by an infinitesimal quantity dV in a single step.5. This process is repeated until the final volume V2 is reached.6. Thus, in a single step the small quantity of work done is given by, dW = –pexdV7 Since the expansion is reversible, therefore, the pressure of the gas is greater by a very

small amount dP than pex.

Hence, p – pex = dP or px = P – dPDerivation :Step - 1

∴∴∴∴∴ dW = – (P – dP).dV = –pdv + dv.dvNeglecting, dP.dV, as it is very small

∴∴∴∴∴ dW = –P.dV


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The total amount of work done during expansion from V1 to V2 will be the sum of allsuch infinitesimal amount of work done of all the steps.Step - 2 : Mathematically, the total amount of work is obtained by integrating the differentialequation between limits V1 to V2.This work is maximum work because the expansion is reversible. Thus,

2

1

– dW∫ =

V2

V1

– P.dV∫ Wmax =

V2

V1

– PdV∫

Step - 3 : The ideal gas equation for n moles is, PV = nRT or P = nRT

V

∴∴∴∴∴ Wmax =

V2

V1

dV– nRT

V∫

Wmax =

V2

V1

dV–nRT

V∫

= –nRT ln v2v1

(V)

= –nRT (ln V2 – ln V1)

= –nRT ln 2

1

V

V

= –2.303 nRT log102

1

V

VAt constant temperature, according to Boyle’s law,

P1V1 = P2V2 or 2

1

V

V = 1

2

P

P

∴∴∴∴∴ Wmax = –2.303 nRT log101

2

P

P

(j) Path dependence nature of work :Work is a path function and not a state function. Since the amountof work done is different in different paths although the initialand final states are the same.Example : Suppose a system (gas) goes from initial state (V1)to final state (V2) by three different paths.Part-I : If the system expands freely in vacuum, then W = 0Part-II : If the system expands against constant external

pressure (pex)then W = –pex (V2 – V1)


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Part-III : If the system expands under Isothermal reversible conditions, then,

Wmax = –2.303 nRT log102

1

V

V

(k) Heat (q) :1. Heat is another way by which a system can exchange energy with its surroundings.

2. When the system and its surroundings are at different temperatures, heat either flows

in or out of the system.

3. Example : Consider a flask containing HNO3 kept in a water bath as shown in fig.

if KOH solution is added to the flask, an exothermic reaction takes place. The temperature

of water bath rises.

3 3 2HNO (aq) + KOH(aq) KNO (aq) + H O( )l⎯⎯→

This indicates that energy is transferred asheat from the flask (system) to the water bath(Surroundings).

4. Heat like work is not the property of thesystem and hence, is not a state function.

5. Example : Suppose a system changes by two different pathsPath - I : The change from state A to State B occurs under adiabatic conditions. Thereis no exchange of heat between the system and its surroundings. q = 0

Path - II : If the change from state A to B takes placeunder isothermal conditions. A quantity of heat q entersthe system q ≠ 0.

Thus, heat transfer is different and it depends on path.Hence, heat is a path function and not a state function.

(l) Units of energy and work :1. The unit of energy as heat or work is Joule (J).

1 J = 1 kg m2s–2

1 J = 1 Pa m3

2. W = –pex.ΔV, if pressure is expressed in atm and volume in litres. Then unit of work

done will be litre-atmosphere (L.atm) where

1 atm = 101.3 × 103 kg m–1 s–2

Adiabatic

q = 0Isothermal

q 0

≠

I

II

A

B


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1 L atm⋅ = dm3 × 101.3 × 103 kg m–1 s–2

= m3 × 10–3 × 101.3 × 103 kg m–1 s–2

= 101.3 kg m2 s–2

= 101.3 J

(m) Sign conventions of W and q :+q → Heat is absorbed by the system from the surroundings.–q → Heat is released from the system to the surroundings.+W → Work is done on the system by the surroundings.–W → Work is done by the system on the surroundings.


*1. Define the terms :(a) system (b) surroundings (c) open system (d) closed system (e) isolated system

Ans. (a) system Refer 3.2 (1), (b) surroundings Refer 3.2 (2), (c) open system Refer 3.2 (a) 1,(d) closed system Refer 3.2 (a) 2, (e) isolated system Refer 3.2 (a) 3

2. Define the terms : *(a) extensive properties *(b) intensive properties. Give two examples of each.Ans. (a) Extensive property : The property of a system whose magnitude depends on the amount

of matter present in the system is called extensive property.Examples : Mass, volume, internal energy, heat capacity, work done, enthalpy, internal energyheat of reaction etc.

(b) Intensive property : Any property of a system whose magnitude is independent of theamount of matter present in the system is called its intensive property.Examples : Temperature, density, surface tension, viscosity, refractive index, melting point,boiling point, specific heat.

*3. Explain the term state function. Give two examples of state functions and two examples ofpath functions.

Ans. Definition : Any property of a system whose value depends on the current state of the systemand is independent of the path followed to reach that state is called the state function.


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(a) For a particular state of the system, the state function has fixed value in the present conditionof the system. It does not depend on previous history of the system.

(b) For example, 1g of water is placed in a container and maintained at 25ºC and 1 atm. pressure.The volume of water will be approximately 1cm3.All the quantities, 1g, 25ºC, 1 atm. and 1 cm3 describe the current state of water andare independent of its previous history.

(c) Hence, the properties mass, temperature, pressure and volume are state functions.Other examples : enthalpy (H), internal energy (U), number of moles, entropy (S) etc.ΔH, ΔU, ΔS, ΔP, ΔVHence, the properties mass, temperature, pressure and volume are state function.Other examples - enthalpy (H), internal energy (H), number of moles entropy (S) etc.Examples of path functions : Work (w) and heat (q).

*4. Explain thermodynamic equilibrium.Ans. Refer 3.2 (e).

*5. Distinguish between Isothermal and Adiabatic processes.Ans. Refer 3.2 (j).

*6. What is a reversible process? What are its features?Ans. Refer 3.2 (h).

*7. Show that pressure times volume is equal to work. OR Show that the product of pressureand volume is equal to work.

Ans. (a) The work (W), that is done due to expansion or compression of a gas against an externalopposing pressure (P) is called pressure-volume work.

(b) The product of PV = W, can be explained as -

PV = × VA

f =

32

× dd

f = f.d = W (Where, d = distance)

*8. Derive the expression for work when a gas expands against constant external pressure.Ans. Derivation :

Step - 1: The work done by the gas during expansion is given by

W = opposing × distance = – f d⋅

(negative sign shows that internal energy of the system during expansion decreases).Step - 2 : But, f = pex.A

∴∴∴∴∴ W = –Pex × A × dStep - 3 : But A × d = Change in volume = ΔV = V2 – V1

∴∴∴∴∴ W = –pex.ΔV = –Pex (V2 – V1)

During expansion : work is done by the system on surroundings,


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V2 > V1 ∴∴∴∴∴ W = –VeDuring compression : work is done on the system by surroundings,V2 < V1 ∴ W = +Ve

*9. Classify the following reactions according to work done by the system, on the system andno work done if pressure is constant.

(a) 2 2H (g) + Cl (g) 2HCl(g)⎯⎯→

Ans. no work is done as volume is constant.

(b)2 3(g)3O (g) 2O⎯⎯→

Ans. Work is done on the system as volume decreases.

(c) 2 2 32SO (g) + O (g) 2SO (g)⎯⎯→

Ans. work is done on the system as volume decreases.

(d) 3 2MgCO (s) MgO(s) + CO (g)⎯⎯→

Ans. work is done by the system as volume increases.

(e) 4 3 2 2NH NO (s) N O(g) + 2H O(g)⎯⎯→

Ans. work is done by the system as volume increases.

*10. Explain the concept of maximum work.Ans. Refer 3.3 (g)

*11. A free expansion of a gas results into no work. Explain.Ans. Expansion of a gas against zero opposing force (as in vacuum) is called force expansion.

When Pex = 0, then W = 0

*12. Derive the expression for maximum work.Ans. Refer 3.3 (i)

*13. What are the sign conventions for q and W?Ans. +q → Heat is absorbed by the system from the surroundings.

–q → Heat is released from the system to the surroundings.+W → Work is done on the system by the surroundings.–W → Work is done by the system on the surroundings. [Refer 3.3 (m)]


• Theoretical MCQs :1. When an ideal gas expands in vacuum, the work done is .......

a) R b) 2R c) Zero d)3

R2


163

*2. In a chemical reaction work is done by the system when .......a) number of moles of gaseous reactants is equal to the number of moles of gaseous productsb) total number of moles increases.c) number of moles of gaseous substances decreases.d) number of moles of gaseous products is greater than the number of moles of gaseous

reactants.*3. When a sample of an ideal gas is allowed to expand at constant temperature against an atmospheric

pressure, .......a) surroundings does work on the system b) ΔΔΔΔΔU = 0c) no heat exchange takes place between the system and surroundingsd) internal energy of the system increases.

*4. In what reaction of the following, work is done by the system on the surroundings?

a) Hg (l) Hg (g)⎯⎯→ b) 2 33O (g) 2O (g)⎯⎯→

c) 2 2H (g) + Cl (g) 2HCl (g)⎯⎯→ d) 2 2 3N (g) + 3H (g) 2NH (g)⎯⎯→

• Numerical MCQs :5. The work done during the expansion of a gas from a volume of 4dm3 to 6dm3 against a constant

external pressure of 3atm is ....... (1L atm = 101.32 J) (CBSE 2004)a) +304J b) –304 J c) –6J d) –608 J

*6. A gas does 0.320 kJ of work on its surroundings and absorbs 120 J of heat from the surroundings.Hence, ΔU is .......a) 440 kJ b) 200 J c) 120.32 J d) –200J

*7. A gas expands in volume from 2L to 5L against a pressure of 1atm at constant temperature.The work done by the gas will be .......

a) 3 J b) –303.9 J c) –303.9 L.atm d) 303.9 L.atm8. An ideal gas expands from 1 × 10–3m3 to 1 × 102 m3 at 3K against a constant pressure of

1 × 105 N/m2. The work done is .......a) –900 J b) 270 kJ c) –900 kJ d) +900 kJ.

(Hints : W = –PexΔΔΔΔΔV)

• Advanced MCQs :9. Which among the following statment is false ....... (IIT 2001)

a) Work is a state functionb) Temperature is a state functionc) Change of state is completely defined when initial and final states are specified.d) Work appears at the boundary of the system.

10. 1 mole of each CaC2, Al4C3 and Mg2C3 reacts with water in separate open flask. Numericalvalue of work done by the system is in order.a) CaC2 < Al4C3 = Mg2C3 b) CaC2 < Al4C3 < Mg2C3c) CaC2 = Mg2C3 < Al4C3 d) CaC2 = Mg2C3 = Al4C3


164

3.4. INTERNAL ENERGY (U) :

Concept Explanation :(1) Every substance (system) contains a definite amount of energy which is called internal

energy (U).(2) Internal energy is the sum of kinetic energies of all the particles of the system and potential

energies due to various bonds between the particles.(3) In thermodynamic, the change in internal energy is determined

i.e. ΔU = U2 – U1Where U1 = system’s energy in initial state

U2 = system’s energy in final state(4) When energy is transferred into the system by heating or doing work on it, the internal

energy increases. i.e. ΔU = +ve(5) When energy is transferred from the system by cooling it or by doing work on the surroundings.

The internal energy decreases. i.e. ΔU = –ve

Note : ΔU always carries a sign even if it is positive.

3.5. FIRST LAW OF THERMODYNAMICS :

Concept Explanation :First law of thermodynamics is simply the law of conservation of energy. It can be statedin different ways –(a) Statements of first law of thermodynamics :1. The total internal energy of an isolated system is constant.2. Energy can be converted from one form to another but it cannot be created or destroyed.3. The total quantity of energy of the universe is constant.4. Whenever a quantity of energy of one kind disappears, an exactly equivalent amount of

energy of other kind must appear.5. The total amount of energy of a system and its surroundings must remain constant although

it may change from one form to another.

(b) Mathematical equation of first law of thermodynamicsThe internal energy of the system can be increased in two ways.

1. by supplying heat to the system and2. by doing work on the system.

(a) If a system with internal energy ‘U1’, absorbs heat ‘q’, its internal energy will changeto U1 + q.


165

(b) Now, if work ‘W’ is done on the system, than the internal energy will further increaseand become equal to U1 + q + W.

(c) Let this be find energy of the system, U2Then, U2 = U1 + q + Wi.e. U2 – U1 = q + W

∴∴∴∴∴ ΔU = q + WThis is mathematical expression of first law of thermodynamics.

(d) For an infinitesimal changes, the above equation is written as, dU = dq + dW.

(c) Modified first law of thermodynamics :1. Mass and energy are equivalent and are mutually interconvertible. (E = mc2)2. Thus, the principle of conservation of energy may be replaced by the principle of conservation

of mass plus energy.3. Hence, the first law of thermodynamics is modified as “The sum of mass and energy

of an isolated system (Universe) remains constant”.

(d) First law of thermodynamics for various processes :1. Isothermal process

T = constant, ∴ ΔU = 0∴ ΔU = q + W

0 = q + W or q = –W or W = –qi.e. heat absorbed is entirely used for doing work by the system on the surroundings orthe work done on the system by the surroundings results in the release of heat by thesystem.

2. Adiabatic processq = 0

∴∴∴∴∴ ΔU = q + WΔU = 0 + W or ΔU = W or –ΔU = –Wi.e. work done on the system is equal to increase is internal energy of the system or workdone by the system is due to the expense of system’s internal energy which decreasesin the process.

3. Isochoric processΔU = q + W

= q – pex × ΔV (∵ W = pex × ΔV)∵ ΔV = 0, ∴∴∴∴∴ ΔU = q – pex × 0

ΔU = qvi.e. the increase in internal energy of the system is equal to the heat absorbed by thesystem at constant volume. As ‘U’ is a state function, qv is also a state function.


166

4. Isobaric processpex = constant, ΔV ≠ 0ΔU = q + W

ΔU = qP – pex.ΔV (Q W = – exp ΔV⋅ )

or qP = ΔU + pex.ΔV

Note : Most of the chemical reactions are carried out at constant atmospheric pressure(pex = constant). A special symbol ΔH called enthalpy change is used for the heat changesproduced in such reactions. ΔH is also a state function.

Limitations of the First Law of Thermodynamics :(i) It does not tell about the direction of heat flow.

(ii) It does not tell about the extent of convertibility of one form of energy into another i.e.how much heat energy would be transferred from one form to the other.


*1. State first law of thermodynamics. Justify its mathematical equation.Ans. (a) Statements of first law of thermodynamics :

(1) The total internal energy of an isolated system is constant.(2) Energy can be converted from one form to another but it cannot be created or destroyed.(3) The total quantity of energy of the universe is constant.(4) Whenever a quantity of energy of one kind disappears, an exactly equivalent amount

of energy of other kind must appear.(5) The total amount of energy of a system and its surroundings must remain constant

although it may change from one form to another.(b) Mathematical equation of first law of thermodynamics :

The internal energy of the system can be increased in two ways.(i) by supplying heat to the system and (ii) by doing work on the system.(1) If a system with internal energy ‘U1’, absorbs heat ‘q’, its internal energy will change

to U1 + q.(2) Now, if work ‘W’ is done on the system, then the internal energy will further increase

and become equal to U1 + q + W.(3) Let this be final energy of the system, U2

Then, U2 = U1 + q + Wi.e. U2 – U1 = q + W∴∴∴∴∴ ΔU = q + W

This is mathematical expression of first law of thermodynamics.(c) For an infinitesimal changes, the above equation is written as, dU = dq + dW


167


1. First law of thermodynamics is merely the law of .......a) Conservation of mass b) Conservation of energyc) Conservation of both mass and energy d) none of these

2. The mathematical expression for the first law of thermodynamics is .......a) H = E + PV b) ΔH = ΔU + P ΔV⋅

c) q = ΔU + W d) ΔΔΔΔΔU = q + W

• Numerical MCQs :3. If a gas absorbs 200 J of heat and expands by 500 cm3 against a constant pressure of

2 × 105 N/m2, then change in internal energy is .......a) –300 J b) +100 J c) –100 J d) +300 J

4. Change in internal energy when 4 kJ of work is done on the system and 1 kJ of heat is givenout by the system is ....... (Kerala PMT - 2008)a) 3 kJ b) 4 kJ c) 7 kJ d) 6 kJ

• Advanced MCQs :5. If W = –20 kJ, the negative sign indicates that the

a) Expanding system gain work energy and does work on the surroundings.b) Expanding system loses work energy and does work on the surrounding.c) Expanding system gain work energy and does work on the system.d) Expanding system loses work energy and does work by the surroundings.

3.6. ENTHALPY (H) :

Concept Explanation :(a) The enthalpy of a system is defined as the sum of the internal energy of the system

and its pressure-volume energy.(b) Mathematically, the enthalpy is defined by the equation, H = U + PV(c) Since U, P and V are state functions, H is also a state function.

(a) Relationship between ΔH, ΔU, and P.ΔV :Step - 1 : Change in enthalpy (ΔH) depends on initial and final states of the system andis given by, ΔH = H2 – H1

Step - 2 : But, H1 = U1 + P1V1 and H2 = U2 + P2V2

∴∴∴∴∴ ΔH = U2 + P2V2 – U1 – P1V1

∴∴∴∴∴ ΔH = (U2 – U1) + (P2V2 – P1V1)


168

∴∴∴∴∴ ΔH = ΔU + Δ(PV)At constant pressure, P1 = P2 = P,

∴∴∴∴∴ ΔH = ΔU + P.ΔVStep - 3 : If pex = P, then from Ist law of thermodynamics,qp = ΔU + P.ΔVComparing the above two equations, we get;ΔH = qp

Thus, increase in enthalpy of a system is equal to the heat absorbed at constant pressure(isobaric process).

Note :(i) Since ‘H’ is a state function, qp is also a state function.(ii) q is not a state function but qp and qv are state functions because the path of the

process is defined and therefore, qp and qv can have only specific values.

(b) Relationship between ΔH and ΔU for chemical reactions :1. We know, ΔH = ΔU + P.ΔV2. For reactions involving solids and liquids, ΔV is very small,∴ P.ΔV is neglected, Hence ΔH = ΔU3. For reaction involving gases, ΔV cannot be neglected,∴ ΔH = ΔU + P.ΔV

= ΔU + P (V2 – V1)= ΔU + PV2 – PV1

4. Assuming ideal behaviour of reactant and product gases,PV1 = n1 RT and PV2 = n2 RT

∴ ΔH = ΔU + n2 RT – n1 RT∴ ΔH = ΔU + RT (n2 – n1)∴ ΔH = ΔU + RT.Δn

Where, Δn = number of moles of gaseous products – number of moles of gaseous reactants.

(c) Work done in chemical reactions :The work done by a system at constant pressure and temperature is given by equation,

∴∴∴∴∴ W = – exp ΔV⋅

Assuming Pex = P∴∴∴∴∴ W = –P.ΔV

= –P (V2 – V1)= –PV2 + PV1


169

Assuming ideal behaviour, PV1 = nRT and PV2 = n2 RT∴∴∴∴∴ W = –n2RT + n1RT

= –RT (n2 – n1)W = –RT.ΔnThe above equation gives the work done by a system, in chemical reaction.

1. If n2 > n1, W is negative and work is done by the system.2. If n2 < n1, W is positive and work is done on the system.3. If n1 = n2, W is zero, no work is done.

(d) Conditions under which ΔH = ΔU :1. If reaction is carried out in closed vessel, volume is constant, ΔV = 0.∴∴∴∴∴ ΔH = ΔU + P.ΔVΔH = ΔU

2. If reaction involves only solids and liquids, ΔV is neglected, hence, ΔH = ΔU.3. If n1 = n2, i.e. Δn = 0, then ΔH = ΔU

3.7. ENTHALPIES OF PHYSICAL CHANGES :

Concept Explanation :They are four main types :(a) phase transition.(b) the breaking of individual atoms, molecules into their ions and other fragments.(c) dissolution of a substance into a solvent and(d) dilution of a solution.

(a) Enthalpy of phase transition :These are,

1. Enthalpy of fusion (Δfus H)(a) Definition : The enthalpy change that accompanies the fusion of one mole of a

solid without change in temperature at constant pressure is called its enthalpyof fusion.

(b) It is denoted by the symbol ΔfusH.(c) Example :

2 2H O(s) H O( )l⎯⎯→ ΔfusH = + 6.01 kJ mol–1 at 0ºC & 1 atm.

(1 mole ice)

2 2H O( ) H O(s)l ⎯⎯→ , ΔfreezH = –6.01 kJ mol–1 at 0ºC & 1 atm.


170

2. Enthalpy of vaporization (ΔvapH) :(a) Definition : The enthalpy change that accompanies the vaporization of one mole of liquid

without changing its temperature at constant pressure is called enthalpy of vaporization.(b) It is denoted by ΔvapH,

(c) Example :

2 (l) 2 (g)H O H O⎯⎯→ ; ΔvapH = + 40.7 kJ mol–1 at 100ºC & 1 atm.(1 mole)

2 (l) 2 (g)H O H O⎯⎯→ ; ΔvapH = + 44.0 kJ mol–1 at 25ºC & l atm.(d) The reverse of vaporization is condensation of vapour which is accompanied with

the evolution of heat.

(e) Example : 2 (g) 2 (l)H O H O⎯⎯→ ; ΔconH = –40.7 kJ mol–1 at 100ºC

Standard enthalpies of fusion and vaporization

Substance F.P./K ΔfusHº/kJ mol–1 B.P./K ΔvapHº/kJ mol–1

Acetone (CH3COCH3) 177.8 5.72 329.4 29.1

Ethanol (C2H5OH) 158.7 4.60 351.5 43.5

Methanol (CH3OH) 175.5 3.16 337.2 35.3

Methane (CH4) 90.7 0.94 111.7 8.2

Benzene (C6H6) 278.7 9.87 353.3 30.8

Water (H2O) 273.2 6.01 373.2 40.7

3. Enthalpy of sublimation (ΔsubH) :(a) Definition : The enthalpy change that accompanies the conversion of one mole

of solid directly into its vapour at constant temperature and pressure is calledits enthalpy of sublimation.

(b) It is denoted by ΔsubH.

(c) Example : 2 (s) 2 (g)H O H O⎯⎯→ ; ΔsubH = 51.08 kJ mol–1 at 0ºC & 1 atm.

Note : It should be noted that whether the conversion of solid to vapour takes placedirectly in one step or in two steps; the enthalpy change is the same since enthalpyis a state function. For example :

2 (s) 2 (l)H O H O ,⎯⎯→ ΔfusH = +6.01 kJ mol–1 at 0ºC

2 (l) 2 (g)H O H O ,⎯⎯→ ΔvapH = +45.07 kJ mol–1 at 0ºC

2 (s) 2 (g)H O H O ,⎯⎯→ ΔsubH = (+6.01 + 45.07) kJ mol–1at 0ºC

= 51.08 kJ mol–1 at 0ºC.


171

4. ΔsubH = ΔfusH + ΔvapH

(b) Enthalpy of atomic or molecular changes :1. Enthalpy of ionization (ΔionH) :

(a) Definition : The enthalpy change that accompanies the removal of an electronfrom each atom or ion in 1 mole of gaseous atoms or ions is called an enthalpyof ionization.

(b) It is denoted by ΔionH.

(c) Example : + –(g) (g)Na Na + e⎯⎯→ ; ΔionH = 494 kJ mol–1

(1 mole)(d) The enthalpy change for the removal of one electron from each atom in one mole

is called first ionization enthalpy :(e) The second ionization energy is larger than first e.g.

+ –Ca(g) Ca (g) + e⎯⎯→ ΔionH = 590 kJ mol–1

+ 2+ –Ca (g) Ca (g) + e⎯⎯→ ΔionH = 1150 kJ mol–1

(f) The reverse of ionization enthalpy is the electron-gain enthalpy, which is definedas the enthalpy change when one mole of gas-phase atoms of an element acceptelectrons so as to form gaseous anions.

(g) Example :

– –Cl(g) + e Cl (g)⎯⎯→ ΔegH = –349 kJ mol–1


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(a) Ionization enthalpies (b) Electron gain enthalpies

Atom ΔΔΔΔΔionH/kJ mol–1 ΔΔΔΔΔionH/kJ mol–1Atom ΔΔΔΔΔegH/kJ mol–1

1st 2nd

Na 494 4560 B –28

K 418 3070 C –122

Li 519 7300 O –141, +844

Mg 736 1450 Cl –349

Ca 590 1150 Br –325

Ba 502 966 F –328

Al 577 1820 I –295

Cl 1260 – S –200, +532

Br 1140 –

I 1010 –

2. Enthalpy of atomization (ΔatomH) :(a) Definition - The enthalpy change accompanying the dissociation of all the

molecules in one mole of a gas phase substance into gaseous atoms is calledenthalpy of atomization.

(b) Example :

2Cl (g) Cl(g) + Cl(g)⎯⎯→ ΔatomH = 242kJ mol–1

(1 mole)

(c) Enthalpy of solution (ΔsolnH) :1. Definition - The change in enthalpy when one mole of a substance is dissolved in

a specified quantity of solvent at a given temperature so as to form a solution ofparticular concentration is called enthalpy of solution.

Note : If the solution so obtained is further diluted, there will again be a change inenthalpy. If we go on diluting the solution, a stage will come when further dilutionproduces no thermal effect. This state is called the state of infinite dilution.

2. Definition - The change in enthalpy when one mole of a substance is dissolved ina solvent so that further dilution does not give any change in enthalpy is calledenthalpy of solution.


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3. Example : When 1 mole of HCl(g) is dissolved in 50 moles of water, the enthalpy of

solution is given as 2 2HCl(g) + 50H O( ) HCl(50H O)l ⎯⎯→ , ΔsoluH = –73.26 kJ

When 1 mole of HCl(g) is dissolved in large quantity of water that is when the solutionis infinitely dilute, the enthalpy of solution is given by.

HCl(g) + aq HCl(aq)⎯⎯→ ; ΔsolnH = –75.14 kJ

Causes of enthalpy of solution :The exact value of ΔsolnH for a given substance is the sum of the contributions of three kindsof interactions.1. Solute - solute interactions :

(a) Energy is required to overcome attractive forces between solute particles to dissolvethe solute.

(b) For an ionic solid, energy is called crystal lattice energy.(c) Crystal lattice energy is defined as the enthalpy change accompanying the formation

of one mole of formula units of a crystalline solid from its constituent ions in the gaseousstate.

(d) Crystal lattice enthalpies are always negative.

+ –(g) (g) (s)M + X MX⎯⎯→ , ΔH is negative

(e) Reverse of crystal formation reaction takes place, during formation of solution.(f) The smaller the magnitude of crystal lattice enthalpy, the more readily the dissolution

of solute occurs.

2. Determination of crystal lattice enthalpy :It is determined by Born - Haber cycle afterMax Born and Fritz Haber, who developed it.The lattice energy of NaCl, for example, is thechange in enthalpy, ΔHº, When Na+ and Cl– ionsin the gas phase come together to form 1 moleof NaCl crystal.

3. Born-Haber Cycle for the formulation of NaCl crystal from its elements :(a) The lattice energy of an ionic crystal can be determined by applying Hess’s law.

(b) Enthalpy change for direct formation of NaCl :

21

Na(s) + Cl (g) NaCl(s)2

⎯⎯→ Δf Hº = –411 kJ


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(c) Enthalpy change by indirect steps :

Step : 1 Sublimation Na(s) Na(g)⎯⎯→ ΔsubH = +109 kJStep : 2 Dissociation

2 (g) (g)1

Cl Cl2

⎯⎯→ ; ΔdH = +122 kJ

Step : 3 Ionisation+ –Na(g) Na (g) + e⎯⎯→ ; ΔionH = 496 kJ

Step : 4 – –Cl(g) + e Cl (g)⎯⎯→ ; ΔegH = –348 kJ

Step : 5 Formation of crystal lattice (one mole of solid NaCl)+ –

(g) (g) (s)Na + Cl NaCl⎯⎯→ ; ΔLH = –(Lattice energy)

∴ ΔfH = ΔsubH + ΔdH + ΔionH + ΔegH + ΔLH–411 kJ = 109kJ + 122 kJ + 496 kJ – 348 kJ + ΔLHBy solving this equation we get,Crystal Lattice energy of NaCl (ΔLH) = –790 kJ mol–1

Lattice enthalpies

Substance ΔLH/ Substance ΔLH/ Substance ΔLH/

kJ mol–1 kJ mol–1 kJ mol–1

LiF –1037 NaF –926 KF –821

LiCl –852 NaCl –786 KCl –717

LiBr –815 NaBr –752 KBr –689

LiI –761 NaI –705 KI –649

4. Solvent - Solvent interactions :During the dissolution process, energy is required for the separation of solvent moleculesin order to make room for the solute particles.

5. Solute - Solvent interactions :The solvent molecules surround the solute particles.This process is called solvation. Energy is released in solvation process.

(d) Enthalpy of dilution :1. When already prepared solution is further diluted, then either heat is released or absorbed,

which is called enthalpy of dilution.2. Definition : The enthalpy change that occurs when a solution of one concentration

is diluted to form the solution of another concentration is called enthalpy of dilution.


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3. Example :

Further diluted Soln. : HCl(g) + aq HCl(aq)⎯⎯→ ; ΔsolnH = –75.14 kJ

Ist prepared Soln. :2 2HCl(g) + 50H O HCl (50H O)⎯⎯→ ; ΔsolnH = –73.26 kJ

– – – +

2 (aq)HCl + 50H O + aq HCl⎯⎯→ ; ΔdilH = –1.88Hence, enthalpy of dilution = –1.88 kJ.


*1. Arrange in order of increasing enthalpy; H2O(s), H2O(g), H2O(l)Ans. Enthalpy increases in the order : H2O(s) < H2O(l) < H2O(g)

*2. What is the difference between ΔH and ΔU? What is the sign of ΔH for exothermic andendothermic reactions? Under what circumstances ΔH = ΔU?

Ans. ΔH → Heat of reaction at constant pressureΔU → Heat of reaction at constant volumeΔH = +ve for endothermic reactionΔH = –ve for exothermic reactionRelationship between ΔH and ΔU for chemical reactions.(a) We know, ΔH = ΔU + P.ΔV(b) For reactions involving solids and liquids, ΔV is very small,

∴∴∴∴∴ P.ΔV is neglected, Hence ΔH = ΔU(c) For reaction involving gases, ΔV cannot be neglected,

∴∴∴∴∴ ΔH = ΔU + P.ΔV= ΔU + P (V2 – V1)= ΔU + PV2 – PV1

(d) Assuming ideal behaviour of reactant and product gases,PV1 = n1 RT and PV2 = n2 RT

∴∴∴∴∴ ΔH = ΔU + n2 RT – n1 RT∴∴∴∴∴ ΔH = ΔU + RT (n2 – n1)∴∴∴∴∴ ΔH = ΔU + RT.Δn

Where, Δn = number of moles of gaseous products – number of moles of gaseous reactants.

*3. Define enthalpy. At constant pressure show that ΔH = ΔU + P.ΔUAns. Definition : Enthalpy is a thermodynamic property of a system. It reflects the capacity to do

non-mechanical work and the capacity to release heat. Enthalpy is denoted as H; specific enthalpydenoted as h.

Relationship between ΔH, ΔU, and P.ΔV.


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Step - 1 : Change in enthalpy (ΔH) depends on initial and final states of the system andis given by, ΔH = H2 – H1

Step - 2 : But, H1 = U1 + P1V1 and H2 = U2 + P2V2

∴∴∴∴∴ ΔH = U2 + P2V2 – U1 – P1V1

∴∴∴∴∴ ΔH = (U2 – U1) + (P2V2 – P1V1)∴∴∴∴∴ ΔH = ΔU + Δ(PV)

At constant pressure, P1 = P2 = P,∴∴∴∴∴ ΔH = ΔU + P.ΔV

Step - 3 : If pex = P, then from Ist law of thermodynamics,qp = ΔU + P.ΔVComparing the above two equations, we get;ΔH = qp

Thus, increase in enthalpy of a system is equal to the heat absorbed at constant pressure (isobaricprocess).

*4. It is difficult to calculate ‘U’, however ΔU can be easily determined. Why? Explain givingexamples.

Ans. The absolute value of internal energy (U) cannot be determined because it is impossible todetermine accurately the most of the quantities that contribute to the internal energy of a system.For example it is difficult to determine the kinetic energies of all the molecules, ions and atomsof the system. It is difficult to determine the potential energies associated with the forces betweenthe particles.

*5. Obtain the relationship between ΔH and ΔU for a chemical reaction.Ans. Refer 3.6(b)

*6. What is the expression for work done in a chemical reaction? Explain the meaning of each term.Ans. Work done in chemical reactions : The work done by a system at constant pressure and temperature

is given by equation,∴∴∴∴∴ W = –pex.ΔV

Assuming pex = P∴ W = –P.ΔV

= –P (V2 – V1)= –PV2 + PV1

Assuming ideal behaviour, PV1 = n1RT and PV2 = n2 RT∴∴∴∴∴ W = –n2RT + n1RT

= –RT (n2 – n1)W = –RT.ΔnThe above equation gives the work done by a system, in chemical reaction.(a) If n2 > n1, W is negative and work is done by the system.


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(b) If n2 < n1, W is positive and work is done on the system.(c) If n1 = n2, W is zero, no work is done.

*7. Define and explain each of the following with one example each :(a) enthalpy of fusion (b) enthalpy of vaporization(c) enthalpy of sublimation (d) enthalpy of atomization(e) enthalpy of ionization

Ans. (a) Enthalpy of fusion (Δfus H) :(1) Definition : The enthalpy change that accompanies the fusion of one mole of a

solid without change in temperature at constant pressure is called its enthalpyof fusion.

(2) It is denoted by the symbol ΔfusH.(3) Example :

2 2H O(s) H O( )l⎯⎯→ ; ΔfusH = + 6.01 kJ mol–1 at 0ºC and 1 atm.

(1 mole ice)

2 2H O( ) H O(s)l ⎯⎯→ ; ΔfreezH = –6.01 kJ mol–1 at 0ºC and 1 atm.

(b) Enthalpy of vaporization (ΔvapH) :(1) Definition : The enthalpy change that accompanies the vaporization of one mole

of liquid without changing its temperature at constant pressure is called enthalpyof vaporization.

(2) It is denoted by ΔvapH,(3) Example :

2 2H O( ) H O(g)l ⎯⎯→ ; ΔvapH = + 40.7 kJ mol–1 at 100ºC and 1 atm.

(1 mole)

2 2H O( ) H O(g)l ⎯⎯→ ; ΔvapH = + 44.0 kJ mol–1 at 25ºC and l atm.

(4) The reverse of vaporization is condensation of vapour which is accompanied withthe evolution of heat.

(5) Example :

2 2H O(g) H O( )l⎯⎯→ ; ΔconH = –40.7 kJ mol–1 at 100ºC

[Refer table 3.7 (a)](c) Enthalpy of sublimation (ΔsubH) :

(1) Definition : The enthalpy change that accompanies the conversion of one moleof solid directly into its vapour at constant temperature and pressure is calledits enthalpy of sublimation.

(2) It is denoted by ΔsubH.


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(3) Example :

2 2H O(s) H O(g)⎯⎯→ ΔsubH = 51.08 kJ mol–1 at 0ºC and 1 atm.

(4) ΔsubH = ΔfusH + ΔvapH[Refer table 3.7 (a)3]

(d) Enthalpy of atomization (ΔatomH) :(i) Definition - The enthalpy change accompanying the dissociation of all the

molecules in one mole of a gas phase substance into gaseous atoms is calledenthalpy of atomization.

(ii) Example :

2Cl (g) Cl(g) + Cl(g)⎯⎯→ ; ΔatomH = 242 kJ mol–1

(1 mole)(e) Enthalpy of ionization (ΔionH) :

(i) Definition : The enthalpy change that accompanies the removal of an electronfrom each atom or ion in 1 mole of gaseous atoms or ions is called an enthalpyof ionization.

(ii) It is denoted by ΔionH.(iii) Example :

+ –(g) (g)Na Na + e⎯⎯→ ; ΔionH = 494 kJ mol–1

(1 mole)(iv) The enthalpy change for the removal of one electron from each atom in one mole

is called first ionization enthalpy :(v) The second ionization energy is larger than first e.g.

+ –Ca(g) Ca (g) + e⎯⎯→ ΔionH = 590 kJ mol–1

+ 2+ –Ca (g) Ca (g) +e⎯⎯→ ΔionH = 1150 kJ mol–1

(vi) The reverse of ionization enthalpy is the electron-gain enthalpy, which is defined asthe enthalpy change when one mole of gas-phase atoms of an element accept electronsso as to form gaseous anions?

(vii) Example : – –Cl(g) + e Cl (g)⎯⎯→ ΔegH = –349 kJ mol–1

[Refer table 3.7 (b)1]


• Theoretical MCQs :

1. Which one does not describe a correct definition of enthalpy? (AIIMS 1997)a) H = E + PV b) H = E – PV c) E = H – PV d) H – E – PV = 0


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2. The enthalpy change of a reaction does not depends on .......a) State of reactants and products b) Nature of reactants and productsc) Different intermediate reactions d) Initial and final enthalpy change of reaction

• Numerical MCQs*3. The heat evolved in the following reaction .......

2 2 22H (g) + O (g) 2H O (l)⎯⎯→ , ΔH = –484 kJ to produce 9 g of water is .......

a) 484 kJ b) 121 × 133 J c) 242 kJ d) 968 kJ

*4. In the reaction, 2 2H (g) + Cl (g) 2HCl (g)⎯⎯→ ΔH = –184 kJ, if 2 moles of H2 react with

2 moles of Cl2, then ΔU is equal to .......

a) –184 kJ b) –368 kJ c) Zero d) +368 kJ

*5. For which of the following substances ΔfHº is not zero .......

a) Ca(s) b) He(g) c) P(red) d) CH3OH(l)

*6. Given the reaction, 2 2 3N (g) + 3H (g) 2NH (g)⎯⎯→ , , , , , ΔH = –92.6 kJ. The enthalpy of

formation of NH3 is .......

a) –185.2 kJ mol–1 b) –46.3 kJ mol–1 c) 92.6 kJ mol–1 d) –92.6 kJ

• Advanced MCQs7. The enthalpy of which of the following substances in standard state is zero?

a) Carbon b) CaCO3 c) NH3 d) HNO3

(H.S.C. Board, March 2008)

8. The enthalpies of formation of N2O and NO are 82 and 90 kJ mol–1. The enthalpy of reaction

2 2(g)2N O(g) + O 4 NO(g)⎯⎯→ is ....... (Hints ΔH = (Products) (reactants)H – H∑ ∑ )

a) 8 kJ b) 88 kJ c) –16 kJ d) 196 kJ

9. Two moles of an ideal gas is expanded isothermally and reversibly from 1L to 10L at 300K.The enthalpy change (in kJ) for the process is .......

(Hint : For an isothermal reversible reaction, ΔH = n Cp ΔT = 0 Q ΔT = 0) (IIT 2004)

a) 11.4 kJ b) –11.4 kJ c) 0 kJ d) 4.8 kJ

3.8. THERMOCHEMISTRY :

Concept Explanation :Thermochemistry is the branch of physical chemistry which deals with the enthalpy changesduring a chemical reactions.


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(a) Enthalpy of chemical reactions (Heat of reaction)(a) Definition : The amount of heat released or absorbed in a chemical reaction at

constant temperature and pressure is called enthalpy of chemical reaction orheat of reaction.

(b) Example : aA + bB cC + dD⎯⎯→

∴∴∴∴∴ Enthalpy change (ΔH) = (cHc + dHD) – (aHA + bHB)

or ΔH = products reactantsH – H∑ ∑

(c) Thus, the enthalpy of chemical reaction is the difference between the sum of theenthalpies of products and that of reactants.

1. Exothermic reaction :(a) When productsH∑ is less than reactantsH∑ , ΔH is negative and heat is released.(b) The reactions in which heat is released from the system to the surroundings are called

exothermic reactions.

(c) Example : 3 22KClO (s) 2KCl(s) + 3O (g)⎯⎯→ , ΔHº = –78 kJ2. Endothermic reaction :

(a) When productsH∑ is more than reactantsH∑ , ΔH is positive and heat is absorbed.(b) The reaction in which heat is absorbed by the system from the surroundings are called

endothermic reaction.

(c) Example : 2 2 2N (g) + 2O (g) 2NO (g)⎯⎯→ ; ΔHº = +66.4 kJ3. Standard enthalpy of chemical reactions :

(a) The standard state of a substance is that form of the substance in which it is moststable at a pressure of 1 bar (105 Pa = 0.987 atm) or 1 atm and at 25ºC.

(b) If the reaction involves species in solution, their standard states are 1M concentrations.(c) Example : Standard states of elements : H2(g), Hg(l), Na(s), C(graphite) at 1 atm. pressure

and 25ºC. Standard states of compounds : C2H5OH(l), H2O(l), CaCO3(s), CO2(g)at 1 atm pressure and 25ºC.

(d) The standard enthalpy of a chemical reaction (Standard heat of reaction) is definedas ‘the enthalpy change accompanying the reaction when all the substancesinvolved are in their standard states’.

(e) It is denoted by ΔHº

4. Thermochemical equation :(a) Definition : A balanced chemical equation in which the exact value of enthalphy

change, physical states and number of moles of reactants and products are specified;is known as thermochemical equation.

(b) Rules for writing the thermochemical equation :(i) The equation must be balanced for the number of moles of reactants and products.


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(ii) The numerical value and sign of enthalpy change (ΔHº) must be shown on theright hand side of the equation.

(iii) The physical states of reactants and products must be specified by the letterss (solid), l (liquid), g (gas) or aq (aqueous).

(c) For the reverse reaction the sign of ΔHº must be changed.(d) If the coefficients indicating the number of moles of all the substances are multiplied

or divided by a numerical factor, the value of ΔHº also be multiplied or divided bythe same factor.

Example : 4 2 2 2CH (g) + 2O CO (g) + 2H O( )l⎯⎯→ , ΔHº = –890 kJ

(b) Standard enthalpy of formation (standard heat of formation) ΔfHº :1. Definition : The enthalpy change that accompanies a reaction in which one mole of

a pure compound in its standard state is formed from its elements in their standardstates, is called standard enthalpy of formation.

2. It is denoted by Δf Hº.3. Example :

2 2 21

H (g) + O (g) H O( )2

l⎯⎯→ , Δf Hº = –286 kJ mol–1

(1 mole)

(graphite) 2 4C + 2H (g) CH (g)⎯⎯→ Δf Hº = –74.8 kJ mol–1

4. The standard enthalpy of formation of an element from itself is zero.i.e.ΔfHº (C) = Δf Hº (H2) = Δf Hº (Cl2) = 0

Standard molar enthalpies of formation

Substance ΔΔΔΔΔfHº Substance ΔΔΔΔΔfHº Substance ΔΔΔΔΔfHº Substance ΔΔΔΔΔfHº

kJ/mol kJ/mol kJ/mol kJ/mol

AlCl3(s) –704.2 Cl2O(g) 80.3 MgCO3(s) –1096 H3PO4(s) –1279

Al2O3 (s) –1676 CuCl2(s) –220.1 NH3(g) –46.1 KClO3(s) –397.7

BaCl2 (s) –855 CuO(s) –157 N2H4(l) 50.6 KNO3(s) –494.6

BaCO3(s) –1216 CuS(s) –79.5 N2H4(g) 95.4 NaCl(s) –411.2

Br2(g) 30.9 HF(g) –271 NO(g) 90.2 Na2O(s) –414.2

HBr(g) –36.4 H2O(l) –285.8 NO2(g) 33.2 NaHCO3(s) –950.8


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CaO(s) –635.1 H2O(g) –241.8 N2O(g) 82.0 NaNO3(s) –468

CaCO3(s) –1207 H2O2(l) –187.8 N2O4(g) 9.16 H2S(g) –20.6

CO(g) –110.5 H2O2(g) –136.3 N2O5(g) 11.0 SO2(g) –296.8

CO2(g) –393.5 HI(g) 26.5 NOCl(g) 51.7 SO3(g) –395.7

CS2(g) –116.7 FeO(s) –272 O3(g) 143 ZnO(s) –348.3

HCl(g) –92.3 Fe2O3(s) –824.2 PCl3(l) –320 ZnS(s) –206

ClO2(g) 102.5 MgO(s) –601.7 PCl3(g) –375 ZnCl2(s) –415.1

Compound ΔΔΔΔΔfHº Compound ΔΔΔΔΔfHº Compound ΔΔΔΔΔfHºkJ/mol kJ/mol kJ/mol

CH3CHO(l) 192.3 CCl4(l) –135.4 HCHO(g) –108.6

CH3CHO(g) 166.2 C2H6(g) –84.7 HCOOH(l) –424.7

CH3COOH(l) –484.5 C2H5OH(l) –277.7 C6H12O6(s) –1260

C6H6(g) 82.93 C2H5OH(g) –235.1 CH4(g) –74.8

C6H6(l) 49.0 C2H4(g) 52.3 CH3OH(l) –238.7

C2H2(g) 226.7 C2H4O(g) –52.6 CH3OH(g) –201.2

(c) Standard enthalpy of combustion (ΔcHº)1. Definition : The enthalpy change accompanying a reaction in which 1mole of a substance

in the standard state reacts completely with oxygen or is completely burnt is calledstandard enthalpy of combustion.

2. It is denoted by ΔcHº3. Example :

2 21

C(g) + O (g) CO (g)2

⎯⎯→ , ΔcHº = –283 kJ mol–1

(1mole)

2 2 2 2 25

C H (g) + O (g) 2CO (g) + H O( )2

l⎯⎯→ , ΔcHº = –1300 kJ mol–1

Standard molar enthalpies of combustion

Element/ ΔΔΔΔΔcHº/ Element/ ΔΔΔΔΔcHº/Compound kJ/mol Compound kJ/mol

C (graphite) –393.5 CH3OH (g) –764

C (diamond) –395.4 CH3OH (l) –726

CO (g) –283.0 C2H5OH (l) –1368


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CH4 (g) –890.0 HCOOH (l) –255

C2H6 (g) –1560 CH3COOH (l) –875

C2H4 (g) –1411 HCHO (g) –571

C2H2 (g) –1300 C2H5OH (g) –1409

C6H6 (l) –3268 CH3CHO (l) –1166

C6H6 (g) –3302 H2 (g) –285.8

(d) Bond enthalpy (Bond energy) :1. Definition : The enthalpy change necessary to break a particular covalent bond in

1 mole of gaseous molecules to produce gaseous atoms and / or radicals is calledbond enthalpy.

2. The bond enthalpy is always positive3. Example : 2 (g) (g) (g)H H + H⎯⎯→ ΔHº = 436.4 kJ.

1 mole∴∴∴∴∴ ΔHº(H–H) = 436.4 kJ mol–1

(4) Bond enthalpy for diatomic molecules is the same as enthalpy of atomization.(5) For polyatomic molecules, the average bond enthalpy of a particular bond is calculated.

Example : H2O has two identical O – H bonds.

2 (g) (g) (g)H O OH + H⎯⎯→ , ΔHº = 499 kJ

2 (g) (g) (g)H O H + O⎯⎯→ ΔHº = 428 kJ

2 (g) (g) (g)H O 2H + O⎯⎯→ ΔHº = 927 kJ

∴∴∴∴∴ Average bond enthalpy of O – H bond = 927

2 = 463.5 kJ

∴∴∴∴∴ ΔHº(0 – H) = 463.5 kJ mol–1

(e) Reaction enthalpy from Bond enthalpy :


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1. In chemical reactions, bonds are broken in reactant molecules and bonds are formed inproduct molecules.

2. Enthalpy of a chemical reaction is the difference between the sum of the reactant bondenthalpies and the sum of the product bond enthalpies.

3. (reaction)ΔHº = (reactant bonds) (Product bonds)Hº Hº–∑ ∑

4. Example : 2 2H (g) + I (g) 2HI(g)⎯⎯→

∴∴∴∴∴ ΔHº = [ΔHº(H – H) + ΔHº (I – I)] – [2 × ΔHº (H – I)]

Single bond enthalpies (kJ mol–1)

Multiple bond enthalpies

Bond ΔΔΔΔΔHº kJ/mol Bond ΔΔΔΔΔHº kJ/mol

C = C 620 C = S 477C ≡ C 812 N = N 418C = N 615 N ≡ N 941.4C ≡ N 891 O = O 498.7C ≡ O 754 N = O 607

(f) Hess’s law of constant heat summation :1. Statement : It states that the change in enthalpy for a reaction is the same whether

the reaction takes place in one or a series of steps. OR It states that the enthalpychange for a chemical reaction is the same regardless of the path by which the reactionoccurs.


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2. Example:Conversion of A to C can take place directly in a single step.

A C⎯⎯→ ; ΔHº1

Or in two steps

Step (1) : A B⎯⎯→ , ΔHº2

Step (2) B C⎯⎯→ , ΔHº3

According to Hess’s law, ΔHº1 = ΔHº2 + ΔHº3

3. Example (2) Preparation of methylene chloride can take place by two different paths.

Path - I : 4 (g) 2 (g) 2 2 (g) (g)CH + 2Cl CH Cl + 2HCl⎯⎯→ , ΔHº1= –203.3 kJ

Path - II : Step (1) - 4 2 2CH (g) + 2Cl (g) CH Cl(g) + 2HCl(g)⎯⎯→ ; ΔHº2 = –98.3 kJ

Step (2) - 3 2 2 2CH Cl(g) + Cl (g) CH Cl (g) + HCl(g)⎯⎯→ ; ΔHº3 = –104.0 kJ

4 2 2 2CH (g) + 2Cl (g) CH Cl (g) + 2HCl(g)⎯⎯→ ; ΔHº = –202.3 kJ

Thus , whether the reaction takes place in one step or several steps the enthalpy of reactionin the same.

(g) Applications of Hess’s law :1. It is useful for the determination enthalpies of those reaction, whose values cannot be

experimentally determined correctly.Example :

(graphite) 2 (g) (g)1

C + O CO2

⎯⎯→ ΔHº = ?

Here some C is oxidised to CO2. Hence, its value is determined from the following twoequation.

(a) (graphite) 2 (g) 2 (g)C + O CO⎯⎯→ ; ΔHº = –393.5 kJ.

Hess’s law

CH + 2Cl4 2 CH + Cl4 2

CH Cl + Cl3 2

CH Cl + HCl3

CH Cl + 2HCl2 2


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(b) (g) 2 (g) 2 (g)1

CO + O CO2

⎯⎯→ ; ΔHº = –283.0 kJ.

[reverse eq. (b) and add eq. (a)]

2. It is used for the calculation of enthalpies of those reaction which do no occur directly.Example - Formation of CH4 from the elements.

3. The enthalpy of formation of any compound from its elements can be calculated.4. The enthalpy of combustion can be calculated.5. The thermochemical equation can be added, subtracted or multiplied by a numerical factor

like ordinary algebraic equation.6. It is useful for the determination of enthalpies of extremely slow reactions.


*1. Define enthalpy of chemical reactions.Ans. (a) Definition : The amount of heat released or absorbed in a chemical reaction at constant

temperature and pressure is called enthalpy of chemical reaction or heat of reaction.

(b) Example : aA + bB cC + dD⎯⎯→

∴∴∴∴∴ Enthalpy change (ΔH) = (cHc + dHD) – (aHA + bHB)

or ΔH = products reactantsH – H∑ ∑

(c) Thus, the enthalpy of chemical reaction is the difference between the sum of the enthalpiesof products and that of reactants.

*2. What is meant by standard state of a substance and standard enthalpy of reactions?Ans. Refer 3.8(a) 3.

*3. Define : (a) Exothermic reactions, (b) Endothermic reactionsAns. (a) Exothermic reaction :

(1) When productsH∑ is less than reactantsH∑ , ΔH is negative and heat is released.

(2) The reactions in which heat is released from the system to the suroundungs are calledexothermic reactions.

(3) Example : 3 22KClO (s) 2KCl(s) + 3O (g), Hº = –78kJ⎯⎯→

(b) Endothermic reaction :

(1) When productsH∑ is more than reactantsH∑ , ΔH is positive and heat is absorbed.(2) The reaction in which heat is absorbed by the system from the surroundings are called

endothermic reaction.

(3) Example : 2 (g) 2 (g) 2 (g)N + 2O 2NO⎯⎯→ , ΔHº = +66.4 kJ


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*4. What are the guidelines for writing a thermochemical equation?Ans. Refer 3.8 (a) 4 – b

*5. Define standard enthalpy of formation. How is it useful to calculate standard reaction enthalpy?Ans. Standard enthalpy of formation (standard heat of formation) ΔfHº :

(i) Definition : The enthalpy change that accompanies a reaction in which one mole ofa pure compound in its standard state is formed from its elements in their standardstates, is called standard enthalpy of formation.

(ii) It is denoted by ΔfHº.(iii) Example :

2 (g) 2 (g) 2 (l)1

H + O H O2

⎯⎯→ , ΔfHº = –286 kJ mol–1

(1 mole)

(graphite) 2 (g) 4 (g)C + 2H CH⎯⎯→ , ΔfHº = –74.8 kJ mol–1

(iv) The standard enthalpy of formation of an element from itself is zero. i.e.ΔfHº (C) = ΔfHº (H2) = ΔfHº (Cl2) = 0

[Refer Table 3.8 (1)b]

*6. Write balanced chemical equation that have ΔHº value equal to ΔfHº for each of the followingsubstances.(a) C2H2 (b) KClO3 (c) CH3COOH (d) C12H22O11 (e) CH3CH2–OH

Ans. We know, ΔHº = f (products) f (reactants)m Δ Hº – n Δ Hº∑ ∑

(a) 2 2 22C(s) + 2H (g) C H (g)⎯⎯→

(b) 2 2 31 3

K(s) + Cl (g) + O (g) KClO (s)2 2

⎯⎯→

(c) 2 2 32C(s) + O (g) + 2H (g) CH COOH( )l⎯⎯→

(d) 2 2 12 22 1111

12C(s) + 11H (g) + O (g) C H O (s)2

⎯⎯→

(e) 2 2 3 21

2C(s) + 3H (g) + O (g) CH CH OH ( )2

l⎯⎯→

*7. Explain standard enthalpy of combustion with one example.Ans. Refer 3.8 (c).

*8. Consider the chemical reaction : 2 2 2(g)OF (g) + H O(g) O + 2HF(g)⎯⎯→ ΔHº = –323 kJ.What is ΔHº of the reaction if : (a) the equation is multiplied by 3 (b) direction of reaction is reversed?

Ans. (a) ΔHº = 3 × –323 kJ = – 969 kJ(b) ΔHº = + 323 kJ.

Unique Solutionsä S.Y.J.C. Science - Chemistry - Part I

188

*9. What is meant by bond enthalpy? How is it useful to calculate reaction enthalpy? Explain withone example.

Ans. Refer 3.8 (d, e).

*10. State and explain Hess’s law of constant heat summation.Ans. Hess’s law of constant heat summation :

(a) Statement : It states that the change in enthalpy for a reaction is the same whetherthe reaction takes place in one or a series of steps. OR It states that the enthalpychange for a chemical reaction is the same regardless of the path by which the reactionoccurs.

(b) Example (1) : Conversion of A to C can take place directly in a single step.

A C⎯⎯→ ; ΔHº1 Or in two steps

Step (1) A B⎯⎯→ , ΔHº2

Step (2) B C⎯⎯→ , ΔHº3 [Refer 3.8 (f)]According to Hess’s law, ΔHº1 = ΔHº2 + ΔHº3

(c) Example (2) : Preparation of methylene chloride can take place by two different paths.

Path - I 4 2 2 2CH (g) + 2Cl (g) CH Cl (g) + 2HCl(g)⎯⎯→ , ΔHº1 = 203.3 kJ

Path - II Step (1) 4 2 2 2CH (g) + Cl (g) CH Cl (g) + 2HCl(g)⎯⎯→ ; ΔHº2 = –98.3 kJ

Step (2) 3 2 3 2CH Cl(g) + Cl (g) CH Cl (g) + HCl(g)⎯⎯→ ; ΔHº3 = –104.0 kJ

4 2 2 2CH (g) + 2Cl (g) CH Cl (g) + 2HCl(g)⎯⎯→ ; ΔHº = –202.3 kJ

Thus, whether the reaction takes place in one step or several step the enthalpy of reactionis the same. [Refer 3.8 (f) flow chart diagram]

*11. The heat evolved in a reaction of 7.5g of Fe2O3 with enough CO is 1.164 kJ. Calculate ΔHº

for the reaction, 2 3Fe O (s) + 3CO(g) 2Fe(s) + 3CO(g)⎯⎯→

Ans. gram Mol. Wt. of Fe2O3 = 2 × 56 + 3 × 16 = 160g∵ 7.5g of Fe2O3, heat evolved = 1.164 kJ.

∴∴∴∴∴ 160g of Fe2O3, heat evolved = 160 × 1.164

7.5 = 24.832 kJ.

*12. Calculate using data of table (3.7), the enthalpy of the reaction.[Refer 3.7 Table Text Book.N.101]

3 3 2 3 2 3 2CH COOH(g) + CH CH OH(g) CH COOCH CH (g) + H O(g)⎯⎯→ [Refer 3.8 (e)]

Ans.

H O H H H O H H

H C C O H + H C C O H H C C O C C

H H H H H H

⎯⎯→ H + H –O–H


189

ΔHº = (reactants bonds) (Product bonds)ΔHº ΔHº–∑ ∑

ΔHº = [8 × ΔHºC – H + 2 × ΔHºC – C + 1 × ΔHºC = O + 2 × ΔHºO – H + 2 × ΔHºC – O] –[8 × ΔHºC – H + 2 × ΔHºC – C + 2 × ΔHºC – O + 1 × ΔHºC = O + 2 × ΔHºO – H]

ΔHº = [8 × 414 + 2 × 347 + 1 × 754 + 2 × 464 + 2 × 351] –[8 × 414 + 2 × 347 + 2 × 351 + 1 × 754 + 2 × 464]

∴∴∴∴∴ ΔHº = 0

*13. The standard enthalpy change for the reaction 2H (g) H(g) + H(g)⎯⎯→ is 436.4 kJ mol–1.Calculate standard enthalpy of formation of atomic hydrogen.

Ans. When 2 moles of H atoms are formed, 436.4 kJ energy is requiredQ 2 × 6.022 × 1023 atoms of H are formed = 436.4 kJ

∴∴∴∴∴ 1 atom of H is formed = 23

436.4

2 × 6.022 × 10 = 3.623 × 10–22 kJ.


• Theoretical MCQs :1. The standard heat of formation of diamond is .......

a) Same as that of graphite b) Greater than that of graphitec) Less than that of graphite d) Taken as zero

2. The heat of combustion of compound is always .......a) Positive b) Negative c) Zero d) Uncertain

• Numerical MCQs :3. The enthalpy changes at 298K in successive breaking of O – H bond of HOH are .......

2H O(g) H(g) + OH(g)⎯⎯→ ; ΔH = 498 kJ mol–1

OH(g) H(g) + O(g)⎯⎯→ ; ΔH = 428 kJ mol–1

The bond enthalpy of O – H bond is .......a) 498 kJ mol–1 b) 463 kJ mol–1 c) 428 kJ mol–1 d) 70 kJ mol–1

4. The enthalpies of combustion of carbon and carbon monoxide are –390 kJ mol–1 and –278kJ mol–1 respectively. The enthalpy of formation of carbon monoxide is .......a) 668 kJ mol–1 b) 112 kJ mol–1 c) –112 kJ mol–1 d) –668 kJ mol–1

• Advanced MCQs :5. The formation of water from H2(g) and O2(g) is an exothermic process because .......

(PMT - 1986)a) The chemical energy of H2(g), and O2 (g) is more than that of waterb) The chemical energy of H2(g), and O2 (g) is less than that of waterc) The temperature of H2(g), and O2 (g) is higher than that of waterd) The temperature of H2(g), and O2 (g) is lower than that of water


190

3.9. SECOND LAW OF THERMODYNAMICS :

Concept Explanation :It is based on human experience.It can be stated as :(i) ‘The spontaneous flow of heat is always unidirectional, from higher temperature to lower

temperature.’(ii) Heat cannot be completely converted into an equivalent amount of work without producing

permanent changes either in the system or its surroundings.(iii) No machine has yet been made that has an efficiency unity.

(The efficiency of machine is the ratio of work done to the total heat absorbed by heatengine).If heat absorbed is completely converted into work without producing changes in the system,the efficiency must be unity.

(iv) During the course of every spontaneous (natural) process, the entropy of the universe(system and surroundings) increases.

3.10. SPONTANEOUS PROCESS (IRREVERSIBLE PROCESS) :

Concept Explanation :1. A reaction that occurs under the specified set of conditions is a spontaneous reaction.2. A reaction that does not occur under the specified set of conditions is a non-spontaneous

reaction.3. Definition : A process that takes place on its own without the external influence is

called a spontaneous process.4. Definition : A process that does not take place without the continuous external force

is called a non-spontaneous process.5. A non-spontaneous process can be made to occur. For example a gas can be compressed

into a smaller volume by applying external pressure on it.6. All the natural (spontaneous) processes proceed till equilibrium is reached.Examples of spontaneous processes :1. Water always flows on its own form higher level to lower level.2. Heat always flows from hotter body to a colder body.3. Solid KCl spontaneously dissolves in water.4. Acid-base neutralization is a spontaneous reaction.5. Consider two bulbs connected through stop-cock. One bulb

is filled with a gas and other is evacuated. As soon as thestop-cock is opened the gas expands spontaneously.


191

(a) Energy and spontaneity :1. The stability of a substance is determined by its energy. The substances having more energy

are less stable whereas those that have less energy are more stable.2. In a spontaneous reaction, substances with more energy are converted into the substances

having less energy.3. A chemical reaction must be exothermic in order to take place spontaneously e.g.

neutralization of acid-base.

3 3 2KOH(aq) + HNO (aq) KNO (aq) + H O⎯⎯→ , ΔH = –57 kJ.

4. Although exothermicity favours spontaneity but does not assure it.There are many processes that are endothermic yet spontaneous.(a) Ice melts spontaneously above 0ºC with the absorption of heat from the surroundings.

above20ºC

Ice H O(l)⎯⎯⎯→ , ΔH = +6.0 kJ mol–1

(b) NaCl dissolves spontaneously in water with the absorption of heat.+ –

(s) (aq) (aq)NaCl + aq Na + Cl⎯⎯→ , ΔH = +3.9 kJ mol–1

Note : Energy is not the only criterion for the spontaneity of processes and hence,there must be another factor (thermodynamic property) called Entropy that determinesthe spontaneity of processes.

(b) Entropy :1. Entropy is a thermodynamic property that determines the spontaneity of a processes.2. Definition : Entropy is a thermodynamic state quantity that is a measure of the

randomness or disorder of the molecules of the system. It is denoted by S.3. Example :

(a) When ice melts, the highly ordered arrangement of molecules collapses and moleculesbecome free to move in liquid, hence, entropy increases and ΔS is positive.

(b) When liquid water vaporizes, the gaseous water molecules move freely and randomly.A less disordered state becomes highly disordered and hence, entropy further increases.

4. Thus, greater the disorder of a system, higher is its entropy.


192

(a) Sign of ΔS different process : When solid I2 undergoes sublimation, the disorderof the system increases and hence, entropy increases. Thus, ΔS is positive.

2 2I (s) I (g)ordered state disordered state

⎯⎯→

ΔS positive

(b) One mole of H2 gas is converted to two moles of gas phase atoms. The numberof particles increases. Thus, disorder increases and hence, entropy of the systemincreases.

2H (g) 2H(g)

disordered state more disordered state⎯⎯→

ΔS positive

(c) 2 2 22H (g) + O (g) 2H O( )l⎯⎯→ ΔS negative

more disordered state less disordered state.In the above process, three gas-phase moles are converted into two moles of liquid.The more disordered state becomes less disordered state. The entropy of the systemdecreases. ΔS is negative.

(c) Quantitative definition of entropy :When one substance changes to another in a chemical reaction, sometimes it is difficultto identify the disorder, hence, it is necessary to define entropy quantitatively.

1. Definition - The entropy change of a system in a process is equal to the amount ofheat transferred to it in a reversible manner divided by the temperature at whichthe transfer takes place.

Thus, ΔS = revq

TWhere, ‘qrev

’ is the heat transferred to the system at temperature T.2. The unit of ΔS is JK–1.3. Entropy (ΔS ) is a state function.4. ΔS that is qrev /T measures the change in the disorder.

(d) Entropy and spontaneity (2nd Law of thermodynamics) :1. Consider the following two spontaneous reactions :

2 2 2 22H O ( ) 2H O( ) + O (g)l l⎯⎯→ , ΔS = +126JK–1

2 2 22H (g) + O (g) 2H O( )l⎯⎯→ ΔS = –327JK–1

2. The above two reactions show that during spontancous process, the entropy may increasesor decreases.This discrepaney is explain by the second law of thermodynamics.

3. Second law of thermodynamics (statement) : It states that the total entropy of the systemand its surroundings (universe) increases in a spontaneous process.


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4. 2nd Law of thermodynamics is expressed as – For spontaneous process,ΔSuniverse = ΔStotal = ΔSsys + ΔSsurr > O

5. (a) When ΔStotal > 0; the process is spontaneous(b) When ΔStotal < 0; the process is non-spontaneous(c) When ΔStotal = 0; the process is at equilibrium


*1. What is spontaneous process? Give examples.Ans. Refer 3.10.

*2. Which of the following are spontaneous?(a) dissolution of sugar in hot coffee.(b) separation of Ar and Kr from their mixture.(c) spreading of fragrance when a bottle of perfume is opened.(d) flow of heat from cold object to hot object(e) heat transfer from ice to room at 25ºC.

Ans. (a), (b), (c), (e) → spontaneous processes.(d) → non-spontaneous

*3. Define entropy. Give its units. What does entropy measure?Ans. Definition : Entropy is a thermodynamic state quantity that is a measure of the randomness

or disorder of the molecules of the system. It is denoted by S.Unit of entropy = JK–1

Entropy (S) measures the randomness or disorder of the molecules of the system.


• Theoretical MCQs :1. Entropy change of a system and its surroundings in equilibrium .......

a) Increases b) Decreasesc) Remains constant d) Either increases or decreases

2. In a reversible process, ΔSsys + ΔSsurr is .......a) > 0 b) < 0 c) ≥ 0 d) = 0

• Numerical MCQs :3. 1g. ice absorbs 335 J of heat to melt at 0ºC, the entropy change will be .......

a) 1.2 JK–1 mol–1 b) 335 JK–1 mol–1 c) 22.1 JK–1 mol–1d) 0.8 JK–1 mol–1


194

4. If enthalpy of vaporisation of benzene is 308 kJ mol–1 at boiling point (80ºC). Calculateentropy in changing it from liquid to vapour in J mol–1 K–1 .......(Haryana PMT - 2004)a) 308 b) 0.873 c) 0.308 d) 873

• Advanced MCQs :

5. For a reaction 2 21

CO(g) + O (g) CO (g)2

⎯⎯→ , ΔH and ΔS are –283 kJ and – 87 JK–1.

It was intended to carry out this reaction at 1000K, 1500K, 3000K and 3500K. At whichof these temperatures would this reaction be thermodynamically spontaneous .......

(Kerala PET-2006)a) 1500 K and 3500 K b) 3000 K and 3500 Kc) 1000 K, 1500 K and 3500 K d) 1500 K, 3000K and 3500 K

3.11. GIBBS ENERGY (G) : Concept Explanation :

According to 2nd of thermodynamics, for spontaneous process ΔStotal (ΔSsys + ΔSsurr) must

be positive. Thus, we have determine two entropy change, ΔSsys and ΔSsurr.But chemists are interested only in system (reaction mixture).This problem was solved by American theoretician J. G. Gibbs. He introduced a newthermodynamics function called Gibbs energy (G).1. Definition - Gibbs energy of a system is defined as the maximum amount of energy

available to a system during a process that can be converted into useful work.2. In other words, it is a thermodynamic quantity which is a measure of capacity of a

system to do useful work.3. It is denoted by symbol G is given by G = H – TS.

Where H is the enthalpy of the system, S is its entropy and T is the absolute temperature.4. H, T and S are state functions, hence, G is also a state function.5. The change in Gibbs energy (ΔG) depends on initial and final states of the system and

not on the path connecting the two states.6. The change in Gibbs energy at constant temperature and pressure is defined as –

ΔG = ΔH – TΔS.

(a) Gibbs energy and spontaneity :1. ΔStotal = ΔSsys + ΔSsurr = ΔS + ΔSsurr (The subscript ‘sys’ is dropped, but it refers

to system only).2. According to 2nd law of thermodynamics, for spontaneous process, at constant temperature

and pressure.ΔStotal > 0.


195

3. If ΔH is the enthalpy change of the system, then enthalpy change of the surroundingsis –ΔH.

4. ∴ ΔSsurr = ΔH

–T

5. Hence, total entropy is given by ΔStotal = ΔH

ΔS –T

.

This equation is expressed in terms of properties of system only.6. On rearrangement of above equation,

–T ΔStotal = –T ΔS + ΔH–T ΔStotal = ΔH – TΔSBut, ΔG = ΔH – TΔS

∴ ΔG = – TΔStotal

This equation indicates that ΔG and ΔStotal have opposite signs because T is alwayspositive.

7. Thus, for a spontaneous process, carried out at constant temperature and pressure.ΔStotal > 0 and hence, ΔG < 0. i.e. Gibbs energy of the system decreases. Thus if,(a) ΔG < 0, the process is spontaneous(b) ΔG > 0, the process is non-spontaneous(c) ΔG = 0, the process is at equilibrium

(b) Predicting spontaneity of a process in terms of ΔG :(i) Gibb’s equation is : ΔG = ΔH – T.ΔS(ii) The spontaneous process is favoured by decrease in enthalpy (–ΔH) and increase in entropy

(+ΔS).(iii) A non-spontaneous reactions is favoured by increase in enthalpy (+ΔH) and decrease

in entropy (–ΔS).(iv) Reactions are classified into four categories with respect to dependence of sign of ΔG

and hence, spontaneity on ΔH, ΔS and T.(a) If ΔH and ΔS are both negative the ΔG will be negative and the reaction will be

spontaneous only when the term ΔH is larger in magnitude than the magnitude ofT ΔS. (This is possible only at low temperatures.)

(b) If ΔH and ΔS are both positive then ΔG will be negative and the reaction will bespontaneous only when the TΔS term is larger in magnitude than ΔH. (This is possibleonly at high temperatures.)

(c) If ΔH is negative and ΔS is positive, ΔG will be always negative regardless oftemperature. The reaction is therefore, spontaneous at all temperatures.

(d) If ΔH is positive and ΔS is negative the ΔG will be always positive regardless oftemperature.The reaction is therefore, non-spontaneous at all temperatures. The factors affectingsign of ΔG and spontaneity are summarized in table. (Factor affecting spontaneity)


196

Factor affecting spontaneity

ΔH ΔS ΔG Spontaneity of reactions– + – Reactions are spontaneous at all the temperatures.

(exothermic)

_ _ – or + Reactions become spontaneous at low temperatures(exothermic) when |T S| < | H|⋅

+ + – or + Reactions become spontaneous at high temperatures(endothermic) when T S > H⋅ .

+ – + Reactions are nonspontaneous at all temperatures.(endothermic)

(c) Temperature of equilibrium between spontaneous andnonspontaneous process :At equilibrium, the process is neither spontaneous nor non-spontaneous.

At equilibrium, ΔG = 0, hence, Gibb’s equation can be written as ΔG = ΔH – T ΔS⋅ = 0

∴∴∴∴∴ ΔH = TΔS or T = ΔH

ΔS‘T’ is the temperature at which change over between spontaneous and non-spontaneousbehaviour occurs.

(d) ΔG and equilibrium constant :In a chemical reaction all the substances may not be in their standard states. Hence, changein Gibbs energy (ΔG) is related to Standard Gibbs energy change (ΔGº) as ΔG = ΔGº+ RT ln Q Where, ΔGº = Standard Gibbs energy change when all the substances arein their standard states. Q = Reaction quotient.Reaction quotient (Q) is in terms of initial concentration or partial pressure of the reactantsand final concentrations or pressures of the products.

Example : aA + bB cC + dD⎯⎯→

Qc = c d

a b

[C] [D]

[A] [B]or Qp =

c dC D

a bA B

p × p

p × pWhere the values of concentrations or partial pressures are other than equilibriumconcentrations or partial pressure values.When the reaction reaches equilibrium, the concentration and partial pressure reach theirequilibrium values and at this stage, Q = KAt equilibrium, ΔG = 0 and Q = K, hence Gibb’s equation becomes,0 = ΔGº + RT In K.


197

∴∴∴∴∴ ΔGº = –RT ln K = –2.303 RT log10K.This equation gives the relationship between Standard Gibbs energy change (ΔGº) of thereaction and its equilibrium constant.

3.12. THIRD LAW OF THERMODYNAMICS :

Concept Explanation :1. Define : It states that the entropy of a perfectly ordered crystalline substance is zero

at absolute zero of temperature.2. Thus, S = 0 at T = 0 for any perfectly ordered crystalline substance.3. If crystal contains some impurities, then S > 0 at T = 0.

This entropy of solid greater than zero is called residual entropy of the substance.4. Third law is used to determine the absolute entropy of any substance either in solid, liquid

of gaseous state at any desired temperature.5. Example : If a perfectly ordered crystalline substance with S = 0 at T = 0 is heated from

(zero) 0K to any desired temperature T, the entropy increases due to the increased vibrationalmotions of molecules.

6. The increase in entropy is given by ΔS = Sr – S0.Where Sr is the absolute entropy of the substance at temperature.T and S0 its value at T = 0.Because, S0 = 0, ∴∴∴∴∴ ΔS = Sr.

7. The value of Sr can be determined by measuring heat capacity of the solid at varioustemperatures by using the expression,

ΔS = Sr – S0 = Sr =

T

O

Cp dT

T

⋅

∫

(a) Standard molar entropy :

1. Definition : The absolute entropy of one mole a pure substance at 1 atm. pressureand 25ºC is called Standard molar entropy of the substance.

2. It is denoted by Sº.

(b) Usefulness of Standard molar entropy (Sº) :1. By knowing the values of Sº of all reactants and products, it is possible to calculate ΔSº.

ΔSº = (products) (reactants)m Sº – n Sº∑ ∑

2. Sº is useful to compare entropies of different substances under the same conditions oftemperature and pressure.→ Sº increases with increasing complexity of molecules.→ Heavier substances have larger Sº values than higher ones.


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*1. Predict the signs of ΔH, ΔS and ΔG of the system when a solid melts at 1 atm. and at(a) –55ºC (b) –95ºC (C) –77ºC, if the normal melting point of solid is –77ºC.

Ans.

*2. Explain how entropy changes in the following processes.(a) freezing of liquid (b) sublimation of solid(c) dissolving of sugar in water (d) condensation of vapour

Ans. (a) Freezing of liquid – Entropy decreases as disordered state becomes ordered state. ΔS is negative.

(b) Sublimation of solid – Entropy increases as ordered state of the system becomes disorderedbecause solid substances goes into gas phase. ΔS is positive.

(c) Dissolving of sugar in water – Entropy increases as ordered state of sugar moleculesgoes into disordered state due to separation of molecules of sugar in water. ‘ΔS’ is positive.

(d) Condensation of vapours – Entropy decreases as gas-phase molecules are convertedinto liquid state.The more disordered state becomes less disordered state. ΔS is negative.

*3. Which member of the following pairs have larger entropy? Why?(a) CO2(s) or CO2(g) (b) CH3OH (l) or C2H5OC2H5(l)(c) NO(g) or N2O4(g) (d) Xe(g) or Kr(g)(e) Na(s) or NaCl(s)

Ans. (a) CO2(g) will have larger entropy because gas phase is highly disordered.(b) CH3OH(l) will have larger entropy because molecules are smaller in size.(c) N2O4(g) will have larger entropy because it has greater complexity.(d) Xe has greater atomic mass and hence, greater entropy than Kr.(e) NaCl(s) has greater entropy since it is more complex than Na(s).

*4. Why is it more convenient to predict spontaneity of reaction in terms of ΔGsys rather thanΔStotal?

Ans. (a) 2nd law of thermodynamics says that ΔStotal (ΔSsys + ΔSsurr) must be positive for allspontaneous processes.


199

(b) Therefore, to know the spontaneity of a process, it is necessary to determine two entropychanges, i.e. ΔSsys and ΔSsurr.

(c) However, the chemists are more interested in the system (reaction mixture) than thesurroundings.

(d) Hence, spontaneity of reaction is expressed in terms of ΔGsys.

*5. Define Gibbs energy and change in Gibbs energy. What are the units of Gibbs energy?Ans. Refer 3.11 (1, 5, 6).

Units of ΔG : ΔG has the units of energy (i.e. kJmol–1or Jmol–1) because both ΔH andTΔS are energy terms.ΔG (J mol–1) = ΔH (J mol–1) – TΔS (k JK–1mol–1)

*6. State second law of thermodynamics in terms of entropy and express it mathematically.Ans. Quantitative definition of entropy : When one substance changes to another in a chemical reaction,

sometimes it is difficult to identify the disorder, hence, it is necessary to define entropyquantitatively.(a) Definition - The entropy change of a system in a process is equal to the amount of

heat transferred to it in a reversible manner divided by the temperature at whichthe transfer takes place.

Thus, ΔS = revq

TWhere, ‘qrev

’ is the heat transferred to the system at temperature T.(b) The unit of ΔS is JK–1.(c) Entropy (ΔS ) is a state function.(d) ΔS that is qrev /T measures the change in the disorder.

*7. State third law of thermodynamics. What is its usefulness?Ans. Refer Concept explanation 3.12 (1) (b)

*8. Why is it impossible for substance to have an absolute entropy zero at temperature greaterthan 0 K?

Ans. If a perfectly ordered crystalline substance with S = 0, at T = 0 is heated from 0 K to anydesired temperature T. The entropy increases due to increased vibrational motions of molecules.Hence, it is impossible to have an absolute entropy zero at temperature greater than 0 K.

*9. It is possible for a reaction to be spontaneous yet endothermic. Comment with exampleAns. Energy and spontaneity :

(a) The stability of a substance is determined by its energy. The substances having more energyare less stable whereas those that have less energy are more stable.

(b) Although exothermicity favours spontaneity but does not assure it.There are many processes that are endothermic yet spontaneous.


200

(1) Ice melts spontaneously above 0ºC with the absorption of heat from the surroundings.

above2 (l)0ºC

Ice H 0⎯⎯⎯→ , ΔH = +6.0 kJ mol–1

(2) NaCl dissolves spontaneously in water with the absorption of heat.+ –

(s) (aq) (aq)NaCl + aq Na + Cl⎯⎯→ , ΔH = +3.9 kJ mol–1

*10. Is it possible for a reaction to be nonspontaneous yet exothermic? Explain with example.Ans. Normally exothermic process are spontaneous but there are some processes which are non-

spontaneous inspite of being exothermic. For example : H2O(l) at 1 atm. and 0ºC forms ice.Here both the phases are in equilibrium. This process is non-spontaneous.

1atm,0ºC2 2H O(l) H O(s) ; H < 0

.

*11. Predict the sign of ΔS in the following processes. Give reasons for your answer.

(a) 2 4 2N O (g) 2NO (g)⎯⎯→

Ans. sign of ΔS is +ve as disorder in the product are larger than in the reactants.

(b) 2 3 2 2Fe O (s) + 3H (g) 2Fe(s) + 3H O(g)⎯⎯→

Ans. reaction is in equilibrium, ΔS = 0.

(c) 2 2 3N (g) + 3H (g) 2NH (g)⎯⎯→

Ans. sign of ΔS is –ve, as disorder in the product are less than in the reactants.

(d) 3 2MgCO (s) MgO(s) + CO (g)⎯⎯→

Ans. ΔS is +ve, as disorder increases in the products.

(e) 2 2CO (g) CO (s)⎯⎯→

Ans. ΔS is –ve as disorder decreases in the product.

(f) 2Cl (g) 2Cl(g)⎯⎯→

Ans. ΔS is +ve, as one mole of Cl2(g) gives two moles of Cl(g) atoms and hence, disorder increases.

*12. What can be said about the spontaneity of reactions when(a) ΔH and ΔS are both positive.

Ans. Reactions become spontaneous at high temperature When T ΔS⋅ > ΔH(b) ΔH and ΔS are both negative.

Ans. Reactions become spontaneous at low temperatures When T ΔS⋅ < ΔH(c) ΔH is positive and ΔS is negative.

Ans. Reactions are nonspontaneous at all temperatures.(d) ΔH is negative and ΔS is positive.

Ans. Reactions are spontaneous at all the temperatures.


201

*13. Identify which member of the following pairs has larger entropy. Why?(a) He (g) in a volume of 1L or He (g) in a volume of 5L both at 25ºC.

Ans. He (g) in a volume of 5L will have larger entropy become disorder will be more due to largevolume.(b) O2 (g) at 1 atm or O2 (g) at 10 atm both at the same temperature.

Ans. O2(g) at 10 atm. will have lesser disorder due to high pressure and hence, entropy will be less.(c) C2H5OH (l) or C2H5OH (g)

Ans. C2H5OH(g) will have more disorder and hence, entropy will be greater.(d) 5mol of Ne or 2 mol of Ne.

Ans. 5 mole of Ne will be more, Ne particles and hence, disorder will be more. Therefore, entropywill be more.

*14. What are the signs of ΔS and ΔH for the following reaction? Explain with reasons.

22H(g) H (g)⎯⎯→

Ans. 2 moles of gas-phase atoms are converted into 1mole of H2(g). Number of particles decreaseand disorder is less and hence, entropy of the system decreases. ΔS is negative.The above reaction shows formation of bond, which is exothermic process, hence, ΔH is –ve.

*15. Derive the relationship between ΔG and ΔStotal.Ans. Refer 3.11 (a)

*16. Justify the inclusion of qrev and T in the definition of entropy, ΔS = qrev/T.Ans. (a) When heat is added to a system the molecular motions are increased due to an increase

in the kinetic energies of the molecules.(b) This results in increasing molecular disorder, hence, entropy increases. i.e. ΔS is proportional

to qrev.(c) If a given quantity of heat is added to a system at higher temperature, then the additional

disorder created is less than if the same heat is added at lower temperature.(d) This shows that ΔS is inversely proportional to the temperature.

*17. The increase in entropy of a system alone does not guarantee the spontaneity of a process.Explain.

Ans. (a) There are many spontaneous processes in which system’s entropy decreases.

(b) For example, 2 2 22H (g) + O (g) 2H O( )l⎯⎯→ , ΔS = – 327 JK–1

(c) In the above reactions, 3 moles of gaseous substances are converted to 2 moles of liquidsubstance.

(d) The disorder in the reactants is larger than that in the product, hence, entropy decreases.(e) The above reaction is spontaneous once initiated it proceeds with explosive violence.

*18. The criterion of spontaneity in terms of Gibbs energy is the same as that laid down by secondlaw of thermodynamics. How?

Ans. (a) Gibbs energy (ΔG ) increases in a nonpontanceous process (ΔG > 0).


202

(b) Thus, the criterion of spontaneity according to second law of thermodynamicsΔStotal > 0 remains the same.The way to express the spontaneity of a process is different. In other words, the secondlaw of thermodynamics is not violated.

(c) For a spontaneous process ΔG < 0 that is Gibbs energy of the system decreases.


• Theoretical MCQs :*1. If for a reaction ΔH is negative and ΔS is positive then the reaction is ........

a) spontaneous at all temperatures b) non-spontaneous at all temperaturesc) spontaneous only at high temperatures d) spontaneous only at low temperatures

*2. Which of the following has highest entropy?a) Al(s) b) CaCO3(s) c) H2O(l) d) CO2(g)

*3. For the process, 2 2H O( ) H O(g)l ⎯⎯→ at 100ºC, ΔS is ........

a) positive b) negative c) zero d) unpredictable*4. For the conversion of liquid to solid below the melting point of solid ........

a) ΔSsys is negative and the process is spontaneousb) ΔSsys is positive and the process is spontaneousc) ΔSsurr is positive and the process is non-spontaneousd) ΔSsys is zero and the process is at equilibrium.

*5. Which of the following conditions guarantee that a reaction is spontaneous at constant T and P?a) entropy of system increases b) enthalpy of system decreasesc) entropy of system decreases and that of surroundings increases.d) Gibbs energy of the system decreases.

*6. Which of the following processes is non-spontaneous?a) dissolving KCl in waterb) mixing of iodine vapour and nitrogen gasc) decomposition of NaCl to Na (s) and Cl2 (g)d) freezing of water at 270K.

*7. For which of the following reactions ΔS is negative?

a) 2 2Mg(s) + Cl (g) MgCl (s)⎯⎯→ b) 2 2H O( ) H O(g)l ⎯⎯→

c) 3 2CaCO (s) CaO(s) + CO (g)⎯⎯→ d) 2I (g) 2I(g)⎯⎯→

8. Heat energy cannot completely transformed into work without producing some changesomewhere, is the statement of ........a) Hess’s law b) First law of thermodynamicsc) Kirchoff’s law d) Second law of thermodynamics


203

9. Considering entropy (s) as a thermodynamic parameter, the criterion for the spontaneity ofany process is ........ (CBSE 2004)a) ΔSsys > 0 only b) ΔSsurr > 0 onlyc) ΔΔΔΔΔSsys + ΔΔΔΔΔSsurr > 0 d) ΔSsys – ΔSsurr > 0

• Numerical MCQs :10. If k < 1. What will be the value of ΔGº of the following ....... (Gujarat CET 2007)

a) 1 b) Zero c) Negative d) Positive

11. What is ΔGº of the following reaction 2 21

CO(g) + O (g) CO (g)2

⎯⎯→ (CBSE 1990)

ΔHº = –282.84 kJ mol–1, SºCO2 (g) = 213.8 JK–1 mol–1, SºCO(g) = 197.9 JK–1 mol–1, SºO2

(g) = 205.0 JK–1 mol–1 .......a) –257 kJ mol–1 b) –250 kJ mol–1 c) +25 kJ mol–1 d) +300 kJ mol–1

• Advanced MCQs :

12. For a process 2 2H O ( ) (1bar,373K) H O(g)l ⎯⎯→ (1 bar, 373 K). The correct set of

thermodynamic parameters is ....... (IIT 2007)a) ΔG = 0, ΔS = +ve b) ΔG = 0, ΔS = –vec) ΔG = +ve, ΔS = 0 d) ΔG = –ve, ΔS = +ve

HOURS BEFORE EXAM Energy is capacity to do work. It has many forms such as kinetic energy potential energy,

heat, work, chemical energy and so on.

System - is the part of the universe chosen for thermodynamic consideration.

Surroundings - The part of universe other than system is called surroundings.

Extensive property - depends on the amount of matter present in the system.

Intensive property - does not depend on the amount of the substance.

A state function is a property whose value depends on the present state of the systemand not on the path.

A reversible process is any process in which the system is always in temperature-pressure equilibrium with its surroundings.

Expansion (–W) → work is done by the system.

Compression (+W) → work is done on the system.

First law of thermodynamics - its mathematical expression is ΔU = q + W

Work = force × displacement, but in chemistry we deal with the PV work done, whichis due to volume change.


204

Work done during expansion, W = ex– P ΔV⋅

Maximum work done during expansion, Wmax = 210

1

V– 2.303 n RT log

V Wmax = 1

102

P– 2.303 n RT log

P Internal energy (U) : of a system is the sum of all kinetic and potential energies of

all the particles in the system.

Enthalpy (H) : of a system is defined as H = U + PV.

Change in enthalpy at constant pressure is given by ΔH = ΔU + P ΔV⋅

For reactions involving solids and liquids only, ΔH = ΔE

Work done in a chemical reaction is given by, W = –Δn RT⋅

Enthalpy change (ΔH) in a chemical reaction is given by, ΔH = ΔU + RT Δn⋅

ΔH = (products) (Reactants)H – H∑ ∑

Hess’s law - the change in enthalpy for a reaction is the same whether the reactiontakes place in one or a series of steps.

Spontaneous process - A process which occurs on its own without external help.

Entropy - is a measure of molecular disorder or randomness in a system it determines

spontaneity of a process. ΔS = revq

T Second law of thermodynamics the total entropy of the universe (system + surroundings)

increases in a spontaneous process. ∴∴∴∴∴ ΔStotal > 0

Change in Gibbs energy is given by, ΔG = ΔH – T.ΔS

ΔG is negative for a spontaneous change, positive for a non-spontaneous change andZero for equilibrium.

Gibbs energy change for a reaction is given by, ΔΔΔΔΔG = ΔΔΔΔΔGº + RT ln Q.

Third law of thermodynamics : the entropy of a pure perfectly ordered crystallinesubstance is zero at 0 K.

The standard entropy of a reaction is given by, ΔΔΔΔΔSº = º º

(products) (Reactants)S – S∑ ∑

Conversions :1 L.atm = 101.32J = 1.013 × 109erg = 24.206 cal1 cal = 4.184 J and 1 J = 0.2390 Cal1 atm = 1.01325 × 105 Pa (Nm–2)1 J = 1 Nm = 1kg m2 s–2

1 Pa = 1 Nm–2 = 1 kg m–1 s–2

1 L = 10–3 m3 = 1dm3 = 103 cm3


205

*1. Three moles of an ideal gas are expandedisothermally from a volume of 300 cm3 to2.5 L at 300 K against a pressure of 1.9atm. Calculate the work done in L atm andjoules.

Given :n = 3 moles,V1 = 300 cm3 = 0.300 L,V2 = 2.5 LT = 300 K,pex = 1.9 atm.

To find :Work done (W) = ?(in L and J)

Solution :W = –pex (V2 – V1)

= –1.9 atm (2.5 L – 0.3 L)= –1.9 × 2.2 L atm.

W = –4.18 L atmQ 1 L atm = 101.3 J∴∴∴∴∴ W = –4.18 × 101.3

W = –423.4J

*2. One mole of an ideal gas is compressedfrom 500 cm3 against a constant pressureof 1.216 × 105 Pa. The work involved inthe process is 36.50 J. Calculate the finalvolume.

Given :n = 1 mole,V1 = 500 cm3 = 500 × 10–6m3

pex = 1.216 × 105 Pa (Nm–2),W = 36.50 J.

To find :V2 = ?

Solution :W = –pex (V2 – V1)36.50 J = –1.216 × 105 Nm–2

× (V2 – 500 × 10–6 m3)= –1.216 × 105 Nm–2 × V2 +

1.216 × 105 Nm–2 × 500 ×10–6 m3

= –1.216 × 105 × V2 + 60.8∴∴∴∴∴ 1.216 × 105 V2 = 60.8 – 36.50

V2 = 5

24.3

1.216 × 10= 200 × 10–6m3 = 200cm3

∴∴∴∴∴ V2 = 200 cm3

*3. Calculate the maximum work when24 g of oxygen are expanded isothermallyand reversibly from a pressure of1.6 × 105 Pa to 100 kPa at 298 K.

Given :

O2W = 24g,

P1 = 1.6 × 105 Pa= 1.6 × 102 kPa= 160 kPa

P2 = 100 kPaT = 298 K.

To find :Wmax = ?

Solution :

no. of moles (n) = O 2

W

M =

24

32 = 0.75 mol

Wmax = –2.303 × n × R × T × log10 1

2

P

P= –2.303 × 0.75mol × 8.314 JK–1

mol–1 × 298 K ×log10

160 kPa

100 kPa



206

= –2.303 × 0.75 × 8.314 × 298 ×0.2041

Wmax= –873.4 J

*4. Three moles of an ideal gas arecompressed isothermally and reversiblyto a volume of 2L. The work done is2.983kJ at 22ºC. Calculate the initialvolume of the gas.

Given :n = 3 mol,V2 = 2L,Wmax = 2.983 kJ = 2983 JT = 22ºC + 273 = 295K

To find :V1 = ?

Solution :

Wmax = –2.303 × n × R × T × log10 2

1

V

V

2983 J = –2.303 × 3 mol × 8.314 JK–1

mol–1 × 295K × log10 1

2L

V

2983 = + 16945.2 × log10 1V

2L

log10 1V

2L=

2983

16945.2= 0.1760

∴∴∴∴∴ 1V

2L= AL (0.1760)

= 1.500∴∴∴∴∴ V1 = 1.5 × 2L = 3.0 L

*5. 2.8 × 10–2 kg of nitrogen is expandedisothermally and reversibly at 300K from15.15 × 105 Nm–2 when the work doneis found to be –17.33 kJ. Find the finalpressure.

Given :

N 2W

= 2.8 × 10–2 kgT = 300 KP1 = 15.15 × 105 Nm–2

Wmax = – 17.33 kJ = –17330 JTo find :

P2 = ?Solution :

n = W

M =

–2

–3

2.8 × 10

28 × 10 = 1

Wmax = –2.303 × n × R × T × log10 1

2

P

P

–17330 J = –2.303 × 1mol × 8.314 JK–1mol–1

×300K× log10

5 –2

2

15.15×10 Nm

P

=

5

102

15.15×10 N2303× 1×8.314×300×log

P–⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

17330J =5

102

15.15 × 105744.1 × log

P

17330

5744.1=

5

102

15.15 × 10log

P

3.0170 =5

102

15.15 × 10log

P

Al (3.0170) =5

2

15.15 × 10

P

1.040 × 103 = 5

2

15.15 × 10

P

∴∴∴∴∴ P2= 5

3

15.15 × 10

1.040 × 10= 14.5673 × 102 Nm–2≈ 1456.7 Nm–2

∴∴∴∴∴ P2= 1456.7 Nm–2


207

*6. Calculate the work done in each of thefollowing reactions. State whether work isdone on or by the system.

(a) The oxidation of one mole of SO2 at 50ºC.

2 2 32SO (g) + O (g) 2SO (g)⎯⎯→

(b) Decomposition of 2 moles of NH4NO3 at100ºC.

4 3 2 2NH NO (s) N O(g) + 2H O(g)⎯⎯→

Given :

(a) one mole of SO2 reacts with 1

2 mole of

O2 to form 1mole of SO3

Hence, n1 = 1.5 mol, n2 = 1mole,R = 8.314 JK–1 mol–1

T = 50ºC + 273 = 323 KTo find :

W = ?Solution :

W = –Δn RT= –RT (n2 – n1)

= –8.314 (JK–1mol–1) × 323 (K) × (1 – 1.5) (mol)= –8.314 × 323 × (–0.5) J

∴∴∴∴∴ W = +1343 J∴∴∴∴∴ Work is done on the system.

Given :(b) 2 moles of NH4NO3(s) gives 2 moles of

N2O(g) and 4 moles of H2O(g)Hence, n1 = 0,n = 6,

R = 8.314 JK–1 mol–1

T = 100ºC + 273K = 373K.

To find :W = ?

Solution :W = –RT (n2 – n1)

= –8.314 (JK–1 mol–1)× 373 (K) × (6 – 0) (mol)

= –18610 J= –18.610 kJ

∴∴∴∴∴ W = –18.61 kJ∴∴∴∴∴ Work is done by the system

*7. The enthalpy change for the reaction

2 4 2 2 6C H (g)+H (g) C H (g)⎯⎯→ is –620J

when 100 mL of ethylene and 100 mL ofH2 react at 1 atm pressure. Calculate thepressure-volume work and ΔU.

Given :Initial volume (V1) = 100 mL + 100 mL

= 200 mL.Final volume (V2) = 100 mL, Δ H = –620J

∴∴∴∴∴ ΔV = V2 – V1

= 100 – 200= –100 mL= –0.1 L

To find :Work = ? and ΔU = ?

Solution :

W = ex–p ΔV⋅

= –1 atm × (–0.1L)= +0.1 L atm.

But, 1 L atm = 101.3 J

∴∴∴∴∴ W = +0.1 (L.atm) × 101.3 J

L atm⋅

W = +10.13 J

We know, ΔH = ΔU + P ΔV⋅

or ΔU = ΔH – P ΔV⋅

= –620J – (–10.13 J)ΔU = –609.87 J.


208

*8. Oxidation of propane is represented

as 3 8 2 2 2C H (g)+5O (g) 3CO (g)+ 4H O(g)⎯⎯→ ,ΔHo = –2043 kJ. How much pressure -volume work is done and what is the valueof ΔU at constant pressure of 1 atm whenthe volume change is +22.4 L.

Given :ΔHº = –2043 kJP = 1atmΔV = +22.4L

To find :ΔU = ?

Solution :

W = ex–p ΔV⋅

W = –1atm × 22.4 L = –22.4 L –atm.W = –22.4 × 101.3 J

(Q 1 atm = 101.3J)= –2269.12 J= –2.269 kJ

W = –2.27 kJΔU = qp + W

at constant pressure,ΔHº = qp = –2043 kJ

∴∴∴∴∴ ΔU = –2043 kJ + (–2.27 kJ)ΔU = –2045.27 kJ

FIRST LAW OF THERMODYNAMICS

*9. A sample of gas absorbs 4000 kJ of heat.a) If volume remains constant, what is ΔU?b) Suppose that in addition to absorption

of heat by the sample, the surroundingsdoes 2000 kJ of work on the sample,what is ΔU?

c) Suppose that as the original sample absorbsheat, it expands against atmosphericpressure and does 600 kJ of work on itssurroundings. What is ΔU?

Given :q = 4000KJ, W = 2000 KJW = 600KJ

To find :ΔU = ?

Solution :i) ΔU = exq – p ΔV⋅

∴∴∴∴∴ ΔV = 0,∴∴∴∴∴ ΔU = q = 4000 kJ.ii) ΔU = q + W

= 4000 kJ + 2000 kJΔU= 6000 kJ

iii) ΔU = q + W= 4000 kJ + (–W)(Q work is done on surroundings)= 4000 kJ – 600 kJ

ΔU= 3400 kJ.

ENTHALPY CHANGE

*10. Calculate standard enthalpy of the reaction Fe2O3(s) + 3CO(g) ⎯⎯→ 2Fe(s) + 3CO2(g) from

the following data :

Δf Hº (Fe2O3) = –824.2 kJ mol–1, ΔfHº (CO) = –110.5 kJ mol–1, ΔfHº (CO2) = –393.5 kJ mol–1.

Given :

Δf Hº (Fe2O3) = –824.2 kJ mol–1, Δf Hº (CO) = –110.5 kJ mol–1,

Δf Hº (CO2) = –393.5 kJ


209

To find : ΔHº = ?

Solution :

ΔHº = f (Products) f (reactants)m Δ Hº – n Δ Hº⎡ ⎤∑ ⋅ ∑ ⋅⎣ ⎦

ΔHº = ºf (Fe) f (CO ) f (Fe O ) f (CO)2 2 3

2 × Δ H + 3 × Δ Hº – 1 × Δ Hº + 3 × Δ Hº⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

ΔHº = [2mol × 0 + 3 mol × (–393.5 kJ mol–1)] – [1 mol × (–824.2 kJ mol–1)+ 3mol × (–110.5 kJ mol–1)]

= [–1180.5 kJ] – [–824.2 kJ – 331.5 kJ]= [–1180.5 kJ] – [–1155.7 kJ]

ΔHº = –24.8 kJ.

*11. Calculate the standard enthalpy of formation of C2H6 from the following data :

2 6 2 2 22C H (g) + 7O (g) 4CO (g) + 6H O( )l⎯⎯→ , ΔHº = –3119 kJ

Δf Hº (CO2) = –393.5 kJ mol–1

Δf Hº (H2O) = –285.8 kJ mol–1

Given :ΔHº = –3119 kJ, ΔfHº (CO2) = –393.5 kJ mol–1 , ΔfHº (H2O) = –285.8 kJ mol–1

To find :

f (C H )2 6H = ?

Solution :

ΔHº = f (Products) f (reactants)m Δ Hº – n Δ Hº⎡ ⎤∑ ⋅ ∑ ⋅⎣ ⎦

ΔHº = f (CO f (H O) f (C H f O2 2 2 6 24mol×Δ H +6mol×Δ H – 2mol×Δ H +7mol×Δ H) )

⎡ ⎤⎡ ⎤⎣ ⎦ ⎣ ⎦

–3119 kJ = –14(mol) × (–393.5) (kJ mol ) + 6(mol)⎡⎣

f (C H )2 6– 2(mol) × Δ H + 7(mol) × 0⎡ ⎤⎣ ⎦

–3119 kJ = [ ] f (C H )2 6–1574kJ + (–1714.8kJ) – 2(mol) × Δ H⎡ ⎤

⎣ ⎦

∴∴∴∴∴ f (C H2 62(mol) × Δ H ) = –3288.8 kJ + 3119 kJ

= –169.8 kJ

∴∴∴∴∴ f (C H2 6H )Δ =

169.8 (kJ)–

2 (mol)

f (C H )2 6HΔ = –84.9 kJmol–1.


210

*12. How much heat is evolved when 12g ofCO react with NO2 according to the

following reaction, 4CO (g) +

2 2 22NO (g) 4CO (g) + N (g)⎯⎯→ =

ΔHº = –1198 kJ.Given :

Mol. Wt of CO = 12 + 16 = 284CO = 4 × 28 = 112g.

To find : Heat evolved = ?

Solution :When 112 g of CO reacts with NO2,Heat Consumed = –1198 kJ

∴∴∴∴∴ 12g of CO reacts with NO2 heat

consumed =–12 × 1198

112ΔH = –128.4 kJ.

∴∴∴∴∴ Heat evolved = 128.4 kJ.

*13. 38.55 kJ of heat is absorbed when 6.0gof O2 react with CIF according to the

reaction, 2CIF(g) + O2(g) ⎯⎯→

2 2Cl O(g) + OF (g) . What is the standard

enthalpy of the reaction?Given :

ΔHº = 38.55KJTo find :

Heat abosrbed = ?Solution :

When 6g of O2 reacts with CIF,heat absorbed = 38.55 kJ.

∴∴∴∴∴ 32g of O2 reacts with ClF,

heat absorbed =32 × 38.55

6= +205.6 kJ

∴∴∴∴∴ ΔΔΔΔΔHº = + 205.6 kJ.

*14. Calculate the enthalpy change for the

reaction 2H O(g) 2H(g) + O(g)⎯⎯→ and

hence, calculate the bond enthalpy ofO–H bond in H2O from the following data :ΔvapH (H2O) = 44.0 kJ mol–1

ΔfH (H2O) = –285.8 kJ mol–1

ΔaH (H2) = 436.0 kJ mol–1, ΔaH (O2)= 498.0 kJ mol–1 where ΔaH is theenthalpy of atomization.

Given :

i) 2 2H O( ) H O(g)l ⎯⎯→ ;

ΔvapH = +44 kJ mol–1

ii) 2 2 21

H (g) + O (g) H O( )2

l⎯⎯→

ΔfH = –285.8 kJ mol–1

iii) 2H (g) 2H(g)⎯⎯→

ΔaHH2 = 436 kJ mol–1

iv) 21

O (g) O(g)2

⎯⎯→

a O2H Δ = 249 kJmol–1

To find :ΔH = ?and Bond enthalpy of O – H bond = ?

Solution :Treatment : Reverse eq. (i) + Reverse eq.(ii) + eq. (iii) + eq. (iv)

2 2H O(g) H O( )l⎯⎯→ ;ΔvapH = –44KJ mol–1

2 2 21

H O( ) H (g) + O (g)2

l ⎯⎯→ ;

ΔfH = +285.8 KJ mol–1

2H (g) 2H(g)⎯⎯→ ;

H 2aH = 436.0KJ mol–1


211

21

O (g) O(g)2

⎯⎯→ ;

O2Ha = 249.0kJ mol–1

2H O(g) 2H(g) + O(g)⎯⎯→ ;

ΔH = (–44.0 + 285.8 + 436.0 +249.0)KJ mol–1

ΔH = 926.8k]H–O–H molecule contains two (O–H)bonds

∴∴∴∴∴ Bond enthalpy of O–H bonds = 926.8

2 =

463.4 kJ mol–1 ≈ Bond enthalpy ofO – H bond = 4634 kJmol–1 in H2O.

*15. Calculate the total heat required to melt180 g of ice at 0ºC, heat it to 100ºC andthen vaporize it at that temperature.ΔfusH(ice) = 6.01 kJ mol–1 at 0ºC,ΔvapH(H2O) = 40.7 kJ mol–1 at 100ºC.Specific heat of water = 4.18 J g–1 K–1.

Given :(i) number of moles of (ice) H2O(s)

=Wt

Mol.Wt. =

180

18 = 10 moles.

(ii) ΔfusH (ice) = 6.01 kJ–1 at 0ºC∴∴∴∴∴ For 10 moles, ΔfusH ice = 60.1 J at 0ºC

...(i)(iii) ΔvapH (H2O) = 40.7 kJ mol–1 at 100º C∴∴∴∴∴ For 10 moles, ΔvapH(H2O) = 407 kJ

...(ii)(iv) Heat absorbed to raise the temperature

from 0ºC to 100ºCsp. heat of water = 4.18 Jg–1 K–1

To find :Total heat required = ?

Solution :

heat heat(m.c. t)

180g

ice liquid vapours

0ºC 100ºC 100ºC

Δ⎯⎯⎯⎯→ ⎯⎯⎯→

q = m.c. Δt = 180 × 4.18 × 100

1000= 75.24 kJ ...(iii)

∴∴∴∴∴ Total heat required= 60.1 kJ + 407 kJ + 75.24 kJ= 542.34 kJ.

Total heat required = 542.34 kJ.

*16. 6.24 g of ethanol are vaporized by supplying5.89 kJ of heat energy. What is enthalpyof vaporization of ethanol?

Given :ΔHº = 5.89kJ (Heat of vaporisation),mass of ethanol = 6.24g

To find : ΔvapH = ?

Solution :gram molecular weight of C2H5OH

= 2 × 12 + 1 × 5 + 16 + 1 = 46g∴∴∴∴∴ 6.24 g ethanol are vaporised= 5.89 kJ of heat energy

∴∴∴∴∴ 46g of ethanol are vaporised

=46 × 5.89

6.24

= 43.5 kJ mol–1

46g of ethanol are vaporised= 43.5 kJ mol–1.


212

*17. Enthalpy of fusion of ice is 6.01 kJ mol–1. The enthalpy of vaporization of water is45.07 kJ mol–1. What is enthalpy of sublimation of ice?

Given :ΔfusionH = +6.01kJmol–1 ΔvapH = 45.07 kJ mol–1

To find : ΔsubH = ?

Solution :

2 2H O(s) H O( )l⎯⎯→ ; ΔfusH = +6.01 kJ mol–1

2 2H O( ) H O(g)l ⎯⎯→ ; ΔvapH = +45.07 kJ mol–1

2 2H O(s) H O(g)⎯⎯→ ; ΔsubH = (+6.01 + 45.07) kJmol–1

∴∴∴∴∴ ΔΔΔΔΔsubH = 51.08 kJ mol–1.

BOND ENERGY

*18. Calculate ΔHº of the reaction 4 2 2 2CH (g) +O (g) CH O(g)+ H O(g)⎯⎯→ from the followingdata :Bond C–H O = O C = O O – HΔHº/kJ mol–1 414 499 745 464

Given :ΔHº(C – H) = 414kJmol–1, ΔHº(O – H) = 464 kJmol–1

ΔH (O – O) = 499 kJmol–1, ΔHº(C = O) = 745 kJmol–1

To find : ΔHº of the reaction = ?

Solution :

H H

| |

H – C – H + O = O H – C = O + H – O – H

|

H

⎯⎯→

ΔHº = (reactant bonds) (product bonds)ΔHº – ΔHº∑ ∑

ΔHº = [4(mol) × ΔHº(C – H) + 1(mol) × ΔHº(O = O)] – [2(mol) × ΔHº(C – H)

+ 1(mol) × ΔHº(C = O) + 2 (mol) × ΔHº(O – H)]ΔHº = [4 (mol) × 414 (kJ mol–1) + 1 (mol) × 499 (kJ mol–1)] –

[2 (mol) × 414 (kJmol–1) + 1 (mol) × 745 (kJmol–1) + 2 (mol) × 464 (kJmol–1)]ΔHº = [1656 kJ + 499 kJ] – [828 kJ + 745 kJ + 928 kJ]ΔHº = 2155 kJ – 2501 kJ

= –346 kJ∴∴∴∴∴ ΔΔΔΔΔHº = –346 kJ.


213

*19. Calculate C–Cl bond enthalpy from the following data :

3 2 2 2CH Cl (g) + Cl (g) CH Cl (g) + HCl(g)⎯⎯→ ΔHº = –104 kJ

Bond C–H Cl–Cl H–ClΔHº/kJ mol–1 414 243 431

Given :ΔHº(C – H) = 414kJmol–1, ΔHº(Cl – Cl) = 243 kJmol–1

ΔHº (H – Cl) = 431 kJmol–1

To find : ΔHº(C – Cl) = ?

Solution :

H H

| |

H – C – Cl + Cl – Cl H – C – Cl + H – Cl

| |

H Cl

⎯⎯→

ΔHº = (reactant bonds) (Product bonds)ΔHº – ΔHº∑ ∑

ΔHº = º º º(C–H) (C–Cl) (Cl–Cl)3(mol) × ΔH + 1(mol) × ΔH + 1(mol) × ΔH⎡ ⎤

⎣ ⎦ –º º º(C – H) (C – Cl) (H – Cl)2(mol) × ΔH + 2(mol) × ΔH + 1(mol) × ΔH⎡ ⎤

⎣ ⎦

ΔHº = –1 º –1(C–Cl)3(mol) × 414 (kJmol ) + 1(mol) × ΔH + 1(mol) × 243(kJmol )⎡ ⎤

⎣ ⎦

– –1 º –1(C–Cl)2(mol) × 414(kJmol ) + 2(mol) × ΔH + 1(mol) × 431(kJmol )⎡ ⎤

⎣ ⎦

–104 kJ = º(C–Cl)1242kJ + (1mol) × ΔH + 243kJ⎡ ⎤

⎣ ⎦ – ºC–Cl828kJ + 2(mol) × ΔH + 431kJ⎡ ⎤

⎣ ⎦

–104 kJ = º º(C–Cl) (C–Cl)1485kJ + 1(mol) × ΔH – 1259kJ + 2(mol) × ΔH⎡ ⎤ ⎡ ⎤

⎣ ⎦ ⎣ ⎦

–104 kJ = º º(C–Cl) (C–Cl)1485kJ + 1(mol) × ΔH – 1259kJ – 2(mol) × ΔH

–104 kJ = º(C–Cl)226kJ – 1(mol) × ΔH or º

C–Cl1(mol) × ΔH = 226 kJ + 104 kJ

∴∴∴∴∴ ºC–ClHΔ = 330 kJmol–1.

PROBLEMS (HESS’S LAW)

*20. Calculate the standard enthalpy of the reaction 2 2 62C (graphite) + 3H (g) C H (g)⎯⎯→ from

the following ΔHº values :

a) 2 6 2 2 2C H (g) + 7/2O (g) 2CO (g)+3H O ( )l⎯⎯→ , ΔHº = –1560 kJ

b) 2 2 2H (g)+1/2O (g) H O ( )l⎯⎯→ ; ΔHº = –285.8 kJ


214

c) 2 2C(graphite) + O (g) CO (g)⎯⎯→ , ΔHº = –393.5 kJ

Given :

i) 2 6 2 2 2C H (g)+7/2O (g) 2CO (g)+3H O ( )l⎯⎯→ , ΔHº = –1560 kJ

ii) 2 2 2H (g) + 1/2O (g) H O( )l⎯⎯→ , ΔHº = –285.8 kJ

iii) 2 2C(graphite) + O (g) CO (g)⎯⎯→ , ΔHº = –393.5 kJ

To find : ΔHº(reaction) = ?

Solution :

Required equation : (graphite) 2 2 62C + 3H (g) C H (g)⎯⎯→ ; ΔHº = ?

Treatment : Reverse eq. (i) + 3 × eq. (ii) + 2 × eq. (iii)

2 2 2 6 27

2CO (g) + 3H O( ) C H + O (g)2

l ⎯⎯→ , ΔHº = + 1560 kJ

2 2 23

3H (g) + O (g) 3H O( )2

l⎯⎯→ , ΔHº = –857.4 kJ

2 22C(graphite) + 2O (g) 2CO (g)⎯⎯→ , ΔHº = –787.0 kJ

2 2 62C(graphite) + 3H (g) C H (g)⎯⎯→ , ΔHº = +1560 kJ – 857.4 kJ – 787 kJ

ΔΔΔΔΔHº = –84.4 kJ.

*21. Given the following equations calculate the standard enthalpy of the reaction,

2 2 32Fe(s) + 3/ 2 O (g) Fe O (s)⎯⎯→ , ΔHº = ?

a) 2 3 2 32Al(s) + Fe O (s) 2Fe(s) + Al O (s)⎯⎯→ , ΔHº = –847.6 kJ

b) 2 2 32Al(s) + 3/2 O (g) Al O (s)⎯⎯→ , ΔHº = –1670 kJ

Given :

i) 2 3 2 32Al(s) + Fe O (s) 2Fe(s) + Al O (s)⎯⎯→ , ΔHº = –847.6 kJ

ii) 2 2 32Al(s) + 3/2 O (g) Al O (s)⎯⎯→ , ΔHº = –1670 kJ


Solution :

Required equation : 2 2 33

2Fe(s) + O (g) Fe O (s)2

⎯⎯→

Treatment : Reverse eq. (i) + eq. (ii)


215

2 3 2 32Fe(s) + Al O (s) 2Al(s) + Fe O (s)⎯⎯→ ; ΔHº = +847.6 kJ

2 2 33

2Al(s) + O (g) Al O (s)2

⎯⎯→ ; ΔHº = –1670 kJ

2 2 33

2Fe(s) + O (g) Fe O (s)2

⎯⎯→ ; ΔHº = +847.6 kJ – 1670 kJ

ΔΔΔΔΔHº = –822.4 kJ

*22. Given the following equations and ΔHº values at 25ºC,

a) 2 2Si(s) + O (g) SiO (s)⎯⎯→ ; ΔHº = –911 kJ

b) 22C(graphite) + O (g) 2CO(g)⎯⎯→ ; ΔHº = –221 kJ

c) Si(s) + C(graphite) SiC(s)⎯⎯→ ; ΔHº = –65.3 kJ

Calculate ΔHº for the following reaction, 2SiO (s) + 3C(graphite) SiC(s) + 2CO(g)⎯⎯→ .

Given :

i) 2 2Si(s) + O (g) SiO (s)⎯⎯→ ; ΔHº = –911 kJ

ii) 22C(graphite) + O (g) 2CO(g)⎯⎯→ ; ΔHº = –221 kJ

iii) Si(s) + C(graphite) SiC(s)⎯⎯→ ; ΔHº = –65.3 kJ


Solution :

Required equation : 2 (g)SiO (s) + 3C(graphite) SiC(s) + 2CO⎯⎯→

Treatment : Reverse eq. (i) + eq. (ii) + eq. (iii)

2 2SiO (s) Si(s) + O (g)⎯⎯→ ; ΔHº = +911 kJ

22C(graphite) + O (g) 2CO(g)⎯⎯→ ; ΔHº = –221 kJ

Si(s) + C(graphite) SiC(s)⎯⎯→ ; ΔHº = –65.3 kJ

2SiO (s) + 3C(graphite) SiC(s) + 2CO(g)⎯⎯→ ; ΔHº = +911 kJ –221 kJ –65.3 kJ∴∴∴∴∴ ΔΔΔΔΔHº = +624.7 kJ

*23. Given the following equations and ΔHº values at 25ºC,

a) 3 3 2 3 22H BO (aq) B O (s) + 3H O( )l⎯⎯→ ; ΔHº = +14.4 kJ

b) 3 3 2 2H BO (aq) HBO (aq) + H O( )l⎯⎯→ ; ΔHº = –0.02 kJ

c) 2 4 7 2 3 2H B O (s) 2B O (s) + H O ( )l⎯⎯→ ; ΔHº = 17.3 kJ calculate ΔHº for the following

reaction 2 4 7 2 2H B O (s) + H O ( ) 4HBO (aq)l ⎯⎯→


216

Given :

a) 3 3 2 3 22H BO (aq) B O (s) + 3H O( )l⎯⎯→ ; ΔHº = +14.4 kJ

b) 3 3 2 2H BO (aq) HBO (aq) + H O ( )l⎯⎯→ ; ΔHº = –0.02 kJ

c) 2 4 7 2 3 2H B O (s) 2B O (s) + H O( )l⎯⎯→ ; ΔHº = 17.3 kJTo find :

ΔHº(reaction) = ?Solution :

Required equation : 2 4 7 2 2H B O (s) + H O( ) 4HBO (aq)l ⎯⎯→

Treatment : eq (iii) + 2 × reverse eq (i) + 4 ×eq (ii)

2 4 7 2 3 2H B O (S) 2B O (s) + H O( )l⎯⎯→ ; ΔHº = 17.3 kJ

2 3 2 3 32B O (s) + 6H O( ) 4H BO (aq)l ⎯⎯→ ; ΔHº = –28.8 kJ

3 3 2 24 H BO (aq) 4HBO (aq) + 4H O( )l⎯⎯→ ; ΔHº = –0.08 kJ

2 4 7 2 2H B O (s) + H O( ) 4HBO (aq)l ⎯⎯→ ; ΔHº = +17.3 kJ – 28.8 kJ – 0.08 kJ

ΔΔΔΔΔHº = –11.58 kJ

ENTROPY

*24. Calculate ΔStotal and hence, show whether the following reaction is spontaneous at

25ºC. (s) 2 2HgS + O (g) Hg( ) + SO (g)l⎯⎯→ ;ΔHº = –238.6 kJ and ΔSº = +36.7 JK–1

Given :ΔHº = –238.6 kJ ΔSº = +36.7 JK–1

To find : ΔS–

total = ?Solution :

The heat evolved in the reaction is 238.6 kJ. The same quantity of heat is absorbed by thesurroundings. Hence, entropy change of the surroundings is given by.

ΔSsurr =ΔHº

–T

= –(238.6) (kJ)

298 (K) Log calculations

ΔSsurr = +0.8007 kJ K–1 2.3777= + 800.7 JK–1 –2.4742

ΔStotal = ΔSsys + ΔSsurr AL 1.9035ΔSsys = ΔSº = +36.7 JK–1 0.8007ΔStotal = 36.7 (JK–1) + 800.7 (JK–1)ΔStotal = 837.4 JK–1

ΔΔΔΔΔStotal is positive, the reaction is spontaneous at 298 K.


217

*25. Determine whether the reactions with thefollowing ΔH and ΔS values arespontaneous or nonspontaneous. Statewhether they are exothermic orendothermic.

i) ΔH = –110 kJ and ΔS = +40 JK–1 at400 K.

ii) ΔH = +50 kJ and ΔS = –130 JK–1 at250 K.

Given :i) ΔH = –110 kJ and ΔS = +40 JK–1 at

400 Kii) ΔH = +50 kJ and ΔS = –130 JK–1 at

250 K.To find :

ΔG = ?Solution :

i) ΔG = ΔH – T ΔS⋅

= –110 kJ – (400 K × 0.04 kJk–1)= –110 – 16.0= –126 kJ

ΔΔΔΔΔG = –126 kJ., spontaneous andexothermic

Given :ΔH = +50 kJ, ΔS = –130 JK–1 = –0.130kJk–1, T = 250 K

To find : ΔG = ?

Solution :

ii) ΔG = ΔH – T ΔS⋅

= 50 kJ – (250k × (–0.130 kJ K–1)= 50 kJ + 32.50 kJ.= 82.50 kJ

ΔΔΔΔΔG = +82.5 kJ, nonspontaneousand endothermic.

*26. For a certain reaction, ΔHº = –224 kJ andΔSº = –153 JK–1. At What temperaturewill it change from spontaneous tononspontaneous? (Spontaneous below1464 K.)

Given :ΔHº = 224 kJ,ΔSº = 153 JK–1

To find : T = ?

Solution :The temperature at which the reactionchange from spontaneous tononspontaneous is given by,

T = ΔHº

ΔSº= –3 –1

– 224 (kJ)

–153 × 10 (kJ K )

= 1.4640 × 103 K= 1464 K

As both ΔΔΔΔΔHº and Δ Δ Δ Δ ΔSº are negative,therefore, the reaction will bespontaneous below 1464 K.

*27. Determine whether the following reactionwill be spontaneous or nonspontaneousunder standard conditions.

2 + 2+Zn(s)+Cu (aq) Zn (aq) +Cu(s)⎯⎯→

ΔHº = –219kJ, ΔSº = –21JK–1.(spontaneous)

Given :ΔHº = –219kJ, ΔSº = –21JK–1

To find : T = ?

Solution :

T = ΔHº

ΔSº= –3 –1

–219 (kJ)

–21× 10 (kJ K )


218

= 10.428 × 103 K= 10428 K

As both ΔΔΔΔΔHº and ΔΔΔΔΔSº are negativetherefore, the reaction will bespontaneous below 10428K.

*28. Determine whether the following reactionis spontaneous under standard conditions.

2 2 2 22H O( ) + O (g) 2H O ( )l l⎯⎯→ ΔHº

= +196 kJ, ΔSº = –126 JK–1

Does it have a cross-over temperature?Given :

ΔHº = +196 kJ, ΔSº = –126o JK–1

To find : State the reaction condition = ?

Solution :As ΔHº is positive and ΔSº is negativetherefore, the reaction will benonspontaneous at all temperatures.No, it has no cross over temperature.

*29. What is the value of ΔSsurr for the followingreaction at 298 K?

6CO2(g) + 6H2O(l) ⎯⎯→ C6H12O6(s) +6O2(g), ΔGº = 2879 kJ mol–1, ΔSº = –210JK–1 mol–1.

Given :ΔGº = 2879 kJmol–1,ΔSº = –210 JK–1 mol–1

= –0.210 kJ K–1mol–1

T = 298 KTo find :

ΔSsurr = ?Solution :

ΔGº = ΔHº – T ΔSº⋅

2879 kJmol–1

= ΔHº – 298 K × (–0.210kJ K–1 mol–1)

= ΔHº + 62.58 kJ mol–1

∴∴∴∴∴ ΔHº = (2879 – 62.58) kJ = 2816.42 kJ.

ΔSsurr =ΔHº

–T

ΔSsurr =2816.42

–298

= –9.45 kJ K–1

ΔΔΔΔΔSsurr = –9.45 kJ K–1

*30. Calculate ΔSsurr when one mole of methanol(CH3OH) is formed from its elementsunder standard conditions if Δf Hº(CH3OH) = –238.9 kJmol–1

Given :The heat evolved in the reaction is 23.89kJ. The same quantity of heat is absorbedby the surroundings. Hence, entropychange for the surroundings is given byT = 25ºC = 289K

To find : ΔSsurr = ?

Solution :

ΔSsurr = ΔHº

–T

=–238.9 (kJ)

–298 (K)

= 0.8017 kJ K–1

= 801.7 JK–1

ΔΔΔΔΔSsurr = 801.7 JK–1

GIBBS FREE ENERGY

*31. Calculate Kp for the reaction,

2 4 2 2 6C H (g) + H (g) C H (g)⎯⎯→ ,ΔGº

= –100 kJ mol–1 at 25ºC.Given :

ΔGº = –100kJmol–1 = –100000 Jmol–1

T = 25ºC + 273K = 298K


219

R = 8.314 JK–1 mol–1

To find : Kp = ?

Solution :ΔGº = –2.303 RT log10 Kp

log10 Kp =–ΔGº

2.303RT

=–1

–1 –1

–100000 Jmol–

2.303×8.314(JK mol )×298K

log10 kp =100000

5705.8

= 17.5260∴∴∴∴∴ Kp = AL.[17.52]

Kp = 3.356 × 1017

*32. Kp for the reaction, MgCO3(s) ⎯⎯→

MgO(s) + CO2(g) is 9 × 10–10. CalculateΔGº for the reaction at 25ºC.

Given :Kp = 9 × 10–10,T = 25ºC + 273K = 298 K.

To find : ΔG = ?

Solution :ΔGº = –2.303 RT + log10 Kp

= –2.303 × 8.314 JK–1 mol–1

× 298 K × log10(9 × 10–10)

= –2.303 × 8.314 JK–1 mol–1

× 298 K × [–10 + 0.9542]= –2.303 × 8.314 JK–1 mol–1

× 298 K × [–9.0458]

= +2.303 × 8.314 × 298 × 9.0458 J.= 51613 J

ΔΔΔΔΔGº= 51.613 kJ.

*33. Calculate ΔG for the reaction at 25ºC,

2 3CO(g) + 2H (g) CH OH(g)⎯⎯→ ,

ΔGº = –24.8 kJ mol–1 if Pco = 4 atm, H2p

= 2 atm, CH OH3p = 2 atm

Given :ΔGº = –24.8 kJmol–1, Pco = 4 atm,

H 2p = 2 atm, CH OH3

p = 2 atm, T = 25ºC

+ 273K = 298K, R = 8.314 JK–1mol–1

To find : ΔG = ?

Solution :

QP = CH OH3

2CO H2

P

P × P = 2

2

4 × 2

= 2

4 × 4 =

2

16 =

1

8ΔG = ΔGº + 2.303 RT log10 Qp

ΔG = –24.8 (kJmol–1) + 2.303× 8.314 (JK–1mol–1)

× 298 (K) × log10

1

8ΔG = –24.8 (kJmol–1) – 2.303

× 8.314 (JK–1mol–1)

× 298 (K) × log108

ΔG = –24.8 × 103Jmol–1

– [2.303 × 8.314 (JK –1mol–1)

× 298(K) × 0.9031ΔG = –24.8 kJmol–1 – 5.15 × 103 kJmol–1

ΔG = –29.95 × 103kJmol–1

= –29.95 kJmol–1

ΔΔΔΔΔG = –29.95 kJmol–1.


220

NUMERICALS FOR PRACTICE*1. Two moles of an ideal gas are expanded isothermally from a volume of 15.5 L to the volume

of 20 L against a constant external pressure of 1 atm. Estimate the amount of work inL atm and J. (Ans. : W = –4.5 L. atm and –455.8 J)

*2. Calculate the constant external pressure required to compress one mole of an ideal gas froma volume of 20 L to 8 L when the work obtained is 44.9 L atm.

(Ans. : Pex = 3.74 atm)

*3. 100 mL of ethylene (g) and 100 mL of HCl(g) are allowed to react at 2 atm pressure according

to the reaction, 2 4 2 5C H (g) + HCl(g) C H Cl(g)⎯⎯→ . Calculate the pressure-volume work

in joules. (Ans. : W = 20.26J)

*4. 3 moles of an ideal gas are expanded isothermally and reversibly from 10 m3 to 20 m3 at 300K. Calculate the work done. (R = 8.314Jk–1mol–1) (Ans. : –5.187kJ)

*5. 4.4 × 10–2 kg of CO2 are compressed isothermally and reversibly at 293 K from the initialpressure of 150 kPa when the work obtained is 1.245 kJ. Find the final pressure.(R = 8.314Jk–1mol–1) (Ans. : final pressure 250kPa)

*6. 280 mmol of a perfect gas occupies 12.7 L at 310 K. Calculate the work done when the gasexpands (a) isothermally against a constant external pressure of 0.25 atm (b) isothermally andreversibly (c) into vacuum, until its volume has increased by 3.3 L.

(Ans. : (a) W = –83.6J, (b) Wmax –166.9J, (c) W = 0)

*7. For a particular reaction, the system absorbs 6 kJ of heat and does 1.5 kJ of work on itssurroundings. What are the changes in internal energy and enthalpy of the system?

(Ans. : Internal energy 4.5kJ and enthalpy 6 kJ)

*8. In a particular reaction, 1 kJ of heat is released from the system and 4 kJ of work is doneon the system. What are the changes in internal energy and enthalpy of the system?

(Ans. : Internal energy = > 3kJ and enthalpy = < 1 kJ)

*9. An ideal gas expands from a volume of 6 dm3 to 16 dm3 against a constant external pressureof 2.026 × 105 N m–2. Find ΔH if ΔU is 418 J. (Ans. : ΔH = qp2444J)

*10. Calculate the work done in the following reaction when 2 moles of HCl are used at constant

pressure and 423 K. 2 2 24HCl(g) + O (g) 2Cl (g) + 2H O(g)⎯⎯→ . State whether work is done

on the system or by the system. (Ans. : W is positive, work done on the system is 1758J)


221

*11. CO reacts with O2 according to the reaction 2 22CO(g) + O (g) 2CO (g)⎯⎯→ ΔH = –566kJ

How much pressure volume work is done and what is the value of ΔU for the reaction of7.0g of CO at 1 atm pressure, if the volume change is –2.8L? (Ans. : –70.47kJ)

*12. Calculate standard enthalpy of formation of benzene from the following data :

6 6 2 2 215

C H (l) + O (g) 6CO (g) + 3H O(l),2

⎯⎯→ ΔHº = ––3267 kJ

Δf Hº(CO2) = –393.5 kJ mol–1, Δf Hº(H2O) = –285.8 kJ mol–1

(Ans. : ΔformHº 48.6 kJ mol–1)

*13. Estimate the standard enthalpy of combustion of acetylene from the following data :Δf Hº(CO2) = –393.5 kJ mol–1, Δf Hº(H2O) = –285.8 kJ mol–1

Δf Hº(C2H2) = 227.3 kJ mol–1

(Ans. : ΔHº = –1300 kJ; ΔcHº = –1300 kJ/ mol)

*14. Calculate the standard enthalpy of the reaction,

2 2 24CO(g) + 2NO (g) 4CO (g) + N (g)⎯⎯→ from the following data.

ΔfHº(CO) = –110.5 kJ mol–1, ΔfHº(NO2) = 33.2 kJ mol–1

ΔfHº(CO2) = –393.5 kJ mol–1 (Ans. : ΔHº –1198.4kJ)

*15. Given the reaction, 2 2 2 2CH O(g) + O (g) CO (g) + H O(g)⎯⎯→ , ΔH = –527 kJ

How much heat will be evolved in the formation of 30 g of CO2?(Ans. : Heat evolved = 359.4 kJ)

*16. Calculate the enthalpy change for the reaction 4 2 3CH (g) + Cl (g) CH Cl(g) + HCl(g)⎯⎯→

The bond enthalpies are :Bond C – H Cl – Cl C – Cl H – ClΔHº/kJ mol–1 414 243 330 431

(Ans. : ΔHº –104kJ)

*17. Calculate the N – N bond enthalpy in the reaction :

2 4 2 3N H (g) + H (g) 2NH (g)⎯⎯→ , ΔHº = –184 kJ if ΔHº(N–H) = 389 kJ mol–1,

ΔHº (H – H) = 435 kJ mol–1 (Ans. : N – N bond enthalpy is 159 kJ/mol–1)

*18. Calculate the standard N – H bond enthalpy from the following data :

2 2 3N (g) + 3H (g) 2NH (g)⎯⎯→ , ΔHº = –83 kJΔHº (N ≡ N) = 946 kJ mol–1, ΔHº (H – H) = 435 kJ mol–1

(Ans. : ΔHº(N – H) 389 kJ/mol–1)


222

*19. Calculate ΔHº for the reaction 2 2 22ClF(g) + O (g) Cl O(g) + OF (g)⎯⎯→ from the following

data :

(a) 2 3F (g) + ClF(g) ClF ( )l⎯⎯→ ΔHº = 139.2 kJ

(b) 3 2 2 22ClF (l) + 2O (g) Cl O(g) + 3OF (g)⎯⎯→ ΔHº = + 533.4 kJ

(c) 2 2 21

F (g) + O (g) OF (g)2

⎯⎯→ ΔHº = 24.7 kJ

(Ans. : ΔHº = +205.6 kJ)

*20. Calculate ΔHº for the reaction between ethene with water to form ethanol from the followingdata :

(a) 2 5 2 2 2C H OH(l) + 3O (g) 2CO (g) + 3H O(l)⎯⎯→ ΔHº = –1368 kJ

(b) 2 4 2 2 2C H (g) + 3O (g) 2CO (g) + 2H O( )l⎯⎯→ ΔHº = –1410 kJ

Is ΔHº calculated the enthalpy of formation of liquid ethanol?(Ans. : ΔHº = – 42 kJ, No it is not)

*21. Calculate the standard enthalpy of formation of CH3COOH (l) from the following data :ΔfHº (CO2) = –393.3 kJ mol–1, ΔfHº (H2O) = –285.8 kJ mol–1

ΔfHº (CH3COOH) = –875 kJ mol–1

(Ans. : ΔHº – 483.2 kJ mol–1)

*22. For a certain reaction ΔH = –25 kJ and ΔS = –40 JK–1. At what temperature will it changefrom spontaneous to nonspontaneous? (Ans. : T = 625 K)

*23. Determine ΔStotal and decide whether the following reaction is spontaneous at 298 K.

2 3 2Fe O (s) + 3CO(g) 2Fe(s) + 3CO (g)⎯⎯→ , ΔHº = –24.8 kJ, ΔSº = 15 JK–1

(Ans. : ΔStotal = 98.2 JK–1)

*24. Determine whether the reaction, 2 4 2N O (g) 2NO (g)⎯⎯→ is spontaneous at 298 K from

the following data, ΔfHº = (N2O4) = 9.16 kJ mol–1, ΔfHº (NO2) = 32.2 kJ mol–1

ΔSº = 175.8 J K–1. At what temperature will the reaction become spontaneous?(Ans. : T = 325.6 K)

*25. Kp for the reaction, 2 2 32SO (g) + O (g) 2SO (g)⎯⎯→ is 7.1 × 1024 at 298 K. Calculate ΔGº

for the reaction. (R = 8.314 J K–1 mol–1) (Ans. : ΔGº –141.8 KJmol–1)


223

*26. Calculate kp for the reaction at 513 K, 22NOCl(g) 2NO(g) + Cl (g)⎯⎯→ , ΔGº = 17.38

kJ mol–1. (Ans. : kp = 0.017)

*27. Calculate ΔG for the reaction at 298 K, 2 2 32SO (g) + O (g) 2SO (g)⎯⎯→ , ΔGº = –142 kJ

mol–1 if partial pressures of SO2, O2 and SO3 are 50 atm, 75 atm and 2 atm respectivelyat 298 K. (Ans. : qp = 2.13 × 10–5, ΔG – 168.7 kJ mol–1)

HIGHER ORDER THINKING SKILLS1. Which of the following processes are accompanied by increase of entropy. (NCERT)

(a) Dissolution of iodine in a solvent

(b) HCl is added to AgNO3 and a precipitate of AgCl is obtained

(c) A partition is removed to allow two gases to mixAns. Increase of entropy : (a) and (c)

2. Predict the sign of entropy change for each of the following changes of state. (NCERT)

(a) Hg( ) Hg(g)l ⎯⎯→

Ans. ΔS = +ve because liquid changes to more disordered gaseous state.

(b) 3 3AgNO (s) AgNO (aq)⎯⎯→

Ans. ΔS = +ve ; because aqueous solution has more disorder than solid.

(c) 2 2I (g) I (s)⎯⎯→

Ans. ΔS = –ve ; because gas is changing to less ordered solid.

(d) (graphite) (diamond)C C⎯⎯→

Ans. ΔS = +ve ; because graphite has more disorder more disorder than diamond.

3. How is the concept of coupling of reactions useful in explaining the occurrence of a non-spontaneous reaction? (NCERT)

Ans. When a non-spontaneous reaction is coupled with a highly spontaneous reaction, having highnegative ΔG value, the overall free energy of the two combined reactions becomes negative.Thus, both the coupled reaction take place spontaneonsly.


224

4. Comment on the following statements. (NCERT)(a) An exothermic reaction is always thermodynamically spontaneous.

Ans. (i) They are accompanied by decrease of energy.(ii) The heat released is absorbed by the surroundings so that the entropy of the surroundings

increases and therefore, ΔStotal is positive. But, if ΔS system is –ve and T is very high,then ΔG (ΔG = ΔH – T.ΔS) may be positive and the reaction may be non-spontaneous.

(b) Reaction with ΔGº < 0 always have an equilibrium constant greater than 1.Ans. ΔGº < 0. Shows that process is spontaneous, equilibrium constant may not have value > 1

(ΔGº = –RT in K).

4Introduction :→→→→→ Electrochemistry is a branch of physical chemistry, which deals with the relationship

between chemical energy and electrical energy.→→→→→ The energy accompanying spontaneous chemical reactions is utilised to produce electric

current. The cells used are called electrochemical cells. e.g. Daniel cell, Dry cell etc.Here chemical energy is converted into electrical energy.

Chapter

SYLLABUS4.1 REDOX REACTIONS

4.2 CONDUCTANCE IN ELECTROLYTIC

SOLUTIONS

(a) Conductivity

(b) Molar conductivity of an electrolyte

(c) Variation of conductivity with concentration

(d) Variation of molar conductivity with

concentration

(e) Kohlrausch law of independent migration of

ions

(f) Application of Kohlrausch law

(g) Relation between molar conductivity and

degree of dissociation of week electrolytes

(α)

(h) Measurement of conductivity


Theoretical MCQs

Numerical MCQs

Advanced MCQs

4.3 ELECTROCHEMICAL CELLS

(a) Electrochemical cells

(b) Types of cells

(c) Distinguish between electrolytic cell and

electrochemical cell

4.4 ELECTROLYTIC CELLS

(a) Electrolysis of molten NaCl

(b) Faraday’s laws of electrolysis

(c) Quantitative aspects of Faraday’s laws of

electrolysis


Theoretical MCQs

Numerical MCQs

Advanced MCQs

4.5 GALVANIC OR VOLTAIC CELLS

(a) Salt Bridge

(b) Representation or formulations of galvanic

cells

Electrochemistry

"The Sun, with all the planets revolving around it, and depending on it,

can still ripen a bunch of grapes as though it had nothing else in the Universe

to do." - Galileo

226


(c) Writing of cell reactions

(d) Daniell cell

(e) Types of electrodes

4.6 REFERENCE ELECTRODES

(a) Standard hydrogen electrode

(b) Calomel electrode


4.7 ELECTROCHEMICAL SERIES

4.8 ELECTRODE POTENTIALS AND CELL

POTENTIAL

(a) Standard potentials

(b) Nernst equation

(c) Cell potentials and Gibbs energy changes

for cell reactions

(d) Standard cell potentials and equilibrium

constants

(e) Measurement of cell potentials


Theoretical MCQs

Numerical MCQs

Advanced MCQs

4.9 COMMON TYPES OF CELLS

(a) Dry cell

(b) Lead accumulators

4.10 FUEL CELLS

4.11 CORROSION


Theoretical MCQs

Advanced MCQs

Hours Before Exam

Numericals with Solution

Numericals for Practice

Higher Order Thinking Skills (HOTS)

→→→→→ When a non-spontaneous redox reaction is made to take place by supplying electrical

energy. The process is called electrolysis. The cells used are called electrolytic cells.

Here electrical energy is converted into chemical energy.→→→→→ Electrochemistry is also related with the physical properties such as resistance and

conductance of electrolytic solutions.

→→→→→ The study of electrochemical cells is very important in science and technology, becauseit makes possible the manufacture of many important chemicals such as NaOH. It is

used in the manufacturing of soaps, detergents, papers etc.

→→→→→ The metals like Al and Cu are useful for electrical wiring. They are prepared by electrolysisfrom their ores.

227

Chapter - 4 Electrochemistry

4.1 REDOX REACTIONS :

Concept Explanation :1. A reaction in which oxidation and reduction reactions occur simultaneously, is called redox

reaction. e.g..2+ 2+Zn(s) + Cu (aq) Zn (aq) + Cu(s)⎯⎯→

2. In the above reaction, one species (Zn) loses electrons and undergoes oxidation. e.g..2+ –Zn(s) Zn (aq) + 2e (oxidation)⎯⎯→

3. The other species (Cu2+) accepts the electrons and undergoes reduction. e.g..2+ –Cu (aq) + 2e Cu(s) (reduction)⎯⎯→

4. Thus, redox reactions are electron transfer reactions.5. Oxidation is defined as the loss of one or more electrons from a substance.6. Reduction is defined as the gain of one or more electrons by a substance.7. Oxidizing agent (oxidant) is a species that accepts electron(s) and causes other substance

to lose electron(s).8. Reducing agent (Reductant) is a species that donates electron(s) and causes other

substance to accept electron(s).9. Oxidation is also defined as a process in which oxidation number of an element increases.

10. In reduction; oxidation number of an element decreases.11. The total increase and decrease in oxidation number are equal in all balanced redox reactions.12. Thus, redox reaction form the basis of all electrochemical processes.

4.2 CONDUCTANCE IN ELECTROLYTIC SOLUTIONS :

Concept Explanation :1. The substances which allow the flow of electricity through them are called conductors.2. The flow of electricity through a conductor involves the transfer of electron from one

point to the other.

228


3. The mechanism of the transfer of electron is not the same for all the conductors, hence,conductors are classified into two types :(a) Electronic conductors :

Definitions : The conductors through which the conduction of electricity occursby a direct flow of electrons under the influence of applied potential are knownas electronic conductors. e.g.. Al, Cu etc.

(b) Electrolytic conductors :(1) Definition : The conductors in which the conduction takes place by the

migration of positive and negative ions are known as electrolytic conductors.e.g.. Solution of electrolytes.

(2) The current flow in these conductors is accompanied with the chemical changesat the electrodes.

(3) A measurement of conductivity of solution provides an information regarding thenature of solutions.

(4) If the conductivity of the solution is same as that of water, they are called non-electrolytes. e.g.. sucrose, urea etc.

(5) If the conductivity of the solution is much higher than that of water, they arecalled electrolytes. e.g. KCl, CH3COOH etc.

(6) The solutions of all the electrolytes of the same concentration (molarity) do nothave the same conductivity.

(7) Thus, electrolytes are classified as strong electrolytes and weak electrolytes.

Distinguish between electronic and electrolytic conductors.

Electronic conductors

1. The flow of electricity occurs by themigration of electrons through theconductor.

2. The conduction do not involve the transferof matter.

3. The conduction process do not involve thechemical changes.

4. The resistance of the conductor increasesand conductivity decreases with increasingtemperature.

Electrolytic conductors

1. The electron transfer takes place bymigration of positive and negative ionstowards the electrodes.

2. The conduction involves the transfer ofmatter.

3. The current flow is always accompaniedwith chemical changes.

4. The resistance decreases and conductivityincreases with increasing temperature.

229


(a) Conductivity :1. (a) According to Ohms law the electrical resistance of a conductor i.e. the passage of

current is equal to the potential difference, (V) divided by the electric current (I).

∴∴∴∴∴ R = V

I...(i)

(b) The electrical conductance (G) of a solution is the reciprocal of its resistance.

G = 1

R...(ii)

(c) The electrical resistance (R) of a conductor is linearly proportional to its length (l)and inversely proportional to its area of cross section (a),

∴∴∴∴∴ Ra

l∝ or R = a

lρ or ρ = a

Rl

...(iii)

The proportionality constant (ρ) is called resistivity of the conductor.(d) If l = 1m and a = 1m2 then R = ρ(e) Thus, the resistivity of a conductor (ρ) is defined as the resistance of the conductor

that is 1m long and 1m2 in cross-sectional area.(f) From eq. (ii) and (iii);

G = 1 a

×lρ =

ak ×

lThe proportionality constant k is called conductivity of the conductor.

or k = Ga

l =

1×

R a

l...(iv)

(g) If l = 1m and a = 1m2 then k = G. Thus, conductivity (k) is defined as theconductance of unit cube of a solution of an electrolyte.

(h) Combinations of eq. (iii) and (iv) gives, k = 1

ρThus, conductivity (k) is the inverse of resistivity (ρ).

2. Unit of various terms :(a) Unit of (Resistance) R : (SI unit)

R =V

I=

Volt

Ampere= ohm (Ω)

(b) Unit of electrical conductance (G) :

G =1

R=

1

ohm= ohm–1 (Ω–1) or mho

= (reciprocal ohm) Siemens (s) ...(SI unit)

1S = 1Ω–1 = A V–1 = C V–1s–1

CCoulomb (C) = A.s or A =

s⎡ ⎤⎢ ⎥⎣ ⎦

230


(c) Unit of resistivity the conductor (ρ) :

ρ = a

Rl

=2m

ohmm

= ohm.m (SI unit) = Ωm

(d) Unit of conductivity of the conductor (k) :

k =1

×R a

ρ = 2

1 m×

ohm m= ohm–1. m–1

= S. m–1 (SI unit)

Note : Common unit of conductivity (k) are Ω–1 cm–1 or S cm–1.

(b) Molar conductivity of an electrolyte () :In 1880’s the German physicist George Kohlrausch introduced the concept of molarconductivity ().Suppose, a cell consists of two parallel plates (electrodes) of 1 sq.cm. area of cross section,and 1 cm apart from each other. If 1 cm3 of solution containing, 1 mole of an electrolyteis placed between the plates, then the conductance shown by the solution is equal to itsmolar conductivity.

1. Definition : It is defined as the conducting power of all the ions produced by dissolvingone mole of an electrolyte in solution.

2. Relation between conductivity (k) and molar conductivity () := k × V

V = volume of solution containing one mole of the electrolyte or = 1000

k ×C

C = molar concentration, i.e. mol–1 (or mol dm–3)3. Unit molar conductivity () :

(a) In SI units :

=k

C=

–1

–3

Sm

molm= S m2 mol–1

(b) If concentration is expressed in mol L–1 and volume (V) in cm3

π =1000 × k

C=

3 –1 –1 –1

–1

1000(cm L ) × k ( cm )

C(molL )

Ω

= Ω–1 cm2 mol–1

= S cm2 mol–1

231


(c) Variation of conductivity (k) with concentration :1. The conductivity of an electrolytic solution always decreases with decrease in

concentration (i.e. on dilution). Reason :(a) The conductivity is the conductance of unit volume of solution (say 1 cm3) and depends

on the number of ions present in unit volume.(b) On dilution i.e. on decreasing the concentration of solution, the number of ions per

unit volume decreases. Hence, conductivity of solution decreases.2. The conductivity of an electrolyte with concentration differs for strong and

weak electrolyte :(a) For strong electrolytes, the conductivity increases rapidly with increasing concentration.(b) For weak electrolytes, the conductivity is very low in dilute solution and increases

gradually with increase in concentration.(c) In case of strong electrolytes, the increase in number of ions per unit volume increases

rapidly with concentration, because they are completely dissociated.(d) In case of weak electrolytes, the increase in number of ions per unit volume is not

so large because they are partially dissociated.

(d) Variation of molar conductivity with concentration :1. Unlike conductivity (k), the molar conductivity of both strong and weak electrolytes

increases with dilution that is with decrease in concentration. Reason :(a) The molar conductivity is the conductance of all the ions produced from 1 mole of

the electrolyte.(b) As solution is diluted, the total number of ions increases due to increase in degree

of dissociation. Hence, molar conductivity increases.

(c) We know, = k

C = k.V

Although k decreases on dilution, but the increase in V is more than compensatedby the decrease in the value of k.

2. The molar conductivities is of strong and weak electrolytes show different behaviouron decrease in concentration that is dilution :(a) The molar conductivity of strong electrolytes increases rapidly and soon approaches

a maximum limiting value.(b) The limiting value of molar conductivity is the value at zero concentration (0) or

at infinite dilution.(c) The molar conductivity of strong electrolytes approaches close to 0 value in just

0.001 M or 0.0001 M solutions.

232


(d) The molar conductivity of weak electrolytes also increases on dilution. But atconcentration of 0.001 M or 0.0001 M, it is very much than the maximum limitingvalue 0.

(e) Kohlrausch, on the basis of experiments, showed that the molar conductivity (), ofstrong electrolytes varies linearly with the square root of concentration ( )C andestablished the relation = 0 – a C

(where a = constant for a particular solvent at a particular temperature.)

(f) The molar conductivity of weak electrolytes doesnot show such linear variation with C .

The figure shows how values of strong andweak electrolytes show different behaviour withdecrease in concentration.

Note :

(a) 0 is the molar conductivity at zero concentration or at infinite dilution. The infinitedilution means the solution is so dilute that further dilution does not increase themolar conductivity. In such a solution, the ions are far apart and do not interactwith one another.

(b) The 0 value of strong electrolytes can be obtained by extrapolating linear part

of versus C graph to zero concentration.

(c) But this method cannot be used for weak electrolyte because versus C curve

does not approach linearity.(d) The 0 value of weak electrolytes can be calculated by Kohlrausch law.

(e) Kohlrausch law of independent migration of ions :1. Statement : The law states that at infinite dilution, each ion migrates independently

of its co-ion and makes its own contribution to the total molar conductivity of anelectrolyte irrespective of the nature of other ion with which it is associated.

2. At infinite dilution (i.e. at zero concentration), 0 is the sum of molar conductivity of cation

( 0+λ ) and anion ( 0

–λ ) that is

0= 0+λ and 0

–λ

for NaCl, 0= 0 0+ –Na Cl

+λ λ

233


(f) Application of Kohlrausch law :1. Molar conductivity of infinite dilution (0) for weak electrolyte : The law can be

used to calculate the molar conductivity of any electrolyte at zero concentration. [However,this law is particularly useful to calculate 0 of weak electrolytes for which the extrapolationmethod (by graph) is not useful.]Example :0value of weak electrolyte, CH3COOH can be calculated from the 0values of strongelectrolytes HCl, CH3COONa and NaCl.0 (HCl) + 0 (CH3COONa) – 0 (NaCl)

=0 0 0 0 0 0

+ – – + + –H Cl CH COO Na Na Cl3+ + + – –λ λ λ λ λ λ

= 0 0+ –H CH COO3

+λ λ = 0 (CH3COOH)

Thus, 0 (CH3COOH) = 0 (HCl) + 0 (CH3COONa) – 0 (NaCl).The 0 values of strong electrolytes can be calculated by extrapolation method and 0

of weak electrolytes can be evaluated.

(g) Relation between molar conductivity () and degree of dissociationof weak electrolytes (ααααα) :

1. α = 0

Where is the molar conductivity of the weak electrolyte at the given concentrationC, and 0 is its molar conductivity at zero concentration.

2. The dissociation constant of a weak electrolyte (K) is given as,

K =2C

1 –

αα =

2

0

0

C

1 –

⎛ ⎞⎜ ⎟⎝ ⎠

=

2

0

0

0

C

–

⎛ ⎞⎜ ⎟⎝ ⎠

=2

02

00

× × C( – )

∴ K =2

0 0

C

( – )

(h) Measurement of conductivity :The determination of conductivity (k) and molar conductivity () of a solution consistsof measurement of resistance of the solution by Wheatstone Bridge principle usingconductivity cell.

234


Wheatstone bridge principle

1. Conductivity cell : It consists of a glass tubewith two platinum plates coated with platinumblack. The cell is to be dipped in a solutionwhose resistance is to be measured.

2. Cell constant :(a) The conductivity (k) of an electrolytic solution is given by,

k = 1

×R a

l...(i)

(b) For any given cell, the ration a

l is fixed quantity and is called the cell constant denoted

by b.(c) Definition – Cell constant is defined as ‘the ratio of the distance between the

electrodes divided by the area of cross section of the electrode.’

Thus, cell constant = b = a

l...(ii)

from equation (i) and (ii)

k = b

R...(iii)

Unit of cell constant : m–1 or cm–1.3. Determination of cell constant :

235


(a) The cell constant is determined by using 1M, 0.1M or 0.01M KCl solution.(b) The conductivity (k) of KCl is known at various temperature.(c) The resistance of KCl solution is determined by Wheatstone bridge principle.(d) AB is the uniform slide wire.(e) Rx is the variable known resistance placed in one arm of Wheatstone bridge.(f) The conductivity cell dipped in a solution of KCl of unknown resistance placed

in the other arm.(g) D is a current detector, F is the sliding contact. A.C. represents the source of attempting

current.(h) The sliding contact is moved along AB until no current flows. Suppose a balance point

is obtained at C.(i) According to Wheatstone bridge principle,

R(Solution)

R(AC) =Rx

R(BC)∴∴∴∴∴ Resistance (R) ∝ length (l)

∴∴∴∴∴R(Solution)

(AC)l =Rx

(BC)l

∴∴∴∴∴ R(solution) =(AC)

Rx(BC)

l

lThe cell constant (b) = k.R (solution)

= Conductivity of solution × resistance offeredby the solution.

The conductivity (k) of KCl is known, here b can be calculated.4. Determination of conductivity of the given solution :

KCl solution is replaced by the given solution and its resistance is measured by Wheatstonebridge principle as described earlier.

∴∴∴∴∴ Conductivity (k) = b

R(given solution)

Molar conductivity of the given solution = () = k

CSince concentration c of solution is known, hence, can be calculated.


*1. Define the term (a) resistivity, (b) conductivity, (c) molar conductivityAns. (a) Resistivity : The resistivity of a conductor is defined as the resistance of the conductor

that is 1m long and 1m2 in cross-sectional area.(b) Conductivity : Conductivity (k) is defined as the conductance of unit cube of a solution

of an electrolyte.

236


(c) Molar conductivity : It is defined as the conducting power of all the ions produced bydissolving one mole of an electrolyte in solution.

*2. Give SI units of (a) resistivity, (b) conductivity, (c) molar conductivity.

Ans. (a) Resistivity : Ra

l∝ or R = a

lρ or ρ = a

Rl

The proportionality constant ρ is called resistivity of the conductor.

(b) Conductivity : G = 1 a

×lρ =

ak ×

l

The proportionality constant k is called conductivity of the conductor.

or k = Ga

l =

1×

R a

l

(c) Molar Conductivity : In SI units :

=k

C=

–1

–3

Sm

molm = S m2 mol–1

*3. Explain the terms conductivity and molar conductivity. How are they interrelated?

Ans. G = 1 a

×lρ =

ak ×

l

The proportionality constant k is called conductivity of the conductor or k = Ga

l =

1×

R a

l

(a) The conductivity (k) is defined as the conductance of unit cube of a solution of an electrolyte.(b) Molar conductivity is defined as the conducting power of all the ions produced by dissolving

one made of an electrolyte in solution.Relation between conductivity (k) and molar conductivity () : = k × V

V = volume of solution containing one mole of the electrolyte or = 1000

k ×C

c = molar concentration, i.e. mol–1 (or mol dm–3)

*4. Why does conductivity of a solution decreases on dilution of the solution?Ans. (a) The conductivity is the conductance of unit volume of solution (say 1 cm3) and depends

on the number of ions present in unit volume.(b) On dilution i.e. on decreasing the concentration of solution, the number of ions per unit

volume decreases. Hence, conductivity of solution decreases.

237


*5. How does molar conductivity of an electrolyte vary with concentration?Ans. Refer 4.2 (d) 1.

*6. How does the variation in molar conductivity of an electrolyte with concentration differs for strongand weak electrolytes? OR How is the molar conductivity of strong electrolytes at zero concentrationdetermined by graphical method? Why is this method not useful for weak electrolytes?

Ans. Refer 4.2 (d) 2.

*7. Explain the different behaviour of strong and weak electrolytes towards the variation ofconductivity with concentration.

Ans. Refer 4.2 (c) 2.*8. State and explain Kohlrausch law of independent migration of ions. How is it useful to determine

the molar conductivity of weak electrolytes at zero concentration?Ans. Refer 4.2 (e) and (f).

*9. What is Cell constant? What is its units? How is it determined?Ans. Refer 4.2 (h) 2 and 3


• Theoretical MCQs :*1. The SI unit of molar conductivity is ........

a) S cm2 mol–1 b) S dm2 mol–1 c) S m2 d) S m2 mol–1

2. The molar conductance of solution of an electrolyte is measured in ........a) ohm cm mol–1 b) ohm–1 cm–1 mol–1

c) ohm cm–1 mol–1 d) ohm–1 cm2 mol–1

• Numerical MCQs3. The distance between two electrodes of a cell is 3.0 cm and area of each electrode is 6.0

cm. The cell constant is ........a) 2.0 b) 1.0 c) 0.5 d) 18

• Advanced MCQs4. Conductivity (units Siemen’s S) is directly proportional to the area of the vessel and the

concentration of the solution in it and is inversely proportional to the length of the vessel, thenthe unit of constant of proportionality is ...... (A.I.E.E.E. 2002)a) S m mol–1 b) S m2 mol–1 c) S–2 m2 mol d) S2 m2 mol–2

2 –1

2 –3

Area×Conc. conductivity×length S×mHint:Conductivity= k× or k = k = =Sm mol

length area×conc. m ×mol∴

⎡ ⎤⎢ ⎥⎣ ⎦

238


5. Electrolyte KCl KNO3 HCl NaOAc NaCl

2 –1(Scm mol )

∞

149.9 145.0 426.2 91.0 126.5

Calculate ∞ HOAc using appropriate molar conductance of the electrolytes listed at infinitedilution of H2O at 25ºC ...... (A.I.E.E. 2005)a) 517.2 b) 552.7 c) 390.7 d) 217.5

Hint : HOA c∞

= NaOAc HCl NaCl–∞ ∞ ∞

+

= 91.0 + 426.2 – 126.5= 390.7

6. The molar conductivites 0NaOAc and 0

HCl at infinite dilution in water at 25ºC are 91.0 and

426.2 S cm2/mol respectively. To calculate0HOAc the additional value required in ......

(A.I.E.E.E.2006)

a) 0NaOH b) 0

NaCl c) 0H O2

d) 0KCl

0 0 0 0HOAc NaOAc HCl NaCl[Hint : = + – ]

4.3 ELECTROCHEMICAL CELLS :

Concept Explanation :(a) Electrochemical cells :1. The reactions which occur in electrochemical cells are called electrochemical reactions.

These reactions are redox reactions.2. An oxidation half reaction occurs at one electrode and a reduction half reaction occurs

at the other electrode.3. Cathode : It is defined as an electrode at which reduction occurs as electrons are

gained by some species.4. Anode : It is defined as an electrode at which oxidation occurs as electrons are lost

by some species.

(b) Types of cells :There are two types of cells :

1. Electrolytic cells – A cell in which nonspontaneous reaction is forced to occur bypassing a direct current into the solution, is called an electrolytic cell.

2. Electrochemical cells or voltaic or galvanic cells - A cell in which a spontaneouschemical reaction produces electricity is called electrochemical cell.

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(c) Distinguish between electrolytic cell and electrochemical cell.

4.4 ELECTROLYTIC CELLS :

Concept Explanation :1. The process which occurs in the electrolytic cell is called electrolysis.2. Electrolysis : It is defined as a ‘process in which an electric current is used to bring

about a nonspontaneous chemical reaction.’

(a) Electrolysis of molten NaCl :Construction :

1. It consists of a vessel with two graphite (carbon)electrodes dipped in fused NaCl (electrolyte).

2. The electrode connected to the negativeterminal of a battery is called cathode.

3. The electrode connected to the positiveterminal of a battery is called anode.Working :

1. When an electric current is passed throughfused NaCl then its electrolysis takes place.

2. Sodium is deposited at cathode and chlorineis liberated at anode.

Electrolytic cell

1. It is a device in which electrical energyis converted into chemical energy.

2. The redox reaction is non-spontaneous andtakes place only when electrical energy issupplied.

3. It is used to bring about electrolysis.4. In this cell cathode has negative polarity

and anode has positive polarity.5. No salt bridge is used in this cell.

6. Example : Voltmeter used for electrolysisor water, electroplating, cells used forpurification of metals.

Electrochemical cell (Voltaic cell)

1. It is a device in which chemical energyis converted into electrical energy.

2. It is based upon the redox reaction whichis spontaneous.

3. It is used to produce electricity.4. In this cell cathode has positive polarity

and anode has negative polarity.5. To set up this cell a salt bridge or porous

pot is used.6. Example : Daniel cell, Dry cell, Nickel

Cadminim cell, Lead storage cell, Fuel cell.

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3. + –2NaCl 2Na + 2Cl⎯⎯→ (Dissociation)

4. Reduction half reaction at cathode : Na+ ions migrate to the cathode. At cathode, eachNa+ ion accepts an electron which is supplied by the battery.

+ –2Na ( ) + 2e 2Na( )l l⎯⎯→ (reduction) ...(i)

5. Oxidation half reaction of anode : Cl– ions migrate to anode. At anode, each Cl– iongives one electron to the anode and is converted into neutral Cl atom in the primary process.Two Cl atoms then combine to form Cl2 gas in the secondary process.

– –2Cl ( ) Cl(g) + Cl(g) + 2el ⎯⎯→ (Oxidation)

2Cl(g) + Cl(g) Cl (g)⎯⎯→ (secondary process)

– –22Cl ( ) Cl (g) + 2el ⎯⎯→ ..(ii)

6. Net cell reaction : on adding eq. (i) and (ii)+ –

22Na ( ) + 2Cl ( ) 2Na( ) + Cl (g)l l l⎯⎯→

7. Results of electrolysis :(a) A pale green Cl2 gas is released at anode and (b) A molten silvery-white Na depositsat cathode.

(b) Faraday’s Laws of Electrolysis :First Law : It selects that the amount of substance that undergoes oxidation or reductionat each electrode during electrolysis is directly proportional to the amount of electricitythat passes through the cell.Second law : It states that when the same amount of electricity is passed through differentcells containing different electrolytes and arranged in series, the amounts of substancesoxidized or reduced at the respective electrodes are directly proportional to their chemicalequivalent masses.Electrical units of charge :

1. Coulomb (C) : Coulomb is the smaller unit of charge or electricity. It is defined asthe amount of charge (electricity) that passes a given point when 1A current flowsfor 1s.

2. Faraday (F) : Faraday is the bigger unit of charge. It is defined as the amount ofelectric charge on one mole of electrons.The charge of one electron is 1.602 × 10–19C. Hence, the charge of Avogadro’s numberof electrons that is of 1 mole of electrons will be 1.602 × 10–19(C) × 6.022 × 1023 = 96473C/mol e– ≈ 96500 C/mole– = 1F.Thus, Faraday is the charge of one mole of electrons and we write 1F = 96500 C/mol e–.

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(c) Quantitative aspects of Faraday’s laws of electrolysis :1. Faraday’s first law of electrolysis : The mass of the reactant consumed or of the product

formed at an electrode during electrolysis is calculated by using the stoichiometry of thehalf reaction and the molar mass of the substance. The following procedure is used inthe calculations.(a) Quantity of electricity passed : The quantity of electricity (charge) passed through a

cell is calculated by measuring the quantity of electric current (I) passed through the celland the time (t) of the passage of current. The quantity of electricity Q is given by Q(C)= I(A) × t(s)

(b) Number of electrons passed in moles : The charge of one mole electrons is

96500 C. Hence, number of moles of electrons actually passed = –

Q (C)

96500(C/mol e )(c) Number of moles of product formed : The balanced equation of the half reaction

(stoichiometry) indicates the number of moles of electrons used and number of molesof product formed. From these quantities the number of moles of product actuallyformed can be calculated as Number of moles of the product formed = number molesof electrons actually passed × mole ration

Where mole ratio = moles of product formed in half reaction

moles of electrons required in half reaction(d) Mass of product formed : Mass of product formed = number of moles of product

formed x its molar mass.From the above steps, the following general formulae can be used to calculate themass of the substance produced.(i) Moles of the substance produced

= –

I(A) × t(s)× mole ratio from the stoichiometry

96500 (C/mol e )(ii) Mass of the substance produced

= –

I(A) × t(s)× mole ratio × molar mass of the substance

96500(C/mol e )

2. Faraday’s second law of electrolysis : Two cells containing different electrolytes areconnected in series. The same quantity of electricity that is the same number of moles ofelectrons are actually passed through them.Moles of A produced in one cell = moles of electrons actually passed × mole ratio ofA half reactionMoles of B produced in other cell = moles of electrons actually passed × mole ratio ofB half reaction.

Hence, moles of A produced

moles of B produced = mole ratio of A half reaction

mole ratio of B half reaction

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*1. Sketch the cell for electrolysis of molten MgCl2. Indicate anode, cathode with their signs. Showthe direction of electron and ion flow. Write electrode reactions and net cell reaction.

Ans.

Electrolysis of molten MgCl22+ –

2MgCl Mg + 2Cl⎯⎯→ (Dissociation)

At anode : – – –2Cl ( ) Cl(g) + Cl (g) + 2el ⎯⎯→ (Oxidation)

2Cl(g) + Cl(g) Cl (g)⎯⎯→

– –22Cl ( ) Cl (g) + 2el ⎯⎯→ ...(i)

At cathode : 2+ –Mg ( ) + 2e Mg( )l l⎯⎯→ (reduction) ...(ii)

Net cell reaction : on adding eq. (i) and (ii);2+ –

2Mg ( ) + 2Cl ( ) Mg( ) + Cl (g)l l l⎯⎯→

*2. Define anode and cathode.Ans. Cathode : It is defined as an electrode at which reduction occurs as electrons are gained by

some species.Anode : It is defined as an electrode at which oxidation occurs as electrons are lost by somespecies.

*3. Predict the half cell reactions that occur when fused KCl is electrolyzed in a cell with inertelectrodes. What is the overall cell reaction?

Ans. + –2KCl 2K + 2Cl⎯⎯→ (Dissociation)

At cathode + –2K ( ) + 2e 2K( )l l⎯⎯→ (reduction) ...(i)

At anode – –22Cl ( ) Cl (g) + 2el ⎯⎯→ (oxidation) ...(ii)

over all reaction + –22K ( ) + 2Cl ( ) 2K( ) + Cl (g)l l l⎯⎯→

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*4. State Faraday’s laws of electrolysis.Ans. (a) First Law : It states that the amount of substance that undergoes oxidation or reduction

at each electrode during electrolysis is directly proportional to the amount of electricitythat passes through the cell.

(b) Second law : It states that when the same amount of electricity is passed through differentcells containing different electrolytes and arranged in series, the amounts of substancesoxidized or reduced at the respective electrodes are directly proportional to their chemicalequivalent masses.

*5. How will you calculate the number of moles of electrons actually passed and mass of thesubstance produced during electrolysis of a salt solution using reaction stoichiometry?

Ans. Refer 4.4 (c).

*6. How many electron will have a total charge of 1 coulomb?Ans. 1.602 × 10–19C charge is present on = 1 electron

∴∴∴∴∴ 1 C charge is present on = –19

1

1.602 × 10 = 6 × 1020 electron

*7. How many Faradays would be required to plate out 1.00 mole of free metal from the following cations?(a) Mg2+ (b) Cr3+ (c) Pb2+ (d) Cu+

Ans. (a) 2+ –Mg + 2e Mg(s)⎯⎯→

The reaction indicates that 1 mole of Mg is produced by the passage of 2 moles of electrons.

The charge of 2 mole electrons is 2 Faradays.

(b) 3+ –Cr + 3e Cr(s)⎯⎯→ ; 3 Faradays

(c) 2+ –Pb + 2e Pb(s)⎯⎯→ ; 2 Faradays

(d) + –Cu + e Cu(s)⎯⎯→ ; 1 Faraday

*8. Define Coulomb and Faraday.Ans. Refer 4.4 (b) 1 and 2.

*9. Describe the electrolysis of molten NaCl using inert electrodes.Ans. Refer 4.4 (a)

*10. What is the difference between electrolytic cell and voltaic cell?Ans. Refer 4.3 (c).*11. Sketch the cell for the electrolysis of molten NaCl. Indicate cathode, anode and their signs.

Show the flow of electrons and ions.Ans. Refer 4.4 (a)

12. Why is cathode in an electrolytic cell considered to be negative and anode positive?Ans. (a) In electrolytic cell, cathode is connected to the negative terminal of the battery and anode

in connected to the positive terminal of the battery,(b) During electrolysis, at negative electrode, reduction reaction takes place hence, it acts as

cathode, while at positive electrode oxidation reaction takes place, hence, it acts as anode.

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Multiple Choice Questions (MCQs)• Theoretical MCQs :

*1. In the electrolysis of molten Al2O3 with inert electrodes .........a) Al is oxidised at anode to Al3+ b) O2 gas is produced at anodec) O– is reduced at cathode c) O is oxidised at anode

*2. The number of electrons that have a total charge of 965 coulombs is .........a) 6.022 × 1023 b) 6.022 × 1022 c) 6.022 × 1021 c) 3.011 × 1023

*3. The time required to produce 2F of electricity with a current of 2.5 amperes is .........a) 13.4h b) 1200 min c) 50000s d) 1.5h

4. When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriateelectrodes are .........

cathode anodea) Pure Zn Pure Cub) Impure sample Pure Cuc) Impure Zn Impure sampled) Pure Cu Impure sample

5. The electroplating with chromium is undertaken because .........a) electrolysis of chromium is easierb) chromium can form alloys with other metalsc) chromium gives a protective and decorative coating to the base metald) of high reactivity of chromium metal.

6. The storage battery acts as .........a) Electrolytic cell b) Voltaic cell c) Fuel cell d) both a and b

7. An example of a simple fuel cell is .........a) Daniel cell b) H2 – O2 cell c) Lead storage cell d) Dry cell

8. The cell reaction in Daniell cell is .........

a) 2+ 2+Zn + Cu Zn + Cu⎯⎯→ b) 2+ 2+Zn + Cu Zn + Cu⎯⎯→

c) 2+ 2+Zn + Cu Zn + Cu⎯⎯→ d) 2+ 2+Zn + Cu Zn + Cu⎯⎯→

• Numerical MCQs :*9. The number of Faradays required to produce 0.5 mol of free metal from Al3+ is .........

a) 3 b) 2 c) 6 d) 1.5*10. The number of coulombs necessary to deposit 1 g of potassium metal (molar mass 39 g mol–1)

from K+ ions is .........a) 96500 C b) 1.93 × 105 C c) 1237 C d) 2474 C

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*11. During electrolysis, 2A current is passed through an electrolytic solution for 965s. The numberof moles of electrons passed will be .........a) 0.02 b) 0.01 c) 200 d) 0.037

*12. The same quantity of electricity is passed through two cells one containing AlCl3 solutionand the two other containing ZnCl2 solution. If 0.04 mole of Al is produced in the first cell,the number of moles of Zn produced in the second cell will be .........a) 0.08 b) 0.0267 c) 0.06 d) 0.02

• Advanced MCQs :13. What is the ratio of the weights liberated at the cathode when the same current is fused through

two solutions of ferric and ferrous salts arranged in series for a given time interval?(MHT - CET 2001)

a) 3 : 2 b) 1 : 3 c) 2 : 3 d) 3 : 1

Fe

Fe

W from ferric 56/3 2Hint : = = = 2 : 3

W from Ferrous 56/2 3

⎛ ⎞⎜ ⎟⎝ ⎠

14. For the electrochemical cell, M / M+ || X–/X, Eº(M+/M) = 0.44V and Eº (X/X–) = 0.33VFrom this data, one can deduce that, ......... (IIT 2000)

a) + –M + X M + X⎯⎯→ is the spontaneous reaction

b) + –M + X M + X⎯⎯→ is the spontaneous reaction

c) Ecell = 0.77Vd) Ecell = –0.77V

[Hint : (b), Eºcell = +0.11V, here it is spontaneous]15. The amount of electricity required to deposit 1 mol of aluminium from a solution of AlCl3 will

be .......... (AIIMS 1992)a) 0.33F b) 1F c) 3F d) 1 Ampere

16. The change required for the reduction of 1 mol of –4MnO to MnO2 is ..........

(AIIMS 2006)a) 1F b) 3F c) 5F d) 6F

Hint : –7+ 4+

4 2Mn O MnO⎛ ⎞

⎯⎯→⎜ ⎟⎝ ⎠i.e. – –

4 2MnO + 3e MnO⎯⎯→

–– –

4 2 2MnO + 2H O + 3e MnO + 4OH⎛ ⎞

⎯⎯→⎜ ⎟⎝ ⎠ Thus, for reduction of 1 mol of –4MnO to MnO2

Charge required = 3F

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4.5 GALVANIC OR VOLTAIC CELLS :

Concept Explanation :1. In this cell, a spontaneous chemical reaction occurs, to produce electricity.2. Each galvanic cell is made up of two half cells.

3. Each half-cell consists of a metal rod immersed in a solution of its own ions of known

concentration.e.g. a rod of Zn metal is dipped in 1M solution of Zn2+ ions. It forms a half-cell.

4. The metal rods (electrodes) are connected by a wire.

5. The solutions of both the half-cells are connected by a salt bridge.6. The anode is a negative electrode and cathode is a positive electrode.

(a) Salt Bridge :1. It is a U-shaped glass tube containing a saturated solution of an

electrolyte such as KCl or NH4NO3 and 5% agar gel.

2. The ions of the electrolyte do not react with the ions of electrodesolutions.

3. Both the ends of the U tube are plugged with glass wool.Functions of the salt bridge :

1. It provides an electrical contact between the two solutions and

thereby completes the electrical circuit.2. It prevents the mixing of electrode solutions.

3. It maintains electrical neutrality in both the solutions by the flow

of ions.

(b) Representation or formulation of galvanic cells :A galvanic cell is represented by a short notation or diagram which includes electrodes,aqueous solutions of ions and the other species that may or may not be involved in the

cell reaction. The following conventions are used to write the cell notation :

1. The metal electrodes or the inert electrodes (used as a conducting device in case of gaselectrodes) are placed at the ends of the cell formula. The anode (–) is written at the

extreme left and cathode (+) is placed at the extreme right of the cell notation.

2. The insoluble substances if any or gases are placed in the interior positions adjacent tothe metal electrodes.

247


3. The aqueous solutions of ions are placed at the middle of the cell formula.4. A single vertical line is used between two phases to indicate the phase boundary, such

as that between the solid electrode and the aqueous solution of ions. It represents thedirect contact between them.

5. A double vertical line is used between two solutions to indicate that they are connectedby a salt bridge.

6. The additional information such as concentration of solutions and the gas pressure is also given.7. The ions in the same phase are separated by a comma. For example Fe3+, Fe2+ | Pt8. A single half cell is written in the order, aqueous solution of ions first and after that the

solid electrode. For example, Zn2+(1M) | Zn. This order is reversed when the electrodeacts as an anode in the galvanic cell.The following examples will illustrate these conventions :

1. The cell composed of Mg (anode) and Cu (cathode) consists of two half cells, Mg2+(1M)|Mg

and Cu2+(1M)|Cu. It is formulated as

Mg(s) | Mg2+(1M) || Cu2+(1M) | Cu

2. The notation for the cell consisting of H+(1M) | H2(g 1atm) | Pt and Ag+(1M) | Ag(s)

half cells with hydrogen gas electrode as anode and Ag electrode as cathode will be,

Pt | H2(g, 1atm) | H+(1M) || Ag+(1M) | Ag(s)

(c) Writing of Cell reaction :1. The cell reaction corresponding to the cell diagram is written on the assumption that the

right hand electrode is cathode (+) and left hand electrode is anode (–).2. In all the voltaic cells the electrons flow spontaneously from left to right that is from negative

electrode to positive electrode through external circuit. This implies that electrons are releasedat left hand side electrode (anode) and consumed at right hand side electrode (cathode).In other words oxidation half reaction takes place at left hand side electrode and reductionhalf reaction at right hand side electrode.

3. The overall cell reaction is the redox reaction which is the sum of oxidation half reactionat anode and reduction half reaction at cathode.

4. While adding the half reactions the electrons must cancel. For this purpose it may benecessary to multiply one or both the half reactions by a suitable numerical factor(s).

5. No electrons appear in the overall cell reaction because all the electrons lost by the speciesundergoing oxidation are gained by the species that undergoes reduction.

6. It is important to note that the individual half reactions may be written with one or moreelectrons. For example, half reaction for H2 gas whether written as

248


2H+(aq) + 2e– → H2(g) or as H+(aq)+e– ½H2(g)makes no difference. However, in writing the overall cell reaction, obviously, the electronsmust be balanced. The following examples will illustrate the above arguments :

(i) Consider the cell,

The oxidation half reaction at anode is,

The reduction half reaction at cathode is,

The overall cell reaction is obtained by multiplying the reduction half reaction by 2to balance the electrons and then adding to oxidation half reaction.Thus,

2+ –Pb(s) Pb (1M) + 2e⎯⎯→ (oxidation half reaction)

+ –2Ag (1M) + 2e 2Ag(s)⎯⎯→ (reduction half reaction)

+ 2+Pb(s) + 2Ag (1M) Pb (1M) + 2Ag⎯⎯→ (overall cell reaction)

(ii) Consider the cellPt(s) | H2 (g, 1atm) | H+(1M) || Cr3+ (1M) | Cr(s)The oxidation half reaction at anode is

+ –2H (g, 1atm) 2H (1M) + 2e⎯⎯→

The reduction half reaction at cathode is 3+ –Cr (1M) + 3e Cr(s)⎯⎯→

To write the overall cell reaction, oxidation half reaction is multiplied by 3 and reductionhalf reaction is multiplied by 2 and then the modified half reactions are added together.Thus,

+ –23H (g, 1atm) 6H (1M) + 6e⎯⎯→ (Oxidation half reaction)

3+ –(s)2Cr (1M) + 6e 2Cr⎯⎯→ (reduction half reaction)

3+ +23H (g, 1atm) + 2Cr (1M) 6H (1M) + 2Cr(s)⎯⎯→ (Overall cell reaction)

(d) Daniell Cell :Principle : It is an electrochemical cell in which a spontaneous chemical reaction occursto produce electricity.

249


Construction :1. It consists of two half cells.2. One half-cell is a beaker containing a strip of

metallic Zn that is immersed in 1M ZnSO4

solution.3. The second half-cell is a beaker containing a

strip of metallic Cu that is immersed in 1MCuSO4 solution.

4. These two half-cells are connected by a saltbridge.

Formulation (Representation) :

Cell Reaction :1. Zn electrode is anode (–) : Here Zn atoms are oxidised to Zn2+ ions which go into

the solution and the electron are left on the metal strip.2+ –Zn(s) Zn (1M) + 2e⎯⎯→ (Oxidation half reaction) ...(i)

2. The electrons produced at anode are transferred through the wire to Cu electrode.3. Cu electrode is a cathode (+), where Cu2+ ions are reduced to metallic Cu; which is

deposited on the Cu strip.2+ –Cu (1M) + 2e Cu(s)⎯⎯→ (Reduction half reaction) ...(ii)

4. Salt bridge maintains electrical neutrality of both the solutions. Two Cl– ions from the saltbridge migrate into the anode solutions for every Zn2+ ions formed. Two K+ ions fromthe salt bridge migrate into the cathode solution to replace every Cu2+ ion reduced.

5. Overall cell reaction : It is obtained by adding eq. (i) and eq. (ii)2+ 2+Zn(s) + Cu (1M) Zn (1M) + Cu(s)⎯⎯→

Observations :The following are the experimental observations when Daniell cell operates :

1. The potential of the cell is 1.1V, initially.

250


Salt

bri

dge

2. The mass of Cu electrode increases due to the deposition of metallic Cu on it. Theconcentration of Cu2+ ions decreases in the solution around Cu electrode as they are consumedin the reduction reaction.

3. The zinc atoms are oxidized to Zn2+ ions which pass into the solution. This results in increasingthe concentration of Zn2+ ions in the anode solution and decreasing the mass of Zn electrode.

Note : Always remember the following pairs in alphabetical order :Electrode ElectrodeLeft RightAnode CathodeOxidation ReductionNegative Positive

(e) Types of Electrodes :The electrodes are classified according to their composition in the following types :

1. Metal - metal ion electrodes : It consists of a metal strip immersed into the solutionof its own ions. It is formulated as M+n(aq) | M(s)

The half reaction for the electrode is ( )+n –M aq + ne M(s)⎯⎯→

For example, Zn2+(aq) | Zn for which the half reaction is 2+ –Zn (aq) + 2e Zn(s)⎯⎯→

2. Non metal-non metal ion electrode (Gas electrode) : It consists of a gas about aninert metal electrode immersed in a solution containing ions of the gas. For example,

(i) Hydrogen gas electrode +2 H2

H (aq) | H (g, P ) | Pt

The electrode reaction is+ –

2 H2

1H (aq) + e H (g, P )

2⎯⎯→

(ii) Chlorine gas electrode –2 Cl 2

Cl (aq) | Cl (g, P ) | Pt

The electrode reaction is – –2 Cl2

1Cl (g,P ) + e Cl (aq)

2⎯⎯→

(iii) Oxygen gas electrode –2 O2

OH (aq) | O (g, P ) | Pt

The electrode reaction is – –2 O 22

1O (g, P ) + H O( ) + 2e 2OH (aq)

2l ⎯⎯→

251


3. Metal - Sparingly soluble salt electrode :It consists of a metal coated with one of itssparingly soluble salts immersed in a solutionof soluble salt that contains the same anion asthat of the sparingly soluble salt.Example :Silver-silver chloride electrode that consists ofa silver wire coated with AgCl and immersedin a solution of a soluble salt such as KCl orHCl or any other soluble salt that contains Cl–

ions. It is represented asCl–(aq) | AgCl(s) | Ag

The electrode reaction is written as – –AgCl(s) + e Ag(s) + Cl (aq)⎯⎯→

4. Redox electrode : It consists of metal wire (Pt) serving as an inert electrode immersedin a solution containing the ions of the same substance in two valency states. For example,(a) Fe2+ (aq), Fe3+ (aq) | Pt

The order of writing the two ions is immaterial. The reduction reaction for the electrode

is 2+ – 3+Fe (aq) + e Fe (aq)⎯⎯→

(b) Cu+(aq), Cu2+(aq) | Pt

Reduction reaction is 2+ – +Cu (aq) + e Cu (aq)⎯⎯→

(c) 3+ 4+Sn (aq), Sn (aq) | Pt

Reduction reaction is 4+ – 3+Sn (aq) + e Sn (aq)⎯⎯→

4.6 REFERENCE ELECTRODE :

Concept Explanation :Definition : An electrode whose potential is arbitrarily taken as zero, is exactly knownas known as reference electrode.

Primary reference electrode :1. Standard hydrogen electrode (SHE) is used as a primary reference electrode. In this electrode

H2 gas at 1 atm. pressure, is in contact with 1M solution of H+ ions.2. The half-cell reaction of SHE can be written either as oxidation or reduction.3. The potential of SHE is considered as zero volt.

252


Note : SHE is not very convenient electrode, therefore, several other electrodes areused as a secondary reference electrode.

Secondary reference electrode :1. The potentials of these electrodes are exactly determined with respect to SHE. These electrode

are most convenient e.g. standard calomel electrode (SCE), silver–silver chloride electrode,glass electrode.

2. These electrodes are used to determine the potentials of all other electrodes.

(a) Standard Hydrogen Electrode (SHE) :1. Definition : An electrode in

which pure and dry hydrogengas is bubbled at 1 atmpressure around a platinisedplatinum plate immersed in 1MH+ ion solution, is calledstandard hydrogen electrode.

2. It is a primary referenceelectrode.

3. By convention, SHE is arbitrarilyassigned, a potential of exactlyzero volt at all temperatures.Construction :

1. It consists of a platinum plate coated with platinum black.2. The platinum plate is seated into a thin glass tube.3. The glass tube contains little Hg and a Cu-wire.4. The glass tube is surrounded by outer glass jacket.5. The platinum plate is immersed in a 1M H+ ion solution.6. Pure and dry H2 gas at 1 atm. pressure is bubbled through the side arm, over the surface

of platinum plate.7. Platinum does not take part in the reaction. It serves only as the site of electron transfer.

Representation :+

2H (1M) | H (g,1atm) | Pt

Electrode reaction (working) :1. The platinum black is capable of absorbing large quantities of H2 gas. It allows the change

from gaseous to ionic form and the reverse process to occur.

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2. At anode,

+ –2H (g, 1atm) 2H (1M) + 2e ; Eº = 0.0000V⎯⎯→ (oxidation)

3. At cathode

+ –22H (1M) + 2e H (g, 1atm) ; Eº = 0.000V⎯⎯→ (reduction)

Application of SHE :1. When an electrode of unknown potential is combined with SHE to form a cell. Then the

measured cell potential is the potential of unknown electrode.2. Example :

2+ +2Zn | Zn (1M) || H (1M) | H (g,1atm) | Pt

By experiment, cell potential is found to be 0.763VEºCell = EºSHE – EºZn

0.763V = 0 – EºZn

∴∴∴∴∴ EºZn = –0.763V

Difficulties in setting SHE :1. It is difficult to obtain pure and dry H2 gas.2. The pressure of H2 gas cannot be maintained at exactly 1 atm. throughout the measurement.3. The concentration of H+ ion solution cannot be exactly maintained at 1M. Due to the bubbling

of H2 gas into the solution, evaporation of water may take place. This results in changingthe concentration of solution.

(b) Calomel electrode :1. It is a metal-sparingly soluble salt

electrode.2. It is a secondary reference electrode.

Construction :1. It consists of a glass tube provided with

a bent side tube and another side tube ‘B’.2. A little Hg is placed at the bottom of the

dry glass tube.3. A rubber bung (stopper) carrying a thin

glass tube with a platinum wire is theninserted, taking care that the platinum wiredips into Hg.

B

Glass tube

Bent sidetube

KCl solution

Paste ofHg + Hg Cl2 2

Hg Pt

Glass wool plug

Calomel Electrode

254


4. The Hg is then covered with a layer of Hg and Hg2Cl2 (Calomel) paste.5. The glass tube is then filled with KCl solution of definite concentration.

Potential of Calomel Electrode :The potential of the Calomel electrode depends on as the concentration of KCl solution.

Concentration of KCl Reduction potential at 298 K.

1. 0.1M (Decimolar) 0.337 V

2. 1 M (Molar) 0.280 V

3. Saturated Calomel electrode 0.242 V

Formulation :The electrode is represented 2asKCl(sat) | Hg2Cl2 | Hg(l)

Electrode reactions :If the electrode is cathode (+) in thegalvanic cell, the half reaction that occurson it will be reduction.

– –2 2Hg Cl (s)+ 2e 2Hg( )+ 2Cl (sat)l⎯⎯→

If it serves as anode (–) in the galvaniccell the half reaction will be oxidation.

– –2 22Hg( ) +2Cl (sat) Hg Cl (s) +2el ⎯⎯→

Application of calomel electrode :It is used as a secondary reference electrode to determine the standard potentials of theelectrodes. For example, to determine the standard potential of Zn2+ (1M) | Zn electrode,the following cell is constructed :Zn | Zn2+ (1M) || KCl (sat) | Hg2Cl2(s) | HgAs shown in the figure. The emf of this cell is measured experimentally from the measuredcell potential, the potential of the given electrode can be calculated as Ecell = Ec –EºZn.Hence, EºZn = Ecell – Ec where Ec is the potential of calomel electrode.

Advantages of calomel electrode :1. It is easy to construct and transport and convenient to handle.2. The potential of the electrode is reproducible and remains constant.3. No separate salt bridge is required for its combination with other electrodes.

255



*1. A voltaic cell consisting of Fe2+(aq) | Fe(s) and Bi3+ (aq) | Bi(s) electrodes is constructed.When the circuit is closed mass of Fe electrode decreases and that of Bi electrode increases.(a) Write cell formula (b) Which electrode is cathode and which is anode?(d) Write electrode reactions and overall cell reaction.

Ans. When the circuit is closed, the mass of Fe electrode decreases that of Bi electrode increases,it means that oxidation takes place at Fe electrode and reduction takes place at Bi electrode.(a) Cell formula :

(b) Fe electrode, is an anode since oxidation takes place.Bi electrode, is a cathode since reduction takes place.

(c) Electrode reactions2+ –Fe(s) Fe (1M) + 2e × 3⎡ ⎤⎯⎯→⎣ ⎦ (Oxidation half-reaction)

3+ –Bi (1M) + 3e Bi(s) × 2⎡ ⎤⎯⎯→⎣ ⎦ (Reduction half-reaction)

3+ 2+3Fe(s) + 2Bi (1M) 3Fe (1M) + 2Bi(s)⎯⎯→ (Overall cell reaction)

*2. What is Salt Bridge? What are its functions in a galvanic cell?Ans. 1. It is a U-shaped glass tube containing a saturated solution of an electrolyte such as KCl

or NH4NO3 and 5% agar gel.2. The ions of the electrolyte do not react with the ions of electrode solutions.3. Both the ends of the U tube are plugged with glass wool.

Functions of the salt bridge :1. It provides an electrical contact between the two solutions and thereby completes the electrical

circuit.2. It prevents the mixing of electrode solutions.3. It maintains electrical neutrality in both the solutions by the flow of ions. [Refer 4.5 (a)]

*3. What are the conventions used to write cell diagram (cell formula)?Ans. Refer 4.5 (b).

*4. Describe the construction of Daniell cell. Write electrode half reactions and net cell reactionin Daniell cell.

Ans. Refer 4.5 (d).

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*5. Describe the following types of electrodes giving one example, with reference to formulation,electrode reaction and Nernst equation for electrode potential.(a) metal-sparingly salt electrode, (b) gas electrode.

Ans. Refer 4.5 (e).

6. What is reference electrode? Give one example.Ans. Refer 4.6

7. Describe the construction and working of Standard Hydrogen Electrode.Ans. Refer 4.6 (a)

8. Describe the construction and working of Calomel Electrode.Ans. Refer 4.6 (b)

9. Why is anode in galvanic cell considered to be negative and cathode positive electrode?Ans. Negative polarity :

(a) According to latest IUPAC convention, the electrode where de-electronation or oxidationtakes place is called anode, for example zinc electrode, 2+ –Zn(s) Zn (aq) + 2e⎯⎯→

(b) The electrons, thus produced are accumulated on the electrode hence, it acquires negativecharge. Due to above reasons the anode in galvanic cell in considered to be negative.

Positive polarity :(c) According to latest IUPAC convention, the electrode where electronation or reduction

takes place is called cathode, for example copper electrode,2+ –Cu (aq) + 2e Cu(s)⎯⎯→

(d) Here, electrons are removed from the electrode and Cu2+(aq) ions are deposited on theelectrode, hence, it acquires positive charge. Due to above reason the cathode in galvaniccell is considered to be positive.

4.7 ELECTROCHEMICAL SERIES (ELECTROMOTIVE SERIES) Concept Explanation :

1. Definition : A series (table) of the arrangement of electrodes (metal or non-metalsin contact with their ions) with the electrode half reactions in decreasing order ofstandard potentials, is called electrochemical series. OR A series in which electrodesor half-cells of elements are arranged in the decreasing order of their standardreduction potential, is called electrochemical series.

2. The E0 value is a measure of the tendency of the species to be reduced.3. The greater E0 value means greater tendency of the species to accept electrons and undergo

reduction.4. The positive E0 values show the tendency of half reactions to occur in the forward direction.5. The negative E0 values indicate the tendency of the half reactions to occur in the reverse

direction.

258


Applications of electrochemical series :1. Relative strength of oxidising agents in terms of Eº values :

(a) The species on left side of half reactions are oxidising agent.(b) The substances in the upper left side of half reactions are strong electron acceptors,

as they have large positive Eº values and have greater tendency to be reduced. Hence,they are stronger oxidising agents e.g. F2, Au+, Ce4+ etc.

(c) Thus, the species located in the upper left side of half reaction can be chosen asoxidising agents.

(d) The strength as oxidisng agent decreases from top to bottom.2. Relative strength of reducing agents in terms of Eº values :

(a) The species on the right side of half reactions are reducing agent.(b) The substances in the bottom right side of half reaction are strong electron donors,

as they have large negative Eº values and have greater tendency to be oxidised. Hence,they are stronger reducing agents e.g. Li, K, Ca etc.

(c) Thus, the species located at the bottom right side of half-reaction can be chosen asreducing agents.

(d) The strength as reducing agent decreases from bottom to top.3. Identifying the spontaneous direction of reaction :

(i) E0cell is calculated from the E0 value given in the series for half-reactions.

(a) If E0cell is positive, then overall cell reaction is spontaneous.

(b) If E0cell is negative, then it is nonspontaneous.

(ii) Example : + 2+Mg + 2Ag Mg + 2Ag⎯⎯→

By convention, the cell is represented as 2+ +Mg | Mg || Ag | AgGiven, E0 Mg2+ | Mg = –2.37V ; E0

Ag+| Ag = + 0.799V

E0cell = E0

red (Ag) – E0red(Mg)

= R.H.E. L.H.E.= +0.799V – (–2.37V)= 3.169V

Since, emf of the cell comes out to be positive. Hence, the reaction is spontaneous,that is Ag+ can oxidize Mg.

(v) In general, any oxidising agent (the species on the left of half reaction) can oxidizeany reducing agent (the species on the right of half reaction) that appears below itbut cannot oxidize the species to located above it in the electrochemical series. Thisis called diagonal rule.

4. Calculate of E0 cell : The electrochemical series can be used to calculate the standardcell potential from the Eº values for half reactions given in it. For example, consider the

cell, 2+ 2+Zn(s) | Zn (1M) || Pb (1M) | Pb

259


The E0cell is given by, E0

cell = E0cathode – E0

anode

= E0pb – E0

Zn

E0Pb = –0.126V and E0

Zn = –0.763VE0

cell = –0.126V – (–0.763V)= 0.637V

4.8 ELECTRODE POTENTIALS AND CELL POTENTIALS :

Concept Explanation :Electrode Potentials :1. Definition : Electrode

potential is defined as‘the difference ofelectrical potentialbetween metal electrodeand solution around itunder equilibriumcondition.

2. When a metal plate isdipped in its salt solution,the metal atoms tend togo into the solution aspositive ions leavingelectrons on the metalplate.

3. This tendency is called De-electronation or oxidation. It tries to make the plate negativelycharged.

4. As the same time, the positive ions present in the solution tend to deposit on the metalplate by removing electrons from the metal plate.

5. This tendency is called electronation or reduction. It tries to make the plate positivelycharged.

6. At equilibrium the two tendencies are matched.7. If de-electronation is faster than electronation, the mutual electrode acquires negative charge.8. If electronation is faster metal layer electrode acquires positive charge.9. Thus, an electrical double layer is formed at the interface between metal plate and solution.

10. There is a difference of electrical potential metal solution at equilibrium, which is calledelectrode potential.

260


Cell Potentials :(a) When a galvanic cell operates, a spontaneous redox chemical reaction occurs inside it.

(b) The half-cell reaction at anode is oxidation. The potential associated with this, is called

oxidation potential.(c) The half cell reaction at cathode is reduction. The potential associated with this, is called

reduction potential.(d) The cell potential is the algebraic sum of the oxidation potential and reduction potential.(e) Formula : Ecell = Eoxi (anode) E red(cathode)

(f) Definition : The cell potential or electromotive force (emf) of the cell is defined as

‘the difference of potential between the electrodes corresponding to an external flow

of electrons from anode to cathode. (OR) ‘The difference of potential between the

two electrodes of the cell in open circuit (i.e. when no current is allowed to flow

in the circuit) is called cell potential or emf of a cell.’

Note : The flow of current through the circuit is determined by the push of electronsat the anode and attraction of electrons at the cathode. These two forces constitutethe driving force or electrical pressure that sends electrons through the circuit.This driving force is called emf or cell potential or cell voltage. It is measured in Volts.

(a) Standard Potentials :1. Definition : It is defined as the potential developed when a particular electrode is

dipped in the solution of its ions at unit activity. (i.e. 1 molar concentration or 1 atm.pressure in case of gas electrode) and at 25ºC (298K).

2. It is designated as E0.(a) Standard oxidation potential (E0

oxi) : It is defined as the potential developed dueto oxidation when the particular electrode is dipped in the solution of its ions at unitactivity (1 molar concentration) and at 298 K.

(b) Standard reduction potential (E0red) : It is defined as the potential developed due

to reduction when the particular electrode is dipped in the solution of its ions at unitactivity (1 molar concentration) and at 298 K. Eºred = –Eºoxi

Note : According to IUPAC, standard reduction potential is now called standard electrodepotential (Eº). It is determined relative to the SHE molar standard conditions.

Standard cell potential or standard emf of a cell is the algebraic sum of the standard oxidationpotential of anode (left hand electrode) and standard reduction potential of cathode. (Righthand electrode)

261


0 0 0cell oxi redE = E + E (old convention)

(L.H.E.) (R.H.E.)

(Anode) (Cathode)But according to latest IUPAC conventions, the emf of the cell is the difference betweenthe reduction electrode potentials of the two electrodes (or half-cells) and it is given as

0 0 0cell red redE = E – E

(R.H.E.) (L.H.E.)

(Cathode) (Anode)

(b) Nernst Equation :The cell potential and electrode potential depends on temperature, concentration of solutesand partial pressures of gases. This dependence of potential is given by Nernst equation.

0 RTE = E – nQ

nFl

Where Q is reaction quotient given by Q = [products]

[reactants]

E =0 RT [products]

E – nnF [reactants]

l

=0

102.303RT [products]

E – lognF [reactants]

Where Eº = standard potential of electrode or celln = moles of electrons used in the reactionF = Faraday = 96500 C/mol e–

(products) and (reactants) represent the concentrations of products and reactantsrespectively.T = temperature in KR = gas constant = 8.314 J K–1 mol–1

At 25ºC, 2.303 RT

F = 0.0592 V mol e–

Hence, at 25ºC, the Nernst equation becomes

E = 0

100.0592 [products]

E – logn [reactants] ...(i)

The first term in Nernst equation represents standard state electrochemical conditions.The second term is the correction for non-standard state electrochemical conditions. Thecell potential or electrode potential is equal to the standard potential if the concentrations

262


of reactants and products are unity. Thus, if we substitue (products) = (reactants) = 1in equation (i), we see that,

E = 100.0592

Eº – log 1 = Eºn

Calculation of cell potential using Nernst equation :

Consider the cell 2+ +H2

Zn | Zn (aq) || H (aq)(g, P ) | Pt

The Oxidation at anode, reduction at cathode and overall cell reaction are as follows :

2+ –Zn(s) Zn (aq) + 2e⎯⎯→ (Oxidation at anode)

+ –2 H 2

2H (aq) + 2e H (g,P )⎯⎯→ (reduction at cathode)

+ 2+2 H 2

Zn(s) + 2H (aq) Zn (aq) + H (g, P )⎯⎯→ (Overall cell reaction)

The emf of the cell is given by Ecell =

2+H 2

cell 10 + 2

[Zn ] × P0.0592Eº – log

n [H ] At 25ºC

The concentration of solute are molar concentrations and gas pressure is in atm.

Calculation of electrode potential : Consider the electrode Zn2+(aq) | Zn. The reduction

reaction of the electrode is 2+ –Zn (aq) + 2e Zn(s)⎯⎯→ . The potential of the electrode

at 25ºC is given by Nernst equations.

Eel = el 10 2+

0.0592 1Eº – log

2 [Zn ]

=2+

el 100.0592

Eº + log [Zn ]2

The following conventions are followed while calculating the cell potentials (emf) :1. In the formula of a galvanic cell, the right hand side electrode is cathode (+) and left hand

side electrode is anode (–).2. All standard potentials are reduction potentials that is they refer to the reduction reactions.3. The cathode has a higher standard potential than the anode. Hence, to formulate a cell

from the given electrodes, the electrode with higher standard potential must be writtenon the right hand side and that with lower potential on the left hand side.

4. For a spontaneous reaction as written (from left to right) the cell potential should be positive,for this purpose, the cathode is written on the right hand side and anode on the left handside.

263


(c) Cell potentials and Gibbs energy changes for cell reactions :1. When a cell produces a current, the current can be used to do work - to run a motor

for example [it means that electrical work is done by the system (cell)]. Hence,thermodynamic principles can be used.

2. Electrical work done = amount of charge flowing × cell potential3. For every one mole of electron transferred in the cell reaction, the quantity of electricity

that flows through the cell is one Faraday. (1 F = 96500C)4. Hence, if n moles electron are transferred in any cell reaction, the quantity of electricity

flowing = nF Faradays.5. ∴∴∴∴∴ Electrical work done = nF × Ecell

= nFEcell ...(i)6. Maximum electrical work is done by the cell when the process occurs reversibly.7. Electrical work done by the system (cell) results in the corresponding decrease in free

energy (ΔG) of the system.∴∴∴∴∴ Electrical work done = –ΔG ...(ii)8. From equation (i) and (ii);

–ΔG = nFEcell or ΔG = –nFEcell

Thus, the standard Gibbs energy change (ΔGº) for cell reaction and standard emf of thecell (Eºcell) are related as, ΔGº = –nFEºcell.

E0cell values are intensive properties :We have, ΔGº = –nFEºcell.Gibbs energy (ΔGº) is an extensive property since it depends on the amount of substances.If stoichiometric equation of redox reaction is multiplied by 2 that is the amount of substancesoxidised and reduced are doubled, ΔGº doubles. The number of electron transferred ‘n’also doubles.Hence, the ratio,

E0cell =

Gº–

nF

Δ

Becomes,2 Gº

–2nF

Δ =

Gº–

nF

Δ

Thus, E0cell remains constant. This indicates that the electrical potential is an intensiveproperty which does not depend on the amount of substances. Therefore, whenever, thehalf reaction is multiplied by a numerical factor to balance the no. of electrons, thecorresponding Eº value is not to be multiplied by that factor.

264


(d) Standard cell potentials and equilibrium constants :The relation between standard free energy change of cell reaction and emf of the cellis given by the equation.ΔGº = –nFEºcell

The relation between standard Gibbs energy change for a reaction and its equilibrium constantis given by the equation.ΔGº = –RT ln K

Combining these two equations we obtain,–nFEºcell= –RT ln K

Eº =RT

n KnF

l

= 102.303 RT

log KnF

= 100.0592

log K at 25ºCn

(e) Measurement cell potential (EMF) :(Using potentiometer) (Poggendorff’s compensation method)

1. Circuit and apparatus :(a) The potentiometer consists of

a thin conducting wire AB ofuniform cross section. It isstretched over a meter scale.

(b) A two-volt accumulator(battery) C is connected inseries with a rheostat and withthe terminals of wire AB.

(c) S is the standard cell of knownemf.

(d) X is a cell, the emf of whichis to be determined.

(e) The standard cell ‘S’ is connected to one end A of wire AB and through a galvanometerG, to a sliding contact (jockey) J.

(f) The jockey (J) can be moved along wire AB.(g) A special double switch K is provided to permit the cell S or X to be connected in

the circuit.

265


(h) In connecting the accumulator and the cell ‘S’ or cell X to wire AB, it is essentialthat the positive poles should be connected to the same end (A) of the wire.

(i) Thus, the cell ‘S’ or X will send a current through the circuit in the direction oppositeto that supplied by accumulator C.

2. Measurement of emf :(a) The standard cell S is placed in the circuit by means of the switch K.(b) The sliding contact J is moved along AB and its position is adjusted to a point say

D. So that no current passes through the galvanometer.(c) The length AD is measured. The emf of the cell S, Es ∝ l(AD)(d) Now, the cell X is placed in the circuit by means of the switch K.(e) The sliding contact is again moved along AB and the point of balance on the wire

is determined as before. Let this be at D′.(f) The length AD′ is measured. The emf of the cell X, Ex ∝ l(AD′)

3. Calculations : Dividing Ex, by Es, we can write,

x

s

E

E =1(AD )

1(AD)

′

Hence Ex = s1(AD )

E1(AD)

′

Knowing all the quantities on the right hand side of the equation the emf of the cell Xcan be calculated.


*1. The galvanic cells give voltages not more than 2.5V. Suggest a galvanic cell that will havemaximum potential which is greater than 2.5V.

Ans. 2+ 2+Mg | Mg (1M) || (1M)Cu | CuE0

red(Mg) = –2.37V, E0red(Cu) = +0.337V

E0cell = E0

red – E0red

R.H.E. L.H.E.= +0.337 – (–2.37)= +0.337 + 2.370

E0cell = 2.707 volts

*2. Define the terms : (a) Oxidation potential, (b) Reduction potential, (c) Cell potential.Ans. (a) Oxidition potential : It is defined as the potential developed due to oxidation when the

particular electrode is dipped in the solution of its ions at unit activity (1 molar concentration)and at 298 K.

266


(b) Reduction potential : It is defined as the potential developed due to reduction whenthe particular electrode is dipped in the solution of its ions at unit activity (1 molarconcentration) and at 298 K. E0

red = – E0oxi

(c) Cell potential : The cell potential or electromotive force (emf) of the cell is defined as‘the difference of potential between the electrodes corresponding to an external flow ofelectrons from anode to cathode. (OR) ‘The difference of potential between the twoelectrodes of the cell in open circuit (i.e. when no current is allowed to flow in the circuit)is called cell potential or emf of a cell.’

*3. What conditions are required for a cell potential to be called standard cell potential?Ans. Following conditions are required for the cell potential to be called as standard cell potential :

(a) The concentration of the solutions of cathode and anode must be 1 molar.(b) In case of gas electrode, the pressure of the gas must be 1 atmosphere.(c) The working temperature of the cell must be 25ºC (298K). The standard cell potential

is calculated as :E0

cell = E0red(cathode) – E0

red(anode)RHE LHE

*4. Formulate a cell for each of the following reactions :

(a) 2+ 4+ –Sn (aq) + 2AgCl(s) Sn (aq) + 2Ag(s) + 2Cl (aq)⎯⎯→

Ans. 2+ 4+ –(s)Sn (aq) | Sn (aq) || Cl (aq) | AgCl | Ag

(b) 2+ –2Mg(s) + Br ( ) Mg (aq) + 2Br (aq)l ⎯⎯→

Ans. 2+ –2Mg | Mg (aq) || Br ( ) Br (aq) | Ptl

*5. Formulate a cell from the following electrode reactions :

(a) – –2Cl (g) + 2e 2Cl (aq)⎯⎯→

(b) – –22I (aq) I (s) + 2e⎯⎯→

Ans. – –2 2Pt | I (s) | I (aq) || Cl (aq) | Cl (g) | Pt

*6. Write Nernst equation and explain the terms involved in it. What part of the equation representsthe correction factor for non-standard state conditions?

Ans. Refer 4.8 (b)

*7. Write Nernst Equation for the following reactions :

(a) 3+ 3+ 2+Cr(s) + 3Fe (aq) Cr (aq) + 3Fe (aq)⎯⎯→

Ans. Ecell = 3+

0cell 10 3+ 3

0.0592 [Cr ]E – . log

3 [Fe ]

267


(b) 3+ –Al (aq) + 3e Al(s)⎯⎯→

Ans. Eel = 0

el 10 3+

0.0592 1E – log

3 [Al ]

*8. Consider the following E0 values and half reactions :– –

2I (s) + 2e 2I (aq)⎯⎯→ E0 = 0.535V

2+ –Cu (aq) + 2e Cu(s)⎯⎯→ E0 = 0.337V

2+ –Cd (aq) + 2e Cd(s)⎯⎯→ E0 = –0.403V

(a) Which of the metals or non-metals or ions is the strongest oxidising agent and which is thestrongest reducing agent?

Ans. I2(s) will be the strongest oxidising agent.Cd(s) will be the strongest reducing agent.

(b) The half reactions can be used to construct three galvanic cells. Which will have the highestcell potential?

Ans. 2+ –2Cd(s) | Cd (aq) || I (aq) | I (s) | Pt

*9. Predict whether,(a) Ag+ can oxidize Pb to Pb2+ under standard state conditions.

E0Ag = 0.799 V and E0

Pb = –0.126 VAns. The half-reactions are;

(i) + –Ag (aq) + e Ag(s)⎯⎯→ ; E0Ag = 0.799V

(ii) 2+ –Pb (aq) + 2e Pb(s)⎯⎯→ E0Pb = –0.126V

for the oxidation of Pb, the half reaction is reverse of equation (ii)

(iii) 2+ –Pb(s) Pb (aq) + 2e⎯⎯→ E0Pb = 0.126V

now add equation (i) and (iii),+ –Ag (aq) + e Ag(s) (reduction)⎯⎯→ E0

Ag = 0.799V2+ –Pb(s) Pb (aq) + 2e (oxidation)⎯⎯→ E0

Pb = 0.126V+ 2+Pb(s) + 2Ag (aq) 2Ag(s) + Pb (aq)⎯⎯→ E0 = 0.925V

E0 for the overall reaction is positive. Hence, the overall reaction is spontaneous. Thisindicates that Ag+ can oxidize Pb.

(b) Cu2+ can oxidize Cd to Cd2+ under standard state conditions.E0

Cu = 0.337 V, E0Cd = –0.403V

Ans. (i) 2+ –Cu (aq) + 2e Cu(s)⎯⎯→ ; E0Cu = 0.337V

(ii) 2+ –Cd (aq) + 2e Cd(s)⎯⎯→ ; E0Cd = –0.403V

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∴∴∴∴∴ for oxidation Cd(s), the equation is;

(iii) 2+ –Cd(s) Cd (aq) +2e⎯⎯→ ; E0Cd = +0.403V

Now add eq. (i) and (iii);

2+ –Cu (aq) + 2e Cu(s) (reduction)⎯⎯→ ; E0Cu = 0.337V

2+ –Cd(s) Cd (aq) + 2e (oxidiation)⎯⎯→ ; E0Cd = 0.403V

(overall) 2+ 2+Cd(s) + Cu (aq) Cu(s) + Cd (aq)⎯⎯→ ; E0Cd = 0.740V

ΘΘΘΘΘΘΘΘΘΘ∵ E0 for the overall reaction is positive. Hence, the overall reaction is spontaneous. This indicatesthat Cu2+ can oxidize Cd.

*10. How can Nernst equation be used to show that electrode potential is equal to the standardelectrode potential by putting appropriate values in Nernst equation? Use Cd2+|Cd half cellfor illustration.

Ans. The cell potential or electrode potential is equal to the standard potential if the concentrationsof reactants and products are unity.

2+ –Cd (aq) + 2e Cd(s)⎯⎯→

E = 010

0.0592 [products]E – log

n [reactants]

E = 010 2+

0.0592 [Cd(s)]E – log

2 [Cd ]The concentration of solid phase [Cd(s)] is taken to be unity.If [Cd2+] = 1 as in 1 M solution

E = 0

100.0592

E – . log 12

= E0

*11. How are ΔG0, E0cell and equilibrium constant related for a particular reaction?

Ans. The relation between standard free energy change of cell reaction and emf of the cell is givenby the equation.

ΔGº = –nFE0cell

The relation between standard Gibbs energy change for a reaction and its equilibrium constantis given by the equation.

ΔGº = –RT ln KCombining these two equations we obtain,

–nFE0cell = –RT ln K

E0 =RT

n KnF

l

269


= 102.303 RT

log KnF

= 100.0592

log K at 25ºCn

*12. How does the equation ΔG0 = –nFEº explain that an electrical potential is an intensive property?Ans. E0cell values are intensive properties

We have, ΔG0 = –nFE0cell

Gibbs energy (ΔG0) is an extensive property since it depends on the amount of substances.If stoichiometric equation of redox reaction is multiplied by 2 that is the amount of substancesoxidised and reduced are doubled, ΔGº doubles. The number of electron transferred ‘n’ alsodoubles.

Hence the ratio, E0cell =

0G–

nF

Δ becomes =

02 G–

2nF

Δ =

0G–

nF

Δ

Thus, E0cell remains constant. This indicates that the electrical potential is an intensive propertywhich does not depend on the amount of substances. Therefore, whenever, the half reactionis multiplied by a numerical factor to balance the no. of electrons, the corresponding E0 valueis not to be multiplied by that factor.

*13. Arrange the following oxidizing agents in order of increasing strength under standard stateconditions. The standard potentials for the reduction half reactions are given.Ag+(aq) (0.8V), Al3+(aq) (–1.66V), F2(g) (2.87V), Cl2(g) (1.36V), I2(s) (0.54V), Cd2+(aq) (–0.4V).

Ans. Oxidising strength decreases as :F2(g) (2.87V) > Cl2(g) (1.36V) > Ag+(aq)(0.8V) > I2(s)(0.54V) > Cd2+(aq) (0.4V) > Al3+(aq)(–1.66V)

*14. Arrange the following reducing agents in order of increasing strength under standard stateconditions, the standard potentials for the reduction half reactions being given.Al(s) (–1.66V), Cl–(aq) (1.36V), Cu(s) (0.34V), Fe(s) (–0.44V), Br–(aq) (1.09V), Ni(s) (–0.26V)

Ans. Increasing strength as reducing agent :Cl–(aq) (1.36V) < Br–(aq) (1.09V) < Cu(s) (0.34V) < Ni(s) (–0.26V) < Fe(s) (–0.44V) <Al(s) (–1.66V)

*15. Which species in each of the following pairs is better oxidizing agent under standard stateconditions? (standard potentials are given) Give reasons for your answer.(a) Br2(l) (1.09V) or Au3+(1.4 V)

Ans. The greater E0 value means greater tendency of the species to accept electrons and undergoreduction. Hence, they are better oxidising agents.Au3+ (1.4 V) has greater E0 value than Br2(l) (1.09 V),Hence, it is a better oxidising agent.

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(b) H+(aq) or Ag+(aq) (0.8 V)Ans. Ag+(aq) (0.8 V) has greater Eº value than H+(aq)

Hence, Ag+ is a better oxidising agent.

(c) Pb2+(aq) (–0.13 V) or Co2+ (–0.28 V)Ans. Pb2+(aq) (0.13 V) has greater Eº value than Co2+(–0.28 V)

Hence, Pb2+ is a better oxidising agent.

(d) Cl2(g) (1.36 V) or Cr3+ (–0.74 V)Ans. Cl2(g) (1.36 V) has greater Eº value than Cr3+ (–0.74 V)

Hence, Cl2 is a better oxidising agent.

*16. Which species in each of the following pairs is better reducing agent under standard stateconditions. E0 values are given. Give reasons for your answer.

(a) K(s) (–2.93 V) or Mg(s) (–2.36V)Ans. E0 values for K(s) is more negative, it has less tendency to accept electron, on the contrary

it has greater tendency to donate electron and hence, K is a better reducing agent.(b) Co2+(aq) (1.81 V) or In(s) (–0.14 V)

Ans. E0 values for In is negative, it has less tendency to accept electron, on the contrary it hasgreater tendency to donate electron and hence, it is a better reducing agent.

(c) Ce3+(aq) (1.61 V) or Ti2+ (–0.37 V)Ans. E0 value for Ti2+ is less hence, it is a better reducing agent.

(d) Hg(l) (0.86 V) or Ni(s), (–0.23 V)Ans. E0 value of Ni(s) is less hence, it is a better reducing agent.

*17. What is redox electrode? Give an example with formulation, electrode reaction and Nernstequation for electrode potential.

Ans. Redox electrode :(a) The electrode at which oxidation reduction takes place is called redox electrode.(b) This electrode consists of platinum wire dipped in a solution containing the ions of the

same metal in two different oxidation states. e.g. Fe2+, Fe3+ ion or Sn2+, Sn4+ ionsFormulation : Fe2+(aq), Fe3+(aq) | PtElectrode reaction :

Reduction reaction : 3+ – 2+Fe (aq) + e Fe (aq)⎯⎯→

Nernst equation for electrode potential : In the above reduction reaction n = 1

2+0

3+ 2+ 3+ 2+ 10 3+Fe , Fe Fe , Fe

0.0592 [Fe ]E = E – log

n [Fe ]

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*18. What are standard electrode potentials and standard emf of the galvanic cell?Ans. Standard electrode potential :

(a) Definition : It is defined as the difference of electrical potential between metal electrodeand the solution around it, when all the substance involved in electrode reaction are intheir standard states, having 1 M concentrations, the gases at 1 atm pressure at 25ºC.The standard electrode potential is represented as E0.

(b) Standard electrode potential can be standard oxidation potential or standard reduction potential.(c) If electrode half reaction is oxidation, than the potential developed is called standard oxidation

potential. (E0oxi)

(d) If electrode half-reaction is reduction, then the potential developed is called standard reductionpotential. (E0

red)Standard emf of galvanic cell :

(a) Definition : The standard emf of the cell is the difference between the standard potentialof cathode (RHE) and the standard potential of anode (LHE), when all the substancesinvolved in the cell reaction are in their standard state that is solutions are at 1Mconcentration, gases at 1 atm pressures and solids and liquids are in pure form at 25ºC.

(b) E0cell = E0

(cathode) – E0(anode)

RHE LHE

*19. With electrode reaction and over all cell reaction when lead accumulator behaves as an electrolyticcell.

Ans. (a) During recharging lead accumulator behaves as an electrolytic cell.(b) The cell must be recharged when its potential falls to 1.8V.(c) To recharge the cell a potential greater than 2V is applied.(d) All the cell reaction that occur during discharging are reversed and H2SO4 is regenerated

and its concentration is increased.(e) During this process the roles of anode and cathode are reversed. Now, PbO2 electrode

is anode (+) and Pb is cathode (–).(f) Oxidation reaction at anode (+) : It is the reverse of reduction at cathode during discharging.

+ 2– –4 2 2 4PbSO (s) + 2H O( ) PbO (s) + 4H (aq) + SO (aq) + 2el ⎯⎯→

(g) Reduction reaction at cathode (–) : it is the reverse of oxidation at anode during discharging.– 2–

4 4PbSO (s) + 2e Pb(s) + SO (aq)⎯⎯→

(h) Overall cell reaction during recharging :

4 2 2 2 42PbSO (s) + 2H O( ) Pb(s) + PbO (s) + 2H SO (aq)l ⎯⎯→

*20. ΔG0 for a redox reaction depends on the number of electrons transferred. Explain.Ans. (a) In redox reaction of the cell, the amount of electricity passed depends on the number of

moles of electron transferred (n).

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(b) Standard free energy change = ΔG0 = –nF E0cell.

(c) If the stoichiometry of the redox reaction changes, (n) also changes, Since free energychange is an extensive property, therefore ΔG0 also changes with the number of electrons(n) transferred.

*21. Write shorthand notation for the cell for each of the following reactions :

(a) 2+ –Cu (aq) + 2Ag(s) + 2Br (aq) Cu(s) + 2AgBr(s)⎯⎯→

Ans. 2+ –Cu (aq) + 2Ag(s) + 2Br (aq) Cu(s) + 2AgBr(s)⎯⎯→

(i)– –2Ag(s) + 2Br (aq) 2AgBr(s) + 2e

(oxidation half-reaction at anode)⎯⎯→

(ii) 2+ –Cu (aq) + 2e Cu(s) (reduction half-reaction at cathode)⎯⎯→

Shorthand notation of the cell :Ag(s) | AgBr(s) | Br–(aq)(1M) || Cu2+(aq) (1M) | Cu

(b) Sn2+(aq) is oxidized by Br2(l)

Ans. 2+ 4+ –2Sn (aq) + Br ( ) Sn (aq) + 2Br (aq)l ⎯⎯→

(i) 2+ 4+ –Sn (aq) Sn (aq) + 2e (oxidation half-reaction at anode)⎯⎯→

(ii) – –2Br (l) + 2e 2Br (aq) (reduction half-reaction at cathode)⎯⎯→

Shorthand notation of the cellPt | Sn2+(aq), Sn4+(aq) || Br–(aq) | Br2(l) | Pt


• Theoretical MCQs :*1. Daniell cell operates under non-standard state conditions. If the equation of the cell reaction

is multiplied by 2 then .......a) E and Eº remain unchanged b) E is doubledc) n remains unchanged in Nernst equation d) Q is halved in Nernst equation

*2. Consider the cell 2 2Pt | Cl (g) | HCl(aq) || HBr(aq) | Br ( ) Ptl . If concentration of HCl is

increased, the cell potential will .......a) increase b) decreasec) remain the same d) become maximum

3. Standard cell potential is .......a) measured at a temperature of 25ºCb) measured when ion concentrations of aqueous reactants are 1.00 Mc) measured the conditions of 1.00 atm. for gaseous reactants.d) all of the above

273


4. 2+ –Zn Zn + 2e⎯⎯→ , Eº = +0.76V, 3+ –Cr 3e Cr+ ⎯⎯→ , Eº = –0.74V the anode in this

cell is .......a) Zn b) Cr c) Zn2+ d) Cr3+

5. Write the cell diagram for the reaction below : + –2Cl (g) + 2Ag(s) 2Ag (aq) + 2Cl (aq)⎯⎯→

a) + –2Ag | Ag (aq) | Cl (g), Cl (aq) | Pt b) + –

2Ag | Ag (aq), Cl (aq) | Cl (g) | Pt

c) – +2Pt Cl (g) | Cl (aq) || Ag (aq) | Ag c) + –

2Ag | Ag (aq) || Cl (g), Cl (aq) | Pt6. Chlorine cannot displace ;

a) Fluorine from NaF b) Iodine from NaIc) Bromine from NaBr d) Iodine from KI

*7. E0 of an electrode half reaction is related to ΔGº by the equation, Eº = –ΔGº/nF. If the amount

of Ag+ in the half reaction + –Ag + e Ag⎯⎯→ is tripled then .......a) n is tripled b) ΔGº reduces to one thirdc) Eº reduces to one third d) all the above

8. Calculate the potential (in volts) for the following voltaic cell at 25ºC.3+ 2+Cr | Cr (0.10M) || Cu (0.0010M) | Cu

a) 1.25V b) 1.33V c) 1.41V d) 1.57V9. Nernst equation is given by .......

a) E = 0

100.0592 [Reduced state]

E – logn [oxidised state] b) E= 0

102.303 [oxidised state]

E + logn [Reduced state]

c) E = 010

0.0592 [oxidised state]E + log

n [Reduced state]d) E =

010

2.303 [Reduced state]E – log

n [oxidised state]

• Numerical MCQs*10. The reaction, – 2+

22Br (aq) + Sn (aq) Br ( ) + Sn(s)l⎯⎯→ with the standard potentials, E0Sn=

–0.114V, 0Br2

E = +1.09V, is .......

a) spontaneous in reverse direction b) spontaneous in forward directionc) at equilibrium d) non-spontaneous in reverse direction

*11. The strongest oxidizing agent among the species In3+ (E0 = –1.34V), Au3+ (E0 = 1.4V), Hg2+

(E0 = 0.86V), Cr3+ (E0 = –0.74V) is .......a) Cr3+ b) Au3+ c) Hg2+ d) In3+

12. The standard reduction potential of Li+ | Li, Ba2+ | Ba; Na+ | Na and Mg2+ | Mg are –3.05,–2.73, –2.71 and –2.37V respectively. Which of the following is the strongest oxidising agenta) Na+ b) Li+ c) Mg2+ d) Ba2+

(Hint : The species having the maximum value of standard reduction potential will be the strongestoxidising agent. Greater the reduction potential (less negative) stronger the oxidising agent.)

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13. Which among the following is the strongest reducing agent.2+ –Fe + 2e Fe(–0.44V)⎯⎯→

2+ –Ni + 2e Ni (–0.25V)⎯⎯→

2+ –Sn + 2e Sn (–0.14V)⎯⎯→

3+ 2+Fe + e Fe (0.77V)⎯⎯→

a) Fe b) Fe2+ c) Ni d) Sn(Hint : smaller the reduction potential stronger is the reducing agent)

*14. The standard potential for the cell reaction 2+ +2T (s) + Hg (1M) 2T (1M) + Hg( )l l l⎯⎯→

where EºTl = –0.34V, EºHg = 0.86V is .......a) 0.52V b) –0.52V c) –1.2V d) +1.2V

*15. The following reaction occurs in a galvanic cell + 2+2Cu Cu + Cu⎯⎯→ . If 0+Cu | Cu

E = 0.16V,

and 0+Cu | Cu

E = 0.52V the standard cell potential will be .......

a) 0.68V b) 0.36V c) –0.36V d) –0.68V*16. Consider the following half reaction and choose the correct alternative .......

i) – –2Cl (g) + 2e 2Cl (aq)⎯⎯→ ; Eº = 1.36V

ii) – –2Br ( ) + 2e 2Br (aq)l ⎯⎯→ ; Eº = 1.07V

iii) – –2I (s) + 2e 2I (aq)⎯⎯→ ; Eº = 0.53V

a) Br2 cannot oxidize I– b) Cl2 can oxidize Br– but not I–

c) I2 can oxidize Cl– d) Br2 can oxidize I– but not Cl–

*17. Maximum standard emf will be delivered by the cell consisting of the half cells .......a) F– | F2 and Br– | Br2 b) F– | F2 and Li+ | Lic) Li+ | Li and Ca2+ | Ca d) Br– | Br2 and Au+ | Au

*18. ΔGº for the reaction + +

21

Ag (aq) + H (g) H (aq) + Ag(s)2

⎯⎯→ , where standard potential for

silver half cell reaction is 0.8V, will be .......a) –77.2 kJ b) +77.2 kJ c) 154.4 kJ d) –38.6kJ

*19. The value of constant in Nernst equation E = 0 constant

E –n

ln Q at 25ºC is .......

a) 0.0592mV b) 0.0592V c) 25.7mV d) 0.0296V20. The standard reduction potential of Cu2+ and Ag+ in V are + 0.34 and + 0.80, respectively.

Determine the value of E in volts for the following cell at 25ºC.

2+ +Cu | Cu (1.00M) ||| Ag (0.0010M) | Ag

a) 0.37V b) 0.55V c) –0.28V d) 0.28V

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*21. Consider the cell, + –2 2Pt | H (g) | H (aq) || I (aq) | I (s) . If the standard cell potential is 0.54V

then the standard potential for cathode half reaction will be .......a) 0V b) –0.54V c) +0.54V d) 1.08V

• Advanced MCQs22. For the electrochemical cell,

M / M+ || X–/X, E0(M+/M) = 0.44V and E0 (X/X–) = 0.33VFrom this data, one can deduce that, (IIT 2000)

a) + –M + X M + X⎯⎯→ is the spontaneous reaction

b) + –M + X M + X⎯⎯→ is the spontaneous reaction

c) Ecell = 0.77Vd) Ecell = –0.77V

(Hint : (b), E0cell = +0.11V, here it is spontaneous)

23. The 03+ 2+M /M

E values for Cr, Mn, Fe and Co are –0.41, +1.57, 0.77 and +1.97V respectively.

For which one of these metals the change in oxidation state from +2 to +3 is easiest? (AIEEE 2004)

a) Cr b) Mn c) Fe d) Co

[Hint : Give values are reduction potentials 2+ 3+ –M M + e⎯⎯→ means oxidation. Oxidation

potentials of Cr will be highest and hence, most easily oxidized]24. The standard reduction potential for Fe2+ | Fe and Sn2+ | Sn electrodes are –0.44 and –0.14

volt respectively. For the cell reaction. 2+ 2+Fe + Sn Fe + Sn⎯⎯→ , the standard emf is ........

(IIT 1990)a) +0.30V b) –0.58V c) +0.58V d) –0.30V

25. The standard reduction potential values of the three metallic cations, X, Y, Z are 0.52, –3.03

and –1.18V respectively. The order of reducing power of the corresponding metals is ..........

(IIT 1998)

a) Y > Z > X b) X > Y > Z c) Z > Y > X d) Z > X > Y

26. For a spontaneous reaction ΔG, equilibrium constant K and E0cell will be respectively ..........

(AIEEEE 2005)

a) –ve, > 1, +ve b) +ve, > 1, –ve c) –ve, < 1, –ve d) –ve, > 1, –ve

[Hint : –ΔGº = RT log K, –veΔG implies that log K will be the i.e. K > 1]

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4.9 COMMON TYPES OF CELLS :


The voltaic cells (electrochemical cells) are classified as :

1. Primary voltaic cells - The voltaic cells that cannot be recharged are called primary

voltaic cells e.g. Dry cell.

2. Secondary voltaic cells - The voltaic cells that can be recharged by reversing the direction

of current flow and thereby regenerating the original reactants (chemicals) are called

secondary voltaic cells. e.g. Lead accumulators.

(a) Dry cell (Leclanche’ cell) :

It is a primary voltaic cell which cannot be recharged. It is called dry cell because the

electrolyte is a viscous aqueous paste and not a liquid solution.

1. Construction :

(a) The container of cell is made up of zinc which also acts as anode (negative electrode)

(b) The Zn container is lined from inside with a porous paper to separate it from the

other materials of the cell.

(c) An inert graphite rod in the centre of

the cell, immersed in an electrolyte paste

acts as cathode (positive electrode).

(d) Graphite rod is surrounded by a paste

of MnO2 and carbon black.

(e) The rest of the cell is filled with

electrolyte which is a moist paste of

NH4Cl and ZnCl2 (some starch is

added to the paste to make it thick so

that it cannot be leaked out).

(f) The cell is sealed at the top to prevent the drying of the paste by evaporation of moisture.

2. Cell reaction :(a) Oxidation of anode - When the cell operates that is current is drawn from the cell,

metallic zinc is oxidized to Zn2+ ions.2+ –Zn(s) Zn (aq) + 2e⎯⎯→

Brass cap

Zn container

Paper spacer

Paste of MnO +carbon2

Graphite rod

Paste of NH CI + ZnCl4 2

(+)

(-)Dry cell

277


(b) Reduction at cathode - The electron liberated in oxidation reaction at anode flow

along the container and migrate to the cathode through the external circuit. At cathode

NH4+ ions are reduced.

+ –4 3 22NH (aq) + 2e 2NH (aq) + H (g)⎯⎯→

The H2 gas formed is oxidised by MnO2 present at the cathode.

2 2 2 3 2H (g) + 2MnO (s) Mn O (s) + H O( )l⎯⎯→

Thus, MnO2 prevents the collection of H2 gas on the cathode. Hence, net reduction

process is the combination of these two reactions.+ –

4 2 2 3 3 22NH (aq) + 2MnO (s) + 2e Mn O (s) + 2NH (aq) + H O( )l⎯⎯→

(c) Net Cell reaction - The net cell reaction is the sum of oxidation at anode and reduction

at cathode.

2+ –Zn(s) Zn (aq) + 2e⎯⎯→ (oxidation)

+ –4 2 2 3 3 22NH (aq) + 2MnO (s) + 2e Mn O (s) + 2NH (aq) + H O( )l⎯⎯→ (reduction)

+ 2+4 2 2 3 3 2Zn(s) + 2NH (aq) + 2MnO (s) Zn (aq) + Mn O (s) + 2NH (aq) + H O( )l⎯⎯→

(overall)The ammonia produced at cathode combines with Zn2+ to form a soluble compoundcontaining complex ion

2+ 2+3 3 4Zn (aq) + 4NH (aq) [Zn(NH ) ] (aq)⎯⎯→

(d) Representation :

(e) Cell potential = 1.5V(f) Cause of irreversibility of dry cell : Zn2+ ions produced at anode combine with ammonia,

NH3 produced at cathode and form very stable complex ions .2+ 2+

(aq) 3(aq) 3 4 (aq)Zn + 4NH [Zn (NH ) ]⎯⎯→

Since Zn2+ ions are internally removed, the reaction at anode can't be reversed, hence

overall discharging reaction can't be reversed and results in poor life of the cell.

(g) Disadvantages :(i) Dry cell can't be recharged.

(ii) If the current from the dry cell is removed rapidly, gaseous H2 can be consumed

rapidly as a result the voltage drops rapidly.

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(iii) Zn anode corrodes due to its action with H+ from +4NH .

+ 2+(S) (aq) (aq) 2(g)Zn + 2H Zn + H⎯⎯→

(h) Applications :(i) Dry cell is used as a source of electric power in radios, flashlights, torches, clocks, etc.

(ii) Since they are available in small size, they can be used conveniently.

3. Alkaline dry cell :(a) The alkaline dry cell is a modified form of Leclanche's dry cell.

(b) If the current from the dry cell is removed rapidly, gaseous H2 can be consumed rapidly

as a result the voltage drops rapidly.(c) Zn anode corrodes due to its action with H+ from +

4NH .

+ 2+(S) (aq) (aq) 2(g)Zn + 2H Zn + H⎯⎯→

(d) To avoid these difficulties, an alkaline dry cell is developed. In this, alkali like NaOH or

KOH is used instead of NH4Cl.(e) Reaction in alkaline dry cell :

(i) Oxidation at zinc anode :– –

(S) (aq) (S) 2 (1)Zn + 2OH ZnO + H O + 2e⎯⎯→

Reduction at graphite cathode :–

2(S) 2 (1) 2 3(S) – (aq)2MnO + H O + 2e Mn O + 2OH⎯⎯→

The overall cell reaction :

(S) 2(S) (S) 2 3(S)Zn + 2MnO ZnO + Mn O⎯⎯→

(f) The voltage of alkaline dry cell is 1.54 V which is more than Leclanche's dry cell which

has voltage of 1.5 volt.

(g) The alkaline dry cell can be represented as, Zn|ZnCl2(aq)NaOH(aq)MnO2(s)|C +(h) Advantages of alkaline dry cell over Leclanche's cell :

(i) It has higher emf of 1.54 V.

(ii) Cell potential is independent of concentrations of alkali electrolytes.(iii) No gases are formed as in Leclanche's dry cell where H2 gas is formed.

(iv) There is no decline in voltage even if high electric current is withdrawn.

(v) The alkaline dry cell has longer life than acidic dry cell, since Zn corrodes very slowlyin alkali medium than in acidic medium.

279


(b) Lead Accumulators (Lead storage battery) :1. Principle :

(a) It is a reversibleelectrochemical cell.

(b) In this cell electrical energyis not generated but it ispreviously stored fromexternal source. Hence, it iscalled secondary cell orstorage battery oraccumulator.

(c) It can be recharged and reused.2. Construction :

(a) A group of lead plates packed with spongy lead acts as anode (negative electrode)(b) Another group of lead plates packed with PbO2 acts as cathode (positive electrode)(c) The electrodes are immersed in an aqueous solution of 38% (by mass) of H2SO4

of density equal to about 1.28 g/mL or 4.963 M which serves as an electrolyte.3. Notation of the cell (Representation)

Working of the cell :(i) Cell reactions during discharge (cell functions as galvanic cell) :

(a) Oxidation reaction at anode (negative electrode) : When the cell providescurrent, spongy lead is oxidized to Pb2+ ions.

2+ –Pb(s) Pb (aq) + 2e⎯⎯→ (oxidation)2+ 2–

4 4Pb (aq) + SO (aq) PbSO (s)⎯⎯→ (precipitation)

2– –4 4Pb(s) + SO (aq) PbSO (s) + 2e⎯⎯→ (overall oxidation at anode)...(i)

(b) Reduction reaction at cathode (positive electrode) : The electrons travel throughthe external circuit from anode to cathode. Here PbO2 is reduced to Pb2+ ions.

+ – 2+2 2PbO (s) + 4H (aq) + 2e Pb (aq) + 2H O( )l⎯⎯→ (reduction)

2+ 2–4 4Pb (aq) + SO (aq) PbSO (s)⎯⎯→ (precipitation)

+ 2– –2 4 4 2PbO (s) + 4H (aq) + SO (aq) 2e PbSO (s) + 2H O( )l+ ⎯⎯→

(overall reduction at cathode) ...(ii)

(-) (+)

38% H SO2 4

Acid proof container

Pb plates packed with PbO2

Pb plates packed with spongy lead

280


(c) Net cell reaction during discharge : On adding equation (i) and (ii);

2 2 4 4 2Pb(s) + PbO (s) + 2H SO (aq) 2PbSO (s) + 2H O( )l⎯⎯→

(d) The potential of the cell depends on concentration (density) of H2SO4. Densitydischarging H2SO4 is consumed and its concentration (density) decreases.

(e) The emf of lead storage cell is 2 Volts.

(ii) Recharging of the cell (cell now functions as electrolytic cells)(a) The cell must be recharged when its potential falls to 1.8 V.(b) To recharge the cell a potential greater than 2V is applied.(c) Thus, all the cell reactions are reversed.

discharging2 2 4 4 2charging

Pb(s) + PbO (s) + 2H SO (aq) 2PbSO (s) + 2H O( )l

4. Application :(a) It is used in laboratory as a source of direct current.(b) It is used in automobiles.(c) It is used in invertors.

5. Nickel-Cadmium (NICAD) cell.(a) Nickel-Cadminum (NICAD) cell is a secondary dry cell.(b) It can be recharged.(c) It consists of a cadmium electrode in contact with an alkali and acts as anode while nickel

(1V) oxide, NiO2 in contact with an alkali acts as cathode. The alkali used is moist pasteof KOH.

(d) Reactions in the cell :(i) Oxidation at cadmium anode :

– –(S) (aq) 2(S)Cd + 2OH Cd(OH) + 2e⎯⎯→

(ii) Reduction at NiO2(s) cathode :– –

2(S) 2 (1) 2(S) (aq)NiO + 2H O + 2e Ni(OH) + 2OH⎯⎯→

The overall cell reaction is the combination of above two reactions.

(S) 2(s) 2 (1) 2(S) 2(S)Cd + NiO + 2H O Cd(OH) + Ni(OH)⎯⎯→

(e) Since the net cell reaction doesn't involve any electrolytes but solids, the voltage is independentof the concentration of alkali electrolyte.

(f) Since the reaction products at both the electrodes are solids, they adhere to the electrodesurface. Hence, these electrode reactions can be reversed during the charging process.Therefore, this is a chargeable dry cell.

(g) Since no gases are produced, this cell can be sealed.(h) The cell potential is about 1.4 V.(i) This cell has longer life than other dry cells.

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4.10 FUEL CELLS : Concept Explanation :

1. Definition - The galvanic cells in whichthe energy of combustion of fuels isdirectly converted into electricalenergy are called fuel cells.

2. In these cells, one of the reactants is afuel such as H2 gas or CH3OH

3. In fuel cells, reactants are not placedwithin the cell, but they are continuouslysupplied to the electrodes.

4. Example : H2–O2 fuel cell.

(a) Hydrogen - Oxygen fuel cell :1. In this cell fuel used is H2 gas and O2 gas is an oxidizing

agent.2. Here, the energy of combustion of H2 is converted into electrical energy.

(b) Principle : It is an electro-chemical cell which converts the energy of combustion of H2

and O2 directly into electrical energy.(c) Construction :1. The anode consists of a porous carbon rod containing a small amount of finely divided

platinum that acts as a catalyst.2. The H2 gas is continuously bubbled through the anode.3. The cathode is also a porous carbon rod impregnated with finely divided platinum catalyst.4. O2 gas is bubbled through the cathode.5. Both the electrodes are immersed in hot KOH solution.

(d) Cell reactions :1. Oxidation at anode (–) - At anode H2 gas is oxidised to H2O

– –2 22H (g) + 4OH (aq) 4H O( ) + 4el⎯⎯→ (oxidation) ...(i)

2. Reduction at cathode (+) - At cathode, O2 is reduced to OH–.

– –2 2O (g) + 2H O( ) + 4e 4OH (aq)l ⎯⎯→ (Reduction reaction) ...(ii)

3. Net cell reaction - an adding eq. (i) and (ii)

2 2 22H (g) + O (g) 2H O( )l⎯⎯→ (Redox reaction)

(e) Representation :

2 2– Pt | H (g) | OH– | O (g) | Pt +

cathode (+)

O gas2

H O2H gas2

Anode(-)

HotKOH

solution

Porous carbonelectrodes with Pt

e

H – O fuel cell2 2

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EMF of the cell (cell potential) :E0

cell = E0red (cathode) – E0

red (anode)= 0.4V – (–0.83V)= 1.23V

(f) Advantages of fuel cells :1. The reacting substances are continuously supplied to the electrodes. Hence, unlike conventional

cells the fuel cells do not have to be discharged when the chemicals are consumed.2. They are non-polluting because the only reaction product is water.3. The fuel cells provide electricity with an efficiency of about 70% which is almost double

of the efficiency of thermal plants which is only about 40%.

(g) Drawbacks of H2–O2 fuel cell :1. In practice voltage is less than 1.23V because of deviations from reversible behaviour.2. H2 gas is hazardous to use and cost of preparing H2 is high.

Application of fuel cells :1. The fuel cells have been used in automobiles on experimental basis.2. The cell was used for providing electrical power in the space programme.3. In spacecrafts it is operated at such a high temperature that water evaporates at the same

rate as it is produced. The vapour is then condensed and pure water is used for drinkingfor astronauts.

4. In future the fuel cells are also likely to be used as power generators for hospitals, hotelsand homes.

5. The applications of fuel cells using methanol as fuel might also become available in smallelectronic products such as cell phones and laptop computers.

4.11 CORROSION : Concept Explanation :

1. Definition - ‘The process ofdestructive attack on a metalsurface by its environmentthrough chemical orelectrochemical reaction whichleads to loss of metal in the formof its compounds (oxides,sulphides, carbonates, sulphates etc.) is called corrosion.

2. In case of iron the corrosion is called rusting (which is hydrated ferric oxide Fe2O3. H2O.3. Corrosion is an electrochemical phenomenon, since on the surface of iron tiny galvanic

cells are formed.

283


4. Rusting occurs in presence of water and air. Water vapours on the surface of the metaldissolve CO2 from air to from H2CO3. This acts as an electrolyte. H2CO3 ionizes to smallextent producing H+ ions.

5. At a particular spot on the surface of iron oxidation takes place and it behaves as ananode.

At anode : 2+ –2Fe(s) 2Fe (aq) + 2e⎯⎯→ (oxidation) Eº = 0.44V

6. Electron released at anodic spot move through the metal, reach another spot and reduceO2 in presence of H+ ion and this spot behaves as a cathode.

At cathode : + –2 2O (g) + 4H (aq) + 4e 2H O( )l⎯⎯→ (reduction) Eº = 1.23V

7. Net cell reaction : The overall reaction is

+ 2+2 22Fe(s) + O (g) + 4H (aq) 2Fe (aq) + 2H O( )l⎯⎯→ ; Eºcell = 0.44 + 1.23 = 1.67V

8. When Fe2+ ion migrate from anode region, they come in contact with O2 dissolved insurface portion by water droplet. They are further oxidised to Fe3+ ions.

2+ + 3+2 24Fe (aq) + O (g) + 4H (aq) 4Fe (aq) + 2H O( )l⎯⎯→

9. Fe3+ ions form an insoluble hydrated oxide which is deposited as red-brown material calledrust.

3+ +2 2 3 22Fe (aq) + 4H O( ) Fe O .H O(s) + 6H (aq)l ⎯⎯→

(Rust)

Protection of metals from corrosion :Following methods are used :

1. Coating metal surface by paint : It is the most common method to shield the Fe fromO2 and moisture to prevent rusting.

2. Galvanizing :(a) Iron can be protected from rust formation by coating it with another metal, which

is more electropositive, such as Zn. The process of deposition of thin layer of Znon Fe is called galvanizing.

(b) Standard potential for the half reaction are :

2+ –Fe (aq) + 2e Fe(s)⎯⎯→ ; Eº = –0.44 V

2+ –Zn (aq) + 2e Zn(s)⎯⎯→ ; Eº = –0.763V

Thus, Zn is stronger reducing agent than ‘Fe’ and Zn can be more easily oxidisedthan Fe. This shows that during corrosion, Zn is oxidized instead of Fe.

(c) Even if Fe is oxidised to Fe2+, Zn immediately reduces Fe2+ to Fe. Thus, as longas Fe and Zn are in contact, Zn protects Fe from oxidation even if Zn layer is scratched.

284


3. Cathodic Protection : The process is which metal in protects from corrosion byconnecting it to a more easily oxidizable much is called Cathodic Protection.(a) The iron object (pipe or tank) buried in the moist soil to be protected from corrosion

is connected to a more active metal either directly or through a wire.(b) The iron object acts as cathode and the protecting metal (sacrificial metal) acts as anode.(c) The anode is gradually used up due to the oxidation of metal to its ions due to loss

of electrons.(d) These electrons are transferred through the wire to H+ ions present around the iron

object and thus, protect it from rusting.(e) As iron object acts as cathode, it is called cathodic protection.(f) For protecting iron objects, generally Mg or Zn are used, which are called sacrificial

anode. These metals oxidize more easily than Fe.(g) A galvanic cell is formed in which Mg or Zn acts as anode and Fe object acts as

cathode. The moist soil serves as an electrolyte. The half reaction are :2+ –Mg(s) Mg (aq) + 2e⎯⎯→ ;Eº = 2.36V

+ –2 2O (g) + 4H (aq) + 4e 2H O( )l⎯⎯→ ; Eº = 1.23V

Because Mg oxidizes instead of Fe, the Fe object is protected from rusting.4. Passivation :

(a) The process in which metal surface is made inactive is called passivation.(b) The metal to be protected from corrosion is treated with a strong oxidizing agent such

as conc. HNO3. As a result a thin oxide layer is formed on the surface of metal.5. Alloy formation :

(a) The tendency of Fe to oxidize can be reduced by forming its alloy with other metals.(b) Example : stainless steel is an alloy of Fe and Cr.(c) The formation of a layer of chromium oxide protects iron from rusting.


*1. Sketch and describe the operation of Dry cell.Ans. Refer 4.9 (a)

*2. Describe the construction and working of H2 – O2 fuel cell.Ans. Refer 4.10

*3. What are the advantages and disadvantages? (drawbacks) of fuel cells.Ans. Advantages of fuel cells :

(a) The reacting substances are continuously supplied to the electrodes. Hence, unlike conventionalcells the fuel cells do not have to be discharged when the chemicals are consumed.

(b) They are non-polluting because the only reaction product is water.(c) The fuel cells provide electricity with an efficiency of about 70% which is almost double

of the efficiency of thermal plants which is only about 40%.Drawbacks of H2–O2 fuel cell :

(1) In practice voltage is less than 1.23V because of deviations from reversible behaviour.(2) H2 gas is hazardous to use and cost of preparing H2 is high.

285


*4. Sketch the lead storage cell.Ans. Refer 4.9 (b)

*5. Write electrode reactions and overall cell reaction during the operation of lead storage cell.Ans. Refer 4.9 (b) 3

*6. Predict whether the following reactions occur under standard state conditions.(a) Oxidation of Ag(s) by Cl2(g). E0Ag = 0.8V, E0Cl2 = 1.36V(b) Reduction of Fe3+ to Fe2+ by Au(s).

03+ 2+Fe , Fe

E = 0.77V, E0Au = 1.4V..

Ans. (a) Since the standard reduction potential of Cl2(g) is higher 0

–Cl /Cl2(E = 1.36V) than that

of ( )0+Ag /Ag

Ag(s) E = 0.8V hence chlorine will oxidise Ag(s) to Ag+, the redox reaction

taking place is + –22Ag(s) + Cl (g) 2Ag + 2Cl (aq)⎯⎯→

(b) Since the standard reduction potential of Au(s) is higher ( )03+Au /Au

E = 1.4V than that

of Fe3+, Fe2+ couple ( )03+ 2+Fe , Fe

E = 0.77V , hence Au(s) will not reduce Fe3+ to Fe2+

but Au(s) will oxidise Fe2+ to Fe3+ has greater tendency to accept electrons.


1. In the lead storage battery during discharging .......(a) pH decreases (b) pH remains same(c) pH increases(d) pH increases or decreases depending upon the extent of discharging.

2. The efficiency of H2 – O2 fuel cell is about .......(a) 70% (b) 90% (c) 40% (d) 100%

• Advanced MCQs3. When a lead storage battery is discharged .......... (IIT 1987)

a) SO2 is evolved b) lead suphate is consumedc) Lead is fermed d) sulphuric acid is consumed

4. The half-cell reactions for rusting of iron are .......... (IIT 2005)

+ –2 2

12H + O + 2e H O

2⎯⎯→ , E0 = +1.23V

2+ –Fe + 2e Fe(s)⎯⎯→ , E0 = –0.44VΔGº (in kJ) for the reactiona) –76 b) –322 c) –122 d) –176[Hint : For emf to be +ve, oxidation should occur at iron electrodeEcell : 1.23 + 0.44V = 1.67V

ΔGº = –nF Eºcell = –2 × 96500 × 1.67J = –322kJ]

286


HOURS BEFORE EXAM Electrochemistry is a branch of physical chemistry, which is concerned with the inter

conversion of chemical and electrical energy. The conductance (G) of a solution of electrolytes is the inverse of resistance (R) i.e.

G = 1R

The conductivity (k) of a solution is the conductance of unit cube of solution. SI unit of conductivity is Sm–1. The molar conductivity () is defined as the conductivity (k) divided by molar concentration

(C) i.e. = k

C SI unit of molar conductivity is Sm2mol–1. The molar conductivity of solution is maximum at zero concentration that is at infinite dilution. Kohlrausch law states that the molar conductivity of a solution at zero concentration

is the sum of molar conductivities of cation (λº+) and anion (λº–) i.e.0 = λº+ + λº–

The cell constant (b) of the conductivity cell is the ratio of the distance between the

electrodes (l) divided by the area of cross section (a) of the electrode, i.e. b = al

An electrochemical cell is a device to study chemical reactions electrically.

A galvanic cell is an electrochemical cell in which a spontaneous redox reaction is usedto produce an electric current.

Cell reaction consists of two half reaction, oxidation half reaction that occurs at anodeand reduction half reaction that occurs at cathode.

Daniell cell : .........

E0cell = E0

red – E0oxi

cathode anodeR.H.E. L.H.E.

Reaction in strage battery

discharging2 (s) 2 4 ( ) 4 (s) 2 ( )charging

Pb(s) + PbO + 2H SO 2PbSO + 2H Ol l⎯⎯⎯⎯⎯→←⎯⎯⎯⎯⎯

Gibbs energy changes are related with cell potential as.∴∴∴∴∴ ΔGº = –nFEº

Anode – Zn | Zn (1M) || (1M) Cu | Cu + cathode2+ 2+e–

current

287


Faraday (F) is a charge on 1 mole of electrons.1F = 96500 C/mole e–.

The electrochemical series is the arrangement of electrodes in order of their decreasingstandard reduction potentials.

The cell potentials are giving by Nernst equation,

Ecell = 0

cell 100.0592 [products]

E – log at 298kn [reactants]

The equilibrium constant (K) of the cell is related to E0cell by the equation,

E0cell = 10

0.0592log K at 298k

n

Corrosion is an electrochemical process in which Fe is oxidised to form rust.

Galvanisation is a process in which Fe surface is covered with Zn, in order to preventFe from rusting.


TYPE - I - MOLAR CONDUCTIVITY

*1. A conductivity cell filled with 0.01M KClgives at 25ºC the resistance of 604 ohms.The conductivity of KCl at 25ºC is0.00141Ω–1 cm–1. The same cell filledwith 0.001M AgNO3 gives a resistanceof 6529 ohms. Calculate the molarconductivity of 0.001M AgNO3 solutionat 25ºC.

Given :i) For KCl,

C = 0.01M, R = 604Ω,k = 0.00141Ω–1 cm–1

ii) For AgNO3, C = 0.001M,R = 6529Ω

To find :i) Λ = ? ii) b = ?

Solution :i) The cell constant of the cell is given by

b = k.RFor KCl, k = 0.00141Ω–1cm–1

and R = 604ΩHere, b = 0.00141 (Ω–1cm–1) × 604(Ω)

= 0.8516 cm–1.ii) The conductivity of 0.001M AgNO3 is

given by,

k =b

R where, R = 6529Ω

∴∴∴∴∴ k =–10.8516 (cm )

6529( )Ω= 1.304 × 10–4Ω–1cm–1

iii) The molar conductivity of 0.001M AgNO3

is given by,

Λ =1000k

C

Λ = 3 –1 –4 –1 –1

–11000(cm L )×1.304×10 (Ω cm )

0.001(molL )

ΛΛΛΛΛ = 130.4ΩΩΩΩΩ–1cm2mol–1

288


*2. A conductivity cell filled with 0.1M KClgives at 25ºC a resistance of 85.5 ohms.The conductivity of 0.1M KCl at 25ºC is0.01286 ohm–1cm–1. The same cell filledwith 0.005 M HCl gives a resistance of529 ohms. What is the molar conductivityof HCl solution at 25ºC?

Given :i) For KCl,

C = 0.1M R = 85.5Ωii) k = 0.01286Ω–1 cm–1

For HCl,C = 0.005M R = 529Ω

To find :i) b = ? ii) Λ = ?

Solution :i) Cell constant (b) = k.R

For KCl,k = 0.01286(Ω–1 cm–1)R = 85.5Ω

∴∴∴∴∴ b = 0.01286(Ω–1 cm–1) × 85.5(Ω)= 1.0995 cm–1.

(ii) The conductivity of 0.005M HCl is givenby,

k = b

R where, R = 529Ω

∴∴∴∴∴ k =–11.0995(cm )

529Ω

= 2.078 × 10–3Ω–1cm–1

(iii) The molar conductivity of 0.005 M–HClis given by,

Λ =1000k

C

= 3 –1 –3 –1 –1

–11000(cm L )×2.078×10 ( cm )

0.005(molL )

Ω

Λ = 415.6ΩΩΩΩΩ –1cm2mol–1

*3. The molar conductivity of 0.05M BaCl2solution at 25ºC is 223 Ω–1 cm2

mol–1. What is its conductivity?Given :

C = 0.05M = 223Ω–1 cm2 mol–1

To find :k = ?

Solution :

Λ = 1000k

C or k =

× C

1000

∴∴∴∴∴ k =–1 2 –1 –1

3 –1

223(Ω cm mol ) × 0.005(mol L )

1000(cm L )

∴∴∴∴∴ k = 0.01115 ΩΩΩΩΩ–1cm–1

*4. The conductivity of 0.02M AgNO3 at 25ºCis 2.428 × 10–3 Ω–1 cm–1. What is its molarconductivity?

Given :C = 0.02Mk = 2.428 × 10–3 Ω–1 cm–1

To find : = ?

Solution :

=1000k

C

3 –1 –3 –1 –1

–1

1000(cm L ) × 2.428 × 10 (Ω cm )

0.02(mol L )

∴∴∴∴∴ = 121.4 ΩΩΩΩΩ–1cm2mol–1

*5. A conductivity cell filled with 0.02MH2SO4 gives at 25ºC a resistance of 122ohms. If the molar conductivity of 0.02MH2SO4 is 618 Ω–1 cm2 mol–1. What is thecell constant?

Given : C = 0.02MR = 122Ω

289


= 618 Ω–1 cm2 mol–1

To find : b = ?

Sol. : = 1000k

C or k =

× C

1000

∴∴∴∴∴ k = –1 2 –1 –1618(Ω cm mol ) × 0.02(molL )

1000

∴∴∴∴∴ k =1.236 × 10–2 Ω2cm–1

Now, k = b

R or b = k.R

∴∴∴∴∴ R = 122ΩCell constant (b)

= 1.236 × 10–2 (Ω–1cm–1) × 122(Ω)∴∴∴∴∴ Cell constant (b) = 1.507cm–1

*6. A conductivity cell filled with 0.02MAgNO3 gives at 25ºC a resistance of 947ohms. If the cell constant is2.3 cm–1, what is the molar conductivityof 0.02M AgNO3 at 25ºC?

Given :C = 0.02M, R = 947Ω,b = 2.3 cm–1

To find : = ?

Solution :

b = k.R or k =b

R =

–12.3cm

947Ω

= 2.428 × 10–3Ω–1cm–1.

=1000k

C

=3 –1 –3 –1 –1

–1

1000(cm L )× 2.428×10 (Ω cm )

0.02(mol L )

∴∴∴∴∴ = 121.4 ΩΩΩΩΩ–1cm2mol–1

*7. The molar conductivities at zero

concentrations of NH4Cl, NaOH and NaCl

are respectively 149.7 Ω–1 cm2 mol–1,

248.1 Ω–1 cm2 mol–1 and 126.5 Ω–1 cm2

mol–1. What is the molar conductivity of

NH4OH at zero concentration?

Given :

0(NH4Cl) = 149.7Ω–1 cm2 mol–1

0 (NaOH) = 248.1 Ω–1cm2 mol–1

0 (NaCl) = 126.5Ω–1cm2 mol–1

To find :

0(NH4OH) = ?

Solution :

According to Kohlrausch law of

independent migration of ions.

0(NH4Cl) =0 0

–+ ClNH 4λ + λ

0(NaOH) = 0 0–+ OHNa

λ + λ

0(NaCl) =0 0

–+ ClNaλ + λ

Hence,

0(NH4Cl) + 0(NaOH) – 0 (NaCl)

=000 0 0 0 – – ++–+ OH Cl NaNaClNH4

– λ– λλ λ + λ – λ

=0 0

–+ OHNH4λ + λ

= º(NH4OH)

Thus, 0(NH4OH)

= 0(NH4Cl) + 0(NaOH) – 0(NaCl)

∴∴∴∴∴ º(NH4OH) = 149.7(Ω–1cm2 mol–1)+ 248.1(Ω–1cm2 mol–1) – 126.5 (Ω–1cm2 mol–1)

= 271.3 Ω–1cm2 mol–1

∴∴∴∴∴ The molar conductivity of CH3OOHat zeroconcentration = 271.3 ΩΩΩΩΩ–1cm2 mol–1

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*8. What is the molar conductivity of AgI atzero concentration if the 0 values of NaI,AgNO3 and NaNO3 are respectively 126.9Ω–1 cm2 mol–1, 133.4 Ω–1 cm2 mol–1 and121.5 Ω–1 cm2mol–1?

Given :0(NaI) = 126.9 Ω–1 cm2 mol–1

0(AgNO3) = 133.4 Ω–1 cm2 mol–1

0(NaNO3) = 121.5 Ω–1 cm2 mol–1

To find :0(AgI) = ?

Solution :According to Kohlrausch Law ofindependent migration of ions

0(NaI) =0 0–+ INaλ + λ

0(AgNO3) =0 0 –+ NOAg 3

λ + λ

0(NaNO3) =0 0+ –Na NO3

λ + λ

Hence,0(NaI) + 0(AgNO3) – 0(NaNO3)

=0 0 0 0

– ++ –I AgNa NO3

λ + λ + λ + λ

=0 0

–+ IAgλ – λ

= λº (AgI)Thus,λº(AgI) = λº(NaI) + λº(AgNO3) –λ0(NaNO3)

∴∴∴∴∴ º(AgI) = 126.9 (Ω–1 cm2 mol–1)+ 133.4 (Ω–1 cm2 mol–1)– 121.5 (Ω–1 cm2 mol–1)= 138.8 Ω–1 cm2 mol–1

∴∴∴∴∴ The molar conductivity of AgI at zeroconcentration = 138.8ΩΩΩΩΩ–1 cm2 mol–1

TYPE - II - FARADAY’S LAW

*9. Estimate the mass of copper metalproduced during the passage of 5A currentthrough CuSO4 solution for 100 minutes.The molar mass of Cu is 63.5 g mol–1.

Given :I = 5A, t = 100mini.e. 100 min × 60sec = 6000s,molar mass of Cu = 63.5 g mol–1.

To find :mass of Cu metal produced = ?

Solution :The quantity of electricity passed,Q = I(A) × t(s)

∴∴∴∴∴ Q = 5(A) × 6000(s)Q = 30000C

i) Moles of electrons actually passed

= –

Q(C)

9500(C/mole ) = –

30000(C)

96500(C/mole )

= 0.3108 mole–

ii) The half reaction at cathode is,

2+ –Cu ( ) + 2e Cu(s)l ⎯⎯→

∴∴∴∴∴ mole ratio = moles of Cu produced

moles of electrons required

= –

1 (mol Cu)

2 (mol e )

= 0.5 mol Cu/mole–

iii) Moles of Cu produced= moles of electrons actually passed × mole

ratio= 0.3108 (mol e–) × 0.5 (mol Cu/ mol e–)= 0.1554 mol Cu

iv) mass of Cu formed= moles of Cu formed × molar mass of Cu

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= 0.1554 (mol Cu) × 63.5 (g mol–1 Cu)= 9.87 g Cu

Mass of Cu formed = 9.87 g Cu

*10. How long will it take to produce 2.415 gof Ag metal from its salt solution by passinga current of 3A? How many moles ofelectrons are required? Molar mass of Agis 107.9 g mol–1.

Given :I = 3A, mass of Ag produced = 2.415g,molar mass of Ag = 107.9g mol–1

To find :t = ? no. of moles of electrons = ?

Solution :Reduction half reaction at cathode,

+ –Ag (aq) + e Ag(s)1 mole 1 mole 1 mole

⎯⎯→

no. of moles of Ag formed

=mass of Ag formed

molar mass of Ag

=2.415

107.9mol = 0.02238mol

no. of moles of Ag formed =0.02238molFrom the reaction,

∵ 1 mole of Ag requires 1 mole of electrons∴∴∴∴∴ 0.02238 mole of Ag will require = 0.02238

mol electrons∵ 1 mole of electrons carries a charge of

96500C,∴∴∴∴∴ 0.02238 mole of electron will carry a

charge= 0.02238 × 96500= 2160C

∴∴∴∴∴ Quantity of electricity passed = Q = 2160CQ = I × t

∴∴∴∴∴ t = Q

I =

2160

3 = 720s = 12 min

Times of electrolysis = 12 min

*11. What current strength in ampere will berequired to produce 2.369 × 10–3 kg of Cufrom CuSO4 solution in one hour? Howmany moles of electrons are required?Molar mass of Cu is 63.5 g mol–1.

Given :mass of Cu produced = 2.369 ×10–3 kg

= 2.369gt = 1 hr = 1 × 60 × 60 = 3600smolar mass of Cu = 63.5 g mol–1.

To find :I = ? no. of moles of electrons = ?

Solution :1 faraday = 96500C = 1 mol electrons1 mol Cu = molar mass of Cu = 63.5 gReduction half reaction;

2+ –Cu + 2e Cu1mole 1mole 1mole

⎯⎯→

moles of Cu deposited

= 2.369

63.5 = 0.0373 mol Cu

From the reaction∵ 1 mol of Cu requires 2 mole electrons.∴ 0.0373 mol Cu will require = 2 × 0.0373

= 0.0746 mol electronsmoles of electrons required =0.0746molNow,

∵ 1 mole of electron = 96500C∴ 0.0746 mol electron = 96500 × 0.0746

= 7199C∴ Quantity of electricity required = Q = 7199C∵ Q = I × t

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∴ I = Q

t =

7199

3600 = 2A

Current strength (I) = 2 Ampere

*12. A current of 6 amperes is passed throughAlCl3 solution for 15 minutes using Ptelectrodes, when 0.504g of Al is produced.What is the molar mass of Al?

Given :I = 6A, t = 15 min = 15 × 60 = 900smass of Al produced = 0.504g

To find :molar mass of Al = ?

Solution :Reduction half reaction,

3+ –Al (aq) + 3e Al(aq)⎯⎯→

Quantity of electricity passed = Q= I × t = 6 × 900 = 5400C

no. of moles of electrons = Q

F =

5400

96500= 0.05596mol

From half-reaction,∵ 3 moles of electrons deposit 1 mole Al∴∴∴∴∴ 0.05596 moles of electron will deposit

= 0.05596

3 = 0.01865 mol Al

Now,∵ 0.01865 mole Al weighs = 0.504g

∴∴∴∴∴ 1 mole Al will weigh = 0.504

0.01865 = 27g

Thus, molar mass of Al = 27 g mol–1.

*13. How many moles of electrons are requiredfor the reduction of

(i) 3 moles of Zn2+ to Zn(ii) 1 mole of Cr3+ to Cr?

How many faradays of electricity will berequired in each case?

Given :moles of Zn = 3 moles,moles of Cr = 1 mole

To find :(i) Faraday of electricity required for Zn = ?(ii) Faraday of electricity required for Cr = ?

Solution :(i)

a) 2+ –Zn + 2e Zn⎯⎯→

mole ratio

=moles of Zn produced

moles of electrons

= –

1 (mol Zn)

2 (mol e )= 0.5 mol Zn/ mol e–

b) moles of electrons required

=moles of Zn

mole ratio

= –

3(mol)

0.5 (mol Zn/mol e ) = 6 mol e–

c) 1 mole of electron requires = 1F electricity∴∴∴∴∴ 6 moles of electron requires = 6 F

electricity(ii)

a) 3+ –Cr + 3e Cr⎯⎯→

mole ratio =moles of Cr produced

moles of electrons

= –

1 (mol Cr)

3 (mol e )= 0.333 mol Cr/mol e–

(b) Moles of electron required

=moles of Cr

mole ratio

= –

1 (mol)

0.333 mol Cr/mol e = 3 mole–

= 3 mole–

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(c)∴∴∴∴∴ 1 mole of electron requires = 1 F electricity∴∴∴∴∴ 3 moles of electron requires = 3 F electricity

*14. In the electrolysis of AgNO3 solution 0.7gof Ag is deposited after a certain periodof time. Calculate the quantity of electricityrequired in coulomb. Molar mass of Ag is107.9 g mol–1.

Given :

mass of Ag = 0.7g,molar mass of Ag = 107.9 g mol–1

To find :(i) Q = ? (in coulmb)

Solution :(i) moles of Ag produced

=mass of Ag

molar mass of Ag

= –1

0.7(g)

107.9(g mol )

= 0.0064 mol Ag

(ii) + –Ag (aq) + e Ag(s)⎯⎯→

mole ratio

= –

moles of Ag

moles of e = –

1 (molAg)

1(mole )= 1 mole Ag/mol e–

(iii) moles of electron actually passed

=moles of Ag

mole ratio

= –

0.0064 (mol Ag)

1 (mole Ag)/mol e= 0.0064 mol e–

(iv) Q = moles of electrons actually passed× 96500 (C/mole–)

= 0.0064 (mole–) × 96500 –

C

(mol e )Q = 626 C

*15. Calculate the amounts of Na and chlorinegas produced during the electrolysis offused NaCl by the passage of 1 amperecurrent for 25 minutes. Molar masses ofNa and chlorine gas are 23 g mol–1 and71g mol–1 respectively.

Given :

I = 1A,t = 25 mins i.e. 25 × 60 = 1500s,molar masses of Na = 23g mol–1,molar mass of Cl2 gas = 71g mol–1

To find :(i) mass of Na produced = ?(ii) mass of Cl2 produced = ?

Solution :Half reaction at the electrodes are

(i) + –Na (aq) + e Na(s)⎯⎯→ (red at cathode)

– –122Cl (aq) Cl (g) + e⎯⎯→ (oxi at anode)

(ii) mole ratio of Na

= –

1 mol Na

1 mol e = 1 mol Na/mol e–

(iii) mass of Na produced

= –

I(A) × t(s)

96500(C/mol e ) × mole ratio of Na ×

molar mass of NaI = 1A, t = 25 × 60 = 1500smass of Na = 23g mol–1

∴∴∴∴∴ mass of Na produced

= – –1

–

1(A)×1500(s)×1mol Na/mole ×23g mol

96500(C/mol e )

Na = 0.3575 g(7iv) mole ratio of Cl2

=2

–

0.5 mol Cl

1 mol e= 0.5 mol Cl2/mol e–

294


(v) mass of Cl2 produced

= –

I(A) × t(s)

96500(C/mole )× mole ratio of Cl2

× molar mass of Cl2

= –

1(A) × 1500(s)

96500(C/mole ) × 0.5 mol Cl2/mol e–

× 71 g mol–1

= 0.5517 g

*16. Calculate the mass of Mg and volume ofchlorine gas at STP produced duringelectrolysis of molten MgCl2 by the passageof 2 amperes of current for 1 hour. Molarmasses of Mg and Cl2 are respectively 24g mol–1 and 71 g mol–1.

Given :

I = 2A, t = 1hr i.e. 1 × 3600 sec = 3600secmolar mass of Mg = 24g mol–1

molar mass of Cl2 = 71g mol–1

To find :(i) mass of Mg = ?(ii) volume of Cl2 gas = ?

Solution :Half reaction at the electrodes are

2+ –Mg (aq) + 2e Mg(s)⎯⎯→

...(reduction at cathode)– –

22Cl (aq) Cl (g) + 2e⎯⎯→

...(oxidation at anode)Mass of Mg produced at cathode

= –

I(A) × t(s)

96500(C/mole )× mole ratio of Mg

× molar mass of MgI = 2 A, t = 1hr × 60 × 60

= 3600smolar mass of Mg = 24 g mol–1

Mole ratio of Mg = –

1 mol Mg

2 mol e

Hence, mass of Mg produced

=–1

– –

2(A) × 3600(s) 1 mol Mg× ×24 g mole

96500 (C/mole ) 2 mol e

= 0.8953gMoles of Cl2 produced

= 2–

I(A) × t(s)× mole ratio of Cl

96500(C/mol e )

mole ratio of Cl2

=2

–

1 mol Cl

2mol e

∴∴∴∴∴ moles of Cl2 produced

=2

– –

1 mol Cl2(A) × 3600(s)×

96500 (C/mole ) 2 mol e= 0.0373 mol Cl2

Volume of Cl2 is given by ideal gas equation

V = nRT

Pn = 0.0373 mol,R = 0.08205L atm K–1mol–1

T = 298KP = 1 atm

∴∴∴∴∴V = –1 –10.0373(mol)×0.08205(Latm.K mol )×298(K)

1atm

V = 0.9/20 L = 912.0 cm3

*17. In a certain electrolysis experiment 0.561g of Zn is deposited in one cell containingZnSO4 solution. Calculate the mass of Cudeposited in another cell containing CuSO4

solution in series with ZnSO4 cell. Molarmasses of Zn and Cu are 65.4 g mol–1 and63.5 g mol–1 respectively.

Given :

molar mass of Zn = 65.4 mol–1

molar mass of Cu = 63.5 g/molmass of Zn deposited = 0.561g

295


To find :mass of Cu deposited = ?

Solution :The half reaction for the formation of Znat cathode of ZnSO4

Cell is 2+ –Zn + 2e Zn(s)⎯⎯→

mole ratio = –

–

1 mol Zn

2 mol emoles of Zn deposited

= –1

mass of Zn deposited 0.561(g)=

molar mass of Zn 65.4(g mol )= 0.008578 mol Zn

Now,

molesof mole ratioofCu deposited Cu half reaction

=molesof mole ratio Zn

Zn deposited half reaction

Hence,

–

–

1(mol Cu)moles ofCu deposited 2(mol e )

=1(mol Zn)0.00857 mol Zn

2(mol e )

∴∴∴∴∴ moles of Cu deposited = 0.008578 mol Cu∴∴∴∴∴ mass of Cu deposited= moles of Cu deposited × molar mass of Cu= 0.00857(mol) × 63.5 (g mol–1)= 0.5447g

*18. Two electrolytic cells one containing AlCl3solution and other containing ZnSO4

solution are connected in series. The samequantity of electricity is passed betweenthe cells. Calculate the amount Zndeposited in ZnSO4 cell if 1.2g of Al aredeposited in AlCl3 cell. The molar masses

of Al and Zn are 27g mol–1 and 65.4g mol–1

respectively.Given :

molar masses of Al = 27 g mol–1

molar masses of Zn = 65.4 g mol–1

mass of Al deposited = 1.2gTo find :

mass of Zn deposited = ?Solution :

3+ –Al (aq) + 3e Al(s)⎯⎯→ ;

mole ratio = –

1 mol Al

3 mol e2+ –Zn (aq) + 2e Zn(s)⎯⎯→ ;

mole ratio = –

1 mol Zn

2 mol emoles of Al deposited

=mass of Al deposited

molar mass of Al

= –1

1.2 g

27g mol= 0.0444 mol Al

Now,

molesofZn deposited

molesofAldeposited

=

moleratioof Zn

moleratioof Al

∴∴∴∴∴molesof Zn deposited

0.0444molAl =

–

–1(molZn) / 2(mole )

1(molAl) /3(mole )

∴∴∴∴∴ moles of Zn deposited

=3 mol Zn

0.0444(mol Al) ×2 mol Al

= 0.0666 mol Zn∴∴∴∴∴ mass of Zn deposited= moles of Zn deposited ×molar mass of Zn= 0.0666 (mol) × 65.4 (g mol–1)= 4.36 g

296


*19. Consider the following cell2+ 2+Pb(s)|Pb (0.5M)||Cu (0.001M)|Cu

(a) Write the cell reaction.(b) Calculate Ecell and E0

cell at 25ºC(c) Calculate ΔG and ΔGº for the cell

reaction.Given :

Pb(s) | Pb2+(aq)(0.5M) || Cu2+(aq)(0.001M) | Cu(s)

02+Pb /Pb

E = –0.126V0

2+Cu /CuE = 0.337V

[Pb2+] = 0.5M; [Cu2+] = 0.001MTo find :

E0cell = ?, Ecell = ?, ΔG0 = ?, ΔG = ?

Solution :

(a) –Pb(s) Pb(aq) + 2e⎯⎯→

(Oxidation at LHE)2+ –Cu (aq) + 2e Cu(s)⎯⎯→

(Reduction at RHE)2+ 2+Pb(s) + Cu (aq) Pb (aq) + Cu(s)⎯⎯→

(Overall cell reaction)∴∴∴∴∴ n = 2

(b) (s) E0cell = 0 0

2+ 2+Cu /Cu Pb /PbE – E

= 0.337 – (–0.126)= 0.463V

E0cell = 0.463V

Ecell= 2+

010 2+

0.0592 [Pb ]E cell – log

n [Cu ].

= 100.0592 0.5

0.463 – . log2 0.001

= 0.463 – 0.0296 log10 500

= 0.463 – 0.0296 × 2.6990= 0.463 – 0.0799= 0.3831 VEcell = 0.3831 V

(c) ΔG0 = –nFE0cell

= –2 × 96500 × 0.463= –89360J= –89.36kJ

ΔG0 = –89.36kJΔG = –nFEcell

= –2 × 96500 × 0.3831= 73940J= –73.94kJ.

ΔG = –73.94kJ.

*20. In the electrolysis of water, one of the half

reactions is + –22H (aq)+ 2e H (g)⎯⎯→

Calculate the volume of H2 gas collected at25ºC and 1 atm pressure by passing 2A for1h through the solution. R = 0.08205 L. atmK–1 mol–1. (Hint : Use equation 4.19 andthen PV = nRT.)

Given :+ –

22H (aq) + 2e H (g)⎯⎯→ ,

I = 2A, t = 1 hr i.e. 1 × 3600 sec. = 3600s,R = 0.08205 L.atm K–1mol–1, T = 25ºC, i.e. 25 + 273 = 298K

To find :volume of H2 gas collected V = ?

Solution :+ –

22H (aq) + 2e H (g)⎯⎯→

mole ratio of H2 =2–

1 mol H

2 mol e= 0.5 mol H2/mol e–

moles of H2 produced

297


= ( )2–

I(A) × t(s)× mole ratio of H

96500 C/mol e

=–

2–

2(A) × 3600(s)× 0.5 mol H /mol e

96500(C/mol e )

= 0.0373 mol H2

V = nRT

Pn = 0.0373 mol,R= 0.08205 L atm K–1 mol–1,T = 25ºC + 273 = 298K,P = 1 atm

V = –1 –10.0373(mol)×0.08205(Latm K mol )×298(K)

1atm

V = 0.917L

TYPE - 2 - CELL POTENTIALS AND

GIBBS ENERGY CHANGE

*21. Write the cell reaction and calculate thestandard potential of the cell.

2+ –2Ni(s)|Ni (1M)||Cl (1M)|Cl (g,1atm)|Pt

0 0Cl Ni2

E – 1.36V and E = –0.25V

Given :0

Cl2E = 1.36V, E0

Ni = –0.25V

To find :Cell reaction = ?E0

cell = ?Solution :

2+ –Ni(s) Ni (1M) + 2e⎯⎯→

...(Oxidation at anode)– –

2Cl (g) + 2e 2Cl (1M)⎯⎯→

...(reduction at cathode)

2+ –2Ni(s)+Cl (g) Ni (1M)+ 2Cl (1M)⎯⎯→

...(overall cell reaction)

E0cell = 0

Cl2E – E0

Ni

Cathode – Anode= 1.36V – (–0.25V)

E0cell = 1.36 + 0.25 = 1.61VE0

cell = 1.61V

*22. Write the cell reaction and calculate theemf of the cell,

2+2 2Pb(s)|Pb (1M)||KCl(sat)|Hg Cl (s)|Hg

0 0anode cathodeE = – 0.126V, E = 0.242V

Identify anode and cathode. Name the righthand side electrode.

Given :E0

Pb = –0.126V, E0Cl = 0.242 V

To find :Cell reaction = ?Right hand side electrode = ?

Solution :2+ –Pb(s) Pb (1M) + 2e⎯⎯→

...(Oxidation at anode)– –

2 2Hg Cl (s) + 2e 2Hg( ) + 2Cl (sat)l⎯⎯→


2 2Pb(s)+Hg Cl (s) ⎯⎯→

2+ –Pb (1M)+2Hg( )+ 2Cl (sat)l

...(overall reaction)Right hand electrode is cathodeE0

cell = E0Cl – E0

Pb

Cathode – Anode= 0.242V – (–0.126V)

E0cell = 0.242 + 0.126 = 0.368VE0

cell = 0.368V

298


*23. The following redox reaction occurs in agalvanic cell.

2+ 3+2Al(s)+3Fe (1M) 2Al (1M)+3Fe⎯⎯→

(a) Write the cell notation.(b) Identify anode and cathode.(c) Calculate E0

cell if E0

anode

= –1.66V and E0cathode = –0.44V

(d) Calculate ΔGº for the reaction.Given :

E0Al = –1.66V, E0

Fe = –0.44VTo find :

i) E0cell = ?

ii) ΔG = ?Solution :

i) 3+ 2+Al(s) | Al (1M) || (1M) Fe | Fe(s)ii) Al electrode is anode and Fe electrode is

cathodeiii) E0

cell = E0Fe – E0

Al

= cathode – anode= –0.44V – (–1.66)

E0cell = 1.22(V)

(iv) From equation, n = 6 mol e–

ΔGº = – nFE= –6(mol e–) × 96500 (C/mol e–) × 1.22V= –706380 (VC) = –706380J= –706.38 kJ

(The emf of the cell is positive, thereaction is spontaneous)

*24. Calculate the potential of the following cellat 25ºC.

2+ +Sn(s)|Sn (0.025M)||Ag (0.015M)Ag(s)

E0Sn = –0.136V,

E0Ag = 0.799V

Given :E0

Ag = 0.799V, E0Sn = –0.136V

To find :E0

cell = ?Nernst equation = Ecell = ?

Solution :E0

cell = E0Ag – E0

Sn

= (cathode) (anode)= 0.799V – (–0.136V)

E0cell = 0.799V + 0.136 = 0.935V

E0cell =

– 2+0

10– + 2

0.0592(Vmole ) (Sn )E cell – log

2(mole ) (Ag )...because n = 2mol e–

= 102

0.0250.935V – 0.0296(V) × log(0.015)

= 100.0250.935(V) – 0.0296(V) ×log

0.000225= 0.935V – 0.0296(V) × log10 111.111= 0.935V – 0.0296(V) × 2.0457= 0.935V – 0.0605V

∴∴∴∴∴ Ecell = 0.8745V

*25. Consider the following redox reaction

2+ 2+Mg(s) + Sn (aq) Mg (aq) + Sn(s)⎯⎯→

(a) Write the cell formula(b) Calculate Ecell for the reaction at

25ºC if[Mg2+] = 0.035M, [Sn2+] = 0.025ME0

Mg = –2.37Vand E0

Sn = –0.136V(c) Calculate ΔG for the cell reaction.

Given :2+ 2+Mg(s)+Sn (aq) Mg (aq) +Sn(s)⎯⎯→

To find :i) Cell formula = ? ii) E0

cell = ?iii) Ecell = ? iv) ΔG = ?

299


Solution :

i) 2+ 2+Mg(s)|Mg (0.035M)||(0.025M)Sn |Sn(s)

ii) E0cell = E0

Sn – E0Mg

Cathode Anode= –0.136V – (–2.37V)= –0.136V + 2.37V

E0cell = 2.234V

iii) Ecell = – 2+

0cell 10– 2+

0.0592(V mole ) (Mg )E – .log

2(mol e ) (Sn )

n = 2

= 100.0352.234V – 0.0296(V) × log0.025

= 2.234V – 0.0296(V) × log10 1.4= 2.234V – 0.0296(V) × 0.1461= 2.234V – 0.00432V

E0cell = 2.229V

iv) ΔG = –nFE= –2(mol e–) × 96500 (C/mol e–)

× 2.229V= –430197 (VC) = –430197J

= –430.197kJΔG= –430.197kJ

*26. Write the cell reaction and calculate theemf of the cell at 25ºC.

3+ 2+Cr(s) | Cr (0.0065M) || Co (0.012M) | Co(s)

E0Co = –0.280V, E0

Cr = –0.74VWhat is ΔG for the cell reaction?

Given :

3+ 2+Cr(s)|Cr (0.0065M)||Co (0.012)M|Co(s) E0

Co = –0.280V, EºCr = –0.74VTo find :

i) Cell reaction = ? ii) Ecell = ?iii) ΔG = ?

Solution :3+Cr(s) Cr (0.0065M) + 3e–] × 2⎯⎯→

2+ –Co (0.012)M + 2e Co(s) ] × 3⎯⎯→

2+ 3+2Cr(s) + 3Co 2Cr + 3Co(s)(0.012M) (0.0065M)

⎯⎯→

E0cell = E0

Co – E0Cr

= cathode anode= –0.280V – (–0.74V)= –0.280V + 0.74V

E0cell = 0.460V

Ecell=–

0cell –

0.0592(V mol e )E –

6(mole )3+

10 2+

[Cr ]× log

[Co ]Ecell = 0.460V – 0.00986(V)

× 2

10 3

(0.0065)log

(0.012)

= 0.460V – 0.00986(V)–5

10 –5

4.225 × 10× log

0.1728 × 10= 0.460V – 0.00986(V)

× log10 24.450= 0.460V – 0.00986(V) × 1.3883= 0.460V – 0.01368(V)

Ecell = 0.4463VΔG = –nFE

= 6(mol e–) × 96500(C/mol e–)× 0.4463V

= –258407.7 (VC)ΔG = –258407.7J = –258.407kJ

*27. Construct a cell consisting of Ni2+ | Ni halfcell and H+ | H2 (g, 1atm) | Pt half cell.

(i) Write the cell reaction.(ii) Calculate emf of the cell if [Ni2+]

= 0.1M, [H+] = 0.05Mand EºNi = –0.257

300


Given :(i) Half cell : Ni2+ | Ni, H+ | H2 (g, 1atm)|Pt(ii) [Ni2+] = 0.1M, [H+] = 0.05M,

E0Ni = –0.257V

To find :(i) Cell reaction = ?(ii) E0

cell = ?(iii) Ecell = ?

Solution :Ni(s) | Ni2+(0.1M) || (0.05M)H+

| H2(g, 1atm) | PtE0

Ni = –0.257V

E0cell = 0

H2E – E0

Ni

= cathode Anode= 0.00 – (–0.257 V)

E0cell = 0.257V

2+ –Ni(s) Ni (0.1M) + 2e⎯⎯→

...(Oxidation at anode)+ –

22H (0.05M) + 2e H (g, 1atm)⎯⎯→

Ecell =

2+–H 20

+ 2×

[Ni ]×P0.0592(Vmole )– logE 10cell

n [H ]

Ecell =–

0cell 10– 2

0.0592(Vmole ) 0.1 ×1– × logE

2(mol e ) (0.05)

+Ni(s) + 2H (0.05M) ⎯⎯→

2+2Ni (0.1M) + H (g, 1atm)


Ecell = 100.10.257V – 0.0296(V) × log

0.0025

= 0.257V – 0.0296 (V) × log10 40= 0.257V – 0.0296 (V) × 1.6021= 0.257V – 0.04742VEcell = 0.2095V

*28. Calculate the equilibrium constant for theredox reaction at 25ºC,

2+ 2+Sr(s)+ Mg (aq) Sr (aq)+ Mg(s)⎯⎯→

that occurs in a galvanic cell. Write the cellformula.E0

Mg = –2.37V and E0Sr = –2.89V

Given :

2+ 2+Sr(s) +Mg (aq) Sr (aq)+ Mg(s)⎯⎯→

E0Mg = –2.37V and E0

Sr = –2.89VTo find :

(i) E0cell = ?

(iii) K = ? (equilibrium constant)Solution :

2+ 2+Sr(s) + |Sr (aq) ||Mg (aq)| Mg(s)

E0Mg = –2.37V and E0

Sr = –2.89VE0

cell = E0Mg – E0

Sr

= cathode Anode= –2.37V – (–2.89 V)

E0cell = 0.52V

Ecell = 100.0592 log K

n

log10K =0.52 × 20.0592

= 17.56K = Antilog (17.56)K = 3.690 × 1017

*29. Using Nernst equation, calculate thepotentials for the following halfreactions :

(a) – – –2I (s) + 2e 2I , [I ] = 0.03M⎯⎯→

0I2

E = 0.535V

(b) 3+ – 2+Fe (aq) +e Fe (aq),⎯⎯→

301


2+[Fe ]=0.1M, [Fe3+] = 0.01M,

02+ 3+Fe , Fe

E = 0.771V

Given :

(i) – – –2I (s) + 2e 2I , [I ] = 0.03M⎯⎯→

0I2

E = 0.535V

(ii) 3+ – 2+Fe (aq) + e Fe (aq),⎯⎯→

2+[Fe ]=0.1M, [Fe3+] = 0.01M,

02+ 3+Fe ,FeE = 0.771V

To find :

(i) –I /I 2E

= ?

(ii) 2+ 3+Fe , FeE = ?

Solution :(i) The Nernst equation for electrode potential

is :

Eel = –

0el 10

0.0592(Vmole ) [products]E – log

n [reactants]

–I /I2E =

–0 – 2

– 10I /I2

0.0592(V mole )E – .log [I ]

n

0I2

E = 0.535V, n = 2 mol e–, [I–] = 0.03M

–I /I 2E =

–2

10–0.0592(V mole )

0.535V – .log (0.03)2(mole )

=0.535V – 0.0296(V) . log10 9 × 10–4

=0.535V + 4 × 0.0296(V) × 0.9542=0.535V + 0.1129V

I/I2E = 0.6479V

(ii) 3+ – 2+Fe (aq) + e Fe (aq)⎯⎯→

2+ , 3+Fe FeE = 0

2+ , 3+Fe FeE

– 2+

– 3+

0.0592(V mole ) (Fe )– log

1(mole ) (Fe )

03+2+ FeFe

E , = 0.771V, (Fe3+) = 0.01M,

(Fe2+) = 0.1M

2+ 3+Fe FeE ,

=0.0592(V) 0.10.771V – . log10

1 0.01

= 0.771V – 0.0592(V) × log10 10= 0.771(V) – 0.0592(V) × 1

2+ 3+Fe FeE , = 0.7118V

*30. Consider a galvanic cell that uses the halfreactions.

+ –22H (aq) + 2e H (g)⎯⎯→

2+ –Mg (aq) + 2e Mg(s)⎯⎯→

Write balanced equation for the cellreaction. Calculate E0

cell, Ecell and ΔGº

if concentrations are 1M each and H 2P =

10 atm.E0

Mg = –2.37VGiven :

(i) + –22H (aq) + 2e H (g)⎯⎯→ ,

(ii) 2+ –Mg (aq) + 2e Mg(s)⎯⎯→ ,

0Mg H 2

E = –2.37V, P =10atm

To find :(i) E0

cell= ?(ii) Ecell= ?(iii) ΔGº = ?

Solution :

(i) + –22H (aq) + 2e H (g)⎯⎯→

302


(ii) 2+ –Mg (aq) + 2e Mg(s)⎯⎯→

Reverse equation (ii) and then add in eq. (i)2+ –Mg(s) Mg (aq) + 2e⎯⎯→

+ –22H (aq) + 2e H (g)⎯⎯→

+ 2+2Mg(s)+2H (aq) Mg (aq)+H (g)⎯⎯→

...(Balanced equation)

E0cell = 0

H2E + E0

Mg

cathode Anode= 0.00 – (–2.37V)

E0cell = 2.37V

Ecell

=2+–

H2cell 10 + 2

[Mg ]×P0.0592(V mole )Eº – .log

n [H ]

=–

10 2

0.0592(V mole ) (1)2.37V – log ×10

2 (1)= 2.37V – 0.0296(V) × log10 10

Ecell = 2.34VΔGº = –nFEº

= –2(mol e–) × 96500 (C/mole–)× 2.37(V)

= –457410 (VC)ΔGº = –457410J = –457.410 kJ

*31. Calculate E0cell, ΔGº and equilibrium

constant for the reaction.+ 2+2Cu (aq) Cu (aq) + Cu(s)⎯⎯→

0+Cu /CuE = 0.52V, and

0+ 2+Cu , Cu

E =

0.16V

Given :0

+Cu /CuE = 0.52V, 0

+ 2+Cu , CuE = 0.16V+ 2+2Cu (aq) Cu (aq) + Cu(s)⎯⎯→

To find :i) E0

cell = ? ii) ΔGº = ?iii) K = ? (Equilibrium constant)

Solution :The standard potential of Cu+ | Cu isgreater. Hence, this electrode is cathode.E0

cell = E0cathode – E0

anode

= 0.52V – 0.16VE0

cell = 0.36VΔGº = –nFEº

= –1(mol e–) × 96500 (C/mol e–)× 0.36V

= –34740(VC)ΔGº = –34740J = –34.740kJ

E0cell = 10

0.0592 . log Kn

0.36(V)= 100.0592 . log K

1

log10K = 0.36 × 10.0592

= 6.081

∴∴∴∴∴ K = antilog (6.081)K = 1.205 × 106

*32. Calculate the potential of the following cellat 25ºC.Zn | Zn2+(0.6M) || H+ (1.2M) | H2

(g, 1 atm) | PtE0

Zn = –0.763VGiven :

Zn | Zn2+(0.6M) || H+ (1.2M) | H2(g, 1 atm) | Pt,E0

Zn = –0.763VTo find :

Ecell = ?Solution :

Zn2+ | Zn half cell has lower potential thanhydrogen gas electrode, whose standardpotential is zero. Hence, hydrogen gas electrodecathode and Zn2+ | Zn electrode will be anode.

2+ –Zn(s) Zn (0.6M) + 2e⎯⎯→

...Oxidation at anode

303


+ –22H (1.2M) + 2e H (g, 1atm)⎯⎯→

...reduction at cathode+Zn(s) + 2H (1.2M) ⎯⎯→

2+2Zn (0.6M) + H (g, 1atm)

...(overall cell reaction)0 0 0cell znH2

E = E – E = 0.00 – (–0.763V) =

0.763VThe emf of the cell is given by nernstequation,Ecell

=2+

H0 2cell 10 + 2

[Zn ] × P0.0592(V mole)E – log

n [H ]

n = 2 mol e–, [Zn2+] = 0.6M, [H+] = 1.2M,

H 2P = 1 atm

Ecell

=–

10– 2

0.0592(V mole ) 0.6×10.763(V) – log

2(mole ) (1.2)

Ecell= 0.763(V) – 0.0296(V) × (–0.3803)= 0.763(V) + 0.01125

Ecell = 0.7742(V)

*33. A constant electric current flows for 4hours through two electrolytic cellsconnected in series. One contains twoelectrolytic cells connected in series. Onecontains AgNO3 solution and secondcontains CuCl2 solution. During this time4 grams of Ag are deposited in the firstcell.

(i) How many grams of Cu are deposited inthe second cell?

(ii) What is the current flowing in amperes?Given :

mass of Ag deposited = 4gmolar mass of Cu = 63.5 g mol–1.molar mass of Ag = 107.9 g mol–1

t = 4 hrs = 4 × 60 × 60 = 14400 sTo find :

mass of Cu deposited (WCu) = ?I = ?

Solution :(i) No. of moles of Ag deposited

=mass of Ag

molar mass of Ag = 4

107.9= 0.03707 mol of Ag

Reactions

(a) + –Ag + e Ag⎯⎯→

(half reaction in AgNO3 cell)

(b) 2+ –Cu + 2e Cu⎯⎯→

(half reaction is CuCl2 cell)mole ratio of Ag

=moles of Ag produced

No. of moles of electrons = 1

1 = 1

mole ratio of Cu = 1

= 0.52

moles of Cu produced

moles of Ag produced = mole ratio of Cu

mole ratio of Ag

moles of Cu produced

=mole ratio of Cu

mole ratio of Ag × moles of Ag produced

=0.5

× 0.037071

mass of Cu produced = 0.01854 × 63.5= 1.177gmass of Cu produced = 1.177g

b) From the reaction,∵ 1 mol Ag+ requires = 1 mole electrons∴∴∴∴∴ 0.03707 mol Ag will require = 0.03707 mol Ag∵ 1 mol electrons = 1 Faraday∴∴∴∴∴ 0.03707 mol electron = 0.03707 Faraday∵ 1 Faraday = 96500C∴∴∴∴∴ 0.03707 Faraday = 0.3707 × 96500

304


= Quantity of electricity (Q) 3577CQ = I × t

∴∴∴∴∴ I = 3577

14400 = 0.25A

Current passed = 0.25A

*34. The passage of 0.95A for 40 minutesdeposited 0.7493g of Cu from CuSO4

solution. Calculate the molar mass of Cu.Given :

I = 0.95A, t = 40 min = 40 × 60 = 2400smass of Cu deposited = 0.7493g

To find :molar mass of Cu = ?

Solution :Reduction half reaction;

2+ –Cu (aq) + 2e Cu(s)⎯⎯→

Q = I × t= 0.95 × 2400= 2280C

no. of moles of electron =2280

96500= 0.02362 mol

∵ 2 mol electron deposit = 1 mol Cu∴∴∴∴∴ 0.02362 mol electron will deposit

= 0.02362

2 = 0.01181 mol Cu

Now,0.01181 mol Cu weighs = 0.7493g

∴∴∴∴∴ 1 mol Cu will weigh = 0.7493 × 1

0.01181 = 63.44g

Thus, molar mass of Cu = 63.44 g mol–1.Molar mass of Cu = 63.44 g mol–1.

*35. A quantity of 0.3 g of Cu was depositedfrom CuSO4 solution by passing 4A throughthe solution for 3.8 min. Calculate the valueof Faraday constant.

Given :mass of Cu deposited = 0.3gI = 4At = 3.8 min = 3.8 × 60 = 228s

To find :Value of Faraday = ?

Solution :Q = I × t

= 4 × 228 = 912CReduction half-reaction;

2+ –Cu (aq) + 2e Cu(s)⎯⎯→

No. of moles of Cu deposited

= 0.3

63.5 = 0.004724 mol

From reduction half reaction;1 mol Cu = 2 mole electrons

∴∴∴∴∴ 0.004724 and Cu = 2 × 0.004724= 0.009448 mol electrons

Now,∵ 0.009448 mol electron = 912C

∴∴∴∴∴ 1 mol electrons = 912

0.009448 = 96528C

∵ 1 Faraday charge is equal to charge on 1mol electrons

∴∴∴∴∴ 1 Faraday = 96528CValue of Faraday = 96528C/mole–

*36. Set up the cell consisting of H+(aq)| H2(g)and Pb2+(aq) | Pb(s) electrodes. Calculatethe emf at 25ºC of the cell if [Pb2+] = 0.1M,[H+] = 0.5M and hydrogen gas is at 2atmpressure.E0

Pb = –0.126V.Given :

+ 2+2H (aq) | H (g) and Pb (aq) | Pb(s) ,

E0Pb = –0.126V

305


[Pb2+] = 0.1M, [H+] = 0.5M,

H 2P = 2 atm pressure

To find :(i) Ecell = ?

Solution :Pb(s)|Pb2+(aq)(0.1M)||(0.5M)H+(aq) |H2(g, 2atm)Pt

0PbE = –0.126V

E0cell = 0

H 2E – 0

PbE(cathode) (Anode)

= 0.00 – (–0.126V)E0cell = +0.126V

Ecell= 2+– H0

10 + 22

[Pb ]× P0.0592(V mole )E cell – log

n [H ]i

n = 2 mol e– . [Pb2+] = 0.1M,[H+] = 0.5M,

H 2P = 2 atm

Ecell = –

– 20.0592(Vmole ) 0.1×20.126(V)– .log10

2(mole ) (0.5)

Ecell= 0.126(V) – 0.0296(V) . log10 0.8= 0.126 (V) – 0.0296(V) × (–0.0969)= 0.126 (V) + 0.0029(V)

Ecell =0.1289V

*37. Write balanced equations for the halfreactions and calculate the reductionpotentials at 25ºC for the following halfcells :

i) C1– (1.2M) | Cl2 (g, 3.6atm) E0 = 1.36Vii) Fe2+ (2M) | Fe(s) E0 = –0.44V

Given :Cl– (1.2M) | Cl2 (g, 3.6 atm) E0 = 1.36VFe2+ (2M) | Fe(s) E0= –0.44V

To find :

i) –Cl /Cl2E = ? ii) 2+Fe /Fe

E = ?

Solution :

(i) – –22Cl (1.2M) Cl (g,3.6atm) +2e⎯⎯→

... Eº = 1.36VThe Nernst equation for electrode potentialis,

Eel = –0.0592(Vmole ) products0E – log 10el n reactants

⎛ ⎞⎜ ⎟⎝ ⎠

⋅

–Cl /Cl2E =

–Cl0 2

– 10 2Cl /Cl –2

p0.0592(V mole )E – .log

n Cl⎡ ⎤⎣ ⎦

0–Cl /Cl2

E = 1.36V, n = 2 mol e–,

[Cl–] = 1.2M0

–Cl /Cl2E

=–

10– 2

0.0592(V mole ) 3.501.36(V) – ×log2 (mol e ) (1.2)

= 1.36(V) – 0.0296(V) × log10 2.4305= 1.36(V) – 0.02966 (V) × 0.3856= 1.36 (V) – 0.01143= 1.348 V

ii) 2+ –Fe (2M) + 2e Fe(s)⎯⎯→

... Eº = –0.44V

E = 100.0592 1Eº – log

2 2

= –0.44 – 0.02966 × log.10 0.5= 0.44 – 0.02966 × (–0.3010)= 0.44 + 0.00892

E = –0.431 V

*38. Device galvanic cell for each of thefollowing reactions and calculate ΔGº forthe reaction.

306


i) Zn dissolves in HCl to produce Zn2+ andH2 gas E0

Zn = –0.763Vii) Cr dissolves in dil. HCl to produce Cr3+

and H2 gas E0Cr = –0.74V

Given :Zn dissolves in HCl to produce Zn2+ andH2 gas ,EºZn = –0.763VCr dissolves in dil. HCl to produce Cr3+

and H2 gas,EºCr = –0.74V

Solution :i) Zn(s) | Zn2+(aq) || H+(aq) | H2(g) || Pt

E0Zn = –0.763V

E0cell = 0

H2E – 0

ZnEcathode Anode

= 0.00 – (–0.763V)E0cell = +0.763VΔGº = –nFEº

= –2 × 96500 × 0.763= 147259 JΔGº= 147.259 kJ

(ii) Cr(s) | Cr3+(aq) || H+(aq) | H2(g) | PtE0

Cr = –0.74V

E0cell = 0

H2E – 0

CrE= cathode Anode= 0.00 – (–0.74V)

E0cell = +0.74V

ΔGº = –nFE0

= –6 × 96500 × 0.74= –428460 JΔGº= –428.460 kJ

*39. The equilibrium constant for the followingreaction at 25ºC is 2.9 × 109. Calculatestandard voltage of the cell.

– –2 2Cl (g) + 2Br (aq) Br (l) + 2Cl (aq)

Given :(i) K = 2.9 × 109,

– –2Cl (g) + 2Br (aq) Br( ) +2Cl (aq)l

To find :(i) E0

cell = ?Solution :

– –2 2Cl (g) + 2Br (aq) Br ( )+ 2Cl (aq)l

n = 2, k = 2.9 × 109

E0cell = 10

0.0592 . log kn

E0cell =

910

0.0592 . log (2.9 × 10 )2

E0cell =

0.0592 × 9.46242

E0cell = 0.280V

307


NUMERICALS FOR PRACTICETYPE - I - MOLAR CONDUCTIVITY AND CELL CONSTANT

*1. The distance between the electrodes of a conductivity cell is 0.98 cm and area of cross sectionof the electrode is 1.3 cm2. Calculate the cell constant for the cell. [Ans. 0.7538 cm–1]

*2. The conductivity of 0.005M NaI solution at 25ºC is 6.065 × 10–4 Ω–1. Calculate its molarconductivity. [Ans. 121.3 Ω–1cm2 mol–1]

*3. The molar conductivity of 0.002 M HCl solution is 407.2 Ω–1cm2 mol–1 at 25ºC. Calculateits conductivity. [Ans. 8.144 × 10–3 Ω–1cm1]

*4. A conductivity cell dipped in 0.0005 M NaCl gives at 25ºC a resistance of 13710 ohms. Ifthe electrodes of the conductivity cell are 0.7 cm apart and the area of cross section of theelectrode is 0.82 cm2, what is the molar conductivity of the solution at 25ºC?

[Ans. 124.6 Ω–1cm2 mol–1]

*5. A conductivity cell dipped in 0.01 M AgNO3 solution gives at 25ºC a resistance of 1442 ohms.If the molar conductivity of 0.01m AgNO3 solution is 124.8 Ω–1cm2 mol–1, what is the cellconstant? [Ans. 1.8 cm–1]

*6. A conductivity cell filled with 0.01 M KCl gives at 25ºC a resistance of 484 ohms. The conductivityof 0.01 M KCl at 25ºC is 0.00141Ω–1cm–1. 0.001M of NaCl solution when filled in the samecell gives a resistance of 5490 ohms at 25ºC. Calculate the molar conductivity of NaCl solution.

[Ans. 124.3 Ω–1cm2 mol–1]

TYPE - II - KOHLRAUSCH LAW

*7. The molar conductivities of potassium acetate, HBr and KBr are respectively 114.4Ω–1cm2

mol–1, 428.2 Ω–1cm2 mol–1 and 151.9 Ω–1cm2 mol–1 at zero concentration. Calculate the molarconductivity of acetic acid at zero concentration. [Ans. 390.7 Ω–1cm2 mol–1]

*8. The molar conductivity of 0.01 M solution of acetic is 16.6 Ω–1cm2 mol–1. If its molar conductivityat infinite dilution is 390.7 Ω–1cm2 mol–1, what is its degree of dissociation is 0.01 M solution?Calculate dissociation constant of acetic acid. [Ans. 1.89 × 10–5]

TYPE - III - FARADAY’S LAWS

*9. During electrolysis of molten CaCl2, 0.8 A current is passed through the cell for 1 h. Calculatethe mass of product formed at cathode. (Molar mass of Ca = 40 g mol–1)

[Ans. 0.5968 g]

*10. How many Faradays of electricity are required to produce 5 g of Mg from MgCl2? (molarmass of Mg = 24 g mol–1) [Ans. 0.4166 F]

308


*11. In the electrolysis of AgNO3 solution 0.42 g of Ag is produced at cathode after a certain periodof time. Calculate the quantity of electricity required. (Molar mass of Ag = 108 g mol–1)

[Ans. 375.4 C]

*12. Calculate the mass of copper metal produced during the passage of 2.5A of current througha solution of CuSO4 for 40 minutes. Molar mass of Cu is 63.5 g mol–1. [Ans. 1.975 g.]

*13. What current strength will be required to deposit 2.692 × 10–3 kg of Ag in 7 minutes fromAgNO3 solution? Molar mass of Ag is 108 × 10–3 kg mol–1. [Ans. 5.72 A]

*14. How much time will be required to liberate 3 × 10–2 kg of iodine from KI solution by thepassage of 4A current through it? Molar mass of iodine is 127 g mol–1. [Ans. 1.583 h]

*15. Calculate the amount of Cu and Br2 produced in 1.5 hours at inert electrodes in a solutionof CuBr2 by a current of 3.5A. Molar masses of Cu and Br2 are 63.5 g mol–1 and 159.8g mol–1 respectively. [Ans. 15.65 g]

*16. Write half reactions for the electrolysis of molten BaCl2. How many grams of Ba are producedby current of 1A passed for 15 min? What volume of Cl2 gas will be liberated at 298K and1 atm pressure? R = 0.08205 L. atm K–1 mol–1. Molar mass of Ba is 137 g mol–1.

[Ans. 0.114 L.]

*17. In a certain electrolysis experiment 1.55 g of Ag were deposited in one cell containing aqueousAgNO3 solution. Calculate the mass of Zn deposited in another cell containing aqueous ZnSO4

solution in series with AgNO3 cell. Molar masses of Zn and Ag are 65.4 g mol–1 and 107.9g mol–1 respectively. [Ans. 0.47g.]

*18. In a certain electrolysis experiment 0.79g of Ag was deposited in one cell containing aqueousAgNO3 solution, while 0.231 g of an unknown metal X was deposited in another cell containingaqueous XCl2 solution is series with AgNO3 cell. Calculate the molar mass of X. Molar massof Ag is 107.9 g mol–1. [Ans. 63.11 g mol–1]

*19. In an electrolysis experiment 0.18 g of Al was deposited in a cell containing aqueous AlCl3

solution. Calculate the mass of Cu deposited in another cell containing aqueous CuSO4 solution,in series with AlCl3 cell. Molar masses of Al and Cu are 27 g mol–1 and 65.4 g mol–1 respectively.

[Ans. 0.654 g]

TYPE - IV - CELL POTENTIAL

*20. Write the cell reaction and calculate the standard emf of the cell.

Cd | Cd2+ (1M) || Ag+(1M) | Ag E0Cd = –0.403V and E0

Ag = 0.799V [Ans. 1.202V]

*21. Construct a cell from Ni2+ | Ni and Cu2+ | Cu half cells. Write the cell reaction and calculatethe standard potential of the cell. [Ans. 0.573V]

309


*22. The standard emf of the following cell is 0.463V Cu | Cu2+(1M) || Ag+(1M) | Ag

If the standard potential of Cu electrode is 0.337V, what is the standard potential of Ag electrode?

[Ans. 0.800V]

*23. Calculate equilibrium constant for the reaction.

+ 2+Ni(s) + 2Ag (aq) Ni (aq) + 2Ag(s)⎯⎯→ at 25ºC EºNi = –0.25V and EºAg = 0.799V

[Ans. 2.754 × 1035]

*24. Predict whether Cd2+(aq) ion can oxidize Zn(s) or Cu(s) under standard state conditions. Eºvalues for half reactions of Cd2+, Zn(s) and Cu(s) are –0.403V,–0.763V and 0.337V respectively.

[Ans. It means Cd2+ ion cannot oxidize metallic Cu]

TYPE - V - NERNST EQUATION

*25. Construct a cell containing Zn2+ | Zn half cell and hydrogen gas electrode. What will be the

emf of the cell at 25ºC if [Zn2+] =0.24M, [H+] = 1.6M and H 2P = 1.8 atm? EºZn = –0.763V.

[Ans. 0.786V]

*26. Calculate Eºcell, Ecell and ΔG for the following reaction at 25ºC.

2+ 2+Mg(s) + Sn (0.003M) Mg (0.04M) + Sn(s)⎯⎯→

EºMg = –2.37V and EºSn = – 0.14V [Ans. –429.6 kJ.]

*27. Write the cell reaction and calculate the emf of the cell.

Pt | H2(g, 1atm) | H+(0.5M) || KCl (1M) | Hg2Cl2(s) | Hg(l)

at 25ºC. Ecalomel = 0.28V. [Ans. 0.2978V]

*28. Calculate the emf of the cell Zn(s) | Zn2+(0.008M) || Cr3+ (0.01M) | Cr

at 25ºC, EºZn = –763V, EºCr = –7.4V [Ans. 0.0456]

*29. Using Nernst equation, calculate the potentials of the following half reactions at 25ºC.

(a) – –2Br (l) + 2e 2Br (aq)⎯⎯→ [Br–] = 0.01M and

0Br

2E = 1.08V

(b) 2+ – +Cu (aq) + e Cu (aq)⎯⎯→ [Cu+] = 0.1M, [Cu2+] = 0.05M, 0+ 2+Cu Cu

E , = 0.153V

[Ans. 0.1352V]

310


HIGHER ORDER THINKING SKILLS (HOTS)1. What would happen if no salt bridge is used in an electrochemical cell such as Zn-Cu cell?

Ans. The metal ions (Zn2+ ions) which are produced by the loss of electron are accumulated inone electrode and the negative ions (SO2–

4 ions) are accumulated in the other electrode. Thus,both the solution will develop charges and the current will stop. Further, inner circuit is alsonot completed.

2. What is the free energy change for(a) Galvanic cell

Ans. For galvanic cell, ΔG < 0, since redox reaction is spontaneous.(b) Electrolytic cell

Ans. (b) For electrolytic cell, ΔG > 0, since redox reaction is non-spontaneous.3. The standard reduction potential values of three metallic cations X, Y, Z are 0.52, –3.03,

–1.18V respectively. What will be the order of reducing power of the corresponding metals.Ans. (a) The standard oxidation potentials of the metals X, Y, Z would be –0.52, + 3.03 and +

1.18V respectively.(b) Higher is the oxidation potential, more easily the metal is oxidised and therefore, greater

is the reducing power.(c) Hence, the reducing powers will be in the order Y > Z > X.

4. State the products of electrolysis obtained on the cathode and the avode when a dilute solutionof H2SO4 with platinum electrodes is electrolysed.

Ans. + 2–2 4 4H SO (aq) 2H (aq) + SO (aq)⎯⎯→

+ –2H O H + OH

At cathode : + –2H + e H; H + H H (g)⎯⎯→ ⎯⎯→

At anode : 2 2OH OH + e ; 4OH 2H O( ) + O (g)l⎯⎯→ ⎯⎯→Hence, H2 (g) is liberated at the cathode and O2 (g) is berated at the anode.

5Introduction :There are three most important characteristics of chemical reactions : position of equilibrium,

feasibility of chemical reactions and rates of reactions.

Chapter

SYLLABUS5.1 Rate of Reaction (Average rate and

instantaneous rate)

(a) Chemical kinetics

(b) Rate of reaction

(c) Average rate of reaction

(d) Instantaneous rate of reaction


Theoretical MCQs

Numerical MCQs

Advanced MCQs

5.2 Rate law, Rate constant and order of

reaction and molecularity

(a) Rate law

(b) Rate constant or specific reaction rate

constant (k)

(c) Order of a reaction

(d) Reaction with no order

(e) Elementary reaction

(f) Complex reaction

(g) Rate determining step

(h) Reaction intermediate

(i) Molecularity

(j) Distinquish between order and molecularity


Theoretical MCQs

Numerical MCQs

Advanced MCQs

5.3 Integrated rate laws

(a) Integrated rate laws

(b) Zero order reaction

(c) First order reaction

(d) Pseudo first order reaction

(e) Experimental determination of rate laws

and order


Theoretical MCQs

Numerical MCQs

Advanced MCQs

5.4 Collision theory, activation energy and

catalyst

(a) Collision theory requirements for a

bimolecular chemical reaction to take place

Chemical Kinetics

"If I have seen further than others, it is by standing upon the shouldersof giants"-Isaac Newton

312


In unit 3 of the present text, we have seen how thermodynamic properties, ΔS and ΔG tell

us whether a chemical reaction represented by an equation can occur under the given set of

conditions. The reactions between strong acids and strong bases are thermodynamically favoured

but occur with rapid rates, the study of reaction rates helps to predict how quickly a reaction

mixture approaches equilibrium. Another view to the study of rates gives an information regarding

the mechanisms of chemical reactions. Many reactions occur into a sequence of steps. These

steps can be discovered only after the study of rates at which the reactions take place.

Therefore, chemical kinetics is a branch of science that deals with the study of rates of chemical

reactions, the factors affecting these rates and the mechanisms by which the reactions occur.

In chemical industries the chemical processes are profitable if reactions rates are fast.

(b) Arrhenius equation for collision theory of

bimolecular reactions

(c) Temperature dependence of reaction rates

(d) Arrhenius equation and temperature variation

(e) Arrhenius equation and energy of activation

(f) Determination of activation energy (Ea)

(g) Effect of catalyst on rates of reactions

(h) Effect of catalyst

(i) Difference between catalyst and reaction

intermediate


Theoretical MCQs

Numerical MCQs

Hours before exam

Numericals with Solution

Numericals for Practice

Higher Order Thinking Skills

313

Chapter - 5 Chemical Kinetics

5.1. RATE OF REACTION (Average rate and instantaneous rate) : Concept Explanation :

(a) Chemical kinetics :1. Definition : “Chemical kinetics is a branch of science that deals with the study of

rates of reactions, the factors affecting these rates and the mechanisms by whichthe reactions occur.”

2. Importance of chemical kinetics :(a) The study of reaction rates helps to predict how quickly a reaction mixture approaches

equilibrium.(b) The study of reaction rates gives information regarding the mechanisms of chemical

reactions.(c) Chemical kinetics helps us to determine conditions by which reaction rates could

be altered.(d) It helps us to study the effect of various factors like temperature, pressure, catalyst

and concentration of reactants on the rate of reaction.

(b) Rate of reaction :1. Definition : “The change in molar concentration of any one of the reactants or products

per unit time is called rate of reaction”.2. Explanation : Consider a simple hypothetical reaction,

A → BIn above reaction, the concentration of reactant A decreases and that of product B increaseswith respect to time.

Rate of disappearance (consumption) of A =Decreasein molar conc.of rectant A

Time taken for decrease

=[A]

–t

∆

∆

OR

Rate of formation of B =Increasein molar conc.of product B

time taken for increase

=[B]

+tΔ

∆

Because the concentration of A decreases, ∆[A] is a negative quantity. Therefore, thenegative sign in the expression for rate of disappearance of the reactant makes the ratea positive quantity.

314


In this reaction, the rate of decrease in concentration of reactant, A is same as the rateof increase in concentration of product B. Therefore,

Rate of reaction = [A]

–t

∆

∆ =

[B]

t

∆

∆

In some chemical reactions, the coefficient of reactants and products are not same forexample,A + B → 2CIn this reaction the rate of disappearance of A and B is same but the rate of appearanceof C is twice as compared to A and B.

∴ 2 × rate of = 2 × rate of = rate ofdisappearance of A disappearance of B appearance of C

OR

Rate of Reaction = – [A]

t

∆

∆ =

– [B]

t

∆

∆ =

1

2

[C]

t

∆

∆

3. Units of rate of reaction :Unit of Rate of Reaction for liquid state reaction, can be given as,

Rate of reaction = Change in concentration

timeSince the concentration of reactants or products are normally expressed in moles per litre(Mole/dm3) and time is expressed in minutes or seconds.Thus, the unit of rate of reaction isMol L−−−−−1. time −−−−−1. Or M time−−−−−1. (Where M = molarity)For gaseous reactions, the concentration of reactants and products are expressed in termsof partial pressures. Therefore, in such cases the unit of rate of reaction willbe atm. time-1.

4. Factors affecting the rate of reaction.The rate of reaction depends upon the following factors;(a) Concentration of reactants : Greater the concentration of the reactants greater

is the rate of reaction.(b) Temperature : Increase in temperature increases the rate of reaction.(c) Catalyst : Catalyst generally increases the rate of reaction.(d) Nature of reactants : Reactants in gaseous state react faster as compared to

liquid or solid reactants. In case of solids, powdered solids react faster than solidsin lump form.

(e) Pressure : In case of gaseous reactants an increase in pressure increases the rateof reaction.

315


(c) Average rate of reaction :1. Definition : The average rate of a reaction is defined as the change in concentration

of a reactant or product divided by the time interval over which the change occurs.2. Explanation :

Eg: A + B 2C⎯⎯→

Average Rate of Reaction = – [A]

t

Δ

Δ =

– [B]

t

Δ

Δ =

1

2

[C]

t

Δ

Δ

3. Example:

2 2 3N + 3H 2NH⎯⎯→

Average Rate of Reaction = 2[N ]–

Δt

Δ =

–1

32[H ]

t

Δ

Δ =

1

23[NH ]

t

Δ

Δ

4. Determination of average rate of reaction :Consider the reaction, A → B

(a) The average rate of this reaction

can be determined graphically.

(b) A graph is plotted between different

values of the molar concentration

of the reactant A, on Y-axis and

corresponding values of time on

X-axis.

(c) Let x1 be the molar concentration

of reactant A, at time t1 and Let

x2 be it’s molar concentration at

time t2.

Change in molar conc. = x2 – x1

Change in time = t2 – t1

Average Rate of Reaction = t

xΔ

Δ =

2 1

2 1

–

t – t

x x

(i) The average rate of reaction does not give the true rate of reaction.(ii) As reaction proceeds the molar concentration of reactant decreases hence, the

rate of reaction also decreases.(iii) Therefore, the rate of reaction does not remain constant over a long interval of

time and hence, it is not the true rate of reaction.

316


(d) Instantaneous rate of reaction :1. Definition : “The rate of chemical reaction, measured at a specific instant is called

instantaneous rate of reaction”.2. Explanation :

A + B 2C⎯⎯→

Instantaneous Rate of Reaction = – d[A]

dt =

– d[B]

dt =

1

2

d[C]

dt

Note : The change in sign from Δ to d in the rate equation.

3. Example :

2 2 3N + 3H 2NH⎯⎯→

Average Rate of Reaction = 2– d[N ]

dt =

–1

32d[H ]

dt =

1

23d[NH ]

dt

4. Determination of instantaneous rate of reaction :Consider the reaction : A → BIn order to determine theinstantaneous rate of reaction,the progress of a reaction isfollowed by measuring theconcentrations of a reactant orproduct at different timeintervals. The concentrations ofa reactant or of a product areplotted against time .

In order to find out, the rate of reaction at very small interval of time say dt, a tangentis drawn to the curve at time t1, as shown in graph.Instantaneous rate of consumption of reactant at time t1 = slope of the tangent

=– d[A]

dtInstantaneous rate of formation of product at time t1 = slope of the tangent

=d[B]

dt

Instantaneous rate of reaction at time t1 = – d[A]

dt =

d[B]

dt

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1. Define Chemical kinetics.Ans. Chemical kinetics is a branch of science that deals with the study of rates of reactions, the

factors affecting these rates and the mechanisms by which the reactions occur.

2. What is the importance of chemical kinetics?Ans. (a) The study of reaction rates helps to predict how quickly a reaction mixture approaches

equilibrium.(b) The study of reaction rates gives information regarding the mechanisms of chemical reactions.(c) Chemical kinetics helps us to determine conditions by which reaction rates could be altered.(d) It helps us to study the effect of various factors like temperature, pressure, catalyst and

concentration of reactants on the rate of reaction.

3. Define and explain rate of a reaction.Ans. Refer 5.1 (b) 1 and 2.

4. Give the units of rate of reactions.Ans. Refer 5.1 (b) - (3)

5. What are the factors affecting rate of a reaction?Ans. Refer 5.1 (b) - (4)

*6. Define average rate of reaction. How is it determined graphically?Ans. Refer 5.1 (c).

*7. Define Instantaneous rate of reaction. How is instantaneous rate evaluated (determined)?Ans. Refer 5.1 (d) 1 and 4


*1. The rate of chemical reaction can be expressed in terms of .......a) rate of consumption of catalystb) rate of consumption of reactants onlyc) rate of consumption of reactants and formation of products bothd) rate of formation of products only

*2. The rate of a reaction is expressed in the units .......a) L mol–1t–1 b) mol dm–3t–1 c) Ms d) M–1s–1

3. Chemical kinetics, a branch of physical chemistry, deal with .......a) heat changes in a reaction b) physical changes in a reactionc) rates of reactions d) structure of molecules

4. For a gaseous reaction the unit of rate of reaction is .......a) mol dm–3 time–1 b) mol dm–3

c) atm time–1 d) mols–1

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5. For the reaction, A + B 2C + D⎯⎯→ which one is the incorrect statements?a) Rate of disappearance of A = Rate of disappearance of Bb) Rate of disappearance of A = Rate of appearance of Dc) Rate of disappearance of B = 2 × Rate of appearance of C

d) Rate of disappearance of B = 1

2× Rate of appearance of C

6. The term d

dt

x in the rate expression refers to the .......

a) instantaneous rate of reactionb) average rate of reactionc) increase in the concentration of reactantsd) concentration of reactants

7. For the reaction, A + 2B C⎯⎯→ , the rate of the reaction at given instant of time is representedby .......

a)d[A]

+dt

= 1 d[B]

2 dt⋅ =

d[C]+

dtb)

d[A]–

dt =

1 d[B]–

2 dt⋅ =

d[C]+

dt

c)d[A]

+dt

= 1 d[B]

+2 dt⋅ =

d[C]+

dtd)

d[A]–

dt =

1 d[B]+

2 dt⋅ =

d[C]+

dt

• Numerical MCQs :*8. In the reaction A + 3B 2C⎯⎯→ , the rate of formation of C is .......

a) the same as rate of consumption of Ab) the same as the rate of consumption of Bc) twice the rate of consumption of Ad) 3/2 times the rate of consumption of B

*9. The instantaneous rate of reaction 2A + B C + 3D⎯⎯→ is given by .......

a)dA

dtb)

1

2

d[A]

dtc)

d[B]

dtd)

13

d[D]dt

*10. Consider the reaction 22 2

d[NO ]2NO(g) + O (g) 2NO (g). If

dt⎯⎯→ = 0.052 M/s then

2d[O ]–

dt will be .......

a) 0.052 M/s b) 0.114 M/s c) 0.026 M/s d) –0.026 M/s

• Advanced MCQs :11. Some statements are given below .......

a) Instantaneous rate of a reaction is the rate at a given pointb) Average rate is over a given time intervalc) As the reaction proceeds it becomes faster and faster

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d) The average rate = Increase in conc. of reactant

Time taken Among the above, the false statement (s)

is/are .......a) only C b) A and C c) C and D d) B, C and D

12. The average rate and instantaneous rate of a reaction may be same if the time interval ismade .......a) infinitesimally small b) moderate c) small d) large

5.2. RATE LAW, RATE CONSTANT AND ORDER OF REACTION ANDMOLECULARITY :

Concept Explanation :(a) Rate Law :1. Definition : “The rate law is defined as an experimentally determined equation which expresses

the rate of a chemical reaction in terms of the molar concentration of the reactants.”

2. Explanation : Consider the general reaction aA + bB cC + dD⎯⎯→

According to the law of mass action, Rate = k[A]a[B]b ... (i)But according to the rate law, expression can be written as Rate = k [A]p [B]q ... (ii)In the above rate law equation, the value of p and q are determined experimentally andthey may or may not be equal to the coefficients of a and b in the reaction.

3. Examples :

(a) 2 2 2 22H O (g) 2H O( ) + O (g)l⎯⎯→ : For the above reaction, rate of the reaction is

experimentally found to be proportional to the concentration of H2O2. Hence, the ratelaw for the reaction is Rate = k[H2O2].

(b) 2 2NO (g) + CO(g) NO(g) + CO (g)⎯⎯→ : For the above reaction, rate of the reaction

is experimentally found to be proportional to [NO2]2 and independent of [CO].

Hence, the rate law for the reaction is Rate = k[NO2]2.

4. Applications of rate law :(a) By knowing the rate law and the rate constant, the rate of reaction for any given

composition of the reaction mixture can be estimated.(b) It can be used to estimate the concentration of the reactants and products at any

instant after the start of the reaction.(c) By knowing the rate law, the mechanism of the reaction can be predicted.

(b) Rate constant or specific reaction rate constant (k) :1. Definition : It is a constant of proportionality in the rate law expression and is equal

to the rate of reaction when the molar concentration of each of the reactants is unity.

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2. Characteristics of rate constant :(a) Greater the value of rate constant, faster is the reaction.(b) Each reaction has a definite value of rate constant at particular temperature.(c) Value of rate constant increases with increase in temperature.(d) Value of rate constant does not depend upon the concentration of the reactants.(e) For a given reaction, rate constant has a higher value in the presence of a catalyst

than in the absence of the catalyst.(f) Units of rate constant depend upon order of reaction.

3. Units of rate constant :The units of rate constant : (Mole Lit–1)1 – n sec–1, where n = order of reaction

Zero order Mole Lit–1sec–1

First order sec–1

Second order Mole–1 Lit sec–1

(c) Order of a reaction :1. Definition : “The order of a chemical reaction with respect to each reactant is defined

as the exponent to which the concentration term of that reactant, in the rate lawis raised.”“The overall order of a reaction is defined as the sum of the exponents (powers)to which the concentration terms in the rate law are raised.

2. Explanation :Consider a chemical reaction,

aA + bB products⎯⎯→

According to rate law expression, Rates of Reaction = k [A]p[B]q.Therefore, the order of reaction (n) is written as, n = p + q.The order of reaction with respect to A is ‘p’ and that of with respect to B is ‘q’.The overall order of reaction is p + q.

3. Examples :

(a) 2 2 2 22H O (g) 2H O( ) + O (g)l⎯⎯→ : For the above reaction, rate of the reaction isexperimentally found to be proportional to the concentration of H2O2. Hence, the ratelaw for the reaction is Rate = k[H2O2].The reaction is first order in terms of H2O2 and its overall order is also first.

(b) 2 2NO (g) + CO(g) NO(g) + CO (g)⎯⎯→ : For the above reaction, rate of thereaction is experimentally found to be proportional to [NO2]

2 and independent of[CO]. Hence, the rate law for the reaction is Rate = k[NO2]

2[CO]0.The reaction is second order in terms of [NO2] and zero order in terms of [CO].Overall order of reaction = 2 + 0 = 2.

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4. Characterstics of order of reaction :(a) It is the sum of powers of the concentration terms in the rate law expression.(b) It is an experimentally determined quantity.(c) Order of reaction may be integer, fraction or even zero.(d) It is not related to the stoichiometric coefficients of the reaction, hence, cannot be

predicted from the stoichiometric balanced equation .(e) It is defined only in terms of the concentrations of the reactants and not of products.(f) It may change with change in the experimental conditions.(g) Order is applicable to elementary as well as complex reactions. The molecularity of

the slowest step will be the order of a complex reaction.

(d) Reaction with no order :1. There are certain reactions which do not have specific order. Hence, they are

reactions with no order.2. Examples :

(a) 2 2H (g) + Br (g) 2HBr (g)⎯⎯→

The rate law for this reaction is found to be rate =

1

22 2

m2

k[H ] [Br ]

1 + [HBr] / [Br ]Where m is a constant. The above equation does not correspond to a simple order.Therefore, the term order cannot be applied for the reaction.

(b) Many enzyme catalysed reactions have no proper order,KbE + S ES P + E⎯⎯⎯→

The rate law for this reaction is found to be rate = b

M

K [S] [E]

K + [S]Where E is the enzyme S is the substrate and Kb and KM are constants. This equationdoes not have any simple order.

(e) Elementary Reaction :1. Definition : An elementary reaction is defined as a reaction that takes place in a single

step and cannot be broken down further into simpler chemical reactions.2. Explanation :

(a) Many reactions that follow a simple rate law actually takes place in a series of steps.Such reactions are called complex reactions.

(b) Each step in a complex reaction is called as elementary reaction.(c) This implies that a complex reaction is broken down in a series of elementary reactions.(d) The overall reaction is obtained by adding all the elementary steps in a reaction.(e) The sequence of elementary reactions in a complex reaction help to determine the

mechanism of a chemical reaction.(f) The concept of molecularity is applicable only for a elementary reaction.

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3. Example :

3 2O (g) O (g) + O(g)⎯⎯→

The above reaction is a single step elementary reaction.

(f) Complex reaction :1. Definition : A reaction that follows a simple rate law but occurs in a series of steps

is termed as a complex reaction.2. Explanation :

(a) A complex reaction occurs in multiple steps.(b) Each step in a complex reaction is an elementary reaction.(c) The overall reaction is obtained by adding all the elementary steps in a reaction.(d) The sequence of elementary reactions in a complex reaction help to determine the

mechanism of a chemical reaction.(e) Molecularity of a complex reaction has no meaning.

3. Example :For example, the reaction

2 2 22NO Cl(g) 2NO (g) + Cl (g)⎯⎯→

The reaction takes place in two steps :

(i) k12 2NO Cl(g) NO (g) + Cl(g)⎯⎯→

(slow, unimolecular)

(ii) k22 2 2NO Cl(g) + Cl(g) NO (g) + Cl (g)⎯⎯→

(fast, bimolecular)

2 2 22NO Cl(g) 2NO (g) + Cl (g)⎯⎯→ (overall reaction)

The first step being slower than the second is the rate determining step. The rate lawpredicted by slow step is rate = k1[NO2Cl]It can be seen that Cl is formed in the first step and consumed in the second. Hence,it is the reaction intermediate.

(g) Rate determining step :1. The slowest step in the reaction mechanism is called a rate determining step.2. Explanation :

(a) A complex reaction occurs in multiple steps.(b) Each step in a complex reaction is an elementary reaction.(c) The sequence of elementary reactions in a complex reaction help to determine the

mechanism of a chemical reaction.(d) One of the steps is the slowest step compared to other steps. Such a step is called

rate determining step.(e) The rate of the overall reaction depends on the rate determining step.

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(f) The overall reaction can never occur faster than its rate determining step.(g) The rate determining step can occur anywhere in the reaction mechanism.(h) The rate law of a rate determining step can be obtained from its stoichiometric equation.

3. Example :For example, the reaction

2 2 22NO Cl(g) 2NO (g) + Cl (g)⎯⎯→


(i) k12 2NO Cl(g) NO (g) + Cl(g)⎯⎯→



(fast, bimolecular)


The first step being slower than the second is the rate determining step. The rate lawpredicted by slow step is rate = k1[NO2Cl]It can be seen that Cl is formed in the first step and consumed in the second. Hence,it is the reaction intermediate.

(h) Reaction intermediate :1. Definition : “The additional species other than the reactants or products formed in

the reaction mechanism is called reaction intermediate.”2. Explanation :

(a) The reaction intermediate appears in the mechanism but does not appear in theoverall reaction.

(b) It is always formed in one step of the mechanism and is consumed in thesubsequent step.

(c) The concentration of reaction intermediate does not appear in the rate law, since itsconcentration is very small and undetermined.

(d) It is not present at the start of the reaction neither at the end of the reaction.3. For example, the reaction

2 2 22NO Cl(g) 2NO (g) + Cl (g)⎯⎯→


(i) k12 2NO Cl(g) NO (g) + Cl(g)⎯⎯→



(fast, bimolecular)


The first step being slower than the second is the rate determining step. The rate lawpredicted by slow step is rate = k1[NO2Cl].

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Order

1. It is the sum of the exponents to which theconcentration terms in the rate law are raised.

2. It is purely an experimental propertyindicating the dependence of observedreaction rate on the concentration of reactants.

3. It may be an integer, fraction or zero.

4. It may change with experimental conditions.

5. It is the property of both elementary andcomplex reactions.

Molecularity

1. It is the number of reactant molecules takingpart in an elementary reaction.

2. It is theoretical property indicating thenumber of reactant molecules involved in eachact leading to reaction.

3. It is always an integer and never be a fractionor zero.

4. It does not change with experimentalconditions.

5. It is the property of only elementary reactionsand has no meaning for complex reactions.

It can be seen that Cl is formed in the first step and consumed in the second. Hence,it is the reaction intermediate.

Note : The example given above can be used as an example for Complex reaction,Rate determining step or for Reaction intermediate.

(i) Molecularity :1. Definition : “The molecularity of an elementary reaction is defined as the number

of reactant molecules taking part in the reaction.”2. Explanation :

(a) It is a theoretical concept.(b) It is always an integer and never be a fraction or zero.(c) It does not change with experimental conditions.(d) It is the property of only elementary reactions and has no meaning for complex reactions.

3. On the basis of molecularity, chemical reactions are classified as follows:(a) Unimolecular reaction: It is the reaction, in which only one molecule of reactant is

involved. for e.g. 2Br (g) 2Br (g)⎯⎯→

(b) Bimolecular reaction: It is the reaction, in which two molecules of reactants are involved.

for e.g. 2 2H + I 2HI⎯⎯→

(c) Trimolecular reaction : It is the reaction, in which three molecules of reactants areinvolved.

(j) Distinguish between Order and Molecularity

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*1. Define rate law.Ans. “The rate law is defined as an experimentally determined equation which expresses the rate

of a chemical reaction in terms of the molar concentration of the reactants.”

*2. Explain rate law with example.Ans. Refer 5.2 (a) 1, 2 and 3.

*3. Define rate constant.Ans. It is a constant of proportionality in the rate law expression and is equal to the rate of reaction

when the molar concentration of each of the reactants is unity.

4. Give characteristics of rate constant.Ans. Refer 5.2 (b)

*5. Give the units of rate constants of first and zero order reactions.Ans. The units of rate constant : (Mole Lit –1)1 – n sec–1, where n = order of reaction

Zero order Mole Lit–1sec–1

First order sec–1

*6. Explain the term order of a reaction with example.Ans. Refer 5.2 (c) 1, 2 and 3.

7. Give the characteristics of order of a reaction.Ans. (a) It is the sum of powers of the concentration terms in the rate law expression.

(b) It is an experimentally determined quantity.(c) Order of reaction may be integer, fraction or even zero.(d) It is not related to the stoichiometric coefficients of the reaction, hence, cannot be predicted

from the stoichiometric balanced equation .(e) It is defined only in terms of the concentrations of the reactants and not of products.(f) It may change with change experimental conditions.(g) Order is applicable to elementary as well as complex reactions. The molecularity of the

slowest step will be the order of a complex reaction.

8. Explain reactions with no order.Ans. Refer 5.2 (d)

*9. Explain the term elementary reaction.Ans. (a) Definition : An elementary reaction is defined as a reaction that takes place in a

single step and cannot be broken down further into simpler chemical reactions.(b) Explanation :

(1) Many reactions that follow a simple rate law actually takes place in a series of steps.Such reactions are called complex reactions.

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(2) Each step in a complex reaction is called as elementary reaction.

(3) This implies that a complex reaction is broken down in a series of elementary reactions.

(4) The overall reaction is obtained by adding all the elementary steps in a reaction.

(5) The sequence of elementary reactions in a complex reaction help to determine the

mechanism of a chemical reaction.

(6) The concept of molecularity is applicable only for an elementary reaction.

(c) Example :

3 2O (g) O (g) + O(g)⎯⎯→

The above reaction is a single step elementary reaction.

*10. What do you mean by a complex reaction?Ans. (a) Definition : A reaction that follows a simple rate law but occurs in a series of steps

is termed as a complex reaction.(b) Explanation :

(1) A complex reaction occurs in multiple steps.(2) Each step in a complex reaction is an elementary reaction.(3) The overall reaction is obtained by adding all the elementary steps in a reaction.(4) The sequence of elementary reactions in a complex reaction help to determine the

mechanism of a chemical reaction.(5) Molecularity of a complex reaction has no meaning.

(c) Example : Refer 5.2 (f).

*11. Explain the terms rate determining step.Ans. (a) The slowest step in the reaction mechanism is called a rate determining step.

(b) Explanation :(1) A complex reaction occurs in multiple steps.(2) Each step in a complex reaction is an elementary reaction.(3) The sequence of elementary reactions in a complex reaction help to determine the

mechanism of a chemical reaction.(4) One of the steps is the slowest step compared to other steps. Such a step is called

rate determining step.(5) The rate of the overall reaction depends on the rate determining step.(6) The overall reaction can never occur faster than its rate determining step.(7) The rate determining step can occur anywhere in the reaction mechanism.(8) The rate law of a rate determining step can be obtained from its stoichiometric equation.Example : Refer 5.2 (g).

*12. Explain the term reaction intermediate.Ans. Refer 5.2 (h).

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*13. What is molecularity of an elementary reaction?Ans. (a) Definition : The molecularity of an elementary reaction is defined as the number of

reactant molecules taking part in the reaction.(b) Explanation :

(1) It is a theoretical concept.(2) It is always an integer and never be a fraction or zero.(3) It does not change with experimental conditions.(4) It is the property of only elementary reactions and has no meaning for complex reactions.

(c) On the basis of molecularity, chemical reactions are classified as follows.(1) Unimolecular reaction: It is the reaction, in which only one molecule of reactant is

involved. for e.g. 2Br (g) 2Br (g)⎯⎯→

(2) Bimolecular reaction: It is the reaction, in which two molecules of reactants are involved.

for e.g. 2 2H + I 2HI⎯⎯→

(3) Trimolecular reaction: It is the reaction, in which three molecules of reactants areinvolved.

*14. Distinguish between order and molecularity.Ans. Refer 5.2 (j)*15. The gas-phase reaction between NO and Br2 is represented by the equation.

22NO (g) + Br (g) 2NOBr (g)⎯⎯→

(a) Write the expression for the rate of consumption of reactants and formation of products.

(b) Write the expression for the rate of overall reaction in terms of rates of consumption

of reactants and formation of products.

Ans. 22 NO(g) + Br (g) 2NOBr (g)⎯⎯→

(a) Rate of consumption of No at time t =– d[NO]

dt

Rate of consumption or Br2 at time t = 2– d[Br ]

dt

Rate of formation of NOBr at time t =d[NOBr]

dt

(b) Rate of reaction = – d[NO]

dt = 2– d[Br ]

dt =

d[NOBr]

dt

*16. The decomposition of N2O5 is represented by the equation

2 5 2 22N O (g) 4NO (g) + O (g)⎯⎯→

(a) How is the rate of formation of NO2 related to the rate of formation of O2?

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(b) How is the rate of formation of O2 related to the rate of consumption of N2O5?

Ans. 2 5 2 22N O (g) 4NO (g) + O (g)⎯⎯→

(a) Rate of formation of NO2 at time t = 2d[NO ]

dt

Rate of formation of O2

at time t = 2d[O ]

dtThey are related by rate of reaction as follows

Rate of reaction = 2d[NO ]1

4 dt = 2d[O ]

dt

Hence, rate of formation of NO2 = 4 × Rate of formation of O2

(b) Rate of formation of O2 = 2d[O ]

dt

Rate of consumption of N2O5 = 2 5– d[N O ]

dt

Rate of reaction = 2d[O ]

dt = 2 5d[N O ]–1

2 dt

Hence, rate of formation of O2 = 1

2 × rate of consumption of N2O5.

*17. Nitric oxide reacts with H2 according to the reaction

2 2 22NO(g) + 2H (g) N (g) + 2H O (g)⎯⎯→

What is the relationship among d[NO]/dt, d[H2]/dt, d[N2]/dt and d [H2O]/dt?Ans. Rate of consumption of reactants and rate of formation of products are related as follows

Rate of Reaction = –1 d[NO]

2 dt = 2d[H ]–1

2 dt = 2+d[N ]

dt = 2d[H O]+1

2 dt

*18. The rate law for the gas-phase reaction 2 22NO(g) + O (g) 2NO (g)⎯⎯→ is rate = k[NO2]2[O2].

What is the order of the reaction with respect to each of the reactants and what is the overallorder of the reaction?

Ans. 2 22NO(g) + O (g) 2NO (g)⎯⎯→

Rate = k [NO]2[O2]∴∴∴∴∴ Order of the reaction with respect to NO = 2∴∴∴∴∴ Order of the reaction with respect to O2 = 1

Overall order of one reaction = 2 + 1 = 3

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*19. The reaction – + –2 2 2 3H O (aq) + 3I (aq) + 2H (aq) 2H O( ) + I (aq)l⎯⎯→ is first order in H2O2

and I–, zero order in H+ Write the rate law.

Ans. – + –2 2 2 3H O (aq) + 3I (aq) + 2H (aq) 2H O( ) + I (aq)l⎯⎯→

Since the reaction is first order in H2O2 and I– and zero order in H+,then Rate = k [H2O2]

1 [I–]1 [H+]0

= k [H2O2] [I–]

*20. Identify the molecularity and write the rate law for each of the following elementary reactions :

(a) 3 3NO(g) + O (g) NO (g) + O(g)⎯⎯→

(b) 2H I(g) + I(g) 2HI(g)⎯⎯→

(c) 2 2 2Cl(g) + Cl(g) + N (g) Cl (g) + N (g)⎯⎯→

Ans. (a) 3 3NO(g) + O (g) NO (g) + O(g)⎯⎯→

Molecularity = 2 Rate k = [NO] [O3]

(b) 2H I(g) + I(g) 2HI(g)⎯⎯→

Molecularity = 2 ∴∴∴∴∴ Rate = k [H2I] [I]

(c) 2 2 2Cl(g) + Cl(g) + N (g) Cl (g) + N (g)⎯⎯→

Molecularity = 3 ∴∴∴∴∴ Rate = k [O]2 [N2]

*21. A complex reaction takes place in two steps as follows :

(i) 3 3NO(g) + O (g) NO (g) + O(g)⎯⎯→

(ii) 3 2 2NO (g) + O(g) NO (g) + O (g)⎯⎯→

(a) Write the overall reaction.(b) Identify reaction intermediate.(c) Identify the rate determining step if the predicted rate law is rate = K[NO] [O3]

Ans. (a) Step I : 3 3NO(g) + O (g) NO (g) + O(g)⎯⎯→

Step II : 3 2 2NO (g) + O(g) NO (g) + O (g)⎯⎯→

Overall Reaction : 3 2 2NO(g) + O (g) NO (g) + O (g)⎯⎯→

(b) The reaction intermediates are [NO2] and [O] since they are formed in the first step andconsumed in the next step.

(c) Rate law for above reaction is Rate = k [NO] [O3]

∴∴∴∴∴ The rate determining step is 1st step i.e. 3 3NO(g) + O NO (g) + O(g)⎯⎯→ .

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*22. What is the order for the following reactions?

(a) 2 2 22NO (g) + F (g) 2NO F(g)⎯⎯→ , rate = k[NO2][F2]

(b) 3 2 4CHCl (g) + Cl (g) CCl (g) + HCl(g)⎯⎯→ , rate = 1

23 2k[CHCl ] [Cl ]

Ans. (a) 2 2 22NO (g) + F (g) 2NO F(g)⎯⎯→ , rate = k[NO2][F2]

∴∴∴∴∴ Order = 1 + 1 = 2

(b) 3 2 4CHCl (g) + Cl (g) CCl (g) + HCl(g)⎯⎯→ , rate = 1

23 2k[CHCl ] [Cl ]

∴∴∴∴∴ Order = 1

1 +2

= 3

2

*23. Write the rate law for the following reactions :(a) A reaction that is zero order in A and second order in B.(b) A reaction that is second order in NO and first order in Br2.

Ans. (a) If the reaction is zero order in A and second order is BThen, Rate = k [A]0 [B]2

Rate = k [B]2.(b) If the reaction is 2nd order in NO and first order in Br2, then Rate = k[NO]2[Br]1.

*24. Consider the reaction

+ 4+ 3+ 3+Tl (aq) + 2Ce (aq) Tl (aq) + 2Ce (aq)⎯⎯→ . The rate law for the reaction is rate = k[Ce4+] [Mn2+]. In the presence of Mn2+ the reaction occurs in the following elementary steps :

(a) 4+ 2+ 3+ 3+Ce (aq) + Mn (aq) Ce (aq) + Mn (aq)⎯⎯→

(b) 4+ 3+ 3+ 4+Ce (aq) + Mn (aq) Ce (aq) + Mn (aq)⎯⎯→

(c) + 4+ 3+ 2+Tl (aq) + Mn (aq) Tl (aq) + Mn (aq)⎯⎯→

Identify (a) catalyst (b) intermediate (c) rate determining step.Ans. (a) The catalyst in the above reaction will be Mn+2. Since Mn+2 is present at the start as

well as end of the reactions.(b) The intermediate in the above reaction will be Mn3+ and Mn4+ since both Mn+3 and Mn+4

are formed in one step and consumed is the next step.(c) The rate law given is Rate = k[Ce4+] [Mn2+].

∴∴∴∴∴ The rate determining step will be 4+ 2+ +3 3+Ce (aq) + Mn (aq) Ce (aq) + Mn (aq)⎯⎯→ .

*25. Comment on the relationship between coefficients of the balanced overall equation for a reactionand the exponents to which the concentration terms in the rate law are raised.What do these exponents represent?

Ans. (a) The rate of a reaction at a given temperature depends on the concentration of reactants.

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(b) Consider the general reaction : aA + bB Products⎯⎯→

(c) The dependence of the reaction rate on the concentration of each reactant is given byan equation called as differential rate law, which is determined experimentally.

(d) Rate ∝ [A]x[B]y, therefore, Rate = K[A]x[B]y, where K = rate constant(e) The exponents x and y in the rate law are not necessarily equal to the stoichiometric

coefficients (a and b) of A and B in the balanced reaction.(f) The values of x and y must be experimentally determined.(g) The values of x and y can be negative, zero or fractions.

(h) For Example : 2 2 2 22H O (g) 2H O( ) + O (g)l⎯⎯→

For the above reaction, rate of the reaction is experimentally found to be proportional tothe concentration of H2O2. Hence, the rate law for the reaction is Rate = k[H2O2].


• Theoretical MCQs :*1. A reaction is first order with respect to reactant A and second order with respect to reactant

B. The rate law for the reaction is given by .......a) rate = k [A][B]2 b) rate = [A][B]2

c) rate = k [A]2[B] d) rate = k [A]0[B]2

*2. Molecularity of an elementary reaction .......a) may be zero b) is always integralc) may be semi-integral d) may be integral, fractional or zero.

*3. The unit of rate constant for first order reaction is .......a) min–2 b) s c) s–1 d) min

4. The order of a reaction .......a) is never equal to zero or fractionb) can be predicted from chemical equation of reactionc) is equal to sum of total number of moles of reactants taking part in reactiond) always determined experimentally

5. Catalytic decomposition of PH3 is a reaction of .......a) Zero order b) Second order c) First order d) Third order

6. For a reaction, A + 2B → D + E following mechanism is suggested .......i) A + B → C + D (Slow step) ii) B + C → E (Fast step)The rate expression for the reaction.a) r = k [A] [B]2 b) r = k [A]2

c) r = k [A] [B]1/2 d) r = k [A] [B]

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7. Which of the following statements regarding molecularity of the reaction is wrong?a) it may be either whole number or fractionalb) it is calculated from the reaction mechanismc) it depends on the rate determining stepd) it is number of molecules of reactants taking part in a single step chemical reaction

8. Many reactions proceed in a sequence of steps, so the overall rate of reaction is determinedby .......a) the order of different steps b) the slowest stepc) molecularity of the steps d) the fastest step

• Numericals MCQ :*9. The rate of the first order reaction A ⎯⎯→ products is 0.01 M/s, when reactant concentration

is 0.2 M. The rate constant for the reaction will be .......a) 0.05 s–1 b) 0.05 min–1 c) 0.1 s–1 d) 0.01 s–1

10. For reaction A + B + C → Products, Rate k [A]2 [B] [C]1/2. The order of the reaction is .......a) 2.8 b) 2 c) 3 d) 3.5

11. For a reaction, A + B → Product it is observed that (i) [A] = constant, [B] is doubled, ratebecomes double (ii) [B] = constant, [A] is doubled, rate becomes four times. The order ofthe reaction is .......a) 1 b) 4 c) 2 d) 3

• Advanced MCQs :*12. The reaction between H2(g) and ICl (g) occurs in the following steps :

(i) 2H + ICl HI + HCl⎯⎯→ (ii) 2 2H + ICl I + HCl⎯⎯→

The reaction intermediate in the reaction is .......a) HCl b) HI c) I2 d) ICl

*13. The formation of SO3 from SO2 and O2 takes place in the following steps :

(i) 2 2 32SO + 2NO 2SO + 2NO⎯⎯→ (ii) 2 22NO + O 2NO⎯⎯→

a) NO2 is intermediate b) NO is catalystc) NO2 is catalyst and NO is intermediate d) NO is catalyst and NO2 is intermediate

5.3 INTEGRATED RATE LAWS : Concept Explanation :

(a) Integrated rate laws :Definition : The equations which are obtained by integrating the differential rate lawsand which give a direct relationship between the concentrations of the reactants andtime are called integrated rate laws.

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(b) Zero Order Reaction :1. Definition : “The reaction whose rate is independent of the reactant concentration and

remains constant throughout the course of the reaction is called zero order reaction”.2. Differential rate law for zero order reaction :

Consider a general reaction,A → productsThe rate of disappearances of reactant, A is given by the relation,

Rate = d[A]

–dt

= k [A]0 = k

∴ Rate = kThus, the rate of this reaction is proportional to the zero power of the concentration ofthe reactant.

3. Examples of zero order reaction :(a) The decomposition of gaseous ammonia on a hot platinum surface is a zero order

reaction at high pressure.1130K

3 2 2Pt catalyst2NH (g) N (g) + 3H (g)⎯⎯⎯⎯→

Rate = k [NH3]0 = k(b) Decomposition of N2O to N2 and O2 on Platinum catalyst :

2 2 22N O(g) 2N (g) + O (g)⎯⎯→

(c) Decomposition of PH3 on hot tungsten at high pressure.4. Integrated Rate law for zero order reaction :

Consider the hypothetical zero order reaction; A → Products

The rate law for this zero order reaction will be, Rate = – d[A]

dt = k[A]0 = k

On rearranging, d[A] = – kdtOn integration of the equation between limits [A] = [A]0 at t = 0 and [A] = [A]t att = t, we get

[A]t

[A]0

d[A]∫ =

t

0

–k dt∫

[A]t[A]0

[A] = t0– k(t)

d[A]∫ = – kdt∫[A]t – [A0] = –kt[A]t = –kt + [A0]

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5. Graphical representation of zero order reaction :(a) The differential rate law for zero order

reaction is given by

Rate = –d[A]

dt = k[A]0 = k

The plot of rate versus [A]t is a horizontalline with zero slope.

(b) The integrated rate law for zero orderreaction is given by[A]t = –kt + [A0]The above equation is in the form of linearequation y = mx + C. Hence, [A]t versust is a straight line with slope –k.

6. Half life of zero order reactions :

The rate law expression for zero order reaction is, [A]t = –kt + [A]0

Where [A]0 and [A]t are the concentrations of the reactant at time, t = 0 and after time

t respectively,

Half-life period, t1/2 is the time when the concentration reduces from [A]0 to [A]0/2. i.e.,

at t = t1/2, [A]t = [A]0 / 2.

∴ 0[A]

2 = –kt1/2 + [A]0

∴ kt1/2 = 0

0[A]

[A] –2

= 0[A]

2

∴ t1/2 = 0[A]

2kHence, for a zero order reaction, half-life period is directly proportional to the initialconcentration of the reactant.

7. Explanation of examples of zero order reaction :(i) The decomposition of gaseous ammonia on a hot platinum surface is a zero order

reaction at high pressure.1130K

3 2 2Pt catalyst2NH (g) N (g) + 3H (g)⎯⎯⎯⎯→

The platinum surface gets completely covered by a layer of NH3 molecules.

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The number of NH3 molecules attached on the surface of platinum is very smallcompared to the total number of ammonia molecules.Most of the ammonia molecules remain in the gas phase and these do not react.Consequently the rate of the reaction is independent of the total concentration of NH3and hence, it remains constant. Hence, it is a zero order reaction.

(ii) The decomposition of nitrous oxide to N2 and O2 in the presence of Pt catalyst.

2 2 22N O(g) 2N (g) + O (g)⎯⎯→

The platinum surface gets completely covered by N2O molecules. Most of the moleculesremain in the gas phase.Because only the N2O molecules on the surface react, the reaction rate is independentof the total concentration of N2O. Hence, it a zero order reaction.

(c) First Order Reaction :1. Definition : “The reaction whose rate depends on the single reactant concentration

raised to a first power is called the first order reaction”.2. Differential rate law for first order reaction :

Consider a general reaction, A → products

The rate of disappearances of reactant, A is given by the relation, Rate = d[A]

–dt

= k[A].

Thus, the rate of this reaction is proportional to first power of the concentration of thereactant.

3. Examples of first order reaction :The examples of first order reaction are :(a) Decomposition of H2O2 :

2 2 2 22H O ( ) 2H O( ) + O (g)l l⎯⎯→ Rate = k[H2O2](b) Decomposition of N2O5 :

2 5 2 22N O (g) 4NO (g) + O (g)⎯⎯→ Rate = k[N2O5](c) Isomerisation of cyclopropane to propene :

Rate = k[C3H6]

4. Integrated Rate law for first order reaction :Consider a general reaction, A → productsThe rate of disappearances of reactant, A is given by the relation,

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Rate = d[A]

–dt

= k[A] ...(i)

(a) Where [A] is the concentration of the reactant A that remains unreacted at timet and –d[A] / dt is the rate of the reaction measured at time t at which A is convertedto the products. By rearranging and integrating the equation (i) between the limits[A] = [A]0 at t = 0 and [A] = [A]t at t = t, we write,

[A]t

[A]0

d[A]

A∫ =

t

0

–k dt∫

where [A]0 is the initial concentration of A at t = 0 and [A]t is the concentration of Athat remains unreacted at t.

On performing the integration, we get[A]t[A]0

In[A] = t0–k(t)

Or ln [A]t – ln[A]0 = –kt

andt

0

[A]ln

[A] = –kt ...(ii)

Hence, kt = 0

t

[A]ln

[A]

that is kt = 0

t

[A]1ln

t [A]

Converting ln to log10 the integrated rate law becomes

k = 0

10t

[A]2.303log

t [A] ...(iii)

Note :The rate law equation can also be written in the following alternative forms :

(a) The equation (ii),t

0

[A]ln

[A] = –kt

can be written by taking antilog of both sides as

t

0

[A]

[A] = e–kt

Or [A]t = [A]0e–kt

(b) If we let ‘a’ mol dm–3 (that is M) as the initial concentration of A at t = 0 andx as the concentration of A that decreases during time t from the beginning of the

This is called as exponential formof first order reaction.

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reaction then the concentration of A that remains unreacted at t is given by[A]t = [A]0 – x = a – xwhere [A]0 = a mol dm–3. With these values the equation (iii) can be written as

k = 102.303 a

logt a – x

5. Graphical representation of first order reaction :Equation :The differential rate law for first order reaction,

Rate = d[A]

–dt

= k[A]

Rate = k [A]t (y = mx)

Nature of graph : When rate of a first order reactionis plotted against concentration[A]t, a straight line isobtained.

Slope : The slope of the line gives the value of K.

Equation :Exponential form of first order reaction,[A]t = [A]0e

–kt

Nature of graph : When the concentration of reactantis plotted against time (t), a curve is obtained.

Equation :The integrated rate law expression

k = 0

10t

[A]2.303log

t [A]On rearrangement we get,

kt

2.303 = log10[A]0 – log10[A]t Hence,

log10[A]t = – k

2.303t + log10[A]0.

The above equation is of the form, y = mx + C

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Nature of graph : When log10[A]t is plotted against time(t), we get a straight line

Slope : The slope of the line is equal to – k

2.303

Equation :

k = 0

10t

[A]2.303log

t [A] . The above integrated rate

law expression can be rearranged to give,

010

t

[A]log

[A] = k

2.303 t + 0. The above equation

is of the form, y = mx + C

Nature of graph : When 0

10t

[A]log

[A] is plotted

against time(t), we get a straight line

Slope : The slope of the line is equal to k

2.303

6. Half life of first order reaction 1/ 2(t ) :

The half life of a reaction is defined as the time needed for the reactant concentrationto fall to one half of its initial value.For the first order reaction the integrated rate law is given by the equation

k = 010

t

[A]2.303log

t [A]

where [A]0 is the initial concentration of the reactant at t = 0. Its concentration falls to[A]t at time t from the start of the reaction. The concentration of the reactant falls to

[A]0/2 at time 1/2t , the half life period. Hence, at t = 1/2t , [A]t = [A]0/2. With this condition,

the equation can be written as

k =0

1001/2

[A]2.303log

[A]t2

= 2101/2

2.303log

t

=1/2

2.303× 0.301

t =

1/2

0.693

t

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= 1/2t = 0.693

k...(iv)

Hence,The equation (iv) implies that the half life of a first order reaction is constant and is independentof the reactant concentration.

(d) Pseudo First Order reaction :1. Definition : A reaction which has higher order true rate law but are found to behave

as first order are called pseudo first order reactions.2. Examples :

(a) The hydrolysis of ester in presence of an acid catalyst, can be explained by takingexample of hydrolysis of methyl acetate in presence of dil. HCl.

dil HCl3 3 2 3 3CH COOCH + H O CH COOH + CH – OH⎯⎯⎯⎯→

The rate of this reaction is given by Rate of reaction = k′[CH3COOCH3] [H2O].According to the rate law, the above reaction is expected to have second order.But practically, it is observed that the water present in this reaction is in large excess,therefore, the concentration of water remains practically constant during the courseof reaction.[H2O] = constant = k′′

∴ Rate of reaction = k′k′′ [CH3COOCH3]= k [CH3COOCH3]

This indicates that, the rate of reaction is determined only by the concentration ofmethyl acetate. Hence, this reaction is first order reaction. Such reactions are knownas pseudo-first order reaction.

(b) The hydrolysis of cane sugar (sugar) in presence of an acid catalyst is an exampleof pseudo–first order reaction.

12 22 11 2 6 12 6 6 12 6C H O (aq) + H O( ) C H O (aq) + C H O (aq)sucrose glucose fructose

l ⎯⎯→

Rate = k′ [C12H22O11] [H2O]The above rate law indicates that the reaction is second order. However, [H2O] isin excess.

∴ [H2O] remain constant

∴ k′ × [H2O] = k∴ rate = k [C12H22O11]

Thus, the rate of reaction is dependent on only concentration of sucrose. Hence, the

reactions is pseudo–first order reaction.

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(e) Experimental Determination of rate laws and order :1. Method I : Isolation Method :

(a) In this method all the reactants except one (isolated) are present in large excess.(b) The concentrations of these reactants remain constant throughout the course of the reaction.(c) The dependence of rate on the concentration of the isolated species is experimentally

determined.(d) This will give the rate law and order of the reaction with respect to the isolated reactant

are measured.(e) This procedure is repeated for other reactants in the reaction.(f) Example : Consider an equation

aA + bB cC + dD⎯⎯→

∴∴∴∴∴ Rate law is rate = k [A]x [B]y

In one experiment, A is isolated by taking B in large excess.

∴∴∴∴∴ Initial concentration of B = [B]0 = remain constant

∴∴∴∴∴ Rate = k [A]x [B]0y

= k [A]x [∵ [B]0y × k = k′]

Hence, order x can be measured.

Similarly ‘y’ can also be determined∴∴∴∴∴ Overall order of reaction = x + y.

2. Method II : Method of Initial rates :(a) This method involes the measurement of rate at the begining of the reaction for different

initial concentrations of reactants by using isolation method.

(b) Consider a reaction aA + bB cC + dD⎯⎯→

∴ Rate law is rate = k′ [A]x [B]y

(c) If B is taken in large excess, its concentration remains constant.∴ The initial rate of the reaction with A isolated is given by r0 = k [A]x

0

Hence, log10 r0 = log10k + log10[A0]x

∴ log10 r0 = x log10[A]0 + log10k[y = mx + c]

(d) For series of concentrations of isolated A, the rates r0 are measured. When the graph

of log10 r0 against log10 [A]0 is plotted, a straight line is obtained. The slope of theline gives order x with respect to A.

(e) The procedure is repeated with B to give order ‘y’ with respect to ‘B’

(f) The overall order of the reaction is x + y.

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2. Method III : Use of Integrated law.(a) This is a trial and error methods. In this the concentration of the reactant is measured

at different intervals of time.(b) These data are substituted in the integrated rate law of reactions of first order, second

order and so on.(c) The equation which gives the constant value of the rate constant (k) gives the correct

order of reaction.

For example, for the first order reaction, k = 010

t

[A ]2.303log

t [A]Where [A0] = Initial concentration of A.and [A]t = concentration that remains unreacted at time t.

(d) By substitution of time-concentration data in the integrated rate law equation the valuesof k at different time intervals are calculated.

(e) In these k values are constant then reaction is a first order reaction.


*1 What is zero order reaction? Write the differential rate law for zero order reaction. Deriveits integrated rate law. How is it represented graphically?

Ans. Refer 5.3 (b) 1 to 5

2. Derive an expression for half life of a zero order reaction.Ans. The rate law expression for zero order reaction is, [A]t = –kt + [A]0

Where [A]0 and [A]t are the concentrations of the reactant at time, t = 0 and after time trespectively,Half-life period, t1/2 is the time when the concentration reduces from [A]0 to [A]0 / 2. i.e.,at t = t1/2, [A]t = [A]0 / 2.

∴∴∴∴∴ 0[A]

2= –kt1/2 + [A]0

∴∴∴∴∴ kt1/2 = 00

[A][A] –

2 =

0[A]

2

∴∴∴∴∴ t1/2 =0[A]

2kHence, for a zero order reaction, half-life period is directly proportional to the initial concentrationof the reactant.

3. Decomposition of NH3 on platinum surface at high temperature is a zero order reaction. Explain.Ans. The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction

at high pressure.

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1130K3 2 2Pt catalyst

2NH (g) N (g) + 3H (g)⎯⎯⎯⎯→

The platinum surface gets completely covered by a layer of NH3 molecules.The number of NH3 molecules attached on the surface of platinum is very small comparedto the total number of ammonia molecules.Most of the ammonia molecules remain in the gas phase and these do not react.Consequently the rate of the reaction is independent of the total concentration of NH3 andhence, it remains constant.Hence, it is a zero order reaction.

4 What is a first order reaction. Explain with an example.Ans. Refer 5.3 (c) 1 to 3.

*5 Define integrated rate law. Derive the expression for integrated rate law for first order reaction,A → products.

Ans. Integrated rate laws :Definition : The equations which are obtained by integrating the differential rate laws and whichgive a direct relationship between the concentrations of the reactants and time are called integratedrate laws.Expression for Integrated Rate law for first order reaction : Refer 5.3 (c) 4.

*6. Describe the graphical representation of first order reaction.Ans: Refer 5.3 (c) 5.

*7. Define half life of a reaction. Derive the relationship between half life and rate constant fora first order reaction. OR Show that half life of a first order reaction is independent of initialconcentration of the reactant.

Ans. Refer 5.3 (c) 6.

*8. What is pseudo-first order reaction? Explain with one example.Ans. Refer 5.3 (d).

*9. The reaction, 3 2 5 2 3 2 5CH COOC H (aq) + H O( ) CH COOH(aq) + C H OH(aq)l ⎯⎯→ follows

the first order kinetics, the rate law being rate = k [CH3COOC2H5]. However, the reactionshows that it is second order. Explain.

Ans. The hydrolysis of ester in presence of an acid catalyst, can be explained by taking exampleof hydrolysis of methyl acetate in presence of dil. HCl.

dil HCl3 3 2 3 3CH COOCH + H O CH COOH + CH – OH⎯⎯⎯⎯→

The rate of this reaction is given by Rate of reaction = k′[CH3COOCH3] [H2O].According to the rate law, the above reaction is expected to have second order.But practically, it is observed that the water present in this reaction is in large excess, therefore,the concentration of water remains practically constant during the course of reaction.

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[H2O] = constant = k′′∴ Rate of reaction = k′k′′ [CH3COOCH3]

= k [CH3COOCH3]This indicates that, the rate of reaction is determined only by the concentration of methyl acetate.Hence, this reaction is first order reaction. Such reactions are known as pseudo-first order reaction.

*10. Describe one method to determine rate law and order of a reaction.Ans. Method I : Isolation Method :

(a) In this method all the reactants except one (isolated) are present in large excess.(b) The concentrations of these reactants remain constant throughout the course of the reaction.(c) The dependence of rate on the concentration of the isolated species is experimentally

determined.(d) This will give the rate law and order of the reaction with respect to the isolated reactant

are measured.(e) This procedure is repeated for other reactants in the reaction.

(f) Example : Consider an equation aA + bB cC + dD⎯⎯→

∴ Rate law is rate = k [A]x [B]y

In one experiment, A is isolated by taking B in large excess.∴ Initial concentration of B = [B]0 = remain constant∴ Rate = k [A]x [B]0

y

= k [A]x [∵ [B]0y × k = k′]

Hence, order x can be measured.Similarly ‘y’ can also be determined

∴ Overall order of reaction = x + y.

*11. A certain reaction occurs in the following steps :

(i) 3 2Cl(g) + O (g) ClO(g) + O (g)⎯⎯→

(ii) 2ClO(g) + O(g) Cl(g) + O (g)⎯⎯→

a) Write the chemical equation for overall reaction.b) Identify the reaction intermediate.c) Identify the catalyst.

Ans. (a) Step I : 3 2Cl(g) + O (g) ClO(g) + O (g)⎯⎯→

Step II : 2ClO(g) + O(g) Cl(g) + O (g)⎯⎯→

Net reaction 3 2O (g) + O(g) 2O (g)⎯⎯→

(b) Reaction intermediate is ClO(g) which is formed in the first step and removed in thesecond step.

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(c) Cl(g) acts as a catalyst. It is present at the start and end of the reaction but is not seenin the overall reaction.

*12. What is the rate law for the reaction 2 2NO (g) + CO(g) NO(g) + CO (g)⎯⎯→ . If the reaction

occurs in the following steps?

[(i) 2 2 3NO + NO NO + NO⎯⎯→ (slow) and (ii) 3 2 2NO + CO NO + CO⎯⎯→ (fast)]

What is the role of NO3?

Ans. Step I : 2 2 3NO + NO NO + NO⎯⎯→ (slow)

Step II : 3 2 2NO + CO NO + CO⎯⎯→ (fast)

Overall reaction 2 2NO (g) + CO(g) NO(g) + CO (g)⎯⎯→

NO3 is formed in the first step and consumed in the second step.Hence, NO3 is an intermediate.

*13. The rate law for the reaction 22NO(g) + Cl (g) 2NOCl(g)⎯⎯→ is given by rate = k[NO][Cl2]. The reaction occurs in the following steps :

(i) 2 2NO(g) + Cl (g) NOCl (g)⎯⎯→

(ii) 2NOCl (g) + NO(g) 2NOCl(g)⎯⎯→

(a) Is NOCl2 a catalyst or reaction intermediate? Why?(b) Identify the rate determining step

Ans. Step I : 2 2NO(g) + Cl (g) NOCl (g)⎯⎯→

Step II : 2NOCl (g) + NO(g) 2NOCl(g)⎯⎯→

Overall reaction 22NO(g) + Cl (g) 2NOCl(g)⎯⎯→

(a) Since NOCl2 is formed in the first step and consumed in the second step. It is an intermediate.(b) ∴ The rate law is rate = k [NO] [O3]

NO and O3 are involed in the first step, it will be the rate determining step and must bethe slower step.

*14. The rate law for the reaction 2 2 22H (g) + 2NO(g) N (g) + 2H O(g)⎯⎯→ is given by

rate = k [H2][NO]2. The reaction occurs in the following steps :

(i) 2 2 2H + 2NO N O + H O⎯⎯→

(ii) 2 2 2 2N O + H N + H O⎯⎯→

What is the role of N2O in the mechanism? Identify the slow step.

Ans. Step I : 2 2 2H + 2NO N O + H O⎯⎯→

Step II : 2 2 2 2N O + H N + H O⎯⎯→

Overall reaction 2 2 22H (g) + 2NO(g) N (g) + 2H O(g)⎯⎯→

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(a) N2O is formed in the first step and consumed in the second step it is an intermediate.

(b) The rate law is rate = k [H2] [NO]2

∴ H2 and NO are involed in the first step, it will be the rate determining step and must be

the slower step.



*1. The integrated rate equation for first order reaction A ⎯⎯→ products is given by .......

a) k = 0

t

[A]2.303ln

t [A] b) k = t

0

[A]1– ln

t [A]

c) k =t

100

[A]2.303log

t [A] d) k = t

0

[A]1ln

t [A]

*2. The slope of the straight line obtained by plotting rate versus concentration of reactant fora first order reaction is .......

a) –k b) –k/2.303 c) k/2.303 d) k3. If [A]0 is the initial concentration and [A] is the concentration at time t, then for a first order

reaction, .......

a) [A]t = [A]0 ekt b) [A]t = [A]0 e – kt

c) [A]0 = [A]t ek/t d) [A]t = [A]0 e – k/t

4. For a first order reaction, the plot of log (a – x) against time(t) would give a straight line

with a slope equal to .......

a) 2.303

kb)

k–

2.303c)

2.303–

kd) 2.303k.

5. Half life period of zero order reaction is given by equation t1/2 = .......

a) 0A2k

b)0.693

kc)

0

1

[A] k d) 20

3

2k[A ]6. Some statements, about a zero order reactions, are given below.

A) Its rate is independent of the conc. of reactant.B) The plot of conc. of reactant with time is a straight line with slope = k.C) It is generally influenced by some external agency.D) The equation, [A]t = [A]o – kt, holds good for such a reaction.Among the above, the incorrect statement(s) is/area) B b) B and C c) B and D d) C and D

7. The half life period of a reaction is constant for .......a) zero order b) first order c) second order d) none

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8. For a zero order reaction, the plot of concentration of reactants Versus time is linear with .......a) +ve slope and passing through the originb) – ve slope and passing through the originc) +ve slope and not passing through the origind) – ve slope and not passing through the origin

9. The units of rate constant for a zero order reaction are ........a) litre sec – 1 b) litre mole – 1sec – 1

c) mol litre – 1 sec – 1 d) mol sec – 1

10. The time required for the completion of 3/4th of the first order reaction is .......

a) 2.303

log4k

b)2.303 1

logk 4

c)2.303 4

logk 3

d)2.303 3

logk 4

11. For a zero order reaction, A → Products, the rate equations is .......a) [A]0 – [A]t = k b) [A]t – [A]0 = kt c) [A]t – [A]0 = k d) [A]0 – [A]t = kt

• Numerical MCQs :12. If half life period for a first order reaction is 2 min. The rate constant is .......

a) 0.5 min–1 b) 1.386 min–1 c) 0.3465 min–1 d) 2 min – 1

*13. The half life of a first order reaction is 30 min and the initial concentration of the reactantis 0.1 M. If the initial concentration of reactant is doubled, then the half life of the reactionwill be .......

a) 1800 s b) 60 min c) 15 min d) 900 s

*14. The slope of a graph ln[A]t versus t for a first order reaction is –2.5 × 10–3s–1. The rateconstant for the reaction will be .......

a) 5.76 × 10–3 s–1 b) 1.086 × 10–3 s–1

c) –2.5 × 10–3 s–1 d) 2.5 × 10–3 s–1

15. A first order reaction is 90% complete in 100sec. The rate constant is .......a) 0 sec – 1 b) 0.693 x 10 – 2 s – 1

c) 2.303 x 10 – 2 s – 1 d) 0.01s – 1

• Advanced MCQs :16. For a first order reaction, the plot of rate versus conc. of reactant is .......

a) non-linear and passing through the originb) linear with the negative slopec) linear with positive slope and passing through the origind) linear with positive slope but not passing through the origin

17. Some statements are given below .......a) For a first order reaction, t1/2 = constantb) For a zero order reaction, t1/2 ∝ Initial conc.

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c) For a zero order reaction, Rate ∝ conc. of reactantd) For a first order reaction, the plot of rate versus conc. is linear with slope = - k.Among the above the false statements are .......a) B, C and D b) B and C c) A, B and D d) C and D

5.4 COLLISION THEORY, ACTIVATION ENERGY AND CATALYST : Concept Explanation :

(a) Collision theory requirements for a bimolecular chemical reaction totake place :

1. Collision between reactant molecules :(a) The basic requirement of collision theory is that the reacting species (atoms, ions or

molecules) must come together and collide in order for the reaction to occur.(b) Therefore, the rate of reaction must be equal to the rate of collision.(c) However, it has been found that the rate of reaction is much lower as compared to

the rate of collisions.(d) It follows, therefore, that all the collisions between reactant molecules are not fruitful,

i.e. do not result in formation of product.(e) This implies that the collision is not the sufficient cause for the chemical reaction

to occur.

2. Energy requirement (activation energy) :

(a) The activation energy [Ea] is defined as theminimum kinetic energy required for amolecular collision to lead to a reaction.

(b) This kinetic energy is required to arrangeouter electrons in breaking and making ofbonds.

(c) It can be seen from the graph that the fraction of molecules with kinetic energy moreor equal to activation energy is less, the rate of reaction is less than the rate or frequencyof collisions.

3. Orientation of reactant molecules :

(a) In reactions involving complex molecules, the above conditions are not sufficient forthe reaction to occur.

(b) Hence, apart from sufficient energy the colliding molecules must have properorientations.

(c) If the molecules have improper orientation they do not react even if they possessthe required kinetic energy.

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(d) Consider a reaction.

2 2 2AB + A B AB + A⎯⎯→

4. Potential energy barrier :(a) When molecules react, an intermediate or activated complex or a transition state is

formed.(b) In the formation of activated complex, atoms, molecules require energy to overcome

the repulsion between reactant molecules.(c) This energy is provided from the kinetic energy which

is converted into potential energy in the activatedcomplex.

(d) Due to high energy, activated complex is unstable,short-lived and decomposed into the products.

(e) To form activated complex, the reactant moleculesmust climb the energy barrier.

(f) Hence, only those molecules having sufficient kineticenergy collide with proper orientation result in formationof products.

(b) Arrhenius equation from collision theory of bimolecular reactions :

1. In order for a bimolecular reaction to occur the reactant molecules must collide and thecolliding molecules must have total kinetic energy equal to or greater than the activationenergy of the reaction.

2. The fraction of molecules having kinetic energy equal to or greater than the activationenergy [Ea] is given by

f = – E /RTae ...(i)

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3. Consider a second order reaction

AB + C AC + B⎯⎯→

Then collision rate for the above reaction is given byCollision rate = Z × [AB] × [C]Where Z = collision frequency and [AB] and [C] are concentration of reactants.

4. Reaction rate = P.f × collision rate= P.f.Z [AB] × [C] ...(ii)

Where f = fraction of collisions with suffcient kinetic energy [Ea]and P = fraction of collisions with proper orientations.

5. Reaction rate for a second order reaction is also given byReaction rate = k [AB] [C] ...(iii)

6. Comparing (ii) and (iii) we get;k = P.f.Z

∴ k = P.Z × – E /RTae [Form (i)]

k = – E /RTaA.e

Where A = P.Z is called a frequency factor or pre-exponential factor.7. The Arrhenius equation can be written in following forms.

k = – E /RTaA.e

∴ ln k = ln A – aE

RT

log10k = log10A – aE

2.303RTWhere k = Rate constant at absolute temperature T.Ea = Energy of activationR = gas constantA = Frequency factor or pre-exponential factor.

(c) Temperature dependence of reaction rates :

1. The kinetic energy of the molecule increases with the increase in temperature.2. The fraction of molecules possessing minimum energy (activation energy [Ea]) increases

with increase in temperature.3. Hence, the fraction of colliding molecules that possess kinetic energy Ea also increases,

hence, the rate of reaction increases with increase in temperature.4. The figure shows that the area that respresents the fraction of molecules with kinetic energy

exceeding Ea is greater at higher temperature T2 than at lower temp T1.

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5. Arrhenius equation indicates that larger the value of Ea, smaller is the rate constant andslower is the rate of reaction.

As T increases aE

RT decreases.

This causes an increase in – aE

RT , – E /RTae , k and hence, the rate of reaction.

–E /RTaa aT E /RT – E /RT k rateeincreases decreases increases increases increases increases

⇒ ⇒ ⇒ ⇒ ⇒

(d) Arrhenius equation and temperature variation :

Relation showing variation in rate constant with temperature.1. As per Arrhenius equation

k = A × – E /RTae

Where k = rate constant at temp TA = frequency factorEa = Energy of activation

2. Taking log we get

ln k = ln A – aE

RT


2.303 RT

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3. lf k1 and k2 are the rate constants at temp T1 and T2 respectively, then

log10k1 = log10A – a

1

E

2.303 RT

log10k2 = log10A – a

2

E

2.303 RT

∴ log10k2 – log10K1 = a

102

Elog A –

2.303RT

⎡ ⎤⎢ ⎥⎣ ⎦

– a

101

Elog A –

2.303RT

⎡ ⎤⎢ ⎥⎣ ⎦

∴ log10 2

1

k

k = a

1

E

2.303 RT – a

2

E

2.303 RT

= a

1 2

E 1 1–

2.303 R T T

⎡ ⎤⎢ ⎥⎣ ⎦

∴ log10 2

1

k

k = a 2 1

1 2

E T – T

2.303 R T T

⎡ ⎤⎢ ⎥⎣ ⎦

(e) Arrhenius equation and energy of activation :

1. Arrhenius equation is k = – E /RTaA.e

Where k = Rate constant at temp T.A = frequency factorEa = energy of activation

2. As the activation energy Ea decreases, theenergy barrier decreases and more reactantmolecules can cross the activation energybarrier.

3. The graph shows when Ea decreases, thefraction of molecules with kinetic energy greaterthan or equal to Ea increases.

4. As Ea decreases,

–E /RTaa a aE E /RT – E /RT k rateeincreases increasesdecreases decreases increases increases

⇒ ⇒ ⇒ ⇒ ⇒

(f) Determination of activation energy (Ea) :

1. Consider a first order reaction B C⎯⎯→

2. The rate constants for the above reaction is determined at various temperatures.3. By Arrhenius equation

k = – E /RTaA × e

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RT

∴ log10k = log10 A – aE

2.303RT

logk = a– E

2.303RT + log10 A

(y = mx + c)4. log10 k is plotted against reciprocal of temperature T.

5. The slope of the straight line graph is a– E

2.303R from which energy of activation can be

calculated as follows slope = aE

2.303R∴ Ea = slope × 2.303 R

(g) Effect of catalyst on rates of reactions :

1. A catalyst is a substance, when addedto the reactants, increases the rate ofreaction without itself being consumed.

2. The catalyst enters the reaction butdoes not appear in the balancedequation. This is because it isconsumed in one step and regeneratedin the another.

3. The catalyst provides an alternativepathway (mechanism) of loweractivation energy for the reaction tooccur.

4. The height of the barrier foruncatalysed reaction is greater thanthat in a catalysed reaction.

5. Therefore, the number of moleculesthat possesses minimum kinetic energy[Ea] increases.

6. According to Arrhenius equation.

k = – E /RTaA.e

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At constant temperature (T) andfor the same concentration,

–E /RTaa a aE E /RT – E /RT k rateeincreases increasesdecreases decreases increases increases

⇒ ⇒ ⇒ ⇒ ⇒

7. eq : 2 2 2 22H O ( ) 2H O( ) + O (g)l l⎯⎯→

The activation energy for above reaction is 76kJ mol–1 Hence, it is very slow at room temperature.In presence of I– as catalyst the activation energy decreases to 57 kJ mol–1 and the reactionbecomes considerably faster.

(h) Effect of catalyst :1. Activation energy : It lowers the activation energy of the reaction.2. Rate of forward reaction : A catalyst increases the rate of forward reaction.3. Rate of backward reaction : A catalyst increases the rate of backward reaction to the

same extent as the rate of forward reaction.Hence, a catalyst does not effect the reaction equilibrium it just helps to achieve the equilibriumfaster.

(i) Difference between catalyst and reaction intermediate :


*1. What are the requirements for a bimolecular reaction to take place?Ans. The main requirements for a bimolecular reaction to occur are :

(a) The reactant molecules must collide.(b) The colliding molecules must have total kinetic energy equal to or greater than the activation

energy of the reaction.(c) The colliding molecules must have proper orientations relative to each other.

*2. Write Arrhenius equation and explain the terms involved.Ans. The Arrhenius equation can be written in following forms.

Catalyst

1. A catalyst is present at the start of thereaction.

2. A catalyst is consumed in one step of thereaction but regenerated in next step.

3. It accelerates the rate of reaction.4. The concentration of catalyst and reaction

intermediate may appear in the rate law.

Reaction Intermediate

1. It is produced during the mechanism ofthe reaction.

2. Reaction intermediate is formed in onestep and consumed in the next step.

3. It has no effect on the rate of reaction.4. The reaction intermediate does not

appear in the rate law.

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k = A.e – –E /RTaA.e


RT


2.303RTWhere k = Rate constant at absolute temperature T.Ea = Energy of activationR = Gas constantA = Frequency factor or pre-exponential factor.

*3. Explain with the help of potential energy barrier, how does increase in temperature increasesthe rate of reaction.

Ans. Refer 5.4 (c)

*4. Define Activation Energy.Ans. Activation energy is defined as the minimum kinetic energy required for a molecular collision

to lead to a reaction.

*5. Using Arrhenius equation explain why the rate of reaction increases with increase in temperature.Ans. Refer 5.4 (c).

6. Derive an expression for Arrhenius equation and temperature variation. OR Derive an expressionshowing variation in rate constant with temperature.

Ans. Refer 5.4 (d).

*7. How will you determine activation energy with the help of a graph.(Intext question Textbook page no : 213)

Ans. Refer 5.4 (f).

*8. Explain, with the help of potential energy barrier, how does a catalyst increase the speed ofa reaction.

Ans. Refer 5.4 (g).

*9. Comment on the effect of catalyst on each of the following :a) Activation energyb) Rate of forward reactionc) Rate of backward reaction.

Ans. a) Activation energy : It lowers the activation energy of the reaction.b) Rate of forward reaction : A catalyst increases the rate of forward reaction.c) Rate of backward reaction : A catalyst increases the rate of backward reaction to the

same extent as the rate of forward reaction.Hence, a catalyst does not effect the reaction equilibrium it just helps to achieve the equilibriumfaster.

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*10. How does a catalyst differs from a reaction intermediate?Ans. Refer 5.4 (i).

*11. Explain briefly the collision theory of bimolecular reactions.Ans. Refer 5.4 (a).

12. Derive Arrhenius equation from Collision theory of bimolecular reactions.Ans. Refer 5.4 (b).


*1. The rate constant of a reaction .......a) decreases with increasing Ea b) decreases with decreasing Ea

c) is independent of Ea d) decreases with increasing temperature*2. A catalyst increases the rate of the reaction by .......

a) increasing Ea b) increasing T c) decreasing Ea d) decreasing T*3. The Arrhenius equation is

a) A =– E /RTake e b)

E / RTaAe

k c) k = E /RTaAe d) k = – RT/EaAe

4. The units of pre-exponential factor or frequency factor are same as .......a) Rate of reaction b) Rate constantc) Order of reaction d) Molecularity of reaction.

HOURS BEFORE EXAM Chemical kinetics is concerned with the rates of reactions factors affecting the rates and the

mechanisms by which they proceed. The rate of reaction is defined as the decrease is molar concentration of a reactant or increase

in molar concentration of a product per unit time.

It can be expressed as average rate during a given time interval or the instantaneous rateat a specific time.

The reaction rate depends on five factors :The concentration of reactants, temperature, presence of a catalyst nature of reactant andpressure.

The concentration dependence is given by the rate law. The rate law for a general reaction,aA + bB cC + dD⎯⎯→ , is given by the expression, rate = k[A]x[B]y, where k is the rateconstant of the reaction and x and y represent the order of the reaction with respect to reactantsA and B respectively. The overall order of the reaction is x + y.

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The order of the reaction is defind as the sum of the exponents to which the concentrationterms in the rate law are raised.

The values of x and y and hence, the order of the reaction must be determined by experimentand cannot be predicted from the stoichiometric equation of the reaction.

The values of x and y are not related to the stoichiometric coefficients of the reactants inthe balanced equation of the reaction.

The integrated rate law is a direct relationship between time and concentration of the reactants.It can be used to calculate concentrations at any time t and the time required for an initialconcentration of the reactants to reach any particular value. For the first order reaction the

integrated rate law is K = 010

t

[A]2.303log

t [A] A graph of log10[A]t versus time is a straight line with slope equal to –k/2.303. The zero order reaction are those for which the rate of the reaction is independent of the

concentration of reactants. The integrated rate law for zero order reaction is[A]t = –kt + [A]0.The graph of [A]t versus t is a straight line with a slope equal to –k. The half life of a reactionis the time needed for the reactant concentration to fall to one half of its initial value. Thehalf lives of first and zero order reactions are given by t1/2 = 0.693/k and t1/2 = [A]02k respectively.

According to the collision theory a bimolecular reaction occurs when two properly orientedmolecules with sufficent kinectic energy collide.

The minimum kinetic energy necessary for a molecular collision to form products is calledactivation energy.

According to collision theory the colliding molecules from an intermediate configuration calledactivated complex or transition state.

The potential energy barrier exists between the reactants and products. The reactant moleculesmust surmount this barrier before they are converted into products. The configuration at thetop of the barrier is an activated complex. The height of the barrier is the activation energy.

The temperature dependence of rate constant is describeal by Arrhenius equation,k = Ae–Ea/RT, where Ea is the energy activation and A is the frequency factor. The graphof log10k against reciprocal of temperature is a straight line with slope –Ea/2.303 R from whichactivation energy can be calculated. The rate constant k1 and k2 at two different temperaturesT1 and T2 are related by the equation.

a2 2 110

1 1 2

Ek T – Tlog =

k 2.303R T T⎡ ⎤⎢ ⎥⎣ ⎦

A catalyst is a substance that increases the rate of the reaction without being itself consumedin the reaction. It functions by providing an alternative mechanism of lower activation energyfor the reaction to occur.

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An elementary reaction is one that occurs in a single step. The molecularity of an elementaryreaction is the number of reactant molecules that participate in the reactions. A complex reactionsoccurs in a series of elementary reaction. The order of an elementary reaction is equal toits molecularity.

The elementary reactions are classified as unimolecular bimolecular or trimolecular dependingon the number of reactant molecules involved. The observed rate law for a complex reactiondepends on the sequence of the elementary steps and their relative rates.

The slowest step in the mechanism is called rate determining step.

A chemical species that is formed in one elementary step in the mechanism of complex reactionand consumed in the subsequent step is called a reaction intermediate.

NUMERICALS WITH SOLUTION Type - I - Rate of reaction (Average rate and instantaneous rate)

*1. Consider the reaction 2A + B 2C⎯⎯→ ,

Suppose that at a particular moment duringthe reaction, rate of disappearance of Ais 0.076 M/s,

i) What is the rate of formation of C?ii) What is the rate of consumption of B?iii) What is the rate of the reaction?

Given :Rate of disappearance of A is

d[A]

dt = 0.076 M/s

To find :

i) rate of formation of C is d[C]

dt

ii) rate of consumption of B is d[B]

dtSolution :

2A + B 2C⎯⎯→

i) The reaction shows that rate of formationof C is same as rate of disappearance ofHence,

Rate of formation of A =d[C]

dt

=– d[A]

dt= 0.076 M/s

ii) Rate of consumption of B is 1

2 of rate

of consumption of AHence, Rate of consumption of B

=– d[B]

dt=

–1

2

d[A]

dt

=–1

× 0.0762

= 0.038 M/s

c) Rate of reaction =–1

2

d[A]

dt

=– d[B]

dt=

1 d[C]

2 dt= 0.038 M/s

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*2. Consider the reaction2– – 2–

2 8 3 43I(aq) + S O (aq) I (aq) + 2SO (aq)⎯⎯→

At a particular time t, 2–4d[SO ]

–dt

= 2.2 × 10–2 M/s.What are the values of

i)–d[I ]

–dt

ii) 2–

2 8d[S O ]–

dt

iii)–3d[I ]

–dt

at the same time?

Solution :2– – 2–

2 8 3 43I(aq) + S O (aq) I (aq) + 2SO (aq)⎯⎯→

a) The given reaction shows that rate of

consumption of I– is 3

2 times of rate of

formation of 2–4SO

∴∴∴∴∴ Rate of consumption of I–

= –– d[I ]

dt

=2–4d[SO ]3

2 dt

=–23

× 2.2 × 102

= 3.3 × 10–2 M/s

b) Rate of consumption of 2–2 8S O is

1

2 of

rate of formation of 2–4SO

∴∴∴∴∴ Rate of consumption of 2–2 8S O

=2–

2 8d[S O ]–

dt

=2–4d[SO ]1

2 dt

= –21× 2.2 × 10

2= 1.1 × 10–2 M/s

c) Rate of formation of I3 is 1

2 of rate of

formation of 2–4SO

∴∴∴∴∴–3

d[I ]

dt=

2–4d[SO ]1

2 dt

= –21× 2.2 × 10

2= 1.1 × 10–2 M/s

*3. Concentrations of various species atdifferent times for reaction

2 5 2 22N O (g) 4NO (g) + O (g)⎯⎯→

Time/s [N2O5]/M [NO2]/M [O2]/M

0 0.0300 0 0200 0.0213 0.0174 0.00435400 0.0152 0.0296 0.00740600 0.0108 0.0 384 0.00960

Calculate the average rate ofdecomposition of N2O5 and the averagerates of formation of NO and O2 duringthe time interval 200s to 400s. What is theaverage rate of the reaction during thesame interval.

Given :Time/s [N2O5]/M [NO2]/M [O2]/M

0 0.0300 0 0200 0.0213 0.0174 0.00435400 0.0152 0.0296 0.00740600 0.0108 0.0 384 0.00960

To find :i) Average rate of decomposition of N2O5 = ?ii) Average rate of formation of NO and O2

during the time interval 200s to 400s = ?iii) Average rate of the reaction = ?

Solution :

2 5 2 22N O (g) 4NO (g) + O (g)⎯⎯→

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a) Average rate of decomposition of N2O5

= 2 5– [N O ]

t

Δ

Δ

=– (0.0152 – 0.0213)

400 – 200

=– (–0.0061)

200= 3.05 × 10–5 M/s

b) Average rate of formation of NO2

= 2Δ [NO ]

Δt

=0.0296 – 0.0174

400 – 200

=0.0122

200= 6.1 × 10–5 Ms–1

c) Average of formation of O2

= 2Δ [O ]

Δt

=0.00740 – 0.00435

400 – 200= 1.525 × 10–5 Ms–1

d) Average rate of reaction

= 2 5[N O ]–1

2 t

Δ

Δ

= 2[NO ]1

4 t

Δ

Δ

=2[O ]

t

Δ

Δ

=–51

× 3.05 × 102

= –51× 6.1× 10

4= 1.525 × 10–5

= 1.525 × 10–5 Ms–1

Type - II - Rate law, Rate constantand Order of reaction andMolecularity

*4. A reaction occurs by the following

mechanism :

a) 2 2 2NO (g) + F (g) NO F(g) + F(g)⎯⎯→

(slow, bimolecular)

b) 2 2F(g) + NO (g) NO F(g)⎯⎯→

(Fast, bimolecular)i) What is the overall reaction?ii) What is the molecularity of each of the

elementary steps?iii) What rate law is predicted by the

mechanism?iv) Identify the intermediate.

Given :

i) 2 2 2NO (g) + F (g) NO F(g) + F(g)⎯⎯→

(slow, bimolecular)

ii) 2 2F(g) + NO (g) NO F(g)⎯⎯→

(Fast, bimolecular)To find :

i) Overall reaction,ii) Molecularity of each of the elementary stepsiii) Mechanism predicted by the rate lawiv) Intermediate identification.

Solution :i) The overall reaction is obtained by adding

the two elementary steps of themechanism. Thus,

2 2 2

2 2

NO (g) + F (g) NO F(g) + F(g)

F(g) + NO (g) NO F(g)

⎯⎯→

⎯⎯→

2 2 22NO (g) + F (g) 2NO F(g)⎯⎯→

(overall reaction)

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ii) The step (i) and (ii) both involve tworeactant molecules each. Hence, both thesteps are bimoecular.

iii) The rate law is predicted by thestoichiometric equation of the slow, ratedetermining step (i)

2 2 2NO (g) + F (g) NO F(g) + F(g)⎯⎯→

(slow, rate determining)The predicted rate law is rate = k [NO2] [F2]

iv) The species F is a reaction intermediatebecause it is produced in step (i) andconsumed in step (ii).

*5. For the reaction,

2A + B + C AC + AB⎯⎯→ , it is found

that tripling the concentration of A2 triplesthe rate, doubling the concentration of Cdoubles the rate and doubling theconcentration of B has no effect.

i) What is the rate law?ii) Why the change in concentration of B has

no effect.Given :

i) Copy the reaction,ii) Rate of reaction is tripled = 3A2,iii) Rate of reaction is doubled = 2 C,iv) Rate of reaction = 2B

To find :i) Rate law,ii) Change in concentration of B has no effect

on the rate of reaction.Solution :

2A + B + C AC + AB⎯⎯→

i) The rate law may be represented as

Rate = k [A2]x [B]y [C]z

The concentration of [A2] is tripled.

The new rate law

rate 2 = k 3[A2]x [B]y[C]z

New rate is tripled

∴∴∴∴∴ Rate2 = 3 × rate1

3[A2]x [B]y [C]z = 3 × [A2]

x[B]y [C]z

3x [A2]x = 3 [A2]

x

∴∴∴∴∴ 3x = 31

∴∴∴∴∴ x = 1

ii) Now If [C] is doubled

The new rate law will be rate3 = k [A2]x

[B]y 2[C]z

New Rate is doubled

∴∴∴∴∴ Rate3 = 2 × Rate1

∴∴∴∴∴ k [A2]x [B]y 2[C]z = 2 × k [A2]

x [B]y

[C]z

2z [C]z = 2 [C]z

2z = 21

Z = 1

iii) Now (B) is doubled

∴∴∴∴∴ Rate4 = [A2]x [2B]y [C]z

The rate reaming unchanged

Rate4 = Rate1

k [A2]x [2y] [B]y [C]z = k [A2]

x [B]y [C]z

∴∴∴∴∴ 2y = 1

∴∴∴∴∴ y = 0

(d) Hence, the rate law is

Rate = k [A2]x [B]y [C]z

= k [A2]1 [B]0 [C]1

= k [A2] [C]

*6. The reaction A + B ⎯⎯→ products is first

order in each of the reactants.(a) Write the rate law.

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∴∴∴∴∴ Hence, new rate becomes 9 time by theinitial rate.

(d) If concentration of [A] is doubled and thatof [B] is halved then new rate4

= k 2 [A] 1

[B]2

= k [A] [B]= rate1 [from (i)]Hence, the rate remains same.

*7. Consider the reaction

2A + 2B 2C + D⎯⎯→ .

From the following data, calculate the orderand rate constant of the reaction.

[A]0/M [B]0/M r0/Ms–1

0.488 0.160 0.240.244 0.160 0.060.244 0.320 0.12

Write the rate law of the reaction.Given :

i) 2A + 2B 2C + D⎯⎯→

ii) [A]0/M [B]0/M r0/Ms–1

0.488 0.160 0.240.244 0.160 0.060.244 0.320 0.12

To find :i) Rate law of the reaction,ii) Order and rate constant of the reaction.

Solution :i) Let the order of the reaction be x in A and

y in B, thenRate = k [A]x [B]y

Substituting the data given we getr1 = 0.24 = k [0.488]x [0.160]y ...(i)r2 = 0.06 = k [0.244]x [0.160]y ....(ii)r3 = 0.12 = k [0.244]x [0.320]y ....(iii)

(b) How does the reaction rate change if theconcentration of B is decreased by a factor 3?

(c) What is the change in the rate if theconcentration of each reactant is tripled?

(d) What is the change in the rate ifconcentration of A is doubled and that ofB is halved?

Given :

i) A + B ⎯⎯→ products

ii) The reaction is a first order.To find :

i) Rate law,ii) Rate of reaction = ? when [B] is decreased

by a factor 3,iii) Change in a rate if 3[B],

iv) Change in the rate if 2[A] and 1

[B]2

.

Solution :

A + B ⎯⎯→ products

(a) The above reaction is first order in eachof the reactantsRate1 = k [A] [B] ...(i)

(b) If concentration of [B] is decrease by afactor of 3

Then new Rate2 = k [A] 1

3 [B]

=1k

3 [A] [B]

= 11

× Rate3

...[From i]

Hence, new rate becomes 1

3rd of the initial

rate.(c) If concentration of each reactant is tripled

Then new Rate3= k 3[A] 3[B]= 9 k [A] [B]

∴∴∴∴∴ Rate3 = 9 × rate1

...[From (i)]

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Dividing (i) by (ii) we get;

0.24

0.06=

y

y

k[0.488] [0.160]

k[0.244] [0.160]

x

x

4 =0.488

0.244

x⎛ ⎞⎜ ⎟⎝ ⎠

4 = 2x

∴ 22 = 2x

∴ x = 2Now dividing (ii) by (iii) We get;

0.06

0.12=

k[0.244] [0.160]

k[0.244] [0.320]

x y

x y

1

2=

0.160

0.320

y⎡ ⎤⎢ ⎥⎣ ⎦

1

2=

y1

2⎡ ⎤⎢ ⎥⎣ ⎦

∴ y = 1Hence, the rate law will be rate = k[A]x [B]y

∴ Rate = k [A]2 [B]1

ii) ∴ Order of the reaction = x + y= 2 + 1= 3

In order to find rate constant [k]0.24 = k [0.488]x [0.160]y ...[From (i)]0.24 = k [0.488]2 [0.160]1

k =0.24

0.488 × 0.488 × 0.160

Order and rate constant of the reaction= 6.3M–2 s–1

*8. For the reaction 2A + B ⎯⎯→ products,

find the rate law from the following data

[A]/M [B]/M r/Ms–1

0.3 0.05 0.150.6 0.05 0.300.6 0.20 1.20

What is the rate constant and order of thereaction?

Given :

i) 2A + B ⎯⎯→ products

ii) [A]/M [B]/M r/Ms–1

0.3 0.05 0.150.6 0.05 0.300.6 0.20 1.20

To find :i) Rate law,ii) Rate constant,iii) Order of the reaction.

Solution :Let the order of the reaction be x in A andy in B then rate law is given byRate = k [A]x[B]y

Substituting the given data in aboveequation we get ;r1 = 0.15 = k [0.3]x [0.05]y ...(i)r2 = 0.30 = k [0.6]x [0.05]y ...(ii)r3 = 1.2 = k [0.6]x [0.2]y ...(iii)Dividing (i) by (ii) we get ;

0.15

0.30=

k [0.3] [0.05]

k [0.6] [0.05]

x y

x y

1

2=

0.3

0.6

x⎡ ⎤⎢ ⎥⎣ ⎦

1

2=

1

2

x⎡ ⎤⎢ ⎥⎣ ⎦

∴∴∴∴∴ x = 1Now dividing (ii) by (iii) we get;

0.3

1.2=

k[0.6] [0.05]

k[0.6] [0.2]

x y

x y

1

4=

0.05

0.2

y⎡ ⎤⎢ ⎥⎣ ⎦

363


1

4=

5

20

y⎡ ⎤⎢ ⎥⎣ ⎦

1

4=

1

4

y⎡ ⎤⎢ ⎥⎣ ⎦

∴∴∴∴∴ y = 1Hence, the rate law will be Rate = k [A]x

[B]y

Rate = k [A] [B]Order of the reaction = x + y

= 1 + 1= 2

For calculation rate constant we cansubstitute the values of x and y in Equ. ...(i)r1 = 0.15 = k [0.3]x [0.05]y

= 0.15 = k [0.3]1 [0.05]1

= 0.15 = k × 0.3 × 0.05= 0.15 = k × 0.015

k = 10 M–1 s–1

*9. The rate law for the reaction

– – –2 4 2 2 4 3C H Br +3I C H +2Br +I⎯⎯→

is rate = k [C2H4Br2] [I–]. The rate of the

reaction is found to be 1.1 × 10–4 M/s whenthe concentrations of C2H4Br2 and I– are0.12 M and 0.18 M respectively. Calculatethe rate constant of the reaction.

Given :

i) – – –2 4 2 2 4 3C H Br +3I C H +2Br +I⎯⎯→

ii) Rate of reaction = 1.1 × 10–4 M/s,iii) [C2H4Br2] = 0.12M, [I–] = 0.18M.

To find :Rate constant of the reaction

Solution :

– – –2 4 2 2 4 3C H Br +3I C H + 2Br + I⎯⎯→

is Rate = k [C2H4Br2] [I–]If [C2H4Br2] = 0.12 M, [I–] = 0.18 MThen rate = 1.1 × 10–4 M/s

∴∴∴∴∴ 1.1 × 10–4 = k [0.12] [0.18]

∴∴∴∴∴ k =–41.1× 10

0.12 × 0.18k = 5.1 × 10–3 M–1 s–1

*10. Consider the reaction,

2A + B ⎯⎯→ products.

If the concentration of A2 and B are halved,the rate of the reaction decreases by afactor of 8. If the concentration of A isincreased by a factor of 2.5, the rateincreases by the factor of 2.5 What is theorder of the reaction? Write the rate law.

Given :

i) 2A + B ⎯⎯→ products

ii) [A2] and [B] = 1

2,

iii) Rate of reaction is decreased by 1

8,

iv) [A2] is increased by 2.5,

v) Rate increases by the factor of 2.5

To find :i) Order of the reaction,ii) Rate law

Solution :

2A + B ⎯⎯→ products

i) The rate law may be represented asRate = k [A2]

x [B]y

ii) If concentration of A2 is increased of afactor by 2.5, ThenNew Rate r2 = k 2.5 [A2]

x [B]y

364


∴∴∴∴∴ The rate increases by a factor of 2.5Hence, r2 = 2.5 r1

k [2.5]x [A2]x [B]y = 2.5 k [A2]

x [B]y

∴∴∴∴∴ (2.5)x = 2.5∴∴∴∴∴ x = 1iii) Now if the concentration of A2 and B are

halved then

New rate, r3 = 21 1

k [A ] [B]2 2

x y⎧ ⎫ ⎧ ⎫⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭

∴∴∴∴∴ The rate decreases by a factor of 8,

∴∴∴∴∴ r3 = 1

8 r1

∴∴∴∴∴ k 21 1

[A ] [B]2 2

x yx y⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 1

8k [A2]

x [B]y

∴∴∴∴∴1 1

2 2

x y⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ =

1

8

∴∴∴∴∴+1

2

x y⎛ ⎞⎜ ⎟⎝ ⎠ =

31

2⎛ ⎞⎜ ⎟⎝ ⎠

∴∴∴∴∴ x + y = 3∴∴∴∴∴ 1 + y = 3 [∴ x = 1]∴∴∴∴∴ y = 3 – 1

y = 2(d) Rate law is

Rate = k [A2]x [B]y

= k [A2]1 [B]2

Order of the reaction = x + y= 1 + 2= 3

*11. Consider the reaction C + D products. Therate of the reaction increases by a factorof 4 when concentration of C is doubled.The rate of the reaction is tripled when

concentration of D is tripled. What is theorder of the reaction? Write the rate law.

Given :

i) C + D products⎯⎯→

ii) Rate of reaction increases by 4, when [C]is doubled.

iv) Rate of reaction is tripled, when[D] istripled.

To find :i) Order of the reaction,ii) Rate law

Solution :

C + D products⎯⎯→

i) The rate law may be representated asRate = k [C]x [D]y

If concentration of C is doubles than Newrate r2 = k [2C]x [D]y

∴∴∴∴∴ The rate increases by a factor of 4∴∴∴∴∴ r2 = 4r1∴∴∴∴∴ k2x [C]x [D]y = 4 k [C]x [D]y

∴∴∴∴∴ 2x = 42x = 22

x = 2If concentration of D is tripled, then Newrate r3 = k [C]x [3D]y

∴∴∴∴∴ The rate is also tripled∴∴∴∴∴ r3 = 3r1∴∴∴∴∴ k [C]x 3y [D]y = 3 k [C]x [D]y

∴∴∴∴∴ 3y = 3∴∴∴∴∴ y = 1

Rate law will be rate = k [C]x [D]y

Rate = k [C]2 [D]1

ii) ∴∴∴∴∴ Order of the reaction = x + y= 2 + 1= 3

*12. The reaction

2 2 2F (g) + 2ClO (g) 2FClO (g)⎯⎯→ is

365


iii) k = 0.42 M–2s–1,iv) [H2] = 0.015M, [NO] = 0.025M

To find :Rate of the reaction = ?

Solution :

2 2 22H (g)+2NO(g) 2H O(g)+ N (g)⎯⎯→

i) The above reaction is first order in H2 and

second order in NO

∴∴∴∴∴ Rate law will be

Rate = k [H2] [NO]2 ..(i)

ii) If [H2] = 0.015 M, [NO] = 0.025M

k = 0.42 M–2 s–1

Then rate = k [H2] [NO]2

= 0.42 × 0.015 × [0.025]2

Rate = 3.94 × 10–6 Ms–1

*14. The rate of reaction A ⎯⎯→ products

1.25 × 10–2 M/s when concentration of Ais 0.45M. Determine the rate constant ifthe reaction is (a) first order in A (b) secondorder in A.

Given :i) Rate of reaction = 1.25 × 10–2 M/s,ii) [A] = 0.45 M.

To find :i) Rate constant (k) if reaction is first order

in A.ii) Rate constant (k) if reaction is second

order in A.Solution :

A ⎯⎯→ products

i) If the reaction is first order in A then ratelaw will beRate = k [A]

first order in each of the reactants. Therate of the reaction is 4.88 × 10–4 M/s when[F2] = 0.015 M and [ClO2] = 0.025 M.Calculate the rate constant of the reaction.

Given :

i) 2 2 2F (g) + 2ClO (g) 2FClO (g)⎯⎯→

ii) The reaction is a first order,iii) Rate of reaction = 4.88 × 10–4 M/s,iv) [F2] = 0.015M, [ClO2] = 0.025M.

To find :Rate constant of the reaction = ?

Solution :

2 2 2F (g) + 2ClO (g) 2FClO (g)⎯⎯→

The above reaction is first order in eachof the reactants

∴∴∴∴∴ Rate = k [F2] [ClO2]If [F2] = 0.015 M, [ClO2] = 0.025 Mthen rate = 4.88 × 10–4 M/s

∴∴∴∴∴ 4.88 × 10–4 = k [0.15] [0.025]

k =–44.88 × 10

0.015 × 0.025∴∴∴∴∴ k = 1.3 M–1 s–1

Rate constant of the reaction= 1.3 M–1 s–1

*13. The reaction

2 2 22H (g) + 2NO (g) 2H O(g) + N (g)⎯⎯→

is first order in H2 and second order in NO.The rate constant of the reaction at a certaintemperature is 0.42 M–2s–1. Calculate therate when [H2] = 0.015 M and[NO] = 0.025 M.

Given :

i) 2 2 22H (g) + 2NO (g) 2H O(g) + N (g)⎯⎯→

ii) The reaction is a first order in H2 andsecond order in NO,

366


1.25 × 10–2 = k × 0.45

∴ k =–21.25 × 10

0.45k = 2.778 × 10–2 s–1

ii) If the reaction is second order in A thenrate law will beRate = k [A]2

1.25 × 10–2 = k × (0.45)2

∴ k =–21.25 × 10

0.45 × 0.45[k = 6.173 × 10–2 M–1s–1]

Type - III - Integrated rate laws*15. In acidic solution sucrose is converted to

a mixture of glucose and fructose in pseudofirst order reaction. It has been found thatthe concentration of sucrose decreasedfrom 20 mmol L–1 to 8 mmol L–1 in 38minutes. What is the half life of thereaction?

Given :i) [sucrose] = [A]0 = 20 m mol L–1

ii) [sucrose] = [A]t = 8 m mol L–1

iii) t = 38 miniv) The reaction is a pseudo first order

reaction.To find :

i) (k) Rate constant = ?ii) half life of the reaction (t1/2) = ?

Solution :i) Initial concentration of sucrose

= [A]0 = 20 mmol L–1

Final concentration of sucrose= [A]t = 8 mmol L–1, t = 38 mimSubstituting the above values in the firstorder equation

k =0

10t

[A]2.303. log

t [A]

= 102.303 20

. log38 8

=2.303

38 × 0.3979

k = 0.02411 min–1

ii) 1/2t =0.693

k

=0.693

0.02411

1/2t = 28.74 min

*16. The half life of a first order reaction is1.7 hours. How long will it take for 20%of the reactant to disappear?

Given :i) (t1/2) = 1.7 hours,ii) 20% of the reactant to disappear.

To find :t = ? (hours or min)

Solution :i) For a first order reaction

1/2t =0.693

k

∴ k =1/2

0.693

t

=0.693

1.7k = 0.4076 h–1

ii) For 20% of the reactant to disappear initialconcentration [A]0 of reactant = 100Final concentration remaining at time t =[A]t = 100 – 20 = 80

∴ k = 010

t

[A]2.303. log

t [A]

367


ii) Now if [A]0 = 100, [A]t = 100 – 90 = 10

k = 0

10t

[A]2.303log

t [A]⋅

0.035 = 102.303 100

logt 10

⋅

t = 2.303 × 1.0

0.035t = 65.8 min

iii) For a first order reaction

1/2t =0.693

k

=0.693

0.035

1/2t = 19.8 min

*18. The rate constant of a first order reactionis 6.8 × 10–4 s–1. If the initial concentrationof the reactant is 0.04 M, what is its molarityafter 20 minutes? How long will it take for25% of the reactant to react?

Given :i) (k) Rate constant = 6.8 × 10–4 s–1.ii) [A]0 = 0.04M,iii) t = 20 mins

= 20 × 60 = 1200 sec.iv) 20% of reactant react

To find :i) [A]t = ?,ii) Reactant remaining = ?iii) t = ?

Solution :i) In a first order reaction

If k = 6.8 × 10–4 s–1

Initial conc [A]0 = 0.04 M.

=10

2.303 100log

t 80⋅

0.4076 =10

2.303log 1.25

t⋅

t =2.303

× 0.09690.4076

t = 0.5483 hours

*17. The gaseous reaction 2A 2A⎯⎯→ is first

order in A2. After 12.3 minutes 65,% ofA2 remains undecomposed. How long willit take to decompose 90% of A2? Whatis the half life of the reaction?

Given :i) [A]0 = 100,ii) [A]t = 65,iii) t = 12.3 minsiv) The reaction is of first order.

To find :i) t = ?,ii) t1/2 = ?

Solution :i) [A]0 = 100, [At] = 65, t = 12.3 mins

For a first order reaction

k = 010

t

[A]2.303log

t [A]⋅

Then k = 010

t

[A]2.303log

t [A]⋅

k = 102.303 100

log12.3 65

⋅

k =2.303 × 0.1871

12.3

k = 0.0350 min–1

368


t = 20 mins = 20 × 60 = 1200 ses.

Then k = 2.303

t log10 0

t

[A]

[A]

6.8 × 10–4 =2.303

1200 log10

0

t

[A]

[A]

∴∴∴∴∴ log10 0

t

[A]

[A]=

–468 × 10 × 1200

2.303= 0.3543

∴∴∴∴∴ Taking antilog an both sides we get

0

t

[A]

[A] = AL [0.3543] = 2.26

∴∴∴∴∴t

0.04

[A] = 2.26

∴∴∴∴∴ [A]t = 0.04

2.28[A]t = 0.0177 M

ii) If 25% of reactant reactsThen Reactant reacted= 25% of [A]0

=25

× 0.04100

= 0.01 MReactant remaining = [A]t

= 0.04 – 0.01Reactant remaining = 0.03 M.

Now, k = 2.303

t log10 0

t

[A]

[A]

6.8 × 10–4 = 2.303

t log10

0.04

0.03

iii) t = –4

2.303 × 0.125

6.8 × 10= 423 secs.

t = 7.05 min.

*19. The rate constant of a certain first orderreaction is 3.12 × 10–3 min–1.

i) How many minutes does it take for thereactant concentration to drop to 0.02 Mif the initial concentration of the reactantis 0.045 M?

ii) What is the molarity of the reactant after1.5h?

Given :i) (k) Rate constant = 3.12 × 10–3 min–1,ii) [A]0 = 0.045 M,iii) [A]t = 0.02 M,iv) t = 1.5 hrs.

To find :i) [A]t = ?ii) t = (in mins)

Solution :i) [A]0 = 0.045 M and [A]t = 0.02M

k = 3.12 × 10–3 min–1

Then k = 010

t

[A]2.303log

t [A]⋅

3.12 × 10–3 = 102.303 0.045

logt 0.02

⎛ ⎞⋅ ⎜ ⎟⎝ ⎠

t =10

–3

2.303 × log 2.25

3.12 × 10

= –3

2.303 × 0.3522

3.12 × 10 = 259.9 min

t ≈≈≈≈≈ 260 minsii) If [A]0 = 0.045 M, t = 1.5 hrs. k = 3.12

× 10–3 min–1

k = 010

t

[A]2.303log

t [A]⋅

3.12 × 10–3 = 0

10t

[A]2.303log

1.5 × 60 [A]⋅

∴∴∴∴∴ log10 0

t

[A]

[A]=

–33.12 × 10 × 1.5 × 60

2.303= 0.122

369


Taking Antilog on both sides

∴∴∴∴∴ 0

t

[A]

[A]= Al [0.122]

∴∴∴∴∴ 0

t

[A]

[A]= 1.324

∴∴∴∴∴ [A]t =0[A]

1.324

[A]t =0.045

1.324 = 0.034 M

[A]t = 0.034 M.

*20. The concentration of a reactant in a first

order reaction A products⎯⎯→ , varieswith time as follows :

t/min 0 10 20 30 40 [A]/M 0.0800 0.0536 0.0359 0.0241 0.0161

Show that the reaction is first order.Given : t/min 0 10 20 30 40 [A]/M 0.0800 0.0536 0.0359 0.0241 0.0161

To find :Rate of reaction is a first order.

Solution :i) If [A]0 = 0.08 M, [A]t = 0.0536, t = 10

mins.

k = 010

t

[A]2.303log

t [A]⋅

k = 102.303 0.08

log10 0.0536

⋅

=2.303

10 . (log10 0.08 – log10 0.0536)

=2.303

× (2.9031 – 2. 7292)10

=2.303

10 × 0.1739

k = 0.04 min–1.

(b) If [A]0 = 0.08M, [A]t = 0.0359, t = 20 min.

k = 010

t

[A]2.303log

t [A]⋅

k =10

2.303 0.08log

20 0.0359⎛ ⎞⋅ ⎜ ⎟⎝ ⎠

k = 2.303

20. [log10 0.08 – log10 0.0359]

= ( )2.303× 2.9031 – 2.5551

20

= 2.303 × 0.3480

20

k = 0.04 min–1

Since the value of rate constant usingfirst order rate law equation comesconstant, the reaction is of first order.

*21. From the following data for the liquid phase

reaction A B⎯⎯→ , determine the order

of the reaction and calculate its rateconstant.

t/s 0 600 1200 1800[A]/mol L–1 0.624 0.446 0.318 0.226

Given :t/s 0 600 1200 1800

[A]/mol L–1 0.624 0.446 0.318 0.226

To find :i) Order of the reactionii) rate constant

Solution :i) If [A]0 = 0.624 mol L–1

[A]t = 0.446 mol L–1

t = 600s

370


k = 010

t

[A]2.303log

t [A]⋅

= 102.303 0.624

logt 0.446

⋅

=2.303

600 . (log100.624 – log100.446)

=2.303

× (1.7952 – 1.6493)600

=2.303

× 0.1459600

k = 5.6 × 10–4 s–1

ii) If [A]0 = 0.624 mol L–1

[A]t = 0.318 mol L–1

t = 1200s

k = 010

t

[A]2.303log

t [A]⋅

= 102.303 0.624

logt 0.318

⎛ ⎞⋅ ⎜ ⎟⎝ ⎠

=2.303

1200 . (log10 0.624 – log10 0.318)

=2.303

× (1.7952 – 1.5024)1200

=2.303 × 0.2928

1200

k = 5.61 × 10–4 s–1

Since the values of rate constant using firstorder rate law equation comes constant.The reaction is of first orderThe rate constant k = 5.6 × 10–4 s–1.

*22. Show that the time required for 99.9%completion of a first order reaction is threetimes the time required of 90% completion.

Given :i) [A]0 = 100,

ii) [A]t = 100 – 99.9 = 0.1To find :

t99.9% = 3t90%

Solution :For 99.9% completion of first order reaction[A]0 = 100, [A]t = 100 – 99.9 = 0.1

k = 010

t

[A]2.303log

t [A]⋅

k = 1099.9%

2.303 100log

t 0.1⋅

k = 1099.9%

2.303log 1000

t⋅

k =99.9%

2.303 × 3

t...(i)

For 90% completion of first order reaction[A]0 = 100, [A]t = 100 – 90 = 10

k = 010

t

[A]2.303log

t [A]⋅

k = 1090%

2.303log 10

t⋅

k =90%

2.303

t...(ii)

Equating (i) and(ii) we get;

99.9%

2.303 × 3

t =

90%

2.303

t∴∴∴∴∴ t99.9% = 3 × t90%

Hence, time required for 99.9%completion of a first order reaction is3 times the time required for 90%completion of the reaction.

*23. A flask contains a mixture of A and B. Boththe compounds decompose by first order

371


kinetics. The half lives are 60 min for Aand 15 min for B. If initial concentrationsof A and B are equal, how long will it takefor the concentration of A to be three timethat of B?

Given :i) [A]0 = [B]0 = M,ii) [A]t = 3x,

iii) [B]t = x,

iv) t1/2 for A = 60 mins,v) t1/2 for B = 15 mins,

To find :t = ? when [A] = 3[B]

Solution :

For A : 1/2t = 60 min; For B : 1/2t = 15minLet the initial concentration of[A]0 = [B]0 = Mmol dm–3

After time t, let the concentration of[B]t = x and then [A]t = 3x

kA =1/2

0.693

t =0.693

60= 0.01155 min–1

kB =1/2

0.693

t=

0.693

15= 0.0462 min–1

kB = 0

t

[B]2.303. log10

t [B]

kB = 102.303 M

. logt x

∴ log10 M

x

= Bk × t

2.303

=0.0462 × t

2.303

log10 M

x = 0.02 t. ...(i)

Now

kA = 2.303

t. log10

0

t

[A]

[A]

=2.303

t. log10

M

3x

= 10 102.303 M 1

log + logt 3x

⎛ ⎞⎜ ⎟⎝ ⎠

= ( )1 310 10

2.3030.02t + (log – log )

t

kA =2.303

t ( 0.02 t – 0.4771)

kA = 2.303 × 0.02

=2.303 × 0.4771

t

∴∴∴∴∴ 0.01155 = 0.04606 1.1

–t

∴∴∴∴∴1.1

t = 0.04606 – 0.01155

1.1

t = 0.03451

∴∴∴∴∴ t = 31.8 min.

Type - IV - Collision theory,activation energy and catalyst

*24. The rate constants of a first order reactionare 0.58s–1 at 313 K and 0.045 s–1 at 293K. What is energy of activation for thereaction?

Given :i) (k) Rate constant = 0.58 s–1

ii) T1 = 293k, T2 = 313kiii) K1 = 0.045 s–1, K2 = 0.58 s–1

372


To find :Ea = ?

Solution :At Temp T1 = 293K k1 = 0.045s–1

Temp T2 = 313K k2 = 0.58s–1

By Arrhenius Equation,

210

1

klog

k = a 2 1

2 1

E T – T

2.303R T T

⎛ ⎞⎜ ⎟⎝ ⎠

100.58

log0.045

= aE 313– 293

2.303×8.314 313×293⎛ ⎞⎜ ⎟⎝ ⎠

log100.58 – log100.045

=aE × 20

2.303 × 8.314 × 313 × 293

( )1.7634 – 2.6532

= aE × 20

2.303 × 8.314 × 313 × 293

1.1102 = aE × 20

2.303 × 8.314 × 313 × 293

∴ Ea = 1.1102×2.303×8.314×313×293

20

Ea = 9.746 × 104 J mol–1

Ea = 97.47 kJ mol–1

Energy of activation = Ea = 97.47 kJmol–1.

*25. The energy of activation for a first orderreaction is 104 kJ mol–1. The rate constantat 25ºC is 3.7 × 10–5 s–1. What is the rateconstant at 30ºC?

Given :i) Ea = 104 kJ mol–1,ii) (k1) Rate consant at 25ºC = 3.7 × 10–5s–1,iii) T1 = 25ºC,iv) T2 = 30ºC.

To find :k2 = ? at 30ºC.

Solution :Energy of activationEa = 104 kJ mol–1

= 104 × 103 J mol–1

At Temp T1 = 25ºC= 25 + 273 = 298K,

k1 = 3.7 × 10–5 s–1

At Temp T2 = 30ºC= 30 + 273 = 303K

log10 2

1

k

k= a 2 1

2 1

E (T – T )

2.303R (T T )⋅

=3104 × 10 (303 – 298)

2.303 × 8.314 × 303 × 298

log10 2

1

k

k= 0.3

∴ 2

1

k

k= Antilog (0.3)

2

1

k

k= 1.995

k2 = 1.995 × k1

= 1.995 × 3.7 × 10–5

k2 = 7.4 × 10–5 s–1

Rate Constant at 30ºC = 7.4 × 10–5 s–1.

*26. What is the activation energy for a reactionwhose rate constant doubles whentemperature changes from 30ºC to 40ºC?

Given :i) k2 = 2k1,ii) T1 = 30ºC & T2 = 40ºC.

To find :Ea = ?

Solution :If Temp T1 = 30ºC = 30 + 273 = 303K

373


Then Rate constant = k1

If Temp T2 = 40ºC = 40 + 273= 313K

Then Rate constant k2 = ?Now k2 = 2k1

log10 2

1

k

k=

a 2 1

2 1

E (T – T )

2.303R (T .T )

log10 1

1

2k

k= aE (313 – 303)

2.303 × 8.314 × 313 × 303

log10 2 = aE × 10

2.303 × 8.314 × 313 × 303

Ea = 0.3010 × 2.303 × 8.314 × 313 × 303

10

= 5.466 × 104 J mol–1

= 54.66 kJ mol–1

Theenergy of activation = 54.66 kJmol–1.

*27. The activation energy for a certain reactionis 334.4 kJ mol–1. How many times largeris the rate constant at 610 K than the rateconstant at 600 K?

Given :i) Ea = 334.4 kJ mol–1

ii) T1 = 600k, T2 = 610 kTo find :

2

1

k

k = ?

Solution :T1 = 600K, T2 = 610KEa = 334.4 kJ mol–1

= 334.4 × 103 J mol–1

Now log10 2

1

k

k = a 2 1

2 1

E T – T

2.303R T T⎛ ⎞⎜ ⎟⎝ ⎠

=3334.4 × 10 (610 – 600)

2.303 × 8.314 × 610 × 600

log10 2

1

k

k= 0.4771

∴ 2

1

k

k= Antilog (0.4771)

∴ 2

1

k

k = 3

∴ k2 = 3k1

∴∴∴∴∴ Rate constant increases by 3 times.

*28. The rate of a reaction at 600K is 7.5 ×105 times the rate of the same reaction at400K. Calculate the energy of activationfor the reaction. (Hint : The ratio of ratesis equal to the ratio of rate constants.)

Given :(i) Rate of a reaction = 7.5 × 105 (at 600k)(ii) R2 = R1,(iii) T1 = 400k, T2 = 600k

To find :(i) Ea = ?

Solution :Let T2 = 600K,

T1 = 400K,

Then 2

1

Rate

Rate = 2

1

k

k = 7.5 × 105

Now log10

2

1

k

k = a 2 1

2 1

E T – T

2.303R T . T

⎛ ⎞⎜ ⎟⎝ ⎠

log10 7.5 × 105

= aE (600 – 400)

2.303 × 8.314 × 600 × 400

5.8751 = aE × 200

2.303 × 8.314 × 600 × 400

374


∴ Ea =5.8751× 2.303 × 8.314 × 600 × 400

200

= 1.35 × 105 kJ mol–1

= 135 × 103Jmol–1

Ea = 135 kJ mol–1.

*29. The rate constant of a first order reactionat 25ºC is 0.24 s–1. If the energy ofactivation of the reaction is 88 kJ mol–1,at what temperature would this reactionhave rate constant of 4 × 10–2 s–1?

Given :i) (k2) Rate constant = 0.24 s–1,ii) Ea = 88 kJ mol–1,iii) (k1) Rate constant = 4 × 10–2 s–1,iv) T2 = 25ºC.

To find :T1 = ?

Solution :Let T2 = 25ºC = 25 + 273 = 298KThen k2 = 0.24 s–1

Then If k1 = 4 × 10–2 s–1

Then T1 = ?

Now log10 2

1

k

k = a 2 1

2 1

E (T – T )

2.303R (T . T )

log10 0.24

0.04⎛ ⎞⎜ ⎟⎝ ⎠

=3

2 1

2 1

T – T88 × 10

2.303 × 8.314 T . T

⎛ ⎞⎜ ⎟⎝ ⎠

log10 (0.24) – log10 0.04

= 3

1 2

88 × 10 1 1–

2.303 × 8.314 T T

⎛ ⎞⎜ ⎟⎝ ⎠

( )3

1 2

88 × 10 1 11.3802 – 2.6021 = –

2.303 × 8.314 T T

⎛ ⎞⎜ ⎟⎝ ⎠

0.7781 = 3

1 2

88 × 10 1 1–

2.303 × 8.314 T T

⎛ ⎞⎜ ⎟⎝ ⎠

1 2

1 1–

T T =

0.7781 × 2.303 × 8.314388 × 10

1 2

1 1–

T T= 1.7 × 10–4

1

1 1–

T 298= 1.7 × 10–4

1

1

T= 1.7 × 10–4 +

1

298= 1.7 × 10–4 + 0.003356= 0.00017 + 0.003356

∴1

1

T= 0.003526

T1 =1

0.003526T1 = 283.6 K.

The temp at which rate constant4 × 10–2 s–1 is 283.6 K.

*30. The rate constant for a reaction at 500ºCis 1.6 × 103 M–1 s–1. What is thefrequency factor of the reaction if itsenergy of activation is 56 kJ mol–1.

Given :i) (k) Rate constant = 1.6 × 103 M–1 s–1,ii) T = 500ºC = 500 + 273 = 773k,iii) Ea = 56 kJ mol–1

To find :Arrhenius frequency factor (A) = ?

Solution :At T = 500 + 273 = 773 K

k = 1.6 × 103 M–1 s–1

Ea = 56 kJ mol–1

= 56 × 103 J mol–1

Now by Arrhenius equation;

k = –E /RTaA × e

∴ log10k = log10A a– E

2.303 RT

375


log10k = –7.6959∴ k = Antilog [–7.6959]

= AL[8.3041]k = 2.01 × 10–8 s–1

Rate constant = k = 2.01 × 10–8 s–1.

*32. The half life of a first order reaction is 900min at 820 K. Estimate its half life at 720K if the energy of activation of the reactionis 250 kJ mol–1.

Given :

i) 1/2 2(t ) = 900 mins,

ii) T2 = 820 k,iii) T1 = 720 k,iv) Ea = 250 kJ mol–1,

To find :

1/2 1(t ) = ?

Solution :At Temp T2 = 820K 1/2t = 900 minIf Temp T1 = 720K ( 1/2t ) = ?Ea = 250 kJ mol–1

= 250 × 103 J mol–1

For a first order reaction

1/21

tk

∝

∴( )

( )1/2 1

1/2 2

t

t = 2

1

k

k ...(i)

Now log10 2

1

k

k

⎛ ⎞⎜ ⎟⎝ ⎠ =

a 2 1

2 1

E T – T

2.303R T × T

⎛ ⎞⎜ ⎟⎝ ⎠

log10 1/2 1

1/2 2

(t )

(t )=

3250×10

2.303×8.314 × (820–720)

(820×720)

log10 ( )1/2 1t

900=

3250 × 10 × 100

2.303 × 8.314 × 820 × 720

log10 ( )1/2 1t

900 = 2.2115

log101.6 × 103

= log10A – 356 × 10

2.303 × 8.314 × 773

3.2041 + 356 × 10

2.303 × 8.314 × 773 = log10A

log10A = 3.2041 + 3.784log10A = 6.9881

∴ A = Antilog (6.9881)A = 9.729 × 106 M–1 s–1.

Note : The unit of frequency factor issame as rate constant.

Frequency factor =A = 9.729 × 106 M–1 s–1

*31. A first order gas-phase reaction has anenergy of activation of 240 kJ mol–1. If thefrequency factor of the reaction is 1.6 × 1013

s–1, calculate its rate constant at 600 K.Given :

i) Ea = 240 kJ mol–1,ii) (A) frequency factor = 1.6 × 1013 s–1,iii) T = 600k.

To find :(k) Rate constant = ?

Solution :Energy of activationEa = 240 kJ mol–1

= 240 × 103 J mol–1

Frequency factor = A = 1.6 × 1013 s–1

T = 600KBy Arrhenius equation


2.303RT

= log10 1.6×1013– 3240 × 10

2.303 × 8.314 × 600= 13.2041 – 20.9

376


∴( )1/2 1t

900= Antilog (2.211)

( )1/2 1t

900= 1.626 × 102 × 900

( )1/ 2 1t

900= 1.463 × 105 min

The half life at 720K will be 1.463 × 105 min.

NUMERICALS FOR PRACTICETYPE - I - RATE OF REACTION AVERAGE RATE AND INSTANTANEOUS RATE :

*1. Write the expressions for the following reaction in terms of rate of consumption of reactants

and formation of products. 2 2 2F (g) + 2ClO (g) 2FClO (g)⎯⎯→

Solution :

Rate of consumption of F2 at time t = 2d[F ]–

dt

Rate of consumption of ClO2 at time t = 2d[ClO ]–

dt

Rate of formation of FClO2 at time t = 2d[FClO ]–

dt

Rate of reaction at time t = 2d[F ]–

dt = 2d[ClO ]1

2 dt = 2d[FClO ]1

2 dt*2. Consider the reaction,

2 5 2 22N O (g) 4NO (g) + O (g)⎯⎯→ in liquid bromine. At a particular moment during the

reaction N2O5 disappears at a rate of 0.02 M/s. At what rates NO2 and O2 are formed?What is the rate of the reaction?

Solution :

2 2d[N O ]–

dt = 0.02 M/s

The reaction shows that the rate of formation of NO2 is twice the rate of consumption ofN2O5. Hence,

2d[NO ]

dt = 2 5d[N O ]

–2dt

= –2 × (–0.02 M/s) = 0.04 M/s

377


The rate of consumption of N2O5 is twice the rate of formation of O2.

2 5d[N O ]–

dt= 2d[O ]

–2dt

or 2d[O ]

dt= 2 5d[N O ]1

–2 dt

Hence, 2d[O ]

dt=

1– × (– 0.02 M/s)

2

= 0.01 M/s

Rate of reaction = 2 5d[N O ]1–

2 dt

= 2d[NO ]1

4 dt

= 2d[O ]

dt

=1

– × (– 0.02 M/s)2

= 0.01 M/s

*3. Nitrogen dioxide decomposes to nitric oxide and molecular oxygen as

2 22NO (g) 2NO(g) + O (g)⎯⎯→

The concentration time data for the consumption of NO2 at 300ºC are as follows :

Time/s 0 100 200 300

[NO2]/M 8.4 × 10–3 5.6 × 10–3 4.3 × 10–3 3.0 × 10–3

Calculate the average rate of decomposition of NO2 and the average rates of formation ofNO and O2 during the time interval 200 s to 300 s. What is the average rate of the reactionduring the same interval?

Solution :Average rate of decomposition of NO2

= 2[ΝΟ ]–

t

Δ

Δ =

–3 –33.0 × 10 (M) – 4.3 × 10 (M)–

300(s) – 200(s) = 1.3 × 10–5 M/s

Average rate of formation od NO

=[ΝΟ]

–t

Δ

Δ = 2[ΝΟ ]

–t

Δ

Δ = 1.3 × 10–5 M/s

Average rate of formation of O2

378


= 2[Ο ]–

t

Δ

Δ = 2[ΝΟ ]1

–2 t

Δ

Δ =

–51.3 × 10 M/s

2 = 6.5 × 10–6 M/s

Average rate of reaction = 2[ΝΟ ]1–

2 t

Δ

Δ =

1 [ΝΟ]

2 t

Δ

Δ = 2[Ο ]

t

Δ

Δ = 6.5 × 10–6 M/s

TYPE - II - RATE LAW, RATE CONSTANT AND ORDER OF REACTION AND MOLECULARITY :

*4. Write the rate law for the reaction A + B C⎯⎯→ , from the following data :

Initial [A]/M Initial [B]/M Initial rate/Ms–1

0.4 0.2 4.0 × 10–5

0.6 0.2 6.0 × 10–5

0.8 0.4 3.2 × 10–4

What is the value of k?[Ans : k = 2.5 × 10–3 M–2 s–1]

*5. The rate law for the reaction A + B ⎯⎯→ products is rate = k [A][B]2. The rate of the reaction

at 25ºC is found to be 0.25 M/s when [A] = 1.0 M and [B] = 0.2 M. Calculate the rate constantk at this temperature.

[Ans : 6.25 M–2 s–1]

*6. The rate law for the reaction, 2NOBr (g) 2NO(g)⎯⎯→ is rate = k[NOBr]2. The rate constant

for the reaction at a certain temperature is 1.62 M–2 s–1. If the concentration of NOBr ata certain time is 2.00 × 10–3 M, what will be the rate of reaction?

[Ans : 6.48 × 10–6 M–2 s–1]*7. Consider the reaction . If the concentration of A is doubled at constant [B], the rate of the

reaction increases by a factor 4. If the concentration of B is doubled at constant [A], therate is doubled. What is the rate law? [Ans : rate = k [A]2[B]]

*8. Consider the reaction 2 2 22H (g) + 2NO(g) 2H O(g) + N (g)⎯⎯→ . The rate of the reaction

doubles if the concentration of H2 is doubled. If the concentration of NO is doubled the rateincreases by a factor 4. Write the rate law. [Ans : rate = k [H2][NO]2

*9. What is the order with respect to each of the reactants and overall order of the followingreaction?

(a)2 2 22NO(g) + 2H (g) N (g) + 2H O(g)⎯⎯→ rate = k [NO]2 [H2]

(b) 3 2 4CHCl (g) + Cl (g) CCl (g) + HCl(g)⎯⎯→ rate = k [CHCl3] [Cl2]1/2

[Ans : 1 + 1/2 = 3/2]

379


*10. The reaction, 22NO(g) + Cl (g) 2NOCl(g)⎯⎯→ is first order in Cl2 and second order in NO.

Write the rate law for the reaction.[Ans : rate = k [NO]2 [Cl2]

*11. The rate of the reaction, A + B P⎯⎯→ is 3.6 × 10–2 M/s [A] = 0.2M and [B] = 0.1M. Calculate

k if the reaction is 2nd order in A and first order in B. [Ans : 9.0 M–2 s–1]

*12. The rate of reaction A + 2B C + 2D⎯⎯→ is 6 × 10–4 M/s when [A] = [B] = 0.3M. What

is the overall order of the reaction if the constant at a given temperature is 2 × 10–3 s–1.[Ans : 1]

*13. The rate of a first order reaction, A B⎯⎯→ is 5.4 × 10–6 Ms–1 when [A] is 0.3M. Calculate

the rate constant of the reaction. [Ans : 1.8 × 10–5 s–1]

*14. For the reaction, A + B P⎯⎯→ . If [B] is doubled at constant [A], the rate of the reaction

doubles. If [A] is tripled and [B] is doubled, the rate of the reaction increases by a factorof 6. What is the order of the reaction with respect to each reactant and the overall orderof the reaction?

[Ans : It follows that the reaction is first order with respect to A and B eachand seconder overall.]

TYPE - III - INTEGRATED RATE LAWS :*15. The half life of a first order reaction is 0.5 min at a certain temperature. Calculate (a) the

rate constant of the reaction (b) the time required for 80% reactant to decompose.[Ans : seconds are required for 80% decomposition of the reactant]

*16. The rate constant for a first order reaction is 7.0 × 10–4 s–1. If the initial concentration ofthe reactant is 0.080 M, what concentration will remain after 35 minutes?

[Ans : The concentration that remains after 35min is 0.0184 M]*17. The half life of a first order reaction is 6.0 h. How long will it take for the concentration of

reactant to decrease from 0.8 M to 0.25 M?[Ans : Half life of reaction is 34 min]

*18. The concentration of N2O5 in liquid bromine varied with time as follows :

t/s 0 200 400 600

[N2O5]/M 0.15 0.098 0.064 0.042

Show that the reaction is first order.[Ans : This indicates the reaction obeys the integrated rate equation of first

order reaction. The reaction is a first order reaction.]

380


*19. For the decomposition of ethylene oxide into CH4 and CO, the variation of total pressure (P)of the reaction mixture with time is as given below :

t/s 0 300 600 900

P/mm 120 127.2 134.0 140.3

Show that the reaction is a first order reaction.[Ans : k = 2.0066 × 10–4 s–1]

*20. The kinetics of hydrolysis of methyl acetate in excess of dilute HCl at 298 K was followedby withdrawing 5 mL of the reaction mixture at different time intervals and titrating againststandard alkali. The following results were obtained.

t/min 0 10 20 30 ∞ Volume of alkali/mL 20.1 20.5 20.9 21.3 35.2

Show that the reaction follows first order kinetics and calculate the rate constant.[Ans : k = 2.71 × 10–3 min–1]

*21. The following results were obtained in the decomposition of H2O2 in pressure of I– ions at298 K.

t/min 3.0 6.0 9.0 ∞ Volume of O2 evolved/cm3 6.2 11.8 16.8 60.6

Show that the reaction, –I

2 2 2 21

H O (g) H O(l) + O (g)2

⎯⎯→ is a first reaction. What is the

rate constant?[Ans : k = 0.0361 min–1]

TYPE - IV - COLLISION THEORY, ACTIVATION ENERGY AND CATALYST :

*22. The rate constant for the reaction, 3 4CH CHO(g) CH (g) + CO(g)⎯⎯→ is 0.035 M–1/2 s–1

at 730 K and 0.343 M–1/2 s–1 at 790 K. Calculate the activation energy of the reaction.(R = 8.314 JK–1 mol–1).

[Ans : The energy of activation of the reaction is 182.4 kJ mol–1]*23. For a certain second order reaction energy of activation is 240 kJ mol–1. Calculate its rate constant

at 1023 K if the rate constant at 923 K is 0.0113 M–1 s–1. (R = 8.314 JK–1 mol–1][Ans : The rate constant at 1023 K is 0.24 M–1 s–1 ]

*24. If the rate of a reaction increases by a factor of 2.36, when the temperature is raised from303 K to 313 K. What is the energy activation? R = 8.314 JK–1 mol–1.

[Ans : 67.6 kJ mol–1]*25. For a first order reaction rate constant at 500 K is 7.66 × 10–4 s–1. Calculate frequency factor

if the energy of activation for the reaction is 160 kJ mol–1.[Ans : 3.96 × 1013 s–1]

381


*26. The energy of activation for a certain second order reaction is 85.2 kJ mol–1 and its frequencyfactor is 3.1 × 1011 L mol–1s–1 at 310 K. Calculate the rate constant of the reaction.

[Ans : 1.372 × 10–3 L mol–1 s–1 ]

HIGHER ORDER THINKING SKILLS (HOTS)*1. The following results were obtained in the decomposition of H2O2 in KI solution at 30ºC.

t/s 100 200 300 ∞Volume of O2 collected/cm3 7.3 13.9 19.6 65.0

Show that the reaction is first order. Calculate the rate constant of the reaction.Given :

t/s 100 200 300 ∞Volume of O2 collected/cm3 7.3 13.9 19.6 65.0

To find :i) Reaction is first order,ii) Rate constant

Solution :

i) 2 2 2 21

H O H O + O2

⎯⎯→

Volume of H2O2 decomposed ∝ volume of O2 formedwhen H2O2 is completely decomposed, at t = ∝1

∞∝ Original amount of H2O2 = [A]0

1

t (volume of O2 liberated at time t) ∝ Amount of H2O2 decomposed at time t.

∴∴∴∴∴ Amount of H2O2 remaining undecomposed at time t = 1 1

–t∞

= [A]t

ii) Hence, the first order Rate law empression.

k = 010

t

[A]2.303log

t [A]⋅

k = 10

12.303

log1 1t –

t

∞

∞

⎛ ⎞⎜ ⎟

⋅ ⎜ ⎟⎜ ⎟⎝ ⎠

= 10652.303

log65 – 7.3t⎛ ⎞⋅ ⎜ ⎟⎝ ⎠

382


= 102.303 65

logt 57.7

⎛ ⎞⋅ ⎜ ⎟⎝ ⎠

=2.303

100 (log10 65 – log10 57.7)

=2.303

× (1.8129 – 1.7612)100

=2.303

× 0.0517100

k = 1.2 × 10–3 s–1.

iii) At t = 200s, [A0] = 1

∞ = 65cm3

[A]t =1 1

–t∞

= 65 – 13.9= 51.1

∴ k =0

10t

[A]2.303log

t [A]⋅

= 102.303 65

logt 51.1

⎛ ⎞⋅ ⎜ ⎟⎝ ⎠

= 10 102.303

. (log 65 – log 51.1)200

= 2.303

200 (1.8129 – 1.7084) = 2.303 × 0.1045

200

k = 1.2 × 10–3 s–1.iv) ∴ The values of rate constant using first order rate law equation comes constant it is a first

order reaction constant value of rate constant [k = 1.2 × 10–3 s–1]

*2. From the following data for the decomposition of azoisopropane,

3 2 3 2 2 6 14(CH ) CHN= CH(CH ) N +C H⎯⎯→ estimate the rate of the reaction when total pressure

is 0.75 atm.

Time/s Total pressure/atm

0 0.65200 1.0

383


Given :i) Time/s Total pressure/atm

0 0.65200 1.0

ii) 3 2 3 2 2 6 14(CH ) CHN= CH(CH ) N +C H⎯⎯→

iii) Total Pressure = 0.75 atm.To find :

rate of reaction = ?Solution :

3 2 3 2 2 6 14(CH ) CHN= CH(CH ) N +C H⎯⎯→

At time t = 0 0.65 atm 0 0time t = 200s 0.65 – x x xTotal pressure = 0.65 – x + x + x1.00 atm = 0.65 + xx = 1 – 0.65x = 0.35 atm

∴ Initial pressure = [A]0 = 0.65 atmpressure at t = 200s =[A]t = 0.65 – x

= 0.65 – 0.35= 0.30 atm

Now, for a first order reaction

k =0

10t

[A]2.303log

t [A]⋅ = 10

2.303 0.65log

200 0.30⋅

k =2.303

200 (log 0.65 – log 0.30) =

2.303(1.8129 – 1.4771)

200

=2.303

× 0.3358200

k = 3.87 × 10–3 s–1 3 2 3 2 2 6 14(CH ) CHN= CH(CH ) N +C H⎯⎯→

At time t = 0 0.65 atm 0 0At time = t 0.65 – x x xPtotal = 0.65 – x + x + x Ptotal = 0.65 + x

∴∴∴∴∴ 0.75 = 0.65 + x∴∴∴∴∴ x = 0.10 atm

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∴∴∴∴∴ Pressure at time = t = [A]t= 0.65 – x= 0.65 – 0.10= 0.55 atm

Now, rate of first order reaction at time ‘t’ isRate = k [A]t

= k × 0.55= 3.87 × 10–3 × 0.55

Rate = 2.13 × 10–3 atm s–1

Rate of the reaction = 2.13 × 10–3 atms–1

*3. The kinetics of hydrolysis of methyl acetate in excess of dilute HCl at 298 K was followedby withdrawing 5 mL of the reaction mixture at different time intervals and titrating againststandard alkali. The following results were obtained.

t/min 0 10 20 30 ∞Volume of alkali/mL 20.1 20.5 20.9 21.3 35.2

Show that the reaction follows first order kinetics and calculate the rate constant.Given :

i) t/min 0 10 20 30 ∞Volume of alkali/mL 20.1 20.5 20.9 21.3 35.2

ii) HCl3 3 2CH COOCH (aq) + H O ⎯⎯⎯→

3 3CH COOH(aq) + CH OH(aq)To find :

i) Rate of reaction is first orderii) Rate constant (k) = ?

Solution :The hydrolysis reaction is HCl

3 3 2CH COOCH (aq) + H O ⎯⎯⎯→

3 3CH COOH(aq) + CH OH(aq)V0 = volume of alkali required for titration at t = 0

= volume of alkali required for neutralization of HCl initially presentV∞ = volume of alkali required after completion of reaction

= volume of alkali requires for HCl initially present and for acetic acid produced aftercompletion of reaction

Hence,V∞ – V0 = volume of alkali required for acetic acid produced from entire ester = [A]0 = ‘a’ Vt = volume of alkali required at different time intervals

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= volume of alkali requried for acetic acid produced at different time intervals andHCl present initially

Vt – V0 = volume of alkali required for acetic acid produced at different time intervals= x

Hence, a – x = V∞ – V0 – Vt + V0 = V∞ – VtThe rate constant of first order reaction is

k =2.303

t. log10

a

a – x

=2.303

t. log10

0

t

V – V

V – V∞

∞

(i) V∞ = 35.2 mL, V0 = 20.1 mL, Vt = 20.5mL, t = 10 minHence, k

= 102.303 35.2 mL – 20.1 mL

log10(min) 35.2 mL – 20.5 mL

⋅

= 102.303

log 1.02710(min)

= 2.66 × 10–3 min–1

(ii) Vt = 20.9 mL, t = 20 min

k = 102.303 15.1 mL

log20(min) 35.2 mL – 20.9 mL

⋅

= 102.303

log 1.05620(min)

⋅

=2.303

× 0.0236620(min)

= 2.72 × 10–3 min–1

(iii) Vt = 21.3 mL, t = 30 min

k = 102.303 15.1 mL

log30(min) (35.2 – 21.3) (mL)

⋅

= 102.303

log 1 08630(min)

.⋅

k =2.303

× 0.0358330(min)

= 2.75 × 10–3 min–1

All the k values are almost same, the reaction is of first order and meank = 2.71 × 10–3 min–1.

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*4. The rate constants for a first order reaction at various temperatures are as follows :

k/s–1 × 10–3 1.0 1.422 2.0 2.782 3.829

Temperature/ºC 25 30 35 40 45

Plot an appropriate graph of the data and calculate the energy of activation.Given :

k/s–1 × 10–3 1.0 1.422 2.0 2.782 3.829

Temperature/ºC 25 30 35 40 45To find :

i) Ea = ?ii) A graph of log10k vs 1/T.

Solution :

By Arrhenius equation


2.303RT

log10k = a10

– E+ log A

2.303RT

Slope = a– E

2.303R

Ea = –slope × 2.303 RNow slope of the above graph

= 2 1

2 1

y – y

–x x = –3 –3

–2.8471 – (–3.0)

(3.3 × 10 – 3.36 × 10 )

= –3

0.1529

– 0.06 × 10 =

–152.9

0.06= –2548

∴ Ea = –(–2548) × 2.303 × 8.314Ea = 48.79 kJ mol–1

Rate constant log10k Temp Abso-1

Tk lute T

1.0 × 10–3 –30 25 298 3.36 × 10–3

1.422 × 10–3 –2.8471 30 303 3.30 × 10–3

2.0 × 10–3 –26990 35 308 3.25 × 10–3

2.782 × 10–3 –2.55336 40 313 3.20 × 10–3

3.829 × 10–3 –2.4169 45 318 3.14 × 10–3

6General Principles andProcesses of Isolationof Elements

Chapter

SYLLABUS6.1 Concentration of ores

(a) Hydraulic washing or gravity separation(b) Hydraulic classifier(c) Magnetic separation method(d) Froth Floatation method(e) Leaching

Questions and AnswersTheoretical MCQsAdvanced MCQs

6.2 Extraction of crude metal from theconcentrated ore

(a) Calcination(b) Roasting(c) Difference between roasting and calcination(d) Smeltings(e) Flux(f ) Slag(g) Reduction by precipitation or displacement

method i.e. Hydrometallurgy(h) Electrolytic reduction or Electrometallurgy


6.3 Thermodynamic principles ofmetallurgy (Ellingham diagram)

(a) Variation of ΔG with respect to temperature(b) Ellingham diagram(c) Importance of carbon and carbon monoxide

as reducing agents using EllinghamDiagram

(d) Limitations of Ellingham diagramQuestions and AnswersTheoretical MCQsAdvanced MCQs

6.4 Purification of refining of crude metal(a) Liquation(b) Polling(c) Zone-refining(d) Electrolytic refining(e) Vapour phase refining(f) Van-Arkel method for refining Zirconium or

titanium(g) Chromatographic method


6.5 Extraction of Zinc from Zinc blende(a) Concentration of ore(b) Roasting(c) Reduction(d) Electrolytic refining of zinc(e) Physical properties(f) Uses of zinc

Questions and AnswersAdvanced MCQs

6.6 Extraction of Iron from Haemetite(a) Concentration of ore(b) Roasting(c) Smelting

Introduction:The metals which are unaffected by moisture, air, carbon dioxide from air, and non-metals occurin free state. However, such metals are very few like gold, silver, platinum, copper etc. Mostof metals occur in the combined form as carbonates, silicates, phosphates, sulphides, oxides etc.

"An investment in knowledge pays the best interest." -Benjamin Franklin

388


Common terms used in metallurgy1) Mineral : A naturally occurring substance obtained by mining which contains the metal in

free or combined state is called mineral.2) Ore : A mineral containing a high percentage of the metal, from which the metal can be

profitably extracted is called an ore.

Note : Every ore is a mineral but every mineral is not an ore.

Metal Minerals Ores

Aluminium (Al) Bauxite, Al2O3.2H2O Cryolite, Na3AlF6 Bauxite Al2O3.China clay, Al2O3.2SiO2.2H2O

Iron (Fe) Haematite, Fe2O3 Magnetite, Fe3O4 Haematite Fe2O3

Limonite, 2Fe2O3,3H2O Iron pyrite, FeS2

Siderite, FeCO3

Zinc (Zn) Zinc blende, ZnS Zincite, ZnO, Calamine, ZnCO3 Zinc blende, Zns

Magnesium Magnesite, MgCO3 Dolomite, MgCO3, CaCO3 Dolomite MgCO3,Epsum salt MgSO4,7H2O CaCO3

1) Metallurgy : The process of extraction of a metal in a pure state from its ore is called metallurgy.2) Different methods used in extraction are:

a) Pyrometallurgy : A process in which the ore is reduced to the metal at high temperatureusing suitable reducing agent like carbon, hydrogen, aluminium etc. is called pyrometallurgy.b) Hydrometallurgy : A process of extracting metals from aqueous solutions of their saltsusing suitable reducing agents is called hydrometallurgy. c) Electrometallurgy : A processof extraction of metals by electrolytic reduction of molten (fused) metallic compounds iscalled electrometallurgy.

3) Gangue : A sandy, earthy and other unwanted impurities present in the ore are called gangue.4) General principles used in metallurgy : The various steps involved in the extraction of pure

metals from their ores are : a) Concentration of ores b) Conversion of ores into oxides orother desired compounds c) Reduction of ores to form crude metals d) Refining of metals.


6.7 Extraction of Aluminium from Bauxite(a) Purification of bauxite(b) Electrolysis of pure alumina(c) Refining of Aluminium (Hoope’s process)

Questions and AnswersTheoretical MCQs

6.8 Extraction of copper from copper pyrites(a) Concentration of ore(b) Smelting(c) Bessemerisation(d) Refining

Questions and AnswersTheoretical MCQs

Hours before exam Higher Order Thinking Skills

389

Chapter - 6 General Principles and Processes of Isolation of Elements

6.1. CONCENTRATION OF ORES :


The process of removal of unwanted materials (gangue) from the ore is known asconcentration of ore or dressing or benefaction of the ore. The selection of method ofconcentration depends on the physical properties of gangue, metal to be extracted, availablefacilities and the environmental factors. Some of the important procedures are as follows.

(a) Hydraulic washing or gravity separation :This method is based on differences inthe density of mineral and gangue. Thepulverised (powdered) ore is fed on tothe top of vibrating table (Wilfley’s table)with a stream of water flowing acrossthe table. Due to jerks, the less denseparticles are carried away by water andthe heavier particles settle between thewooden cleats or riffles fixed on the table.e.g. Cassiterite, SnO2 is concentrated bythis method.

(b) Hydraulic classifier :This is based on difference in gravities ofthe ore and the gangue particles. It consistof a large conical reservoir fitted with ahooper at the top for the addition ofpowdered ore and an inlet at the bottom.There is provision to remove light gangueparticles from the side near the top andconcentrated ore from bottom. Water jet isintroduced from the inlet at the bottom inupward direction to wash the powdered ore.The lighter gangue particles are washed andthe heavier ore is left behind. This methodis also called levigation.

(c) Magnetic separation method :This method is used when the impurities are of magnetic in nature. e.g. Cassiterite (anore of tin) which contains Wolframite or Ferrous tungstate (FeWO4) which is magnetic

390


in nature. The magnetic separator consist ofleather belt moving over two rollers, one ofwhich encloses magnets in it. The powderedore is dropped over this moving belt at oneend.The impurities which are magnetic in natureare attracted by the magnet and hence, fallsjust below the magnet while the non-magnetic ore falls away from it.

(d) Froth floatation method :This method of concentrationof ore is used for the sulphideores e.g. Galena (PbS), Zincblende (ZnS) etc. It is basedon the principle that metallicparticles are wetted by oilwhile gangue particles arewetted by water. It consistof big tank containing waterfitted with stirrer. Thepowdered ore is added totank along with certain oillike pine oil, eucalyptus oil, fatty acids, xanthates etc. which produces froth when stirred bypassing a current of compressed air. Some amount of aniline or cresol is added which stabilizesthe froth. The ore particles are wetted by froth, being lighter, floats on the surface and skimmedoff while the impurities are wetted by water, becomes heavy and settles at the bottom. It ispossible to separate two sulphide ores by adjusting proportions of oil to water or byusing depressants. e.g. An ore containing ZnS and PbS, NaCN is added which selectivelyprevents ZnS coming to froth but allows PbS to come with the froth.

(e) Leaching :1. Define : It is the process of extracting a soluble material from an insoluble solid by

dissolving out in suitable solvent. Leaching is often used if the ore is soluble in somesuitable solvent.

391


2. For example, in the metallurgy of silver and gold (hydrometallurgy), the powdered oreis leached with dilute solution of NaCN or KCN in presence of air when the respectivemetal form soluble cyanides from which metal is obtained by the displacement reactionusing more electropositive or active metal live Zn.

– – –2 2 24M(s) + 8CN (aq) + 2H O(aq) + O (g) 4[M(CN) ] (aq) + 4OH (aq)

[M = Ag or Au] metalcyanide of Ag or Au⎯⎯→

–– 22 42[M(CN) ] (aq) + Zn (s) [Zn (CN) ] (aq) + 2M (s)⎯⎯→

3. Alumina is obtained from bauxite by leaching it with NaOH (Bayer’s process) or withNa2CO3 (Hall’s process).(a) Bayer’s Process : Bayer’s process is used when ore contains ferric oxide as main

impurity. The ore is digested at about 423K with the concentrated solution of NaOHin an autoclave. The Al2O3 in bauxite dissolves in NaOH to form sodium meta aluminateNaAlO2. The impurities being insoluble remains behind.

2 3 2 (s) (aq) 2(aq) 2 (l)Al O 2H O + 2NaOH 2NaAlO + 3H O⋅ ⎯⎯→

The solution is filtered to remove insoluble impurities and agitated with freshly preparedAl(OH)3, so that aluminium in NaAlO2 gets precipitated as Al(OH)3.

2 2 3NaAlO + 2H O NaOH + Al(OH)⎯⎯→

(b) Hall’s process : In this process, the bauxite ore is fused Na2CO3 is to convert Al2O3

into soluble Sodium Aluminate NaAlO2.

2 3 2 2 3 2 2 2Al O 2H O(s) + 2Na CO 2NaAlO (aq) + 3CO (g) + 2H O( )l⋅ ⎯⎯→

The solution is filtered to remove insoluble impurities. The filtered solutions is neutralizedby passing carbon dioxide gas so that Al(OH)3 is precipitated.

2 2 2 3 2 32NaAlO + CO + 3H O 2Al(OH) + Na CO⎯⎯→

The precipitated of Al(OH)3 is washed, dried and heated to obtain Al2O3.

3 2 3 22Al(OH) Al O + 3H O⎯⎯→Δ

Questions and Answers :• Answer in short :

*1. Identify the ores mentioned below which can be concentrated by magnetic separation method?Fe2O3, FeCO3, ZnO, ZnS, CuFeS2

Ans. Fe2O3, FeCO3, CuFeS2

*2. Which process is generally used for the benefication of sulphide ores?Ans. Froth floatation process.

*3. What is the role of depressants in froth floatation process?Ans. Depressants are used to prevent certain types of particles from forming the froth with bubbles

392


in froth floatation process which helps in the separation of two sulphide ores. For example,an ore containing ZnS and PbS, sodium cyanide is used as depressant. It prevents ZnS fromforming froth. Hence, lead sulphide can be separated.

*4. What is the difference between minerals and ores?Ans. (a) Mineral: A naturally occurring substance obtained by mining which contains the metal in

free or combined state is called mineral.(b) Ore : A mineral containing a high percentage of the metal, from which the metal can

be profitably extracted is called an ore.Hence, every ore is a mineral but every minerals is not ore.

*5. Which are the different methods used in metallurgy?Ans. Metallurgy is of three types (a) pyrometallurgy (b) Hydrometallurgy and (c) Electrometallurgy.

*6. Describe the magnetic separation process.Ans. Refer 6.1 (c)

*7. Explain the working of froth floatation process.Ans. Refer 6.1 (d)

*8. Explain the terms : LeachingAns. Refer 6.1 (e) 1 and 2

*9. Which are the various steps involved in the extraction of pure metals from their ores?Ans. The various steps involved in the extraction of pure metals from their ores are broadly classified

into four groups.(a) Concentration of ores.(b) Conversion of ores into oxides or other desired compounds.(c) Reduction of ores to form crude metals.(d) Refining of metals.

*10. What are the different furnaces used for extraction of metals?Ans. (a) Reverberatory furnace, (b) Blast furnace, (c) Bessemer converter (d) Hearth.


*1. The ores that are concentrated by floatation method are .......a) Carbonates b) Sulphides c) Oxides d) Phosphates

*2. Which of the following metal is obtained by leaching its ore with dilute cyanide solution .......a) Silver b) Titanium c) Vanadium d) Zinc

3. Rocky impurities present in the mineral are called .......a) Flux b) Gangue c) Matte d) Slag

4. In order to separate PbS and ZnS in froth floatation process ....... is used as depressants.a) NaCN b) KCN c) AgCN d) AuCN

393


• Advanced MCQs :5. Which one contains both iron and copper?

a) Cuprite b) Chalcolite c) Copper pyrites d) Malachite6. In the equation – – –

2 2 24M + 8CN + 2H O + O 4 [M(CN) ] + 4OH⎯⎯→ the metal M is .......a) Copper b) Iron c) Gold d) Zinc

6.2. EXTRACTION OF CRUDE METAL FROM THE CONCENTRATEDORE :

Concept Explanation :The extraction of metal from it’s concentrated ore depends on the nature of ore as well asimpurities present in it. Generally the concentrated ore is converted into a form which is suitablefor reduction i.e., it is converted into it’s oxide which can be easily reduced using suitable reducingagent. Thus, it is oxidation-reduction which involves following process.

(a) Calcination :1. Define : It is a process of heating of concentrated ore strongly below it’s melting

point in absence of air or in a limited supply of air.2. Calcination is generally carried out for the ores containing carbonates and hydrated oxide.3. During calcination

(a) Organic matter, moisture, volatile impurities like CO2 are expelled from the ore.(b) Ore becomes porous.(c) Carbonate ores are decomposed to give metal oxides and CO2.(d) The hydrated ores lose water of hydration. e.g.

Δ3 2ZnCO (s) ZnO(s) + CO (g)

(Calamine ore)⎯⎯→

Δ3 2CaCO CaO + CO (g)

⎯⎯→

Δ2 3 2 2 3 22Fe O 3H O(s) 2Fe O (s) + 3H O(g)

Limonite ore⋅

⎯⎯→

(b) Roasting :1. Roasting is carried out mainly for sulphide ores :

It is a process in which the concentrated ores are heated to high temperature,below the melting point of the metal, in a reverberatory furnace in excess of airor oxygen. The impurities like moisture, sulphur, arsenic etc. are expelled out astheir volatile oxide.

394


2. 2 2

2 2 3

4 2 2 5

S + O SO (g)

4 As + 3O 2As O

P + 5O 2P O

⎯⎯→ ↑

⎯⎯→ ↑

⎯⎯→ ↑

3. Metal sulphides are converted to metal oxides e.g.

2 2

2 2

2 2 2 2

2ZnS + 3O 2ZnO + 2SO (g)

2PbS + 3O 2PbO + 2SO (g)

2Cu S + 3O 2Cu O + 2SO (g)

⎯⎯→

⎯⎯→

⎯⎯→

4. Sometimes, the oxidation of sulphide takes place only to the sulphate stage. For example,

2 4

2 4

PbS + 2O PbSO

ZnS + 2O ZnSO

⎯⎯→

⎯⎯→

(c) Difference between roasting and calcination :

(d) Smelting :Smelting is the process of extracting the molten crude metal from it’s concentrated oreat high temperature along with reducing agent like carbon or coke. Using carbon as reducingagent has two fold advantages. Carbon itself is reducing agent and on oxidation forms carbonmonoxide which is also a reducing agent. Smelting is generally carried out in a blastfurnace.For example,

2SnO + 2C Sn + 2COCassiterite

↑⎯⎯→

Roasting

1. Ore is heated to a high temperature in thepresence of air.

2. The process is applied for sulphide ores.3. Metal sulphide ores are converted to oxide

or sulphate.

4. 2 22ZnS + 3O 2ZnO + SO (g)⎯⎯→

Calcination

1. Ore is heated in absence of air or in alimited supply of air.

2. This process is applied for Carbonate andHydrated oxides.

3. Metal carbonate are converted to Oxides.

4. 3 2ZnCO ZnO(s) + CO (g)⎯⎯→Δ

395


ΔZnO + C Zn + CO↑⎯⎯→823K

2 3 2Fe O + 3CO 2Fe + 3CO ↑⎯⎯⎯→During smelting some amount of flux is added to the concentrated ore which combinewith impurities present in the ore and forms fusible slag.

2 3FeO + SiO FeSiO(flux) (slag)

⎯⎯→Δ

(e) Flux :1. Flux is a chemical substance which is added to the concentrated ores during smelting in

order to remove the gangue of impurities by chemical reaction forming a fusible masscalled Slag.

2. Basically the selection of flux material depends on the nature of impurities (gangue) tobe removed. If the gangue is acidic in nature (e.g. SiO2) then basic flux is used.

Δ2 3SiO + FeO FeSiO

(gangue)(flux) (slag)⎯⎯→

3. If the gangue is basic in nature (e.g. CaO), then acidic flux like SiO2 is used.

Δ2 3CaO + SiO CaSiO

(slag)⎯⎯→

(f) Slag : It is a waste product formed by the combination of gangue andflux during extraction of metal by smelting process :It is lighter than molten metal and insoluble in it hence, can easily be skimmed of.It prevents the molten metal from oxidation by air as it forms separate layer over themolten metal. for e.g.

2 3CaO + SiO CaSiO(slag)

⎯⎯→

(g) Reduction by precipitation or displacement method i.e.Hydrometallurgy :The process of extraction of metals by dissolving it in suitable chemical reagent andthe precipitations of metal by more electropositive metal is called Hydrometallurgy.Some metals are reduced by displacement using more reactive metal from their complexes.For example silver and gold are extracted from their complexes by more reactive Zincmetal. In this process, the concentrated ore is dissolved in suitable solution to form theirsoluble complexes. The metal ion is than precipitated by adding Zinc dust.For example, in the metallurgy of silver and gold (hydrometallurgy), the powdered oreis leached with dilute solution of NaCN or KCN in presence of air when the respectivemetal form soluble cyanides from which metal is obtained by the displacement reactionusing more electropositive or active metal live Zn.

396


– – –2 2 24M(s) + 8CN (aq) + 2H O(aq) + O (g) 4[M(CN) ] (aq) + 4OH (aq)

[M = Ag or Au] metalcyanide of Ag or Au⎯⎯→

–– 22 42[M(CN) ] (aq) + Zn (s) [Zn (CN) ] (aq) + 2M (s)⎯⎯→

(h) Electrolytic reduction OR Electrometallurgy :

The process of extraction of metals by electrolytic process is known as electrometallurgy.

This process is employed for the extraction of highly electropositive metals like sodium,potasisum, magnesium, calcium and aluminium. The ores of these metals cannot be reducedby conventionals reducing agents like carbon, carbon monoxide etc. In this method, themetallic components such as oxide, hydroxide, halides are electrolyzed in their fused state.The metal ions are discharged at the cathode. For example, Sodium is extracted by theelectrolysis of it’s fused state NaCl.

+ –NaCl( ) +Na ClMolten

l⎯⎯→

– –2

+ –

At Anode : Cl (g) +2Cl 2eAt Cathode : + 2Na (s)2Na 2e

⎯⎯→

⎯⎯→


*1. Explain the terms (a) Roasting, (b) Smelting, (c) CalcinationAns. (a) Roasting : Refer 6.2 (b), (b) Smelting : Refer 6.2 (e) (c) Calcination : Refer 6.2 (a)

*2. What is pyrometallurgy?Ans. The process of extraction of metal by heating the metal oxide with suitable reducing agent at

high temperature is known as pyrometallurgy or thermal reduction. Depending on the nature ofoxide and metal, the extraction of metal can be carried out by using following reducing agent.Reduction with coke and carbon monoxide : In the metallurgy Al, Fe, Pb, Zn, Mg, Co etc.Reduction with Na, Al, Mg or hydrogen : In the metallurgy of Mn, Cr, As, Mo, W etc.Reduction with water gas (CO + H2) : In the metallurgy of Ni.Self reduction or Auto reduction : In the metallurgy of Pb, Hg, Cu etc.

*3. What is the meaning of hydrometallurgy?Ans. Refer 6.2 (h)

*4. Explain the following terms. (a) Gangue, (b) SlagAns. (a) Gangue : A sandy, earthy and other unwanted impurities present in the ore are called

gangue. e.g. in the extraction of iron, silica is the gangue present in haematite ore.(b) Slag : Slag is a waste product formed by the combination of gangue and flux during

smelting process.5. Explain flux.

Ans. Refer 6.2 (e)

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*6. What is calcination? Write examples with reactions.Ans. Refer 6.2 (a)

*7. What is smelting? Explain with an example.Ans. Refer 6.2 (d)

*8. Explain Electrometallurgy.Ans. Refer 6.2 (h)

*9. At 673K, which is the better reducing agent carbon or carbon monoxide?Ans. At 673, carbon monoxide is the better reducing agent.

*10. Using standard potentials write equation for the net reaction that you would predict inthe following experiments.(a) Zinc metal is added to aqueous sodium triiodide.

Ans. The standard reduction potential of Zn 0+2Zn / Zn

E = –0.76 Volt is lower than that of I3–

( )0 – –3E I /I = 0.536 volts. Hence, Zn reduces I3

– to I– and the reduction reaction is

3 2Zn + NaI ZnI + NaI⎯⎯→ .

(b) Iodine is added to excess aqueous HClO3.Ans. The reduction potential of I2 (E

0 I–/I2 = 0.434V) is lower than that of HClO3 (E0ClO3

–/ClO–

= 0.50V). Hence, I2 reduces HClO3 to HOCl and itself is oxidized.

2 3 22I + HClO + 2H O 4HOI + HOCl⎯⎯→

*11. Why is permanganate not a suitable oxidizing agent for quantitative estimation of Fe2+

in presence of HCl?Ans. (a) Fe2+ can be estimated quantitatively by titrating against a standard solution KMnO4 in

acidic medium.2+ 4+ 3+ 2+

25Fe + MnO 5Fe + Mn + 4H O⎯⎯→ .

(b) For acidic medium H2SO4(aq) is used and not HCl solution.(c) In case HCl is used, we will have to consider reduction potential of following reduction

reactions :

(1) 4– + – 2+ 2MnO + 8H + 5e Mn + 4H O⎯⎯→ (E0red = 1.51V)

(2) – 02 red½Cl (g) + e Cl(aq)E = 1.40V⎯⎯→

(3) The reduction potential Fe3+ is 3+ – 2+Fe +e Fe⎯⎯→ E0red = 0.77V.

(4) Since the reduction potential of Cl2/Cl– and Fe2+/Fe3+ are less than for MnO4–. Both

HCl and Fe2+ will be oxidised by permanganate solution.(5) Therefore, Fe2+ cannot be estimated quantitatively by permanganate in the presence

of HCl.

*12. In electrometallurgy of aluminium, why is the graphite rod used?Ans. The process of electrometallurgy is specifically employed for the extraction of highly

398


electropositive metals like sodium, potassium, magnesium, calcium and aluminium. The oresof these metals cannot be reduced by conventional reducing agents such as carbon, carbonmonoxide, hydrogen etc. Since, graphite is good conductor also the liberated oxygen combineswith graphite and forms carbon dioxide.

*13. Suggest a flux for the removal of silica gangue.Ans. Limestone (CaCO3). Limestone on decomposition gives quick lime which combine with silica

to form calcium silicate (CaSiO3)Δ

3 2CaCO CaO + CO⎯⎯→

2 3CaO + SiO CaSiO(Slag)

⎯⎯→


*1. Electrochemical process (electrolysis of fused state) is used to extract .......a) Iron b) Lead c) Sodium d) Silver

*2. In metallurgy, flux is a substance used to convert .......a) Mineral into silicateb) Fusible impurities to infusible impuritiesc) Infusible impurities to soluble impuritiesd) Soluble impurities into infusible impurities

3. Heating of iron pyrites in air to remove sulphur is called .......a) Fusion b) Calcination c) Roasting d) Smelting

4. In blast furnace iron oxide is reduced to iron by .......a) Carbon monoxide b) Lime stone c) Carbon d) Zinc

5. Which one of the following metal is leached by cyanide process?a) Ag b) Na c) Al d) Cu

• Advanced MCQs :6. The method chiefly used for the extraction of lead and tin from their ores are

respectively ....... (IIT 2004)a) self reduction and carbon reduction b) self reduction and electrolytic reductionc) carbon reduction and self reduction d) cyanide process and self reduction

7. In the process of extraction of Gold, .......

Roasted gold ore O– –22+ + +H O [X]CN OH⎯⎯→

[X] + + AuZn [Y]⎯⎯→

Identify the complexes [X] and [Y]

a) X = [Au (CN)2]–, Y = [Zn (CN)4]

2– b) X = [Au (CN)4]–3, Y = [Zn (CN)4]

2–

c) X = [Au (CN)2]– , Y = [Zn (CN)4]

2– d) X = [Au (CN)4]–, Y = [Zn (CN)4]

–

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6.3. THERMODYNAMIC PRINCIPLES OF METALLURGY(ELLINGHAM DIAGRAM)(a) Variation of ΔG with respect to temperature for the following :1. Formation of metal oxide from metal.2. Formation of CO(g) from carbon3. Formation of CO2(g) from carbon

The change in Gibbs free energy is given by ΔG = ΔH – TΔSWhere, ΔG = Change in Gibbs free energy,ΔH = Heat of reaction,ΔS = Change in entropy.For a reaction to proceed in the required direction, the ΔG must be negative.1. Formation of metal oxide from metal :

(a) Consider the formation of metal oxide, MO(s) from metal M(s).

(s) 2(g) (s)1

M + O MO2

⎯⎯→

(b) For this reaction H is negative (since it is combustion) and ΔS is also negative (asO2 is converted to MO(s)).

(c) Since, ΔG = ΔH – TΔS, at low temperature ΔG may be negative. As temperatureincreases, ΔG increases or becomes less negative.

2. Formation of CO(g) from carbon :(a) The oxidation of carbon, C(s) to CO, can be explained with the help of thermodynamic

concepts.

(s) 2(g) (s)1

C + O CO2

⎯⎯→

(b) For this reaction H is negative (since it is combustion) and ΔS is positive (molesof gaseous products> moles of gaseous reactants) .Therefore the term TΔS becomespositive.

(c) At all temperatures ΔG < 0 and ΔG decreases with the increase in temperature.3. Formation of CO2(g) from carbon :

(a) The oxidation of carbon, C(s) to CO2, can be explained with the help of thermodynamicconcepts.

(s) 2(g) (2g)C + O CO⎯⎯→

(b) For this reaction H is negative (since it is combustion) and ΔS is zero (molesof gaseous products= moles of gaseous reactants) .Therefore the term TΔSbecomes zero.

(c) Hence, ΔG does not vary with temperature.

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(b) Ellingham diagram :1. Ellingham diagram : It is the

plot of Gibbs free energychange (ΔG°) with thetemperature for the reactionof a metal and other elementswith one mole of gaseousoxygen at one atmosphereto form corresponding metaloxides.

The corresponding ΔG° iscalled standard free energychange of combustionreaction.

Ellingham diagram can also beplotted for the formation ofsulphides, halides, etc.

2. Features of Ellingham diagram :(a) The graph of G° against temperature for the formation of metal oxide is a straight

line with positive slope.(b) In case of some metal oxides like MgO, ZnO and HgO, there is a sudden change

in the slopes indicating change in the physical state.(c) In case of metal oxides of Hg and Ag, the graphs are at the upper part of Ellingham

diagram indicating less tendency for oxide formation.(d) For the formation of gaseous CO, the graph is a straight line with a negative slope.

This line intersects the lines of many metal oxides.(e) For the formation of gaseous CO2, the graph is a straight line almost parallel to

temperature axis, indicating very less effect of temperature on ΔG° of combustionreaction.

3. Significance of Ellingham diagram :(a) Ellingham diagram is obtained by plotting standard free energy change ΔG° for the

formation of oxides of metals versus temperature.(b) The positive slope of formation of metal oxide lines indicate that the stability of metal

oxides decreases due to increase in ΔG° with the increase in temperature.

+200

0.0

–200

–400

–600

–800

–1000

–12000 500 1000 1500 2000 2500

Temperature / K

Ellingham diagram for oxide formation

ΔGº /

kJ

mol

–1

2Mg + O 2

2MgO 43Al + O

2

23

Al O2 3

2Zn + O2

2ZnO

C + O CO2 2

2Fe + O 2FeO2

2Hg + O

2HgO

2

4Ag + O 2Ag O

2

2

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(c) The sudden change in the graphs and slopes indicates a phase change from solid toliquid or from liquid to vapour.

(d) The negative slope for CO shows that it becomes more stable with the increase intemperature while almost constant value of ΔG° for CO2 indicates, it is less stablethan CO at higher temperatures. Therefore, at high temperature carbon preferablyforms CO and not CO2.

(e) The variation in the negative values of free energy change ( ΔG°) in decreasingorder is,MgO > Al2O3 > Cr2O3 > FeO > Ag2Ohence, the tendency to undergo oxidation of the corresponding metals will beMg > Al > Cr > Fe > Ag.

(f) The Ellingham diagram enables to select a suitable reducing agent for the reductionof given metal oxide. An element can reduce the oxide of another element aboveit in Ellingham diagram.Eg : Al can be used as a reducing agent for the reduction of Cr2O3.

2 3 2 3Cr O + 2Al Al O + 2Cr⎯⎯→

c) Importance of carbon and carbon monoxide as reducing agents usingEllingham diagram.

1. Carbon is used extensively as a reducing agent for several metal oxides. For example,

2 3 2blast furnance2Fe O + 3C 4Fe + 3CO⎯⎯⎯⎯⎯⎯⎯→

1500KZnO + C Zn + CO⎯⎯⎯⎯→

Thus, carbon forms CO2 as well CO in the reduction reactions.

2. For the formation of 2(g) (s) 2(g) 2(g)CO (C + O CO ) S⎯⎯→ Δ is almost zero and the graph

in Ellingham diagram is horizontal showing that G° does not change with temperature.3. For the formation of

(g) (s) 2(g)CO (2C +O 2CO) S⎯⎯→ Δ

is positive and increases withtemperature while ΔG° valuesbecome more negative with theincrease in temperature and thegraph has negative slope at highertemperature. These two graphs(lines) intersect at 700 K.

4. Therefore, below 700 K, the formation of CO2 is favoured while above 700 K, formationof CO is favoured.

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5. CO can also be used as a reducing agent. For example,

(S) (g) (s) 2(g)MO + CO M + CO⎯⎯→

CO gets oxidised to CO2 by oxygen, hence it is a strong reducing agent.

(g) 2(g) 2(g)2CO + O 2CO⎯⎯→

ΔH and ΔS are negative for this reaction. Hence, even at low temperature, CO canreduce metal oxides and therefore, it is a better reducing agent. For example,haematite, Fe2O3 is reduced by CO at low temperature.

6. At low temperature, CO is a better reducing agent than carbon.

(d) Limitations of Ellingham diagram :1. The diagram simply indicates whether a reaction is possible or not. It does not predict

the kinetics of reduction i.e. the rate or time taken for the reduction to occur.2. The interpretation of ΔGº is based on the assumption that the reactants and products are

in a state of equilibrium and ΔG is based on the equation ΔGº = –RT log KHowever, this is not always true because the reactants or products may be solid.


*1. What are the features of Ellingham diagram? What is its significance?Ans. Refer 6.3 (b) 2, 3

*2. Using an Ellingham diagram, indicate the lowest temperature at which ZnO can be reducedto zinc metal by carbon. Write the overall reaction at this temperature.

Ans. The graph for the formation of CO in the Ellingham diagram is a straight line with downwardslope (ΔS increases and ΔGº decreases with temperature). Below the temperature of 1000K,the formation of CO2 is favoured and above 1000 K formation of CO is formed. For example,

1000KZnO + C Zn + CO⎯⎯⎯⎯→

The reduction of Zinc oxide by carbon occurs above 1000 K temperature.

*3. Explain how ΔGº varies with temperature in the reaction.

(s) 2 (g) (g)2C + O 2CO⎯⎯→

Ans. Refer 6.3 (a) 2.

*4. What is the minimum temperature for reduction of MgO by carbon?Ans. 1873 K.

5. Which is better reducing agent C or CO?Ans. Refer 6.3 (c).

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Charge

Sloping hearth

ScreenImpurities

Furnace Liquation

Puremoltenmetal


1. According to Ellingham diagram, the oxidation reaction of carbon to carbon monoxide maybe used to reduce which one of the following oxides at the lowest temperature?a) Al2O3 b) Cu2O c) MgO d) ZnO

• Advanced MCQs :2. In view of the sign of Δr ΔGº for the following reaction .......

2PbO + Pb 2PbO⎯⎯→

ΔGº < 0

2SnO + Sn 2SnO⎯⎯→

ΔGº > O

3. Which oxidation state is more characteristic for lead and tin? (AIEEE 2011)a) For Pb +2, for tin +2 b) For Pb +4, for tin +4c) For Pb +2, for tin +4 d) For Pb +4, for tin +2

6.4. PURIFICATION OR REFINING OF CRUDE METAL : Concept Explanation :

The crude metals obtained after smelting or by any other process contains impurities ofunreduced metal oxide, nonmetals, some other metals and gases. Removal of these impuritiesfrom crude metal is known as purification or refining which can be carried by severalmethods, depending upon the impurity to be removed. These are :

(a) Liquation :This method is employed for themetals having very low melting pointsuch as tin, Bismuth, lead etc. In thismethod the impure metal is placedon the sloping hearth ofreverberatory furnace and heated ininert atmosphere of carbonmonoxide. The metal melts andflows down which is collected at thebottom of sloping hearth in receiverleaving behind non-fusible impuritieson the hearth.

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(b) Polling :This method is used for the metals which contain oxide as an impurity e.g. copper andtin. In this method, molten metal is agitated vigorously with green pole (bamboo) or logsof wood. The heat of molten metal makes the green logs to liberate hydrocarbon gaseswhich reduces metal oxide into metal e.g. molten impure copper becomes 99.5% pureafter polling process.

(c) Zone-refining :This method is based on the principlethat impurities are more soluble inthe melt than in the solid state ofmetal. The method is used formetals which are required in veryhigh purity e.g. extremely puresilicon, germanium, boron, galliumand indium.The metal to be purified is casted into thin bar. A circular mobile heater is fixed at oneend of impure metal. One zone of bar is melted by a circular mobile heater in the atmosphereof an inert gas like argon. At the heated zone, metal melts as the heater moves slowly,the impurities also move into the adjacent molten part. Thus, impurities are made to movein to one end which is finally cut off and discarded. Mean time molten metal solidifiesas it is away from the heater. Thus, we get extremely pure metal.

(d) Electrolytic refining :This is most common method of refining and is based on the principles of electrolysis.The impure metal is made an anode and cathode is strip of pure metal which are placedin electrolytic bath containing soluble salt of same metal on passing current, anodeundergoes oxidation and passes in solution.An equivalent of metal cation fromsolution get deposited on cathodeand grows in size.The impurities fall down below theanode as anode mud. The reactionstaking place at anode and cathodeare as follows.

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At anode :

–n+ neM M +(impure)

⎯⎯→

At cathode : n+ –M Mne+ ⎯⎯→

Metals like Ag, Cu, Ni, Al and Zn arerefined by this method.

(e) Vapour phase refining :This method is based on the fact that certain metals are converted to their volatile compoundwhich on heating decomposes to give metal while impurities are unaffected during compoundformation. The metal to be refined by this method should :

1. Form a volatile compound with available reagent and2. The volatile compound formed should be easily decomposable so that metal can easily

be recovered.e.g. Nickel is refined by this technique and is known as Mond process. Nickel is heatedin stream of carbon monoxide to form nickel tetra carbonyl complex which on the thermaldecomposition gives Nickel.

330-350 K 450-470K4Ni + 4CO Ni(CO) + 4CONi

impure Nickel tetra purecarbonyl

⎯⎯⎯⎯→ ⎯⎯⎯⎯→

(f) Van-Arkel method for refining Zirconium or titanium :

This method is similar to Mond process and used to obtain ultra pure Zirconium and Titaniummetal. In this method, the metal is converted to unstable volatile compound (e.g. Iodide)taking care that impurities are not affected during compound formation. The compoundis then decomposed to get pure metal.

523 K 1700 K2 4 2Ti(s) + 2I (g) TiI (g) Ti(s) + 2I (g)

Impure pure⎯⎯⎯→ ⎯⎯⎯⎯→

870 K 2075 K2 4 2Zr + 2I ZrI Zr + 2I

impure purevapours⎯⎯⎯→ ⎯⎯⎯⎯→

(g) Chromatographic method :The technique of chromatography is based on the principle that different component ofmixture are differently adsorbed on an adsorbent.

406


The mixture is put in a liquid or gaseous medium (called moving phase). Different componentof mixture are adsorbed at different levels of porous column (adsorbent called stationaryphase). The adsorbed components are removed (eluted) by using suitable solvents (eluant).

The mobile phase and stationary phase are chosen such that component of the samplehave different solubilities in two phases. A component which is quite soluble in stationaryphase takes longer time to travel through it than a component which is not very solublein stationary phase. Thus, component of sample are separated from each other as they movethrough stationary phase. Depending upon the physical states of two phases and also onthe process of passage of moving medium, chromatographic technique is given differentnames such as column chromatography, paper chromatography, thin layer chromatography,gas chromatography etc.


*1. What is polling?Ans. Refer 6.4 (c)

*2. How is zone refining process used to obtain ultra pure metals?Ans. Refer 6.4 (c)

407



*1. Van Arkel method of purification of metals involves converting the metal to a .......a) Volatile compound ` b) Volatile unstable compoundc) Non-volatile stable compound d) Non-volatile unstable compound

*2. Zone refining is a method to obtain .......a) Very high temperature b) Ultra pure Alc) Ultra pure Germanium d) Ultra pure oxides

3. Zone refining has been employed for preparing ultrapure sample of .......a) Cu b) Na c) Ge d) Zn

4. Titanium can be obtained in a state of high purity by .......a) Van Arkel method b) Poling c) Cupellation d) Electrorefining

• Advanced MCQs :5. Which of the following pairs of metal is purified by Van Arkel method? (IIT 2011)

a) Ga and In b) Zr and Ti c) Ag and Au d) Ni and Fe

6.5. EXTRACTION OF ZINC FROM ZINC BLENDE : Concept Explanation :

Zinc is reactive metal hence, does not occur in free (native) form. The ores of Zinc ore :(1) Zinc blende ZnS (2) Calamine ZnCO3 (3) Zincite ZnO(4) Franklinite ZnO.Fe2O3 (5) Willemite Zn2SiO4The chief ore of zinc is zinc blende which occurs in Zawar mines located near Udaypur(Rajasthan).

(a) Concentration of ore :The crushed ore is concentrated by :

1. Gravity separation : The powdered ore is washed with powerful stream of water.The lighter gangue impurity particles are washed away and the heavier ore particlesremain behind.

2. Electromagnetic separation : The impurities of iron oxide from zinc blende are separatedby magnetic separation. This is possbile because iron oxide is magnetic in nature.

3. Froth Floatation process : Froth Floatation process is the important step in concentratingzinc blende. In this process, finely powered ore of zinc blende is treated with watercontaining pine oil in a tank. This oily mixture is stirred by passing compressed air intoit by stirrer. This process leads to froth formation. Sulphide particles wetted by pineoil rise to the surface of tank in the form of foam. Earthy impurities wetted by watersettle down at the bottom which are removed. In this way sulphide particles are separatedfrom other impurities.

408


(b) Roasting :The concentrated ore is roasted in excess of air at about 1200 K to convert zinc sulphideinto zinc oxide.During roasting sulphur present in zinc blende is completely expelled out hence, the roastingis known as dead roasting or sweat roasting.

2 2(g)2ZnS + 3O 2ZnO + 2SO⎯⎯→ ↑

Sometimes ZnS gets converted to ZnSO4.

2 4ZnS + 2O ZnSO⎯⎯→

At 1200 K ZnSO4 decomposes to give ZnO.

4 2 22ZnSO 2ZnO + 2SO + O⎯⎯→

(c) Reduction :Zinc oxide is reduced to zinc by heating with crushed coke at 1673 K in vertical fire

clay retort.

(g)1673KZnO + C Zn + COΔ

⎯⎯⎯→

The vapours of zinc formed arecollected in condenser to formmolten zinc which on coolingsolidifies into zinc spelter. Thewaste is continuously removedfrom bottom by automaticmechanism and fresh charge isadded from top of the retort.Thisprocess gives 97% to 98%Zinc contains impurities of iron,cadmium and lead impurities. It isrefined by electrolytic method.

(d) Electrolytic refining of zinc :To remove impurities the spelter is dissolved in dilute sulphuric acid. It forms solution ofZnSO4. The solution of ZnSO4 should contain about 3% free H2SO4. The solution is filteredand electrolyzed using aluminum cathode and lead anode at a voltage of 3.5V. Pure zincis deposited on the aluminium cathode.

(e) Physical properties :1. It is silvery white metal in pure form. However, becomes grey coloured when exposed

to moist atmosphere.

409


2. It’s atomic radius is 125 pm and ionic radius (Zn2+) is 74 pm.3. The specific gravity of zinc metal is 7.14 g cm–3.4. It melts at 692 K and it’s boiling point is 1193 K.5. It is good conductor of heat and electricity, malleable and ductile in the temperature range

of 373 K to 423 K.

(f) Uses of Zinc :1. It is used for galvanizing iron.2. Zinc is used in the extraction of Gold and silver in cyanide process.3. Zinc dust is used as a reducing agent in the laboratory. It is also used in the manufacture

of drugs, dyestuff, paints and other chemicals.4. It is used as an electrode in dry cells.5. It is also used in the manufacturing of alloys such as brass, german silver etc.


*1. How is zinc extracted from zinc blende?Ans. Refer 6.5 (a, b, c and d)

Multiple Choice Questions :• Advanced MCQs :

1. In electro-refining, the impure metal is made .......a) anode b) cathode c) anode or cathode d) electrolyte

2. Extraction of zinc from zinc blende is achieved by ....... (IIT 2007)a) Electrolytic reductionb) Roasting followed by reduction with carbonc) Roasting followed by reduction with other metald) Roasting followed by self reduction

6.6. EXTRACTION OF IRON FROM HAEMATITE : Concept Explanation :

The common ores of iron are :1. Haematite Fe2O3 (red oxide of iron)2. Limonite 2Fe2O3.3H2O (hydrated oxide of iron)3. Magnetite Fe3O4 (magnetic oxide of iron)4. Siderite FeCO3

5. Iron pyrite FeS2

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Extraction of iron :

(a) Concentration of ore :The ore is crushed in jaw crusher and is concentrated by gravity separation process

(discussed earlier) in which it is washed with water to remove sand, clay etc.

(b) Roasting :The concentrated ore is then calcined. (heated in limited supply of air in reverberatory)Impurities like sulphur, phosphorus and arsenic are converted to their oxide and are removedas they are volatile.

2 2

2 2 3

S + O SO

4As + 3O 2As O

⎯⎯→ ↑

⎯⎯→ ↑

2 2 34FeO + O 2Fe O⎯⎯→

The roasted ore is converted into small lumps.

(c) Smelting :The roasted ore is reduced with coke in the blast furnace. A blast furnace is tall cylindricalfurnace having diameter of 5 to 10 meter and height of 25 meter. It is made of steel linedwith fire bricks. It is narrow at the top and has cup and cone arrangement for the introductionof charge. At the base of furnace it is provided with :

1. tuyers arrangement for the introduction of hot air2. taping hole for withdrawal of molten iron and3. an outlet from which slag can be flown out.

The ore (charge) is mixed with coke and lime stone in the ratio 15 : 5 : 3 and introducedthrough cup and cone arrangement and at the same time blast of hot air is blown upwardwith the help of tuyers.The burning of coke to carbon monoxide supplies most of the heat required for the workingtemperature of the furnace and give temperature up to 2200 K at the bottom of furnace.As the gases move up, they meet the descending charge and the temperature falls gradually.At the bottom of furnace, the reducing agent is carbon itself, but at the top of the furnace,reducing agent is carbon monoxide. Following reaction takes place in the furnace.(i) Zone of combustion OR combustion zone : (5-10 m height from the bottom). The

hot air blown through the tuyers reacts with coke to form carbon monoxide.

21

C + O CO2

⎯⎯→ ΔH = –220 kJ

The reaction is highly exothermic. As a result temperature in the zone of combustionis around 2000 K. A part of carbon monoxide dissociates to give finely divided carbon.

22CO O + 2C⎯⎯→

411


In the blast furance the hot gas rich in CO rises up the furnace and heats the chargecoming down and also reacts with charge. This carbon monoxide acts both as a fueland a reducing agent.

(ii) Zone of reduction : (22-25 m height near the top). Carbon monoxide reduces ferricoxide to spongy iron (porous solid) at about 900 K.

2 3 2Fe O + 3CO 2Fe + 3CO⎯⎯→

A small amount of ferric oxide is also reduced to iron by carbon.

2 3Fe O + 3C 2Fe + 3CO⎯⎯→

(iii) Zone of slag formation :(20 m in height). The mainconstituents of gangue are silica,alumina and phosphates. Theremoval of gangue is effectedby adding limestone which actsas flux.Limestone decomposes to givecalcium oxide (quick lime) at about1200 K.

3 2CaCO CaO + CO⎯⎯→

Calcium oxide combines with silicaand alumina at about 1500 K toform molten slag of calcium silicateand calcium aluminate.

2 3CaO + SiO CaSiO⎯⎯→ 2 3 3 3 212CaO + 2Al O 4Ca AlO + 3O⎯⎯→

(iv) Zone of fusion : (15 m height) MnO2 and Ca3(PO4)2 present in the iron ore arereduced to Mn and P. A part of SiO2 is also reduced to Si. The spongy iron comingdown from the top of the furnace melts and absorbs the impurities like carbon, silicon,manganese, phosphorus and sulphur. The molten iron collects at the bottom of thefurnace. The lighter slag floats on the surface of molten iron. Molten slag and ironare removed through separate outlets. Molten iron is cooled in moulds. The solidsblocks of iron are referred as pigs, hence, the name pig iron. It is also called castiron. It contains upto 4% carbon. The hot waste gases consisting of mostly N2,CO, CO2 escapes through an outlet at the top of the furnace. They are used for

412


preheating the blast of air. The various changes taking place in the blastfurnace as the temperature gradually increases (temperature gradient) are summarizedas follows :

Temp K Change taking place Equation

500 K Ore loses moisture

900 K Reduction of ore by CO 2 3 2Fe O + 3CO 2Fe + 3CO⎯⎯→

1200 K Limestone Decomposes 3 2CaCO CaO + CO⎯⎯→

1500 K Reduction of ore by C 2 3Fe O + 3C 2Fe + 3CO⎯⎯→

1500 K Fusion of iron and slag 2 3CaO + SiO CaSiO⎯⎯→

Formation

2000 K Combustion of coke 2 3 2Fe O + 3CO 2Fe + 3CO⎯⎯→

Cast iron is hard and brittle. It cannot be welded or tempered. It is used in the manufactureof casted material and certain automobile parts.Commercial varieties of iron :(i) Cast or pig iron : It contains 2 to 4.5% carbon along with impurities such as sulphur, silicon,

phosphorous, manganese etc. It is brittle and cannot be welded.(ii) Wrought iron : It is the purest form of iron and contains carbon and other impurities not

more than 0.5%. It is malleable and can easily be welded.(iii) Steel : It contains 0.2 to 1.5% carbon. The mechanical properties of steel can be altered

by the addition of small amount of element like Mn, Cr and Ni. It’s properties areintermediate between cast iron and wrought iron.


*1. How is iron extracted from haematite?Ans. Refer 6.6 (a, b, c)

*2. Write reactions involved at different temperature in the blast furnace.

Ans. Refer 6.6 C (iv) Table

3. Name the principal gangue associated with haematite?Ans. SiO2 (silica) and Alumina (Al2O3)

*4. Which is the effective reducing agent in the extraction of iron from haematite?Ans. Carbon monoxide and carbon (coke).

Hot

gas

es

413


*5. Why is carbon monoxide a better oxidising agent than carbon for the reduction of haematiteat lower temperature?

Ans. At lower temperature, carbon dioxide is thermodynamically more stable than CO. The Δ Gfor the conversion of CO to CO2 is negative at lower temperture. Hence, CO can be usedto reduce Fe2O3 to Fe. It reduces Fe2O3 to Fe and itself gets oxidised to CO2.

2 3 2Fe O + 3CO 2Fe + CO⎯⎯→


*1. In blast furnace, iron oxide is reduced by .......a) Silica b) CO c) C d) Lime stone

*2. In extraction of iron limestone is used for .......a) Formation of slag b) Reduction of Fe Orec) Purification of Fe formed d) Oxidation of Fe Ore

3. The reduction of iron in a blast furnace involves all the steps except .......a) Fusion b) Reduction c) Sublimation d) roasting

• Advanced MCQs :*4. Fe2O3 is reduced to spongy iron near the top of blast furnace by .......

a) CO b) CO2 c) C d) H2*5. Highest carbon content iron is .......

a) Stainless steel b) Wrought iron c) Cast iron d) Mild iron

6.7. Extraction of Aluminium from Bauxite : Concept Explanation :

The minerals of aluminium are :(i) Bauxite : Al2O3.2H2O (ii) Cryolite : Na3AlF6(iii) Feldspar : KAlSi3O8 (iv) Mica : KAlSi2O10(OH)2

Aluminium is normally extracted from bauxite which involves following steps :

1. Purification of bauxite and 2. Electrolysis of alumina.

(a) Purification of bauxite :The bauxite ore can be purified by any one the following method, depending on the natureof impurity present in it.(i) Bayer’s process : This method is widely used for the purification of red bauxite which

contains iron as an impurity.(ii) Hall’s process : This process is also used for red bauxite.

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(iii) Serpeck’s process : This method of purification of bauxite is adopted for white bauxitewhich contains silica as chief impurity.

1. Purification of bauxite by Hall’s process : In this method, the bauxite ore is fusedwith sodium carbonate and lime. The fused mass is dissolved in water and filtered to removeimpurities. The clear solution of meta-Aluminate is then heated to 323 – 373 K and streamof carbon dioxide is passed through solution when aluminium hydroxide precipitates outwhich is filtered, washed, dried and ignited at 1473 K to get pure Al2O3.

Δ

2 3 2 2 3 2 2 2Al O .2H O + Na CO 2NaAlO + CO + 2H O⎯⎯→

2 2 2 3 2 32NaAlO + CO + 3H O 2Al(OH) + Na CO⎯⎯→

1473 K3 2 3 22Al(OH) Al O + 3H O⎯⎯⎯⎯→

(b) Electrolysis of pure alumina :The electrolysis is carried outin an iron tank having liningof carbon which acts ascathode. The anode consist ofnumber of carbon rods whichdip in the fused electrolyte.The purified alumina isdissolved in molten cryolite(Na3AlF6 80-85%). Smallamount of flurospar(5 – 6%) is added. The addition of cryolite and flurospar lowers the melting point of purealumina to 1143 K and is good conductor. The electrolyte is covered with a layer of carbonpowder (coke)During electrolysis, following reactions occur :

3 6 3Na AlF 3NaF + AlF

+3 –3AlF Al + 3F

At cathode : +3 –Al + 3e Al⎯⎯→

At anode : – –F F + le⎯⎯→

2 3 2 3 22Al O + 6F 4AlF + 3O⎯⎯→

22C + O 2CO⎯⎯→

2 22C + O 2CO⎯⎯→

415


The O2 liberated at anode reacts with carbon anode and forms CO & CO2. Thus, anodeis being eaten away (used up) and needs replacement from time to time.Aluminium liberated at cathode is removed periodically from tapping hole provided at thebottom of cell. This electrolytic method of reduction of aluminium from bauxite is knownas Hall-Herault’s process.

(c) Refining of Aluminium (Hoope’s electrolytic process) :The aluminium obtained by Hall’s process is 99% pure. It is further refined by Hoope’sprocess. The process is carried out in an iron tank lined inside with carbon. It has threelayers of molten liquids having different densities.The top layer consist of pure aluminium with carbon electrodes dipping in it and acts ascathode. The middle layer is mixture of cryolite and barium fluoride in molten state whichacts as an electrolyte.The bottom layer is impure aluminium which along with carbon lining acts as anode.On passing electric current aluminium ions from middle layer are discharged at the cathodeas pure aluminium. An equivalent amount of aluminium from bottom layer passes into middlelayer leaving behind the impurities. Thus, this method gives pure aluminium.


*1. What are the steps involved in the extraction of aluminium from bauxite?Ans. Refer 6.7 (a. b and c)

*2. Explain the refining of aluminium.Ans. Refer 6.7 (c)

*3. What is the role of CaF2 in metallurgy of aluminium?Ans. The addition of CaF2 is to lower the melting point of pure alumina and is good conductor.

4. Mention names and formulas of two ores of aluminium.Ans. The minerals of aluminium are :

(a) Bauxite : Al2O3.2H2O (b) Cryolite : Na3AlF6

(c) Feldspar : KAlSi3O8 (d) Mica : KAlSi2O10(OH)2

(any two)


1. In alumino-thermic process, aluminium is used as .......a) Reducing agent b) Oxidising agent c) Catalyst d) Electrolyte

2. A common method used for the extraction of metals from their oxides by reduction is .......a) Al b) Fe c) Cr d) Co

416


6.8. EXTRACTION OF COPPER FROM COPPER PYRITES : Concept Explanation :

1. The chief ores of copper are :(i) Copper glance Cu2S (ii) Copper pyrite CuFeS2

(iii) Malachite Cu(OH)2.CuCO3 (iv) Cuprite (Ruby copper) Cu2O(v) Azurite 2CuCO3.Cu(OH)2

2. Extraction : Copper is mainly extracted from copper pyrite which contains iron sulphide,gangue and smaller quantities of arsenic, selenium, tellurium, silver, gold and platinum :

(a) Concentration of ore :The finely powdered ore is concentrated by froth floatation process. The concentratedore is then subjected to roasting in reverberatory furnace in limited supply of air. Inreverberatory furnace, copper pyrite, CuFeS2 get converted to Cu2S and FeS, while impuritieslike arsenic and antimony are removed as their volatile oxide.

2 2 2 22 CuFeS + O Cu S + 2FeS + SO⎯⎯→

(b) Smelting :The roasted mass is mixed with some powdered coke and sand and is heated stronglyin blast furnace. FeS is preferentially oxidised as iron is more reactive than copper. FeOcombines with silica (flux) to form fusible slag.

2 22FeS + 3O 2FeO + 2SO⎯⎯→

2 3FeO + SiO FeSiO(slag)

⎯⎯→

The lower layer of molten mass contains cuprous sulphide and some traces of iron sulphide,known as copper matte which is removed from bottom tapping hole.

(c) Bessemerisation :The molten copper matte is then transferred to Bessemer converter and a blast of hotair mixed with sand is blown. During this process Cu2S is partially oxidised to Cu2O whichfurther reacts with remaining Cu2S to form copper and sulphur dioxide (auto reduction).

2 2 2 22Cu S + +3O 2Cu O 2SO⎯⎯→ ↑

2 22Cu S + + SO2Cu O 6Cu ↑⎯⎯→

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Traces of FeS present in the matte is oxidised to FeO which combines with silica to form slag.

2 22 FeS + +3O 2 FeO 2SO⎯⎯→ ↑

32FeO + FeSiO (slag)SiO ⎯⎯→

After the reaction is completed,

the converter is tilted and the

molten mass is poured in tosand moulds. As the metal

solidifies, the dissolved SO2

escapes producing blisters onthe metal surface hence, known

as blister copper.

(d) Refining :The blister copper thus, obtained is 99% pure and contains impurities of silver and

gold. It is further refined electrolytically to get 99.99% pure copper. In electrolyticrefining, the electrolyte is acidified CuSO4. The anode is impure copper while the

cathode is pure copper rod. On passing current Cu2+ ions from solution get deposited

at cathode while an equivalent amount of anode dissolves. The noble metals do notdissolve in dilute acid present in the electrolyte and settles down as anode mud while

other impurities pass into solution.


*1. Describe the process of extraction of copper.Ans. Refer 6.8 (b)



*1. During smelting silica is added to roasted copper ore to remove .......a) Cuprous Sulphide b) Ferrous sulphidex c) Ferrous oxide d) Ferrous sulphide

*2. The ore which contains copper and iron both .......a) Malachite b) Chalcopyrite c) Chalcocite d) Azurite

pure and gets deposited at the cathode.C gasO2

S gasO2

Acidic/basic lining

Hot air

Bessemer converter

Slag

418


HOURS BEFORE EXAMMetallurgy : Is an art and science of extraction of metal of their ore commercially andeconomically.Ores : Naturally occurring substances (minerals) from which one or more metals can be extractedeconomically.Minerals : Naturally occurring substances which are obtained from mine and contains one ormore metals in it.

Principle underlying metallurgy are :1) Concentration of ore : Removal of unwanted impurities (gangue) from the ore is called

concentration. An ore can be concentrated by different methods depending upon the nature ofimpurity to be removed.

Pyrometallurgy

The metal is obtained by thereduction of their concentratedore by subjecting them to hightemperature using reducingagent. like coke eg. Iron, Zinc,Copper etc.

Hydrometallurgy

The metal is obtained by dissolvingconcentrated ore in suitablesolvent and then by displacementmethod eg. Silver, Gold etc.

ElectrometalluryThe concentrated ore issubjected to electrolysisusing suitable electrolyte.e.g. Aluminium.

Metallurgy

Hand-picking

When the impuritiesare in the form oflumps.

Hydraulic washing orgravity separation

When ore particles are highter(having low density thangangue) than impurity.e.g. Cassiterite, SnO2

Hydraulic classifier

When impurities arelighter than gangueparticle.

Magnetic Separation method

When ore is non-magnetic andimpurities are magnetic.e.g. wolframite (FeWO4) isremoved from casseterite ore.

Leaching

When ore dissolves in somesuitable solvent but impuritiesremain insoluble. e.g. Silver,Gold ore is concentrated bythis method. Alumina fromBauxite.

Froth floatation

Best suited for sulphide ore.Even two sulphide ore can beselectively concentrated usingdepressant. e.g. PbS can beselectively removed from ZnSusing NaCN as depressent.

419


2) Extraction of crude metal from concentrated ore :

3) Reduction of metal oxide into crude metal :(a) Using reducing agent i.e. pyrometallurgy and process is called smelting. The concentrated ore is

mixed with coke and heated in blast furnace. The metal oxide is reduced to metal.

y 2M O + YC M + COx x y⎯⎯→

eg. 2

ZnO + C Zn + CO

ZnO + CO Zn + CO

⎯⎯→

⎯⎯→

Δ

2 3823K

2 3 21123K

2

Fe O + 3C 2Fe + 3CO

Fe O + 3CO 2Fe + 3CO

FeO + CO Fe + CO

⎯⎯→

⎯⎯⎯→

⎯⎯⎯→

To remove impurities (gangue) flux is added which form slag with it.2 3SiO + FeO FeSiO

gangue flux slag⎯⎯→

2 3CaO + SiO CaSiO⎯⎯→

(b) Reduction by precipitation OR displacement method i.e. Hydrometallurgy. The metal is obtainedfrom their complexes by adding to it more reactive metal. eg. Silver sulphide is dissolved inNaCN solution to form diayanoargenta, a complex ion. From this complex, Silver is obtained byadding Zn dust in sol. when Silver get precipitated.

2 2 2Aq S + 4NaCN 2Na[Ag(CN) ] + Na S⎯⎯→

2 2 42Na[Ag (CN) ] + Zn 2Ag + Na [Zn(CN) ]⎯⎯→

(c) Electrolytic reduction or electrometallurgy : The metals like alkali and alkaline earth metalsand aluminium which cannot be reduced by using suitable reducing agent this method isemployed e.g. Electrolysis of fused NaCl.

+ –NaCl Na + Cl

At anode – –22Cl Cl + 2 e⎯⎯→

At cathode +2Na + 2e 2Na⎯⎯→

Calcination

♦ Carried out at about 200 – 300ºC.♦ Carried out in insufficient or absence of air.♦ To remove organic volatile impurities.♦ Moisture, hydrated water is removed.♦ Carbonates decomposes to form oxide.♦ To make concentrated ore porous.

Roasting

♦ Carried out at high temperature (below the m.p.of metal)

♦ Carried out in excess of air or O2.♦ Carried out in a reverberatory furnace.♦ Sulphide is expelled out as SO2.♦ Metal sulphide is converted into metal oxide.

420


4) Purification or refining of crude metal :Removal of impurity from crude metal to get 99.99% pure metal is known as refining orpurification. The different methods of purification are :

Extraction of metals(a) Zinc

Ores :1. Zinc blende, ZnS (chief ore) 2. Calamine, ZnCO3 3. Zincite, ZnO4. Franklinite, ZnO.Fe2O3 5. Willemite, Zn2SiO4

Concentration of ORE (ZnS)

Froth floatation method

Roasting : 2 22ZnS + 3O 2ZnO + 2SO⎯⎯→

(a) Distillation

It is used for low boilingmetals. The impuremetal is evaporated toobtain pure metal asdistillate e.g. Zn andHg.

Electrolytic refining♦ Based on the prin-

ciple of electrotro-lysis.

♦ Most commonmethod.

♦ Metals like Ag, Cu,Ni Zn are refinedby this method.

(b) Liquation

The impure metal is heatedon the sloping hearth ofrever beratory furnace inan inert atmosphere of CO.The metal melts and getscollected in the receiver inpure form e.g. Tin, Bismuth,lead etc.Vapour phase refining

♦ Certain metals areconverted into theirvolatile compound.

♦ On heating theydecompose to givepure metal whilei m p u r i t i e s a r eunaffected.

♦ Nickel is refined bythis process (Mondprocess).

(c) Polling

This method is used toremove impurity ofoxide. The moltenmetal is agitatedvigorously with greenwooden logs or polewhen the hydrocarbongases liberated fromgreen log reducesmetal oxide to metale.g. copper.Van-Arkelmethod♦ Similar to mond

process.

♦ Zirconium and bythis method.

♦ They are convertedto their volatileunstable iodide.

♦ On heating to hightemperature de-compose to givepure metal.

(d) Zone-refining

This method gives ultrapure metal. e.g. Silicon.The principle is,impurities are moresoluble in the moltenliquid than in the solid.Chromatographicmethod♦ Based on the

principle that diff-erent component ofmixture are differ-ently adsorbed on anadsorbent

♦ Different componentof mixture areadsorbed at differentlevels of porouscolumn

♦ The adsorbed comp-onents are removed(eluted) by usingsuitable solvents

421


Reduction : 1673ZnO + C Zn + CO⎯⎯⎯→

Purification : Electrolytic refining

Physical properties :1. Silvery white metal, turns grey when exposed to most atmosphere.2. Atomic radius. 125 pm Ionic (Zn2+) radius = 74pm.3. Specific gravity is 7.14 g cm3.4. Melting point : 692K, Boiling point : 1193 K.5. Good conductor of heat and electricity.6. Malleable and ductile in the temp. range 373 – 423K.

Uses :1. In galvanising of iron.2. In the extraction of gold and silver (cyanide process).3. As reducing agent in laboratory (Zn-dust).4. In the manufacture of drugs, dyestaft point and other chemicals.5. As an electrode in dry cell.6. As an alloy e.g. brass, german silver.

(b) Iron :1. Ores

(i) Haemetite (chief ore), Fe2O3 (ii) Limonite, 2Fe2O3.3H2O(iii) Magnetite, Fe3O4 (iv) Siderite, FeCO3

(v) Ironpyrite, FeS2

2. Concentration of ore : By gravity separation.3. Calcination : Carried out in reverberatory furnace in limited supply of air. During the process

(i) Moisture is removed.(ii) Sulphur, phosphorous, arsenic are converted to their volatile oxide.(iii) Some FeCO3 present get converted to oxide.

2 2S + O SO⎯⎯→ ↑2 2 34As + 3O 2As O⎯⎯→ ↑

4 2 4 10P + 5O P O⎯⎯→ ↑ 3 2FeCO FeO + CO⎯⎯→ ↑

2 2 34FeO + O 2Fe O⎯⎯→

4. Smelting :(i) Carried out in blast furnace.(ii) Calcined ore is mixed with coke.(iii) Reduction of Fe2O3 takes place in steps.(a) Zone of combustion (2200K) – bottom of furnace.

2 2C + O CO⎯⎯→ ΔH = –393.5 kJ

422


(b) Zone of fusion (1500K)

2CO + C CO⎯⎯→ ΔH = +163.2 kJ(c) Slag formation Zone (1270K)

3 2

2 3

CaCO CaO + CO

CaO + SiO CaSiO (slag)

⎯⎯→

⎯⎯→

(d) Reduction Zone (875 k)

2

2 3 3 2

3 4 2

2 3 2

FeO + CO Fe + CO

3Fe O + CO 2Fe O4 + CO

Fe O + CO 3FeO + CO

Fe O + 3CO 2Fe + 3CO

⎯⎯→

⎯⎯→

⎯⎯→

⎯⎯→

at the lower, hotter part the main reaction is

FeO + C Fe + CO⎯⎯→

The metal obtained is called cast iron or pig iron.Commercial varieties of iron :(1) Cast iron : contains 2 – 4% Carbon, cannot be Welded(2) Wrought iron : Purest form of iron and contains impurities of C, not more than 0.5% can

be welded.(3) Steel : Contains 0.2 – 1.5% Carbon. The properties of steel is altered by adding small amount

of metal like Mn, Ni and Cr. It’s property is intermediate between cast iron andwrought iron.

Metallurgy of Aluminium :1. Ores : (i) Bauxite, Al2O3.2H2O (chief ore) (ii) Cryolite, Na3AlF6

(iii) Feldspar, KAlSi3O8 (iv) Mica, K; AlS2O10 (OH)2.2. Concentration : Bauxite ore is fesed with sodium carbonate and lime.

Fused mass of bauxite + water dissolve Filtered⎯⎯→ ⎯⎯→ ⎯⎯→

+CO (g) filtered2filtrate Al(OH) Al O3 2 3and heated to 1473k⎯⎯⎯⎯⎯→ ↓

3. Electrolysis of pure alumina : It is carried out in big iron tank. The Al2O3 is dissolvedin molten cryolite (Na3AF6) – 80 – 85% and 5 – 6% CaF2 because it is good conductorand also lower the m.p. of Al2O3. The anode is graphite rod which remain dipping in theelectrolyte. The iron tank is lined with carbon which acts as cathode. During electrolysis followingreaction occur.

Metallurgy of : 3 6 3Na AlF 3NaF + AlF

423


3+ –3AlF Al + 3F

At anode : – –22F F + 2e⎯⎯→

At cathode : 3+ –Al + 3e Al⎯⎯→

The fluorine liberated at anode, reacts with Al2O3 to form AlF3.

2 3 2 3 22Al O + 6F 4AlF + 3O⎯⎯→

22C + O 2CO⎯⎯→

2 22CO + O 2CO⎯⎯→

Thus, instead of fluorine, O2 is liberated at anode which react with it and forms CO2. Thusanode is eaten away and need to be replaced from time to time.

4. Purification or refining of aluminium : Aluminium obtained by Hall’s Heraults process is99% pure. It is further purified by Hoope’s process to get 99.99% Aluminium.

Copper :1. Ores of Copper :

(i) Copper glane, Cu2S (ii) Copper pyrite, CuFeS2 (Chief ore)(iii) Malachite Cu(OH)2.CuCO3 (iv) Cuprite (Ruby copper), Cu2O(v) Azurite, 2 CuCO3.Cu(OH)2

2. Concentration of ore : The powdered ore is by froth floatation process.3. Roasting : Roasting is carried out in reverberatory furnace in limited supply of air.

2 2 2 2

2 2

2 2 2 2

2CuFeS + O Cu S + 2FeS + SO

2FeS + 3O 2FeO + 2SO

2Cu S + 3O 2Cu O + 2SO

⎯⎯→

⎯⎯→

⎯⎯→

4. Smelting : Roasted mass is mixed with some coke and silica and heated strongly in blastfurnace. During roasting if at all any Cu2O is formed combines with FeS and forms Cu2Swhile Fe is removed as slag.

2 2

2 3

Cu O + FeS Cu S + FeO

FeO + SiO FeSiO

⎯⎯→

⎯⎯→

5. Bessemerisation : The copper matte is transferred to Bessemer converted white metalliccopper is obtained due to auto oxidation. The copper obtained is known as blister copper becauseSO2 escape through molten copper producing blister on the surface of metal.

2 2 2

2 2 2 2

Cu S + 2Cu O 6Cu + SO

2Cu S + 3O 2Cu O + 2SO

⎯⎯→

⎯⎯→

6. Purification or refining : Blister copper is 99% pure and contains impurities of silver andgold. It is further refined electrolytically using CuSO4 as an electrolyte to get 99.99% purecopper.

424


Higher Order Thinking Skills

1. Why can’t aluminium be reduced by carbon?Ans. Aluminium cannot be reduced by carbon, because it is a stronger reducing agent than carbon.

2. Why the graphite rods in the extraction of aluminium from Al2O3 have to be replaced fromtime to time?

Ans. During the reaction, oxygen is liberated at anode which reacts with carbon electrode to formcarbon monoxide and carbon dioxide, which results in slow corrosion of carbon electrode. Hence,have to be replaced from time to time.

3. Thermite process is quite useful for repairing broken parts of machines. Explain.Ans. In thermite process, oxides of metal like iron is reduced by aluminium which is highly exothermic

and large amount of heat is released during the reaction. As a result of this, metal will be

in molten state. e.g. heat

2 3 2 3(s)Fe O + 2Al Al O + 2Fe + heat

molten

⎯⎯⎯→

The metal is allowed to fall between the broken parts which fills the gaps and machine is repaired.4. Giving examples, differentiate between ‘Roasting’ and ‘Calcination.’

Ans.

5. Is it true that under certain conditions, Mg can reduce SiO2 and Si can reduce MgO? Whatare those conditions?

Ans. Below the m.p. of silicon (1693 K) ΔfGº curve for the formation of SiO2, lies above the ΔfGºcurve for formation of MgO. Therefore, below 1693 K, Magnesium can reduce SiO2 to Silicon.However, above 1693 K, Δf Gº curve for MgO lies above Δf Gº curve for SiO2 andtherefore, at temperature above 1693K, Si can reduce MgO to Mg.

Calcination

1. Moisture and organic impurities are removed.

2. It is used for carbonate and oxide ores.3. It is carried out in absence of air.4. It is carried out in the temperature range of

373K – 473 K. e.g.heat

2 3 2 2 3 2Fe O H O(s) Fe O (s) + .H O(g)x x⋅ ⎯⎯⎯→

heat3 2ZnCO (s) ZnO (s) + CO (g)⎯⎯⎯→

heat

3 3

2

CaCO MgCO (s) CaO (s) + MgO (s)+ 2CO (g)

⋅ ⎯⎯⎯→

Roasting

1. Volatile impurities are removed as their oxidese.g. SO2, As2O3 etc.

2. It is used for sulphide ores.3. It is carried out in presence of air i.e. regular

supply of air.4. It is carried out below the melting point of

metal.

2 22ZnS + 3O 2ZnO + 2SO⎯⎯→ ↑

2 2 2 22Cu S + 3O 2Cu O + 2SO⎯⎯→ ↑

2 22PbS + 3O 2PbO +2SO⎯⎯→ ↑

7Introduction :The elements in which the last electron enters p orbitals of the atoms are called p-Block Element.p – block elements are placed in groups 13 to 18 of the periodic table. Their valence shellelectronic configuration is n2np1–6 (except helium which has 1s2 configuration). The propertiesof p-block elements like that of others are greatly influenced by ionization enthalpy, atomicsizes, electron gain enthalpy and electro negativity. p-block elements include metals, non-metalsand metalloids

Chapter

SYLLABUS7.1 GENERAL INTRODUCTION

(a) Group 15 elements

(b) Occurrence

(c) Electronic configuration

(d) Trends in physical and chemical properties

(e) Chemical properties

(f) Dinitrogen

(g) Compounds of Nitrogen - Ammonia

(h) Compounds of Nitrogen - Nitric acid

(i) Oxides of Nitrogen

(j) Allotropic forms of phosphorus

(k) Compounds of phosphorus

(l) Oxyacids of phosphorus


Theoretical MCQs

Advanced MCQs

7.2 GROUP 16 ELEMENTS

(a) General introduction

(b) Trends in physical and chemical properties

(c) Chemical properties

(d) Dioxygen

(e) Classification of oxides

(f) Ozone O3

(g) Sulphur

(h) Compounds of Sulphur

(i) Sulphuric acid

(j) Oxoacids of sulphur


Theoretical MCQs

Advanced MCQs

p – Block

Elements"You cannot teach a man anything; you can only help him to discoverit in himself." - Galileo

426


7.3 GROUP 17 ELEMENTS (HALOGEN

FAMILY)

(a) General introduction

(b) Trends in physical properties

(c) Chemical reactivity

(d) Anomalous behaviour of fluorine

(e) Compounds of halogen - chlorine

(f) Hydrogen chloride

(g) Interhalogen compounds

(h) Oxyacids and oxoacids of halogens


Theoretical MCQs

Advanced MCQs

7.4 GROUP 18 ELEMENTS (NOBLE GASES)

(a) Introduction and electronic configuration

(b) Trends in physical properties

(c) Uses of noble gases


Theoretical MCQs

Advanced MCQs

Hours Before Exam

Higher Order Thinking Skills (HOTS)

427

Chapter - 7 p – Block Elements

7.1 GENERAL INTRODUCTION :

Concept Explanation :(a) Group 15 elements :

Group 15 elements are representative elements. Metallic character increases down thegroup. N, P are non metals. As, Sb, are metalloids. Bi is metal.

(b) Occurrence :N – (a) Nitrogen occurs in gaseous state as free N2 in the atmoshere to the extent of 78%

by volume and in the combined state as nitrate.(b) It is present as constituent of protein and albumins.(c) In the earth’s crust, it occurs as sodium nitrate NaNO3 (chile salt petre) and potassium

nitrate (Indian salt petre).P – (a) Phosphorus occurs in minerals of the apatite family, Ca9[PO4]6.CaX2 (X=F, Cl or

OH) e.q. Fluorapatite, Ca9[PO4]6.CaF2, which is the main constituent of phosphaterocks.

(b) Phosphorus is an essential constituent of animal and plant matter.(c) It is present in bones as well as in living cells.

As, Sb, Bi – Present as oxides and with sulphide minerals.

(c) Electronic configuration :1. Valence shell configuration is ns2np3, which is configuration of exact half filled state of

np subshell and is externally stable.Hence, these elements are very stable and less reactive.

Electronic configuration of group 15 elements

Element

NitrogenPhosphorusArsenicAntimonyBismuth

Symbol

NP

AsSbBi

AtomicNo.

715335183

Electronic Configuration

1s22s22p3

1s22s22p63s23p3

1s22s22p63s23p63d104s24p3

1s22s22p63s23p63d104s24p64d105s25p3

1s22s22p63s23p63d104s24p6

4d104f145s25p65d106s26p3

Briefrepresentationof electronicconfiguration

[He] 2s22p3

[Ne] 3s23p3

[Ar] 3d104s24p3

[Kr] 4d105s25p3

[Xe] 4f145d106s24p3

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2. Five valence electrons are preceded by 2 and 8 electrons in the case of N and P respectively.In the case of As, Sb, Bi, they are preceded by 18 electrons.Hence, As, Sb, Bi behave differently.

3. The np3 configuration being the exact half filled state is stable on the basis of Hund’srule of maximum multiplicity. This explains the greater stability and less reactivity of theseelements.

(d) Trends in physical and chemical properties :Among the members of group 15, there is regular gradation in properties with increasein atomic number form the characteristics of true non-metal (nitrogen) to a distinct metal(bismuth). This transition from non-metal to metallic character is evident from the steadychanges of physical properties of these elements. Some of the physical properties are shownin Table.

Atomic and physical properties of group 15 elements

1. Atomic and ionic radii : Values increase from N to P considerably due to addition ofnew shells but increase is small from As to Bi due to completely filled inner d or f orbitals.

2. Electronegativity : Value decreases down the group with increase in atomic size.3. Ionisation enthalpy : Value decreases down the group due to increase in atomic size.

Due to stable ns2np3 configuration, ionisation enthalpy of group 15 elements is very high.As seen from the table ΔH1 < ΔH2 < ΔH3.

4. Non metallic and metallic character : Electropositive character and metallic natureincreases down the group. N, P are non metals. As, Sb are metalloids. Bi is metal.

No.

1.

2.3.4.

5.6.

7.

8.

9.

Property

Atomic Number

Atomic Mass g/mol–1

Covalent Radius (pm)Ionic Radius (pm)

ElectronegativityIonisation enthalpyΔH/kJmol–1 I

IIIII

Density (g/cm3)

Melting point/K

Boiling Point/K

N

7

14.0170

171 (N3–)

3.00

1402

28564577

0.879 at

63 K63

77.2

P

15

30.97110

212 (P3–)

2.10

1012

190329101.823

317.1(White phosphorus)

554 (white P)

As

33

74.92120

222 (As3–)

2.20

947

179827365.778

(Grey ∝ form)

1087.5

(Grey ∝ form)

883

Sb

51

121.75140

76 (Sb3+)

1.82

834

159524436.580

904

1653

Bi

83

209.00150

108 (Bi3+)

1.67

703

161024669.808

544

1813

429


5. Physical appearance and density : Nitrogen exists as diatomic molecule in N2 state.N ≡ N.P, As, Sb exist as tetraatomic molecules. Bi is monoatomic solid. Density increases downthe group.

6. Atomic volume and melting points : Atomic volume increases down the group. Butatomic volume of P is more than that of Sb. This is due to weak intermolecular attractionbetween P4 units. In As4 – close packing results in greater intermolecular attraction andsmaller volume. Therefore, melting point of Phosphorus is less than that of Arsenic.

7. Allotropy : Except Bi all the other elements in this group exhibit allotropy. Following arethe common allotropes :(i) Solid Nitrogen :

α – Nitrogen β – NitrogenCubical crystal structure Hexagonal crystal structure

(ii) Phosphorus :

White phosphorus Red phosphorus Black phosphorustetrahedron structure complex chain structure stable layer structure

Arsenic :Grey As Yellow As Black As

Antimony :Metallic Sb Yellow as α Sb Explosive Sb

430


8. Conductivity : Conductivity increases down the group N and P are non conductors ofheat and electricity. As is poor conductor. Sb and Bi are good conductors.

9. Oxidation state : Electron configuration is ns2np3. So common oxidation states are +3,+5 and –3. Stability of +3 oxidation state increases and +5 oxidation state decreases downthe group due to ‘Inert pair effect’ of ns2 electrons. However, N does not exhibit +5 oxidationstate through covalence due to non availability of d orbitals in valence shell. N exhibitsall the oxidation states from –3 to +5.

Formation of PCl5, PF5

3s 3p 3dP (Ground state)

P* (Excited state)

– five unpaired electrons.

Compound NH3 N2H4 NH2OH N2 N2O NO HNO2 NO2 HNO3

Oxidation state –3 –2 –1 0 +1 +2 +3 +4 +5

(e) Chemical Properties :1. Action of air : All elements react with oxygen of air to form oxides. Reactivity of the

elements with oxygen increases down the group. Nitrogen reacts with oxygen only at verytemperature, which is possible by striking an electric arc.

electric (2000K)2 2 arc

N + O 2NO

Nitricoxide

⎯⎯⎯⎯⎯⎯→

2. Action of oxidising agents : Hot, conc. HNO3 and H2SO4. Nitrogen does not react.Phosphorus and Arsenic form Oxyacids.(a) Action of hot and conc. Nitric acid

4 3 3 4 2 2P + 20HNO 4H PO + 20 NO + 4H Oconc. Phosphoric acid

⎯⎯→

hot4 3 3 4 2 2As + 20HNO 4H AsO + 20NO + 4H O

Arsenic acid⎯⎯→

hot3 4 10 2 24Sb + 20HNO Sb O + 20 NO + 10H O

Antimonyoxide⎯⎯→

hot3 3 2 2 2Bi + 6HNO Bi(NO ) + 3H O + 3NO

Bismuth nitrate⎯⎯→

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(b) Action of hot and conc. H2SO4hot

4 2 4 3 4 2 2P + 10H SO 4H PO + 10SO + 4H O Phosphoric acid

⎯⎯→

hot4 2 4 3 4 2 2As + 10H SO 4H AsO + 10SO + 4H O

Arsenic acid⎯⎯→

hot2 4 2 4 3 2 22Sb + 6H SO Sb (SO ) + 3SO + 6H O

Antimonysulphate⎯⎯→

hot2 4 2 4 3 2 22Bi + 6H SO Bi (SO ) + 3SO + 6H O⎯⎯→

3. Action of alkalies : Nitrogen has no action.

4 2 3 2 2P + 3NaOH + 3H O PH + 3NaH POPhosphine Sodium hypophosphite

⎯⎯→

4 3 3 2As + 12 NaOH 4 Na AsO + 6Hsoluble (Sodium arsenite)

⎯⎯→

3 3 22Sb + 6 NaOH 2 Na SbO + 3Hsoluble (Sodium antimonite)

⎯⎯→

4. Action of metals : Group 15 elements react with all metallic elements. Nitrogen forms nitrides.

2 36Li + N 2Li N (Lithium Nitride)⎯⎯→

2 3 23Mg + N Mg N (Magnesium Nitride)⎯⎯→

P, As, Sb, Bi form phosphides, arsenide and antimonate and bismuthide as respectively.

4 3 26Ca + P 2Ca P (Calcium phosphide)⎯⎯→

5. Reactivity towards hydrogen :

All the elements of group 15 react with hydrogen to form gaseous hydrides EH3

(E = N, P, As, Sb, Bi)

Properties of hydrides of group 15 elements

Property

E-H distance (pm)

Boiling point/K

Melting Point/KHEH angle (º)

Δdiss (E-H)/kJmol–1

ΔH–f /kJmol–1

NH3

101.7

239.6

195.3107.8

389

– 46.29

PH3

141.9185.5139.593.632213.4

AsH3

151.9

210.6

156.791.8

297

66.4

SbH3

170.7

256.0

18591.3

255

145.1

BiH3

–

290

––

–

278

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(a) Down the group, bond dissociation energy decreases along with stability of hydrides.(b) Down the group, reducing character increases.(c) Down the group, basicity decreases. NH3 > PH3 > AsH3 > SbH3 > BiH3.(d) Down the group, strong smell and poisonous nature increases of group 15 elements.

6. Tendancy to form hydrogen bonding : Among hydrides of group 15 elements, onlyNH3 can form intermolecular hydrogen bond with itself and with water molecules. .

This is due to polarity of N – H bond caused due to large electronegativity difference betweenN and H. Hydrogen bonding is responsible for solubility and high boiling point of NH3.

7. Anomalous nature of Nitrogen :Nitrogen has(a) small atomic size (b) high ionisation enthalpy(c) high electronegativity (d) non availability of d-orbitals in valence shellSo it shows following properties different than other elements of its group.

(a) Nitrogen is a gas while all other elements of the group are solids at roomtemperature.

(b) Nitrogen exits as diatomic molecule (N2), while phosphorus and the other elementsexit as tetraatomic molecules (As4, Sb4, P4, etc).

(c) Due to small size and high electronegativity, nitrogen is the only element of thegroup which can form hydrogen bonds in its hydride compounds. Other elementsdo not form hydrogen bonding in their hydride compounds.

(d) Nitrogen has tendency to form pπ-pπ multiple bonds. Other elements of the groupdo not form pπ-pπ multiple bonding. Instead, they form multiple bonding throughdπ-pπ overlapping.

(e) It shows wide range of oxidation states.(f) Except NF3, other trihalides are unstable.

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(f) Dinitrogen :1. Preparation :

(a) Commercial method : Liquid air fractionaldistillation

⎯⎯⎯⎯→ Liquid Nitrogen distills over,

B.p. of liq. N2 = 77.2 K.B.p. of liq. O2 = 90.0 K.

(b) Laboratory method :

4 (aq) 2(aq ) 2(g) 2 (l) (aq)NH Cl + NaNO N + 2H O + NaCl⎯⎯→

Impurities of NO, HNO3 are removed by passing over K2Cr2O7/H2SO4 mixture.

(c) From Ammonium dichromate : 4 2 2 7 2 2 2 3(NH ) Cr O N + 4H O + Cr O⎯⎯→Δ

(d) From bleaching powder : 2 3 2 2 23CaOCl + 2NH 3CaCl + 3H O + N⎯⎯→

(e) From Sodium or Barium azide : 3 2 2Ba(N ) Ba + 3N⎯⎯→

3 22NaN 2Na + 3N⎯⎯→ Nitrogen obtained by this method is poor.

2. Physical properties :(a) It is a colourless, odourless and inert gas.

(b) Its isotopes are 147 N , 15

7 N .

(c) It is slightly soluble in water.(d) It has low freezing point (195.3 K) and low boiling point (239.6 K).(e) N ≡ N bond has large bond enthalpy.

3. Chemical properties :At high temperature, Dinitrogen undergoes following reactions.(a) It combines with metals and non metals to form ionic and covalent nitrides.

2 3 23Mg + N Mg N⎯⎯→

2 36Li + N 2Li N⎯⎯→ Ionic nitrides

22Al + N 2AlN⎯⎯→

Fe/Al O + K O Haber's process2 3 22 2 3N + 3H 2NH⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→

Covalent nitrides2 2N + O 2NO⎯⎯→

(b) It reacts with calcium carbide to form Calcium cynamide.1000K

2 2 21AtmCaC + N CaCN + C⎯⎯⎯→

4. Uses :N – (a) Dinitrogen dilutes the action of dioxygen in air and thus, makes combustion much

less rapid.(b) It is used for filling electric lamps and provides an inert atmosphere in certain

metallurgical operations.

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(c) It is used in the manufacture of NH3, HNO3, CaCN2 and other nitrogencompounds.

(d) Liquid dinitrogen is used as a refrigerant to preserve biological materials, fooditems and in cryosurgery.

(g) Compounds of Nitrogen - NH3 :1. Preparation of Ammonia :

(a) Laboratory method :

4 2 2 3 2NH Cl + Ca(OH) CaCl + 2NH + 2H OSlaked lime

⎯⎯→

CaO3 3Quick lime

Moist NH Dry NH⎯⎯⎯⎯⎯→

It cannot be dried over H2SO4, P2O5 or anhydrous CaCl2, as it reacts with them.(b) Manufacturing method :

Haber’s process :

2 2 3N + 3H 2NH ΔHf = – 46.1 kJ mol–1

Optimum conditions : (a) High pressure of 200 atm(b) Temperature 700 K(c) Catalyst finely divided Iron and Molybdenum or Al2O3, K2O(d) Ratio N2 : H2 :: 1:3

(c) By hydrolysis of calcium cyanamide CaCN2.1000 K, 1atm

2 2 2CaC + N CaCN + C⎯⎯⎯⎯⎯→

2 2 3 3CaCN + 3H O CaCO + 2NHSteam

⎯⎯→

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2. Physical properties :(a) It is colourless gas with pungent smell.(b) Freezing point is 198.4 K; Boiling point is 239.7 K.(c) Water soluble and basic in nature.(d) Ammonia molecules are associated due to hydrogen bonding in solid and liquid state.(e) Due to intermolecular hydrogen bonding, melting point and boiling point is high.

[Refer Fig. 7.1 (e) 6](f) It liquifies on cooling under pressure to form liquid ammonia, which is used as cooling

agent.3. Chemical properties :

(a) It reacts with Air.

3 2 2 24NH + 3O 2N + 6H O⎯⎯→Δ

Pt3 2 2500K,9bar

4 NH + 5O 4 NO + 6H O⎯⎯⎯⎯⎯→

(b) It has reducing action on CuO and Cl2.

3 2 22 NH + 3CuO 3Cu + 3H O + N⎯⎯→Δ

3 2 4 28NH + 3Cl 6 NH Cl + Nexcess

⎯⎯→

(c) It reacts with Halogens in excess.

3 2 3NH + Cl NCl + 3HClexcess Nitrogen trichloride (explosive)

⎯⎯→

3 2 4 28NH + 3Br 6 NH Br + N⎯⎯→

3 2 3 32 NH + 3I NH NI + 3HIExplosive (Nitrogen triodide ammonia)

⎯⎯→

3 3 2 2 48NH NI 5N + 9I + 6 NH I⎯⎯→

(d) It reacts with active metals like Na, K.

575k3 2 22Na + 2NH 2NaNH + H⎯⎯⎯→

(e) It reacts with Sodium hypochloride.

glue/3 2 2 2gelatin

2NH + NaOCl NH NH + NaCl + H O hydrazine

⎯⎯⎯→

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(f) It undergoes self ionisation in liquid state.–

+3 4 22 NH NH + N H

(g) It reacts with water to form alkaline solution.+ –

3 2 4NH + H O NH + OH⎯⎯→

(h) Formation of coordination compounds.Due to presence of lone pair of electrons of N; NH3 acts as a Lewis base. It linkswith metal ions by donating lone pair.

2+ 2+(aq) 3(aq) 3 4 (aq)Cu + 4 NH [Cu(NH ) ]

blue deep blue

⎯⎯→

+ –(aq) (aq) (s) pptAg + Cl AgCl

White

⎯⎯→

(s) 3(aq) 3 2 (aq)AgCl + 2NH [Ag(NH ) Cl]

ppt colourless

⎯⎯→

(i)Nessler's reagent

K HgI2 4

Brown pptAmmonium salts

Million's base⎯⎯⎯⎯⎯⎯→

2 4 3 2 22K HgI + NH + 3KOH NH HgOHgI + 7 KI + 2H OIodide of Million's reagent

⎯⎯→

This is a test for ammonia and ammonium salts.4. Uses of Ammonia :

(a) For preparation of fertilisers.(b) For preparation of nitric acid in Ostwald’s process.(c) Liquid NH3 is used as refrigerant.

5. Structure of NH3 :State of hybridisation of N is sp3. N – H bonds are formedby sp3 – S overlap. Lone pair of electrons occupy sp3 hybridorbital. Bond angle is 107.50 to avoid repulsion betweenbonded pair - lone pair. Ammonia molecule is trigonal pyramidal

(h) Compounds of Nitrogen – HNO3 :1. Preparation :

(a) Laboratory method :distill

3 2 4 4 3NaNO + H SO NaHSO + HNONitre

⎯⎯⎯→

(b) Manufacture :Ostwald’s process

Pt/Rh gauge catalyst3(g) 2(g) (g) 2 (g)500 K 9bar

4 NH + 5O 4 NO + 6H O

air Nitric oxide

⎯⎯⎯⎯⎯⎯⎯→

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(g) 2 2(g) (g)2 NO + O 2 NO Nitrogen dioxide

2 2 3 (g)(g) l3NO + H O 2HNO + NO

Nitric acid

⎯⎯→

NO is by product. Which is reused. On concentration by distillation 68% by mass Nitric acid is obtained. By dehydration with conc. H2SO4, 98% concentration is obtained. It is aqua fortis i.e. strong water, because it attacks all metals.

2. Physical properties :(a) Pure nitric acid is colourless liquid.

(f.p. 231.4 K and b.p. 355.6 K)(b) In gaseous state molecule of HNO3

has a planar structure.(c) The pure acid is unstable.

In aqueous solution nitric acid behaves as strong acid and undergo ionisation forming

nitrate ions. + –3 2 3 3HNO (aq) + H O(l) H O (aq) + NO (aq)⎯⎯→

3. Chemical properties :(a) It is a strong acid in aqueous solution.

+ –3(aq) 2 ( ) 3 (aq) 3 (aq)HNO + H O H O + NOl ⎯⎯→

(b) Action on metals : Except Gold and Platinum all metals are attacked by nitric acid.It is strong oxidising agent.Nature of products depend on concentration of acid and temperature.

3 3 2 23Cu + 8HNO 3Cu(NO ) + 2 NO + 4H Odil.

⎯⎯→

3 3 2 2 2Cu + 4HNO Cu(NO ) + 2 NO + 2H Oconc.

⎯⎯→

3 3 2 2 24Zn + 10 HNO 4 Zn(NO ) + N O + 5H Odil.

⎯⎯→

3 3 2 2 2Zn + 4 HNO Zn(NO ) + 2NO + 2H Oconc.

⎯⎯→

Metals like cromium and aluminium are covered with a layer of oxide.Hence, they do not react with Nitric acid.

(c) Action on non metals :

3 2 2 2C + 4HNO CO + 4 NO + 2H Oconc.

⎯⎯→

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2 3 3 2 2I + 10HNO 2HIO + 10 NO + 4H Oiodic acid

⎯⎯→

8 3 2 4 2 2S + 48HNO 8H SO + 48NO + 16H O⎯⎯→

4 3 3 4 2 2P + 20HNO 4H PO + 20 NO + 4H Ophosphoric acid

⎯⎯→

(d) Action of Aqua regia : (HCl + HNO3 : : 3:1) on gold and platinum. conc. conc.

3 23HCl + HNO NOCl + 2H O + 2[(Cl)]⎯⎯→

HCl3 4Au + 3[Cl] AuCl HAuCl Aurochloric acid⎯⎯→ ⎯⎯⎯→

2HCl4 2 6Pt + 4[Cl] PtCl H PtCl Chloroplatinic acid⎯⎯→ ⎯⎯⎯→

(e) Action on organic compounds :(1) It introduces ‘nitro’ group in organic compounds.

conc.H SO2 46 6 2 6 5 2 2C H + HONO C H NO + H O

Benzene conc. Nitrobenzene⎯⎯⎯⎯⎯→

conc.H SO2 46 5 3 2 6 2 2 3 3 2C H CH + 3HONO C H (NO ) CH + 3H O

Toluene conc. 2, 4,6- trinitrotoluene⎯⎯⎯⎯⎯→

conc.H SO2 46 5 2 6 2 2 3 2C H OH + 3HONO C H (NO ) OH + 3H O

conc. 2,4,6-trinitrophenol (picric acid)⎯⎯⎯⎯⎯→

(2) It oxidises sugarHNO3

12 22 11 2 2C H O + 18(O) 6[COOH] + 5H OCane sugar Oxalic acid

⎯⎯⎯→

(f) Brown ring test : Determination of nitrateAq solution of nitrate + Conc. H2SO4 → Cool. Add dilute FeSO4 solution from theside of test tube.Brown ring at the junction confirms presence of nitrate.

– +2 + +33 2NO + 3Fe + 4H NO + 3Fe + 2H O⎯⎯→

2+ 2+2 6 2 5 2[Fe(H O) ] + NO [Fe(H O) NO] + H O

Brown complex⎯⎯→

(g) Uses of nitric acid : Manufacture of dyes and fertilisers. Manufacture of cellulose, varnishes. Manufacture of sulphuric acid in lead chamber process. Preparation of explosives TNT (trinitrotoluene) and cellulose nitrate. Pickling of stainless steel, etching of metals and rocket fuels.

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(i) Oxides of Nitrogen :1. Nitrogen forms a number of oxides in different oxidation states. Oxidation number, formula,

preparation method, physical appearance and chemical nature of these oxides are givenbelow :

Name

Nitrous oxideor Dinitrogen

oxide [Nitrogen(I) oxide]

Nitrogenmonoxide orNitric Oxide

[Nitrogen (II)oxide]

Dinitrogentrioxide

[Nitrogen (III)oxide]

Nitrogen dioxide[Nitrogen (IV)oxide]

Dinitrogentetraoxide

[Nitrogen (IV)oxide]

Dinitrogenpentoxide

[Nitrogen (V)oxide]

Formula

N2O

NO

N2O3

NO2

N2O4

N2O5

Oxidationstate of

Nitrogen

+1

+2

+3

+4

+4

+5

Common method of preparation

Heat4 3 2 2NH NO N O + 2H O⎯⎯⎯→

2 4 2 42 NaNO + 2FeSO + 3H SO →

2 4 3 4 2Fe (SO ) + 2NaHSO + 2H O + 2NO

250 K2 4 2 32 NO + N O 2 N O⎯⎯⎯→

673K3 2 2 22Pb(NO ) 4 NO + 2PbO + O⎯⎯⎯→

Cool2 2 4heat

2 NO N O⎯⎯⎯→←⎯⎯⎯

3 4 10 3 2 54HNO + P O 4HPO + 2 N O⎯⎯→

Physicalappearance

and chemicalnature

colourlessgas, neutral

colourlessgas, neutral,reactive

blue, solid,acidic

brown gas,acidic,reactive

colourlesssolid/liquid,acidic

colourlessionicsolid, unstableacidic

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2. Structures of oxides of Nitrogen(a) N2O : Dinitrogen oxide known as ‘Laughing gas’.

Diamagnetic linear molecule.(b) NO : Nitrogen monoxide

It is oxidising as well as reducing agent. It is paramagnetic in gaseous state, due to presence of unpaired electrons. It dimerises in solid or liquid state. Hence, it is diamagnetic in solid/liquid state. It is most stable oxide of Nitrogen.

(c) N2O3 Nitrogen trioxide known as Nitrogen sesquioxide (blue solid and brown gas)

(d) NO2 Nitrogen dioxide

NO2 is paramagnetic, but it dimerises to N2O4 which is diamagnetic.

2NO NO + [O]⎯⎯→

[O] is responsible for oxidising properties.

(e) N2O4 Dinitrogen tetraoxide.

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Formed by Dimerisation of NO2 molecules. Planar molecule, diamagnetic.N – N bond is formed by sp2 – sp2 overlap. It is diamagnetic.

(f) N2O5 Dinitrogen pentaoxide.

Structure is symmetrical and linear. It is good oxidizing agent.

(j) Allotropic forms of phosphorus :1. White (Yellow) phosphorus :

(a) Yellow, white waxy solid.(b) Poisonous(c) Insoluble in water; so stored under water to protect from air.(d) Discrete tetrahedral structure : P4 molecule(e) Reactive due to angle strain.(f) Dissolves in organic solvents like CS2.(g) Dissolves in caustic soda on boiling to form phosphine.

4 2 3 2 2P + 3NaOH + 3H O PH + NaH POPhosphine (sodium hypophosphite)

⎯⎯→

(h) Chemiluminescence : Its vapours undergo atmospheric oxidation to give faint,green glow.

4 2 4 10P + 5O P O (Phosphorous pentaoxide)⎯⎯→

This property of glowing even in dark is called as chemiluminescence.2. Red phosphorus :

(a) It is prepared by heating white

phosphorous in absence of air

in closed Iron pot to 573K.

Iodine is used as catalyst.

(b) It is reddish, violet solid.

(c) It is non poisonous.

(d) It is insoluble in water, alkali, organic solvents.

(e) It dissolves in conc HNO3.

(f) It does not show chemiluminescence.

(g) It is polymeric, with chains of tetrahedral units of P4 linked together.

442


White and red phosphorus3. Black phosphorus :

(a) It is most stable form of phosphorus, so unreactive.(b) It resembles graphite.(c It is insoluble in carbon di sulphide.(d) It is good conductor of electricity.(e) It has two forms α and β black phosphorus.

(f) 803KSealed tube

Red phosphorus α – formmonoclinic

⎯⎯⎯⎯⎯→

453Khigh pressure

White phosphorus β – form⎯⎯⎯⎯⎯→

(k) Compounds of phosphorus :1. Phosphine :

(a) Preparation of phosphine : From Calcium or Aluminium phosphide,

3 2 2 2 3Ca P +6H O 3Ca(OH) + 2PH⎯⎯→ ↑

3 2 2 3Ca P +6HCl 3CaCl + 2PHCalcium phosphide

⎯⎯→ ↑

2 4 2 4 3 32AlP + 3H SO Al (SO ) + 2PHAluminium phosphide

⎯⎯→ ↑

Lab method :

4 2 3 2 2P + 3NaOH + 3H O PH + 3NaH POSodium hypophoshite

⎯⎯→

4 3 2PH I + KOH KI + PH + H OPhosphonium

iodide

→

(pure phosphine is obtained by this method)

(b) Properties of phosphine :• Colourless, poisonous gas, sparingly soluble in water with rotten fish smell.• It explodes with oxygen in air, forming white smoke of phosphorus pentaoxide.

Properties

1. Colour2. Solubility in CS23. Action of air4. M.P.5. Action of hot NaOH6. Physiological Effect

White phosphorus

White (yellow)SolubleChemiluminescence317 KForms phosphinePoisonous

Red phosphorus

Reddish violetInsolubleNo chemiluminescenceAbove 773 KNo Phosphine formed (No action)Non poisonous

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• Flammability of phosphine is due to presence of traces of P2H4 or P4 vapours.• On heating, decomposes to red phosphrous and hydrogen.• It is powerful reducing agent.

4 3 3 2 2 43CuSO + 2PH Cu P + 3H SOcopper phosphide

⎯⎯→

2 3 3 23HgCl + 2PH Hg P + 6HClmercuric phosphide

⎯⎯→

3 2 5PH + 4Cl PCl + 3HCl⎯⎯→

• It reacts with acids

3 4PH + HBr PH Br (phosphonium bromide)⎯⎯→

(c) Uses of phosphine :• Used for preparation of Holme’s signal for ships to indicate position of rocks in

sea.

2 2 2 2 2CaC + 2H O Ca(OH) + C Hcalcium carbide acetylene

⎯⎯→

3 2 2 2 3Ca P + 6H O 3Ca(OH) + 2PHcalcium phosphide

⎯⎯→

PH3 catches fire and lights up acetylene, which acts as a signal for approaching ships.• Used for preparation of organo phosphorus compounds.• Used for preparation of smoke screens.

2. Halides of phosphorus :Halides are represented as PX3 (X = F, Cl, Br, I)PX5 (X = F, Cl, Br) *PI5 – does not exist due to steric effect.(a) Preparation of PCl3 :

4 2 3P + 6Cl 4PClwhite dry

⎯⎯→

4 2 3 2 2 2P + 8SOCl 4PCl + 2S Cl + 4SOwhite thionyl chloride disulphur dichloride

⎯⎯→

(b) Properties of PCl3 :Colourless, oily liquid, which hydrolyses.

3 2 3 3PCl + 3H O H PO + 3HClphosphorus acid

⎯⎯→

3 2 5PCl + Cl PCl⎯⎯→

3 3 3 3 33CH COOH + PCl 3CH COCl + H PO⎯⎯→

3 33AgCN + PCl P(CN) + 3AgCl⎯⎯→

444


(c) Preparation of PCl5 :excess

4 2 5P + 10Cl 4PCl⎯⎯⎯→

4 2 2 5 2P + 10SO Cl 4PCl + 10SOsulphuryl chloride

⎯⎯→

(d) Properties of PCl5 :It is yellowish white powder, which sublimes below 373 K.

5 2 3PCl + H O POCl + 2HCl⎯⎯→

3 2 3 4POCl + 3H O H PO + 3HCl⎯⎯→

5 3 2PCl PCl + Cl⎯⎯→Δ

3 2 5 3 2 3CH CH OH + PCl CH CH Cl + POCl + HCl⎯⎯→

5 32Ag + PCl 2AgCl + PCl⎯⎯→

5 4 3Sn + 2PCl SnCl + PCl⎯⎯→

(e) Uses of PCl5 :(i) Synthesis of alkyl chlorides, acetyl chloride.(ii) Chlorinating agent.

(f) Structures of PCl3 and PCl5PCl3 : P is sp3 hybridised.Shape is pyramidal.

PCl5 : In gaseous and liquid phase.P undergoes sp3d hybridisation, and is trigonalbipyramidal in shape.Three equatorial P–Cl bonds are equivalent.Two axial bonds P– Cl bonds are longer dueto more repulsion between axial bond pairs.PCl5 is unstable due to unsymmetrical shapein gaseous and liquid phase. In the solid stateit exists as an ionic solid, [PCl4]

+ [PCl6]– in

which the cation [PCl4]+ is tetrahedral and

anion [PCl6]– octahedral.

445


(l) Oxyacids of phosphorus :These acids contain at least one P=O bond and one P–OH bond.Oxoacid of phosphorus with +3 oxidation state tend to disproportionate (undergo oxidationand reduction) into higher and lower oxidation state.

3 3 3 4 34H PO 3H PO + PH[+3] [+5] [–3]

⎯⎯→

• Oxoacids with P–H bond act as strong reducing agent.

P–H bonds do not ionise to give H+ ions.• H attached to O (P — O — H) ionise causing basicity.

1. Formula : H3PO2

Name : Phosphinic acid or hypo phosphorus acidMethod of preparation :

boil2 28P + 3Ba(OH) + 6H O ⎯⎯⎯→ 2 2 2 3Ba(H PO ) + 2PH↓

2 2 2 2 4Ba(H PO ) + H SO ⎯⎯→ 4 3 2BaSO + 2H PO↓

3 2 3 2 3 3 44AgNO + 2H O + H PO 4Ag + 4HNO + H PO⎯⎯→

Oxidation state of P : +1; MonobasicStrong reducing agent due to 2 P–H bonds.

2. Formula : H3PO3

Name : Phosphonic or orthophosphorus acidMethod of preparation :

2 3 2 3 3P O + 3H O 2H PO⎯⎯→

Oxidation state of P : +3; Dibasic (due to 2P–OHbonds). It can act as reducing agent due to P–Hbond. It can undergo disproportionation (selfoxidation and reduction)

3. Formula : H4P2O5

Name : Pyrophosphorus acidMethod of preparation :

3 3 3PCl + H PO acid⎯⎯→

Oxidation state of P : +3; Dibasic (due to 2P–OH bonds).It can act as reducing agent.

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4. Formula : H4P2O6

Name : Hypophosphoric acidMethod of preparation :

Red P + alkali acid⎯⎯→

(Decomposition) 4 2 6 3 3 3H P O H PO + HPO⎯⎯→Oxidation state of P : +4; Tetrabasic

5. Formula : H3PO4

Name : Orthophosphoric acidMethod of preparation :

4 10 2 3 4P O + 6H O 4H PO⎯⎯→

Oxidation state of P : +5; Tribasic

6. Formula : H4P2O7

Name : Pyrophosphoric acidMethod of preparation :

Δ

3 4 4 2 7 22H PO H P O + H O⎯⎯→

Reaction is reversed on heating with water.Oxidation state of P : +5, Tetrabasic

7. Formula : (HPO3)n

Name : Polymetaphoric acid, glacial phosphoric acidMethod of preparation :

heated2 4 nin sealed tube

Phosphorus acid + Br (HPO )⎯⎯⎯⎯⎯→

Oxidation state of P : +5


*1. How does NH3 act as Lewis base?Ans. (i) According to Lewis acid-base theory base is a substance which

can donate a lone pair of electron.

(ii) In NH3, N has a lone pair of electron.

(iii) Therefore, NH3 can donate a lone pair of electron and act as Lewis base.+

–3 4NH + HCl N H + Cl⎯⎯→

H|

H – N – Hi i

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*2. How does ammonia react with white precipitate of AgCl?

Ans. 3(aq) 3 2 (aq)AgCl + 2 NH Ag(NH ) Cl

white ppt. colourless complex

⎯⎯→

*3. Enlist the conditions required to maximize the yield of ammonia.

Ans. 2 2 3N + 3H 2NH ΔHf = – 46.1 kJ mol–1

Optimum conditions : (a) high pressure of 200 atm(b) temperature 700 K(c) catalyst finely divided Iron and Molybdenum or Al2O3, K2O(d) Ratio N2 : H2 :: 1:3

*4. Find the covalence of nitrogen in N2O5 (Dinitrogen pentoxide).Ans.

*5. How does nitrogen differ from other elements of group 15?Ans. Refer 7.1 (e) 7.

*6. Why does R3P = O exist but R3N = O doest not?Ans. Due to non availability of d orbitals, N cannot increase its covalency beyond 4.

*7. Why does nitrogen show catenation properties less than phosphorus?Ans. (a) Nitrogen shows poor catenation tendency as compared to phosphorus due to high N–N

bond enthalpy.(b) Nitrogen forms pπ–pπ bonds, whereas phosphorous can form dπ–pπ bonds .

*8. Why does H3PO2 behave as reducing agent?Ans. The structure of H3PO2 (hypophosphorous acid), since it has P – H

bond it is strong reducing agent.

*9. What are the basicities of H3PO4 and H3PO3?Ans. H3PO4 – tribasic acid - due to three P – OH bonds.

H3PO3 – Dibasic acid - due to two P – OH bonds.

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*10. Why are trihalides less covalent than pentahalides?Ans. (a) In trihalide oxidation state of group 15 elements is +3 and pentahalides it is +5.

(b) The energy required to remove all 5 electrons will be very large. Hence, +5 oxidationstate is less stable than +3 oxidation state.

(c) In +5 oxidation state, due to larger positive charge, the polarising power is more.(d) This is in accordance with Fajan’s principle that greater positive charge of cation, more

is the polarising power and greater is the covalent character.(e) Hence, trihalide are less covalent than pentahalides.

*11. Why is BiH3, the strongest reducing agent amongst all the hydrides of group 15 elements?Ans. As Bi has largest atomic size. BiH3 has the lower Bi–H bond dissociation enthalpy amongst

the hydrides of group15.

*12. Why is N2 rather inert at room temperature?

Ans. 2 (g) (g)N 2 N⎯⎯→ ΔH = – 944.7 kg mol–1. High bond enthalpy of N ≡ N is responsible

for chemical inertness of N2.

*13. Name the elements in the groups 15, 16, 17 and 18.Ans. Group 15 – Nitrogen, Phosphorus, Arsenic, Antimony, Bismuth.

Group 16 – Oxygen, Sulphur, Selenium, Tellurium, Polonium.Group 17 – Fluorine, Chlorine, Bromine, Iodine, Astatine.Group 18 – Helium, Neon, Argon, Krypton, Xenon, Radon.

*14. Which are the differences in the reactivity of first and second element in the group 15?Ans. (a) The 1st and 2nd element group 15 are N and P respectively.

(b) N has small atomic size, greater electronegativity and no d-orbitals, hence, it behavesdifferent than P.

(c) Nitrogen reacts with O2 at a very high temp. while P4 reacts with O2 at room temp.

2000K2 2N + O 2NO

; 4 2 4 10P + 5O P Owhite phosphorus pentaoxide

⎯⎯→

(d) N2 does not react with NaOH, while P

4 reacts as follows :

4 2 3 2 2P + 3NaOH + H O PH + 3NaH POphosphine sodium hypophosphide

⎯⎯→

(e) N2 does not react with conc. HNO3 or conc. H2SO4 while P4 reacts with both as

follows :4 3 3 4 2 2

4 2 4 3 4 2 2

P + 20HNO 4H PO + 20NO + 4H O

P + 10H SO 4H PO + 10SO + 4H O

⎯⎯→⎯⎯→

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*15. Explain the non metallic and metallic character of elements of group 15.Ans. Non metallic and metallic character : Electropositive character and metallic nature

increases down the group. N, P are non metals. As, Sb are metalloids. Bi is metal.

*16. What are the different allotropes of phosphorous?Ans. Allotropy : Except Bi all the other elements in this group exhibit allotropy.

Following are the common allotropes :

Phosphorus :

White phosphorus Red phosphorus Black phosphorustetrahedron structure complex chain structure stable layer structure

*17. Draw the structures of white and red phosphorous.Ans. Refer for diagram. Q.No. 16.

*18. Why +3 oxidation state is stable in heavier element of group 15?Ans. Oxidation state : Electron configuration is ns2np3. So common oxidation states are +3, +5

and –3. Stability of +3 oxidation state increases and +5 oxidation state decreases down thegroup due to ‘Inert pair effect’of ns2 electrons. However, N does not exhibit +5 oxidationstate through covalent due to non availability of d orbitals in valence shell. N exhibits all theoxidation states from –3 to +5.

*19. What happens when As4, Sb, and P4 is treated separately with hot and concentrated nitric acid?Ans. Refer 7.1 (e) 2

*20. How is dinitrogen prepared?Ans. Refer 7.1 (f) 1.

*21. What is the action of nitrogen on (a) Magnesium (b) Hydrogen (c) Oxygen?Ans. (a) Magnesium :

2 3 23Mg + N Mg N⎯⎯→

2 36Li + N 2Li N⎯⎯→

22Al + N 2AlN⎯⎯→

450


(b) Hydrogen :

2 2 3N + 3H 2NH⎯⎯→

(c) Oxygen :

2 2N + O 2NO⎯⎯→

*22. How is ammonia prepared in the laboratory?Ans. Laboratory method :

4 2 2 3 2NH Cl + Ca(OH) CaCl + 2NH + 2H OSlaked lime

⎯⎯→

CaO3 3Quick lime

Moist NH Dry NH⎯⎯⎯⎯⎯→

It cannot be dried over H2SO4, P2O5 or anhydrous CaCl2, as it reacts.

*23. How is ammonia manufactured?Ans. Refer 7.1 g (b).

*24. Ammonia is a Lewis base. Explain.Ans. Due to presence of lone pair of electorn of N; NH3 acts as a lewis base. It links with

metal ions by donating lone pair.2+ 2+

3(aq) 3 4 (aq)Cu + 4 NH [Cu(NH ) ]

deep blue

⎯⎯→

+ –(s) pptAg + Cl AgCl⎯⎯→

(s) 3(aq) 3 2 (aq)AgCl + 2NH [Ag(NH ) Cl]

ppt colourless

⎯⎯→

*25. Explain, the structure of ammonia.Ans. Refer 7.1 (g) 5.

*26. What is the action of concentrated nitric acid on (a) Copper, (b) Zinc?Ans. Refer 7.1 (h) 3 (b).

*27. Name the oxides of nitrogen.Ans. Refer 7.1 (i) 1.

*28. Name and draw the structures of oxides of nitrogen.Ans. Refer 7.1 (i) 1 and 2 (a).

*29. Draw the structure of : N2O3, N2O5.Ans. Refer 7.1 (i) b.

*30. What are the oxidation state of nitrogen in N2O4?Ans. Oxidation state of Nitrogen in N2O4 is 4 (2x – 8 = 0) or x = 4

451


*31. What happens when white phosphorous is heated with concentrated NaOH solution?Ans. Refer 7.1 (j) 1.

*32. How is phosphine prepared?Ans. Refer 7.1 (k) 1.

*33. What are the uses of PCl3 and PCl5?Ans. PCl3 : It is used for chlorinating organic compound.

PCl5 : It is used in synthesis of some organic compounds.Eg. C2H5Cl, CH3COCl. It is also used as a chorinating agent.

*34. What happens when PCl5 is heated?

Ans. It sublimes. 5 3 2PCl PCl + Cl⎯⎯→Δ

*35. Name and draw the structures of different oxyacids of phosphorous.Ans. Refer 7.1 - l (1 to 7).

*36. What is the oxidation state of phosphorous in H4P2O6 and H4P2O7?Ans. Oxidation state of phosphorous in H4P2O6 is + 4 H4P2O7 is + 5

*37. Ammonia is highly soluble in water. Explain.Ans. (a) Nitrogen is highly electronegative element.

(b) Hence, N–H bond in NH3 is polor.(c) NH3 is therefore, capable of forming hydrogen bonding with itself as well as with

water molecules.(d)

(e) Therefore, NH3 is highly soluble in water.


• Theoretical MCQs :*1. PCl5 exists but NCl5 does not due to .......

a) Inertness of N2 b) NCl5 is unstablec) Larger size of N d) Non availability of vacant d-atomic orbitals

*2. The p–p–p angle in white phosphorous is .......a) 120º b) 109º28 c) 90º d) 60º

452


*3. Catalytic oxidation of NH3 gives .......a) dinitrogen pentoxide b) Nitrogen dioxide c) Nitrogen d) Nitric oxide

*4. Structure of NH3 is ..........a) Trigonal bipyramidal b) Tetrahedral c) Pyramidal d) Trigonal

*5. The basicity of phosphorous acid (H3PO3) is .......a) one b) two c) four d) three

*6. Maximum covalency of nitrogen is .......a) two b) four c) three d) five

*7. NH3 is basic while PH3 is .......a) acidic b) neutral c) amphoteric d) basic

8. Which of the following orders regarding the melting points of hydrides of Group 15 is .......a) NH3 > PH3 > AsH3 b) NH3 < PH3 < AsH3

c) NH3 > PH3 < AsH3 d) NH3 < PH3 > AsH3

9. The ammonia molecule is .......a) triangular planarb) regular tetrahedron with bond angles equal to 109º 28′.c) trigonal bipyramidald) trigonal pyramidal

10. Which of the following orders regarding the bond angle H – M – H of the hybrides of Group15 elements is correct?a) H–N–H > H – P – H > H – As – H b) H–N–H < H – P – H > H – As – Hc) H–N–H > H – P – H < H – As – H d) H–As–H < H – P – H > H – As – H

11. Which of the following halides is not kown?a) NCl5 b) PF5 c) AsF5 d) SbCl5

12. Which of the following statements is not correct?a) With the exception of nitrogen, all other elements of Group 15 form pentahalides.b) The pentahalides of elements of Group 15 are trigonal bipyramidal.c) The pentahalides of elements of Group 15 undergo incomplete hydrolysis to give

the corresponding acids.d) The pentahalides of elements of Group 15 undergo complete hydrolysis to give the

corresponding acids.13. The dimerization of NO2 is accompanied with .......

a) an increase in paramagnetism b) a decrease in paramagnetismc) no change in paramagnetism d) a darkening in colour

14. Which of the following acids of phosphorus is a reducing acid?a) H3PO3 b) H3PO4 c) H4P2O7 d) (HPO2)3

453


15. Orthophosphoric acid is a .......a) monobasic acid b) dibasic acid c) tribasic acid d) tetrabasic acid

16. Orthophosphorous acid is a .......a) monobasic acid b) dibasic acid c) tribasic acid d) tetrabasic acid

17. Polymetaphosphoric acid has .......a) linear structure of HPO3 unitsb) branched structure of HPO3 unitsc) cyclic structure of HPO3 unitsd) discrete molecules of (HPO3)2, (HPO3)3 and so on

18. In which of the following acids, P–P bond is present?a) Hypophosphoric acid b) pyrophosphoric acidc) Orthophosphoric acid d) Polymetaphosphoric acid

19. The oxide which on dissolving in water turns blue litmus red is .......a) P2O5 b) As2O3 c) BaO d) Sb2O3

20. Conc. HNO3 oxidizes iodine to .......a) HI b) HIO3 c) NH4I d) HIO4

21. Which of the following acids contains phosphorus in the +4 oxidation state?a) Hypophosphorous acid b) Orthophosphorous acidc) Phosphoric acid d) Hypophosphoric acid

22. Phosphine may be produced by adding to water some .......a) Ca3P2 b) P4O6 c) P4O11 d) HPO3

23. Which of the following statements is not correct?a) The molecule of NO2 is angularb) Low temperatures favours the dimerization of NO2 to N2O4.c) NO2 is soluble in water giving a mixture of HNO3 and HNO2.d) The structure of N2O4 is nonplanar.

24. Which of the following statements is not correct?a) Phosphorus trioxide exists as dimer P4O6.b) Phosphorus pentoxide exists as dimer P4O10.c) Orthophosphorous acid, H3PO3 is a tribasic acid.d) Orthophosphoric acid, H3PO4 is a tribasic acid.

25. Which of the following halides does not exist?a) PI5 b) PBr5 c) PCl5 d) PF5

26. Which of the following forms maximum P–H bonds?a) H3PO2 b) H3PO3 c) H3PO4 d) H4P2O7

27. The solid phosphorus pentachloride exists as .......a) PCl5 b) PCl4

+Cl– c) PCl6– d) PCl4

+.PCl6–

454


28. The most reactive form of phosphorus is .......a) red b) yellow c) violet d) black

29. Which of the following hydrides does not exist?a) NH3 b) PH5 c) AsH3 d) N2H4

30. Which of the following hydrides is thermally least stable?a) NH3 b) PH3 c) AsH3 d) SbH3

31. In the reaction 3 2 3N + 3H 2NH , the yield of ammonia is expected to be maximum

at .......a) low temperature and low pressure b) low temperature and high pressurec) high temperature and low pressure c) high temperature and high pressure

32. Highest oxidation state of nitrogen is achieved in .......a) nitrogen dioxide b) dinitrogen trioxidec) dinitrogen tetroxide d) dinitrogen pentoxide

33. Which of the following statements regarding ammonia is not correct?a) Ammonia is a colourless and pungent smelling gasb) Ammonia can be easily liquefied and solidifiedc) Ammonia can act as Lewis acidd) Ammonium chloride dissolves in liquid ammonia has acidic properties.

34. The gas liberated when copper reacts with dilute HNO3 is .......a) NO b) NO2 c) N2O3 d) N2O5

35. In the brown-ring test of nitrate ion, the compound formed is .......a) [Fe(H2O)5NO]2+ b) [Fe(H2O)5NO]3+

c) [Fe(H2O)4(NO2)2+ d) [Fe(H2O)3(NO)3]2+

36. The aqua regia is .......a) 3 parts conc. HNO3 + 1 part conc. HClb) 1 part conc. HNO3 + 3 parts conc. HClc) 2 parts conc. HNO3 + 2 parts conc. HCld) 2.5 parts conc. HNO3 + 0.5 part conc. HCl

37. Which of the following halides of nitrogen is expected to be most stable?a) NF3 b) NCl3 c) NBr3 d) NI3

38. The order of stability of hydrides of Group 15 is .......a) NH3 > PH3 > AsH3 b) NH3 < PH3 < AsH3

c) NH3 > PH3 < AsH3 d) NH3 < PH3 > AsH3

39. The laughing gas is .......a) N2O b) NO c) NO2 d) N2O3

40. Pyrophosphorous acid is a .......a) monobasic acid b) dibasic acid c) tribasic acid d) tetrabasic acid

455


• Advanced MCQs :41. Which of the following molecules includes nitrogen atom having oxidation state equal to –2?

a) N2 b) NH2OH c) N2H4 d) NH3

42. Given that 2 2 42NO N O ΔH = negative. The condition at which the conversion of NO2

to N2O4 becomes more and more are :a) lower temperature and lower pressure b) lower temperature and higher pressurec) higher temperature and higher pressure d) higher temperature and lower pressure

43. Which of the following statements regarding nitrogen molecule is not correct?a) Amongst the homonuclear diatomic molecules of second period, the bond dissociation

enthalpy is maximum in case of N2 molecule.b) Amongst the homonuclear diatomic molecules of second period, the bond length is minimum

in case of N2 molecule.c) Nitrogen molecule is paramagnetic in nature.d) Only at high temperatures, nitrogen molecule reacts with metals and non metals forming

ionic and covalent nitrides.

7.2 GROUP 16 ELEMENTS :


(a) General introduction :

1. Occurrence : First four elements are called as ‘chalcogens’ i.e. ore forming elements.

Oxygen : (1) Oxygen is the most abundant of all the elements on the earth. Oxygenforms about 46.6% by mass of earth crust.

(2) Dry air contains 21.0% of oxygen by volume.

Sulphur : (1) Sulphur occurs less abundantly. It is only 0.03-0.1%.(2) Sulphur in the combined form exists mainly as sulphates, gypsum

CaSO4.2H2O, epsum salt MgSO4, 7H2O, baryte BaSO4 and sulphideslike zinc blende ZnS, galena PbS, copper pyrities CuFeS2.

(3) It is found that very small quantity occurs as H2S (Hydrogen sulphide)in volcanoes, hot springs, in meteorites, etc.

(4) Onion, garlic, mustard, eggs, proteins, hair and wool, the organic materialsalso contain sulphur.

Selenium and tellurium occur as metal selenides and tellurides in sulphide ores.Polonium is radioactive.Group 16 consists of oxygen (O), sulphur (S), selenium (Se), tellurium (Te) and polonium

456


(Po). The first four elements of the group i.e. oxygen, sulphur, selenium and tellurium arecalled chalcogens, meaning ore forming elements. This is because a large number ofthe metals found in the earth’s crust are oxides and sulphides, such as Cu2O, CuO, Ag2S,ZnS, FeS etc. The metals may occur as selenides and tellurides also.

2. Electronic configuration : It is ns2np4. Each element requires two electrons to attainstable inert gas configuration.

Electronic configuration of group 16 elements(b) Trends in physical and chemical properties :

Physical properties of group 16 elements

Property

Atomic Number

Atomic MassCovalent Radius (pm)Ionic Radius M–2(pm)

Density at 298K (gcm–3)

Melting Point K

Boiling Point K

Ionization enthalpy(I)kJ mol–1

Electronegativity

Electron gainEnthalpy ΔegHkJ mol–1

O

816.00

74140

1.32 (at m.p)

55

901310

3.50–141

S

1632.06

1041842.06

(for rhombic form)

386 (for

Monoclinic form)718

1000

2.5–200

Se

3477.96

1171984.19

(for hexagonalgrey allotrope)

490

958941

2.48–195

Te

52127.60

1372216.25

723

1263870

2.1–190

Po

84210

140230 (approx)

–

527

1233812

2.0–183

Elements

Symbol

OSSeTePo

AtomicNo.

816345284


1s22s22p4

1s22s22p63s23p4

1s22s22p63s23p63d104s24p4

1s22s22p63s23p63d104s24p64d105s25p4

1s22s22p63s23p63d104s24p64d104f14

5s25p65d106s26p4

Briefrepresentationof electronicconfiguration

[He] 2s22p4

[Ne] 3s23p4

[Ar] 3d104s24p4

[Kr] 4d105s25p4

[Xe] 5d106s26p4

Elements

OxygensSulphurSeleniumTelluriumPolonium

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1. Physical state and molecular structure :(a) Oxygen is a diatomic gas. O=O is formed

by pπ – pπ overlap. Molecules are heldtogether by weak Van der Waals forces.

(b) Sulphur and Selenium are solids formed as

polyatomic complex structures.

2. Atomic and ionic radii : It increases down the group as electrons are added to a new shell.

3. Density : It increases down the group due to increase in the attractive van der Waal’s

forces. The crystal structure becomes more compact.

4. Electronegativity : It decreases down the group with increase in atomic size.

5. Melting and boiling point : Values increase down the group. Large difference in values

of m.p and b.p between O and S is due to diatomic nature of oxygen O2 and octatomic S8.

M.P. and B.P. of Po < Te, because of maximum inert pair effect in Po. Vander Waal’s

forces are weaker in Po, causing lower m.p and b.p.

6. Ionisation enthalpy : Ionisation enthalpy increases across the period. But first ionisation

enthalpy for group 16 elements is lower than first ionisation enthalpy for groups 15.

Group 15 elements have stability due to exactly half filled configuration.

ns npGroup 15 Elements ns2np3

(more stable, more symmetrical)Group 16 Elements ns2np4 ns np

(Less stable, less symmetrical)

So removal of electron should disturb the stability. Group 16 elements attain the state ofstable half filled configuration on loss of electron. Hence, first ionisation energy is low.

7. Electron gain enthalpy : – –M + e M + energy⎯⎯→

Electron gain enthalpy for group 16 elements is high and it decreases down the group,due to increase in atomic size. But ΔegH for O < S, due to small atomic size of O, incomingelectron is repelled by np3 electrons.

8. Metallic character : Increases down the group. Po is a metal.9. Catenation : Oxygen has little tendency for catenation. Sulphur has maximum tendency

of catenation. It decrease down the group from S to Po.

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10. Allotropy : Following are allotropes :O = Oxygen, ozone S = α– Sulphur, β–sulphurSe = Crystalline, amorphous Te = Metallic, non-metallicPo = α–form, β–form

11. Oxidation state :

3s 3p 3dSulphur atom(ground state)

sp3 hybridisation, tetrahedral atom exhibits +2 oxidation state

3s 3p 3dSulphur atom(first excited state)

sp3d hybridisation, trigonal bipyramidal atom exhibits +4 oxidation state

3s 3p 3dSulphur atom(second excited state)

sp3d2 hybridisation, octahedral atom exhibits +6 oxidation state

Hybridisation of sulphur

Configuration of group 16 elements is ns2np4. So possible oxidation states are –2, +2, +4,+6. Oxygen exhibits oxidation state – 2 in oxides and –1 in peroxides, O shows + oxidationstate when it combines with F. From S to Po, tendency to show –2 oxidation state decreases;Po shows +2 oxidation state. Except Oxygen, excitation of electrons in the empty d orbitalsmay take place.

(c) Chemical Properties :1. Reactivity decreases down the group.

Action of :

(a) air 2 2S + O SO⎯⎯→

(b) acids 3 2 4 2 2S + 6HNO H SO + 6NO + 2H O⎯⎯→

(c) alkali 2 3 2 23S + 6NaOH Na SO + 2Na S + 3H O⎯⎯→

(d) Non-metals 22S + C CS⎯⎯→

2 4Se + 2Cl SeCl⎯⎯→ 2 2Se + 2F SeF⎯⎯→

(e) metals Cu + S CuS⎯⎯→

459


2. Hydrides, halides and oxides.(a) Molecular formula of hybrides is H2M :

O forms additional hybride H2O2

Hybrides H2O H2S H2Se H2Te

H-M-H 104.5o 92.1o 91o 90o

Central atom M is sp3 hybridised. Due to high electronegativity of O, electron densityaround O is maximum with maximum repulsion between the electrons. So H-O-H

bond angle is maximum. As electronegativity of M decreases down the group, bond

angle H-M-H decreases.(b) Physical state and volatility :


Boiling point K 373 213 232 269

Boiling point of H2O is very high

due to association of H2O moleculesdue to hydrogen bonding.

Hydrogen bonding is not possible in

other hydrides, due to lowerelectronegativity differences

between M and Hydrogen. Boiling

point increases from H2S to H2Tedue to increase in molecular size and

increase in Van der Waals forces.

(c) Thermal stability :


ΔHdiss (M–H) kJ mol–1 463 347 276 238

As atomic size increases down the group. M-H bond becomes weaker and thermal

stability decreases.(d) Acidic Character : Hybrides of group 16 elements are weak diprotic acids.

+ –2 2 3H S + H O H O + HS

– + 2 –2 3HS + H O H O + S

460


acid strength varies as 2 2 2 2H O H S H Se H Te< < < .

The decrease is due to decrease in the bond dissociation energy of M-H bond.Hybrides react with bases as follows

( )2 2H S + NaOH NaHS + H O

Sodiumhydrosulphide⎯⎯→

( )2 2NaHS NaOH Na S H O

SodiumSulphide+ → +

(e) Reducing character : Except H2O, all the hybrides of group 16 act as strong reducingagents. Reducing property increases down the group due to decrease in the thermalstability. H2O is very stable. So it does not act as reducing agent.

(f) Reactivity towards halogens :(1) As F is more electronegative than O, F2O is written as OF2.(2) Stability of halides of S is F– > Cl– > Br– > I–.(3) All sulphur hexahalides are stable, gaseous with octahedral structure. S is sp3d2

hybridized.

(4) Among tetrafluorides SF4 is gas, SeF4 is liquid and TeF4 is solid. They are sp3dhybridised with trigonal bipyramidal structure with lone pair occupying equatorialstructure. This is called as see-saw geometry.

(5) Dihalides are formed by sp3 hybridisation. Hence, they have tetrahedral structure.(6) Monohalides are dimeric.

They undergo disproportionation as 2 2 42Se Cl SeCl + 3Se⎯⎯→ .

(g) Reactivity with oxygen :All the elements form oxides MO2 and MO3. M=S, Se, Te, Po. All are acidic in nature.

(h) Anomalous behaviour of oxygen :Oxygen shows anomalous behaviour due to ;(i) High electronegativity and (ii) absence of ‘d’ orbitals in valence shell.

Some important points of difference are :(1) Oxygen is gas, other members are solids.

Element

O

S

Se

Te

Po

Fluoride

OF2

S2F2, SF2, SF4, SF6

Se2F2, SeF4, SeF6

TeF4, TeF6

–

Chlorides

Cl2O, ClO2, Cl2O7

S2Cl2, SCl2, SCl4

Se2Cl2, SeCl2,SeCl4TeCl2, TeCl4PoCl2, PoCl4

Bromides

Br2O

S2Br2, SBr2

SeBr2, SeBr4

TeBr2, TeBr4

PoBr2, PoBr4

Iodides

I2O5

S2I2

–

TeI4

PoI4

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(2) Oxygen is diatomic, while others are polyatomic.(3) Oxygen is paramagnetic, as can be explained from molecular orbital theory, other

elements are diamagnetic.(4) Oxygen shows oxidation state–2. Others show –2, +2, +4, +6.(5) Oxygen can form hydrogen bonds in it’s compounds due to high electronegativity.

Others do not form.(6) Oxygen forms ionic compounds unlike others.(7) Oxygen forms pπ–pπ multiple bonds with C, N etc. Other elements do not form.(8) Hydride of oxygen i.e. H2O is liquid at room temperature. Other elements form

gaseous hydrides.

(d) Dioxygen :It is most abundant element on the earth. Oxygen is 50% by weight of earth’s crust.Oxygen in the atmosphere is present due to photosynthesis by green plants. It can be

represented as 2 2 2 2H O + CO C (H O) + Ox xx x x⎯⎯→ .

1. Methods of preparation :(a) By thermal decomposition of salts.

3 2 (g)MnO22KClO 2KCl + 3O⎯⎯⎯→

Δ

3 2 2 (g)2KNO 2KNO + O⎯⎯→Δ

4 2 4 2 2 (g)2KMnO K MnO + MnO + O⎯⎯→Δ

2 2 7 2 2 3 2 (g)2K Cr O 2K O + 2Cr O + 3O⎯⎯→Δ

4 22KClO 2KCl + 4O⎯⎯→

(b) By thermal decomposition of metallic oxides and peroxides.

2 (g)2HgO 2Hg + O⎯⎯→Δ

2 2 (g)2Ag O 4Ag + O⎯⎯→Δ

3 4 22Pb O 6PbO + 3O⎯⎯→Δ

2 (s) 2 (g)2PbO 2PbO + O⎯⎯→Δ

2 22BaO 2BaO + O⎯⎯→Δ

2 2 2 22H O 2H O + O⎯⎯→Δ

2 3 4 26MnO 2Mn O + 2O⎯⎯→Δ

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(c) By action of water

2 2 2 22 Na O + 2H O 4 NaOH + OSodium Peroxide

⎯⎯→

4 2 2 2 4 2 4 4 2 4 2 22KMnO +5Na O +8H SO K SO +2MnSO + 5Na SO 8H O+5O⎯⎯→ +

(d) Laboratory method

420 K3 2MnO2

2KClO 2KCl + 3O⎯⎯⎯→

KCl : MnO2 : : 4 : 1.MnO2 is catalyst. It lowers reaction temperature. O2 is collected by downwarddisplacement of water.

(e) Manufacture :1. Electrolysis of acidified water gives H2 at cathode and O2 at anode.2. Fractional distillation of liquid air gives two fractions, liq N2 (b.p. 77.2 K) and liq

O2 (b.p. 90 K). Liquid N2 distills over leaving behind liquid O2.2. Physical properties of dioxygen :

(a) It is tasteless, colourless, odourless gas.(b) It’s solubility in water is 30cm3 L–1 at 293 K.

So animals living in water can get their requirement of oxygen for living.(c) It is absorbed by alkaline pyragollal.

(d) Pressure CoolDioxygen Paleblue liquid light blue solidb.p90.2K m.p 54.4K

⎯⎯⎯⎯→ ⎯⎯⎯→

(e) It is heavier than air.(f) It is paramagnetic (due to presence of unpaired electron in antibonding π – orbital.(g) It’s stable isotopes are 16O, 17O, 18O.

3. Chemical properties of Dioxygen : It is very stable due to high bond dissociation energyof 493.4 kJ mol–1. It reacts at very high temperature.(a) It is not combustible, but it supports combustion.(b) It reacts with metals and non metals to form oxides.

1. Active metals like Na, K, Ca react at room temperature.

22Ca + O 2CaO⎯⎯→

2. Magnesium burns.

22Mg + O 2MgO⎯⎯→

3. Al, Fe react on heating.Δ

2 2 36Al + 3O 3Al O⎯⎯→ .

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4. Non metals like S, P, C, H2, N2.

2 2S + O SO⎯⎯→

5. It oxidises various compounds.

(i)Pt/Rh gauze

3 2 2500K, 9bar4 NH + 5O 4 NO + 6H O⎯⎯⎯⎯⎯→

(ii)723K

2 2 2CuCl24HCl + O 2H O + 2Cl⎯⎯⎯→

(iii)723K

2 2 3V O , 2-3atm pr2 52SO + O 2SO⎯⎯⎯⎯⎯⎯→

(iv) 2 2 2 2CS + 3O CO + 2SO⎯⎯→

(v) Sulphides of Zn, Mg.

2 22ZnS + 3O 2ZnO + 2SO⎯⎯→

(vi) Hydrocarbons to CO2 and water. All reactions are highly exothermic.

6 6 2 2 215

C H + O 6CO + 3H O2

⎯⎯→

2 4 2 2 2C H + 3O 2CO + 2H O⎯⎯→

2 6 2 2 27

C H + O 2CO + 3H O2

⎯⎯→

4. Uses of Oxygen :(a) For sustaining life on earth.(b) As constituent of rocket fuel.(c) Oxygen is used in Oxy hydrogen/acetylene flame for cutting and welding.(d) Coal + O2 mixture is used as explosive in mining.(e) In metallurgical process for removing impurities of metals.(f) For preparation of phenols, sulphuric acid, nitric acid etc.

(e) Classification of oxides :They are binary compounds of oxygen with metals and non metals. Some elements formmore than one type of oxides, different in properties and bonding. They are mainly ofthree types.1. Acidic oxides : They are binary compounds of oxygen and non metal. They form

acidic solution with water. Example : B2O3 (Boron trioxide), SO2, SO3 etc.

e.g. 2 (g) 2 ( ) 2 3(aq)CO + H O H COl ⎯⎯→ Carbonic acid

2 5 (g) 2 ( ) 3(aq)N O + H O 2HNOl ⎯⎯→ Nitric acid

4 10(s) 2 ( ) 3 4 (aq)P O + 6H O 4H POl ⎯⎯→ Phosphoric acid

2 7 (g) 2 ( ) 4 (aq)Cl O + H O 2HClOl ⎯⎯→ Perchloric acid

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2. Basic oxides : They are binary compounds of oxygen with highly electropositivemetals.They form basic solution with water. Example : K2O, MgO, Na2O etc.

e.g. (s) 2 ( ) 2 (aq)CaO + H O Ca(OH)l ⎯⎯→

(s) 2 ( ) 2 (aq)BaO + H O Ba(OH)l ⎯⎯→

3. Amphoteric oxides : They are binary compounds of oxygen and border line elements.They react with both acids and bases to form salt and water.

2 3 3 2Al O + 6HCl 2AlCl + 3H O⎯⎯→

[ ]2 3 2 3 6Al O + 6 NaOH + 3H O 2 Na Al(OH)⎯⎯→

2 2ZnO + 2HCl ZnCl + H O⎯⎯→

2 2 2ZnO + 2 NaOH Na ZnO + H O⎯⎯→

4. Neutral oxides : Oxides which are neither acidic nor basic.e.g. NO, N2O, CO.

5. Basic character of oxides decreases on moving from left to right in periodic table.

(f) Ozone O3 :It is an allotropic form of oxygen, which is highly explosive with pungent odour.

1. Importance of ozone and formation in atmosphere. It is available at a hight of 20 km fromearth. Layer of Ozone is Ozone umbrella or ozonosphere.Ozonosphere protects earth from sun’s harmful ultraviolet radiations by filtering them.Step 1 : Formation of excited O*

(i)

h2 175 nm

O O + O*ground state excited stateLower mesosphere

υ⎯⎯⎯→

Step 2 : Deactivation of O*

(ii) 2 2O* + O O + O⎯⎯→

Step 3 : Formation of O3*

(iii) 2 3O + O O *⎯⎯→

Step 4 : Stabilisation of O3*

(iv) 3 2 2 3O * + M (N or O or inert element) O + M⎯⎯→

2. Laboratory preparation :

Pure, silent electric dischargedry oxygen ozonsied oxygen⎯⎯⎯⎯⎯⎯⎯⎯→ (10% Ozone)

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2 33O 2O⎯⎯→ ; ΔH = 142 kJ mol–1.

It is endothermic reaction, use of silent electric discharge is to prevent decomposition of O3.3. Physical properties of ozone :

(a) In gaseous state it is blue, liquid state-blue black, solid state-violet black.(b) ‘Ozein’ = Greek meaning odour. It has pungent odour.(c) It causes nausea and headache in larger concentrations.(d) It is unstable thermodynamically.

decomposition2Ozone O Heat⎯⎯⎯⎯⎯⎯→

Δ is liberated and entropy increases.

It causes large negative Gibbs’s energy change. i.e (–ΔG)4. Chemical properties of ozone :

(a) It is powerful oxidising agent.

(s) 3(g) 4(s) 2 (g)PbS + 4O PbSO + 4O⎯⎯→

(aq) 2 3(g) (aq) 2 (s) 2(g)2KI + H O + O 2KOH + I + O⎯⎯→

(b) It acts as reducing agent with peroxides.

2 3 2BaO + O BaO + 2O⎯⎯→

2 2 3 2 2H O + O H O + 2O⎯⎯→

(c) It forms ozonides with alkenes/alkynes.

(d) Estimation of ozone.Borate buffer

2pH = 9.2Ozone + excess KI I⎯⎯⎯⎯⎯→ .

Iodine is titrated against sodium thiosulphate to estimate ozone.(e) Depletion of ozone layer by exhaust gases (NO) from jet planes.

3 2 2NO + O NO + O⎯⎯→

(f) Freons in refrigerators also causes depletion of ozone due to free chlorine radicals.5. Structure of Ozone :

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Structure of ozone is resonance hybrid of I and II. O–O–O bond angle is 117º.Bond length is 128 pm.

6. Uses of Ozone :(a) It protects earth from harmful radiations of sun by absorbing them.(b) It sterilises air and water.(c) It is used for bleaching starch, oils, waxes.(d) Used as disinfectant.(e) Used for manufacutre of silk and camphor.

(g) Sulphur :1. Occurrence : It occurs in free as well as combined state.

Sulphides Sulphates

Copper pyrite Cu2S Gypsum CaSO4.2H2O

FeS2 Iron pyrite Epsum MgSO4.7H2O

Zinc blende ZnS Baryte BaSO4

Cinnabar HgS Glauber salt Na2SO4.10H2O

Galena PbS

Volcanic gases H2S

2. Allotropic forms of sulphur :(a) Rhombic sulphur (α - sulphur or octahedral

sulphur) This form is stable at room temp.

It is pale yellow. It is insoluble in water but

soluble in CS2 its crystals are obtained by

evaporating solution of sulphur in CS2.

Molecule is formed by S8 units. It has a

puckered ring structure.

(b) Monoclinic sulphur (βββββ - sulphur or prismatic sulphur) : It is obtained by melting

and cooling sulphur. On removing crust, prismatic sulphur crystals are formed.

Molecule is formed by S8 units with puckered ring structure. It is stable above 369K

and gets converted to rhombic sulphur below 369K. At 369K both α and β sulphur

are stable. This temp. is the transition temp.

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Characteristic properties of common allotropes of Sulphur ααααα and βββββ

Common colour sp. gravity solubility stabilityName and mp K

1. Rhombic S α-sulphur pale yellow 2.06/385.8 soluble in CS2 very stable2. Monoclinic β-sulphur 1.98/393 soluble in CS2 stable above 369K

Below 369K gets

< 369Kβ sulphur α sulphur⎯⎯⎯⎯→

converted to α sulphur

(c) Cyclo - S6, ring has chair form and the moleculardimensions are shown in Fig. At higherTemperature (1000K), S2 is the dominant speciesand is paramagnetic like oxygen.

(d) In sulphur vapour S2 (violet) and S3 (cherry red)units exist. S2 molecules is paramagnetic withtwo unpaired electrons in antibonding π*

(e) Plastic S : γ-sulphur is prepared by pouring boiling sulphur in cold water. It is softrubber like and amorphous, specific gravity 1.95. It consists of open chain S8 molecule.

Plastic sulphur standing Rhombic sulphur. It is insoluble in water as well as CS2.

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(f) Milk of sulphur : It is amorphous and insoluble in CS2. It is used in medicine.Preparation :-

2 5 2 3 23Ca(OH) + 12S 2CaS + CaS O + 3H Omilk of lime Calcium Calcium

Pentasulphide thiosulphate

⎯⎯→Δ

5 2 3 2 22CaS + CaS O + 6HCl 3CaCl + 3H O + 12S (milk of sulphur)⎯⎯→

(g) Colloidal sulphur : It is obtained in different colours depending on size of sulphuratom.Preparation :

2 2 22H S + SO 2H O + 3Scolloidal sulphur

⎯⎯→

2 2 3 2 2Na S O + 2HCl 2NaCl + SO + H O + Scolloidal sulphur

⎯⎯→

2 3 2 2H S + 2HNO 2H O + 2NO + Scolloidal sulphur

⎯⎯→

(h) Compounds of sulphur :Sulphur dioxide

1. Preparation :

a) (s) 2(g) 2(g)S + O SO⎯⎯→ ↑

b) Lab method :

2 4 4 2 2Cu + 2H SO CuSO + SO - + 2H Oturning conc.

⎯⎯→

2 3(aq) (aq) (aq) 2 2Na SO + 2HCl 2NaCl + H O + SO

Sodium sulphite

⎯⎯→ ↑

c) Industrial method : By-product of roasting of pyrites;

2(s) 2(g) 2 3(s) 2(g)4FeS + 11O 2Fe O + 8SO -

iron pyrites

⎯⎯→

SO2 is dried and liquefied under pressure2. Properties :

(a) It is colourless, pungent gas with acidic nature.(b) It liquifies at room temperature under a pressure of two atmospheres. It boils at 263 K.(c) It dissolves in water.

2 2 2 3H O + SO H SO⎯⎯→

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(d) It is a strong reducing agent.

(i) 2 2 2 2 4I + SO + 2H O 2HI + H SO⎯⎯→

(ii) 2 2 7 2 2 4 2 4 2 4 3 2K Cr O + 3SO + H SO K SO + Cr (SO ) + H OOrange Green

⎯⎯→

(e) It reacts with NaOH to form sodium sulphite followed by sodium bisulphite.

2 2 3 22NaOH + SO Na SO + H O⎯⎯→

2 3 2 2 3Na SO + SO + H O 2NaHSO⎯⎯→

(f) It reacts with chlorine and oxygen

2 2 2 2SO (g) + Cl (g) SO Cl (g) Sulphuryl chloride⎯⎯→

V O2 52 2 3723K,2-3 atm pr

2SO (g) + O (g) 2SO (g)⎯⎯⎯⎯⎯⎯⎯→

(g) It acts as an oxidising agent

2 (g) 2(g) 2 ( )2H S + SO 2H O + 3Sl⎯⎯→ ↓

2 2 3 24FeCl + SO + 4HCl 4FeCl + 2H O + S⎯⎯→

(h) In most condition it acts as reducing agent.

3 2 2 2 2 42FeCl + SO + 2H O 2FeCl + H SO + 2HCl⎯⎯→

2 4 2 2 4 4 2 45SO + 2KMnO + 2H O K SO + 2MnSO + 2H SOviolet colourless

⎯⎯→

3. Uses :(a) Sulphur dioxide is used in the manufacture of H2SO4.

(b) It is also used in the refining of petroleum and in sugar industry.

(c) For fumigation, as a germicide and preserving fruits.

(d) As an anti-chlor, disinfectant and preservative.

(e) Industrial chemicals like Sulphuric acid, sodium hydrogen sulphite, calcium hydrogen

sulphite etc. are manufactured from sulphur dioxide.

(f) Liquid SO2 is used as a solvent to dissolve many inorganic and organic compounds.

(g) It is used as a bleaching agent in the presence of moisture. It bleaches due to reduction

and bleaching action is temporary. 2 2 2 4SO + 2H O H SO + 2(H)⎯⎯→

(h) This gas acts a Lewis base due to presence of lone pairs of electrons on sulphur

atom. It also acts as ligand and forms numerous coordination compound.

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4. Structure of SO2 molecule :

Sulphur atom is SP2 hybridised.But bond angle is reduced to 119º from 120º due to repulsion of lone pair of electrons.S and O are bonded by σ bond (sp2 – p overlaps) and π bond due to pπ – pπ and dπ- pπ overlap.But both π bonds are identical due to resonance.

(i) Sulphuric acid :1. Contact process of manufacture :

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Following are important steps :(a) Preparation of SO2.

2 2S(s) + O (g) SOexcess

⎯⎯→ or 2 2 2 3 24FeS + 11O (g) 2Fe O + 8SO⎯⎯→

(b) Oxidation of SO2 to SO3.

V O2 52 2 32SO (g) + O (g) 2SO (g)

; ΔH = –196kCal

(1) SO2 and O2 should be pure and dry to avoid poisoning of catalyst.(2) To shift equilibrium towards right reaction should be carried at high pressure and

low temperature.Optimum pressure / temperature are 2-3 atm / 723 KSO2 : O2 :: 2 : 3 by volume

(c) Dissolution of SO2 in H2SO4.

2 2 4 2 2 7SO + H SO H S O98% oleum

⎯⎯→

SO2 cannot be dissolved in water as it forms fog.(d) Dilution of oleum

2 2 7 2 2 4H S O + H O 2H SO (96 – 98% pure)⎯⎯→

2. Lead chamber process : Following reactions take place

heat2 2(g)S + O SO⎯⎯⎯→

2 22NO + O 2NOIntermediate

⎯⎯→

2 2 3NO + SO SO + NO⎯⎯→

3 2 2 4SO + H O H SO←⎯⎯

Net Reaction : NO2 2 2 2 42SO + O + 2H O 2H SO⎯⎯⎯→

3. Physical properties of sulphuric acidH2SO4 : (1) It is colourless, dense, oily liquid (specific gravity 1.84 at 298K).

(2) Freezing point of sulphuric acid is 283K, boiling point is 611 K.(3) It is highly viscous due to Hydrogen (x) bonding.(4) Reaction with water is highly exothermic with spurting.

So water is never added to acid for dilution, but acid is added to water.4. Chemical properties of sulphuric acid.

(a) It is dibasic acid : In aqueous solution.+ –

2 4 4 1H SO H + HSO Ka > 10

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– + 2– –24 4 2HSO H + SO Ka = 1.2 × 10

or 2 –– +4 2 3 4HSO + H O H O +SO⎯⎯→

Ka1 >>> Ka2.(b) It is strong dehydrating agent due to it’s great affinity for water.

H SO2 42HCOOH CO + H O

Formic acid⎯⎯⎯⎯→

H SO2 46 12 6 2C H O 6C + 6H O

Glucose⎯⎯⎯⎯→

H SO2 42 5 2 5 2 5 22C H OH C H OC H + H O

ether⎯⎯⎯⎯→

H SO2 42 2 2[COOH] CO + CO + H O⎯⎯⎯⎯→

H SO2 412 22 11 2C H O 12C + 11H O

cane sugar⎯⎯⎯⎯→

(c) Oxidising property

2 4 2 2H SO H O + SO + [O]conc.

⎯⎯→

Due to liberation of [O] it acts as a powerful oxidising agent.(1) Oxidation of non metals by conc. H2SO4.

2 4 2 2 2C + 2H SO CO + 2SO + 2H O⎯⎯→

2 4 2 2S + 2H SO 3SO + 2H O⎯⎯→

4 2 4 4 10 2 2P + 10H SO P O + 10SO + 10H O⎯⎯→

4 10 2 3 4P O + 6H O H PO⎯⎯→

(2) Oxidation of metals (by conc. H2SO4)

2 4 4 2 2Zn + 2H SO ZnSO + 2H O + SO⎯⎯→

2 4 4 2 2Cu + 2H SO CuSO + 2H O + SO⎯⎯→

(3) Oxidation of Halogen acids

2 4 2 2 22HBr + H SO Br + SO + 2H O⎯⎯→

2 4 2 2 22HI + H SO I + SO + 2H O⎯⎯→

(4) Oxidation of some salts

4 2 4 2 4 3 2 22FeSO + 2H SO Fe (SO ) + SO + H O⎯⎯→

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4 6 2 4 2 2 4 4 4 2 4K Fe(CN) + 6H SO + 6H O 2K SO + FeSO + 3(NH ) SO + 6CO⎯⎯→

3 2 4 4 4 2 23KClO + 3H SO KHSO + HClO + 2ClO + H O⎯⎯→

(5) Action as a sulphonating agent.

6 6 2 6 5 2 2C H + HOSO OH C H SO OH + H OBenzene Sulphonic acid

⎯⎯→

(6) Action of PCl5

2 5 2 3HOSO OH + PCl ClSO OH + POCl + HCl chlorosulphuric acid

⎯⎯→

2 5 2 3ClSO OH + PCl ClSO Cl + POCl + HClsulphuryl chloride

⎯⎯→

(7) Metals below hydrogen in electro chemical series displace hydrogen from diluteacid. e.g.

Mg, Zn, Fe e.g. 2 4 4 2(g)Mg + H SO MgSO + H⎯⎯→

Cu, Pb, Hg, Bi and noble metals do not react with dilute acid

(8) It reacts with CaF2.

2 2 4 4CaF + H SO CaSO + 2HF.⎯⎯→

5. Uses of Sulphuric acid :(a) Sulphuric acid is used for making dyes, detergents, explosives, fertilizers.(b) It is used in chemicals like HNO3, HCl, H3PO4.(c) It is used in lead storage batteries.(d) Sulphuric acid is used as dehydrating agent.(e) It is used in refining in petroleum industry.(f) Also used as a pickling agent in removing layers of metal oxides before electroplating,

galvanising and soldering.

(j) Oxoacids of sulphur :1. They are grouped in five series.

(a) H2SO2 Sulphoxylic acid series

(b) H2SO3, H2S2O2, H2S2O4, H2S2O5 Sulphurous acid series

(c) H2SO4, H2S2O3, H2S2O7 Sulphuric acid series

(d) H2SO5, H2S2O8 Peroxy sulphuric acid series

(e) H2S2O6, H2SnO6 Thionic acid series

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(oil of vitriol)

(oleum)

2. Structures of oxoacides.

Formula Name Structure Oxidation Character- No. of S istic Bond

(a) H2SO2 Sulphoxylic acid H–O–S–OH +2

(b) H2SO3 Sulphurous acid

O||

HO – S – OH. . +4

(c) H2S2O2 Thio sulphuous acid

..HO – S – OH

||S

–2, +4 –S = S– bond

(d) H2S2O4 Dithionous acid

HO – S – S – OH|| ||O O

+3 S – S– bond

(e) H2S2O5 Pyrosulphurous acidHO – S – O – S – OH

|| ||O O

+4

(f) H2SO4 Sulphuric acid

O||

HO – S – OH||O

+6

(g) H2S2O3 Thiosulphuric acid

S||

HO – S – OH||O

–2, +6 S=S bond

(h) H2S2O7 Pyrosulphuric acid

O O|| ||

HO – S – O – S– OH|| ||O O

+6 S–O–S bond

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(Marshall’s acid)

acid (caro’s acid)

Peroxygroup

Peroxygroup

S–S linkage

(i) H2SO5Peroxy monosulphuric

O||

HO – O – S – OH||O

+6

(j) H2S2O8

Peroxy disulphuric acid

O O|| ||

HO – S – O – O – S – OH|| ||O O

+6

(k) H2S2O6Dithionic acid

O O|| ||

HO – S – S – OH|| ||O O

+5


*1. What is the oxidation state of oxygen in compounds (a) O2F2 and (b) H2O2?Ans. (a) Oxidation state of O in O2F2 is +1.

(b) Oxidation state of O in H2O2 is –1.

*2. Name the aerosole which cause the depletion of ozone.Ans. (a) Freon in refrigerants causes depletion of ozone.

*3. Why is sulphur a solid and dioxygen a gas?Ans. (a) Oxygen is a diatomic gas. O=O is formed by pπ – pπ overlap. Molecules are held together

by weak van der Waals forces.(b) Sulphur and Selenium are solids formed as polyatomic complex structures.

*4. Classify the following oxides as acidic, basic and amphoteric :SO2, CaO, Al2O3, BaO, NO2.

Ans. acidic basic amphotericSO2 CaO Al2O3NO2 BaO

*5. In which form of sulphur does it possess an open chain structure?Ans. Plastic sulphur has open chain structure.

*6. Which is the most stable form of sulphur? Draw its structure?Ans. Rhombic sulphur is the most stable form of sulphur. 89 + 8 [Refer 7.2 (b) 1]

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*7. Why does sulphur in vapour state exhibit paramagnetic behaviour?Ans. In vapour state, sulphur partly exists as S2 molecule with two electrons in antibonding π* orbitals.

So it exhibits paramagnetism.

*8. Which hydrogen bond is stronger? S–H–O or O–H–SAns. O–H–S bond is stronger than S–H–O. It is due to larger electronegativity difference between

–O–H than –S–H.

*9. List the names of the allotropes of sulphur. OR What are the allotropic forms of sulphur?Ans. (a) Rhombic sulphur (or alpha (α) sulphur)

(b) Monoclinic sulphur (or beta) (β) sulphur)(c) Plastic sulphur (or gamma (γ) sulphur)(d) Milk of sulphur(e) Colloidal sulphur(f) Cyclo–S6 sulphur

*10. Explain, the nature of S–O bond in SO2. OR Explain, the structure of SO2 molecule.Ans. Refer 7.2 (h) 4

*11. Why is Ka2 smaller than Ka1 for H2SO4 in water?

Ans. + –2 4 4 1H SO H + HSO Ka > 10

– + –2 –24 4 2HSO H + SO Ka = 1.2 × 10

+ –24 2 3 4HSO + H O H O + SO–

⎯⎯→

Ka2 << Ka1 because (i) removal of H+ from HSO4–2 is not favourable ii) –

4HSO ions are

hydrated and stabilised.

*12. Mention the conditions to maximize the yield of H2SO4 by contact process.Ans. In contact process following reaction takes place.

2 2 32SO (g) + O (g) 2SO ΔH = –196.6 kJ

(a) As forward reaction is exothermic yield of SO3 is more at low temperature. Optimumtemperature is 723 K.

(b) As forward reaction proceeds with decrease into volume, yield of SO3 is increased bycarrying out reaction under high pressure of 2-3 atmospheres.

(c) SO2 : O2 :: 2 : 3 by volume.(d) Catalyst is V2O5.(e) Pure and dry SO2 and O2 are used.

*13. Why is H2O a liquid and H2S a gas?Ans. (a) Oxygen is more electronegative than sulphur. Hence, O–H bond is more polar.

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(b) Therefore, H2O molecule are associated through intermolecular hydrogen bonding.

Oδ–

H

H ......δ+

O

H

H ......δ+

hydrogen bonding

O

H

δ–Hδ+

*14. Explain the anomalous behaviour of oxygen.Ans. Refer 7.2 (c) 2 (h).

*15. Oxygen is only diatomic species in group 16 elements. Explain.Ans. Oxygen is a diatomic gas. O=O is formed by pπ – pπ overlap. Molecules are held together

by weak Van der Waals forces.

*16. H2S is less acidic that H2Te, why?Ans. Acidic Character : Hydrides of group 16 elements are weak diprotic acids.

+ –2 2 3H S + H O H O + HS

– + 2 –2 3HS + H O H O + S

acid strength increases 2 2 2 2H O H S H Se H Te< < <

The decrease is due to decrease in the bond dissociation energy of M-H bond.

*17. Why are the first ionization energies of the higher elements of group 16 lower than thoseof group 15 elements?

Ans. Refer 7.2 (b)6.*18. What are the oxidation states of the elements in group 16?

Ans. Configuration of group 16 elements is ns2np4. So possible oxidation states are –2, +2, +4, +6.Oxygen exhibits oxidation state – 2 in oxides and –1 in peroxides, O shows + oxidation statewhen it combines with F. From S to Po, tendency to show –2 oxidation state decreases; Poshows +2 oxidation state. Except Oxygen, excitation of electrons in the empty d orbitals maytake place. Hence, the show higher oxidation state +4 and +6. The stability of +6 oxidationstate decreases but stability of +4 oxidation state increases down the group due to inert paireffect.

*19. What is the action of following elements on sulphur?(a) NaOH (b) Fluorine (c) Carbon (d) Copper

Ans. (a) NaOH : alkali 2 3 2 23S + 6NaOH Na SO + 2Na S + 3H OSodium Sulphite Sodium Sulphide

⎯⎯→

(b) Fluorine : Nonmetals 2 2Se + 2F SeF⎯⎯→

(c) Carbon : Nonmetals 2 6S + 3Cl SCl⎯⎯→

(d) Copper : metals Cu + S CuS⎯⎯→

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*20. How is oxygen prepared by thermal decomposition of certain metallic oxides?Ans. Refer 7.2 (d) 1.

21. What are different types of oxides? Give example.Ans. Refer 7.2 (e) 1 to 5.

*22. Explain : Protective umbrella as ozone for UV from sun.Ans. Ozone is available at a height of 20 km from earth. Ozonosphere protects earth from sun’s

harmful ultraviolet radiations by filtering them.Step 1 : Formation of excited O*(i) h

2 175nmO O + O*

Lower mesosphere ground state excited State

υ⎯⎯⎯→

Step 2 : Deactivation of O*

(ii) 2 2O* + O O + O⎯⎯→

Step 3 : Formation of O3*

(iii) 2 3O + O O *⎯⎯→

Step 4 : Stabilisation of O3*

(iv) 3 2 2 3O * + M (N or O or inert element) O + M⎯⎯→

Thus, O3 is naturally maintained in atmosphere.

*23. Why water is liquid and hydrogen sulphide is a gas?Ans. Refer 7.2 (c) 2. (b)

*24. Name the different oxyacids of sulphur and draw their structures.Ans. Refer 7.2 (j) 2.

*25. Ozone acts as oxidizing and reducing agents. Explain with examples.Ans. (a) It is powerful oxidising agent.

(s) 3(g) 4(s) 2 (g)PbS + 4O PbSO + 4O⎯⎯→

(aq) 2 3(g) (aq) 2 (s) 2(g)2KI + H O + O 2KOH + I + O⎯⎯→

(b) It acts as reducing agent with peroxides.

2 3 2BaO + O BaO + 2O⎯⎯→

2 2 3 2 2H O + O H O + 2O⎯⎯→

*26. How is ozone estimated?Ans. Estimation of ozone.

Borate buffer2pH = 9.2

Ozone + excess KI I⎯⎯⎯⎯⎯→ .

Liberated Iodine is titrated against sodium thiosulphate to estimate ozone.

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*27. How does sulphur occur?Ans. Refer 7.2 (g) 1.

*28. How is sulphur dioxide manufactured?Ans. Refer 7.2 (h) 1 b and c.

*29. How is sulphuric acid manufactured by contact process?Ans. Refer 7.2 (i) 1

*30. Write the uses of sulphuric acid.Ans. Refer 7.2 (i) 5


*1. Oxygen molecule shows .......a) Ferri magnetism b) Ferro magnetism c) Dimagnetism d) Paramagnetism

*2. When SO2 is passed through acidifed K2Cr2O7 solution .......a) The solution turns green b) The solution is decolourizedc) Reduction of SO2 takes place d) Green ppt of Cr2(SO4)3 is formed

*3. FeSO4 forms brown ring with .......a) NO3 b) NO2 c) NO d) N2O3

*4. The most stable allotropic form of sulphur is .......a) Rhombic sulphur b) Flowers of sulphurc) Plastic sulphur d) Mono-clinic sulphur

*5. General electronic configuration of group 16 elements is .......a) ns2np3 b) ns2np4 c) ns2np2 d) ns2np5

*6. Ozone is .......a) A compound of oxygen b) An allotropic oxygenc) An isotope of oxygen d) An isobar of oxygen

*7. Maximum covalency of sulphur is .......a) Four b) Six c) Three d) Two

*8. Oxygen exhibit – 1 oxidation state in .......a) OF2 b) H2O2 c) HCIO d) H2O

9. Which of the following orders of melting points of hydrides of Group 16 elements is true?a) H2S > H2Se > H2Te b) H2S < H2Se < H2Tec) H2S > H2Se < H2Te c) H2S < H2Se > H2Te

10. Which of the allotropes of sulphur represents α–sulphur?a) Rhombic sulphur b) Monoclinic sulphurc) Plastic sulphur d) Colloidal sulphur

11. Which of the following hydrides of Group 16 has the lowest boiling point?a) H2O b) H2S c) H2Se d) H2Te

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12. Which of the following acids does not exist in the free form?a) H2SO2 b) H2S2O3 c) H2S2O7 d) H2SO4

13. The oxidation state of sulphur in peroxodisulphuric acid, H2S2O8 is .......a) +5 b) +6 c) +7 d) +8

14. Which of the following oxo-acids of sulphur has more than one oxidation state of oxygen?a) H2S2O3 b) H2S2O4 c) H2S2O6 d) H2S2O8

15. In the manufacture of sulphuric acid by the contact process SO3 is not added to water directlyby form H2SO4 because .......a) the reaction does not go to completion b) the reaction is exothermicc) the reaction is quite slow d) SO3 is insoluble in water

16. The hybridization of sulphur in SO2 is .......a) sp b) sp2 c) sp3 d) dsp2

17. Marshall’s acid is .......a) H2SO5 b) H2S2O5 c) H2S2O7 d) H2S2O8

18. Oxygen atoms in H2O2 has .......a) sp hybird orbitals b) sp2

hybrid orbitalsc) sp3 hybird orbitals d) pure p orbitals

19. Which of the oxidation state(s) is/are shown by the element sulphur?a) –2, +2, +4 and +6 b) +2 only c) –2, +2 and +4 d) –2 only

20. Which of the following statements regarding sulphur is not true?a) Rhombic sulphur is stable at room temperatureb) Monoclinic sulphur is stable at room temperaturec) Both rhombic and monoclinic sulphur are soluble in CS2d) Both rhombic and monoclinic sulphur have a puckered-ring structure of eight sulphur atoms

21. Which of the following equations represents the oxidising action of sulphur dioxide?

a) 3+ 2+ 2– +2 2 42Fe + SO + 2H O 2Fe + SO + 4H⎯⎯→

b) 23Fe + SO 2FeO + FeS⎯⎯→

c) – 2+ + 2–4 2 2 42MnO + 2H O + 5SO 2Mn + 4H + 5SO⎯⎯→

d) 2– + 3+ 2–2 7 2 4 2Cr O + 2H + 3SO 2Cr + 3SO + H O⎯⎯→

22. The most favourable conditions for the formation of SO3 via the equation.

2 2 31

SO + O SO2

ΔH = –95kJ mol–1 are

a) low temperature and low pressure b) low temperature and high pressurec) high temperature and low pressure d) high temperature and high pressure

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23. The elemental sulphur is written as .......a) S b) S2 c) S4 d) S8

24. Colloidal sulphur is obtained when .......a) sulphur is heated in absence of airb) sulphur is heated in presence of airc) sodium thiosulphate is treated with iodined) H2S is passed through water containing HNO3

25. Burning sulphur in air produces .......a) H2S b) H2S2 c) SO2 d) SO3

26. Which of the following is an anhydride of H2SO4?a) H2S b) H2S2 c) SO2 d) SO3

27. Which of the following is an anhydride of H2SO3?a) H2S b) H2S2 c) SO2 d) SO3

28. Which of the following statements regarding sulphuric acid is not true?a) Concentrated H2SO4 acts as a dehydrating agentb) Hot concentrated H2SO4 acts as a powerful oxidizing agentc) The reaction of PCl5 with sulphuric acid produces sulphurylchlorided) The structure of sulphuric acid in vapour phase is square planar

29. Which of the following statements regarding ozone is not true?a) Ozone is an allotrope of oxygenb) The ozone layer protects the earth’s surface from an excessive concentration of harmful

ultraviolet radiationc) The conversion of oxygen into ozone is an exothermic processd) Ozone is much more powerful oxidizing agent than molecular oxygen

7.3 GROUP 17 ELEMENTS (HALOGEN FAMILY) :

Concept Explanation :(a) General Introduction :1. Group 17 elements are known as halogens (halo = salt, gene = born) from Greek. They

are electronegative and show close resemblance in properties.Occurrence : Fluorine occurs as (i) fluorspar CaF2 (ii) Cryolite Na3AlF6 (iii) Fluoropatite3Ca3(PO4)2.CaF2.Chlorine as (i) NaCl in seawater (ii) Carnalite KCl.MgCl2.6H2O. Bromine as bromidesof alkali and alkaline earth metals. It is also oxidation of Br– ions from sea water. Iodineas iodide in (a) Brine (b) sodium iodate in chile salt petre (iii) thyroxin in thyroid glands.At is radio active.

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2. Electronic configuration : Valence shell configuration is ns2 np5. They require ONE electronto attain stable inert gas configuration.

(b) Trends in physical properties :1. Physical state : They exist as diatomic molecules. Fluorine and chlorine are gases at room

temperature. Bromine is liquid and iodine is solid. As the size of the atoms increase downthe group, intermolecular Vanderwaal’s forces increase; and molecules are closer.

2. Atomic and ionic radii : They have smallest atomic radii in the respective period. Atomicand ionic radii increase down the group as valence electron is added in the new shell.

3. Ionisation enthalpy : They show least tendency to lose electron due to small size, soionisation enthalpy is very high. It decreases down the group as atomic size increases.I forms I+ ion.

4. Electronegativity : Halogen posses highest electronegativity in the respective period dueto smallest size. Electronegativity decreases down the group with increase in atomic size.Fluorine is most electronegative.

5. Electron gain enthalpy : – –X + e X + E⎯⎯→ Δ . As halogens aquire stable inert gas

configuration on gaining electron, large amount of energy is evolved with very high electrongain enthalpy. Value decreases down the group. But fluorine shows less electron gain enthalpythan Chlorine, due to it’s small size. The incoming electron is repelled by 2p electronsin small fluorine atom.

6. Colour :

Halogen Fluorine Chlorine Bromine Iodine

Colour pale yellow greenish yellow red violet

Element

Fluorine

Chlorine

Bromine

Iodine

Astatine

Symbol

F

Cl

Br

I

At

Atomic No.

9

17

35

53

85

ElectronicConfiguration

1s2 2s2 2p5

1s2 2s2 2p6 3s2 3p5

1s2 2s2 2p6 3s2 3p6

3d10 4s2 4p5

1s2 2s2 2p6 3s2 3p6

3d10 4s2 4p6 4d10 5s2 5p5

1s2 2s2 2p6 3s2 3p6

3d10 4s2 4p6 4d10

4f14 5s2 5p6 5pd10 6s2 6p5

Brief representation ofelectronic configuration

[He]2s2 2p2

[Ne]3s2 3p5

[Ar]3d10 4s2 4p5

[Kr]4d10

5s2 5p5

[Xe]4f14

5d106s26p5

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Outer electron is excited due to absorption of light energy. With increase in atomic sizedown the group, excitation energy decreases. So fluorine absorbs violet light of higherenergy and transmits yellow light of lower energy. Iodine absorbs yellow light of lowerenergy and transmits violet light of higher energy.

7. Non-metallic character : Halogens are non metals due to high electronegativity. Non metalliccharacter decreases down the group with increase in atomic size, Iodine forms I+ andI+3 under certain conditions.

8. Density : Density increases down the group due to increase in the magnitude of Van derwaal’s intermolecular attractive forces.

9. Melting point and boiling point : Halogens are characterised by law values of mp and bp,which increase down the group.

10. Oxidation states : All halogens show – 1 oxidation state due to their configuration ns2 np5.Except F, other halogens show oxidation state of +1, +3, +5, +7 due to excitation of outerns or np electron into empty d orbitals.

Energy for excitation is compensated by energy released on bond formation. Cl and Bralso show +4 and +6 oxidation states in acids and oxyacids.

(c) Chemical reactivity :Halogens are highly reactive due to :(i) Very high electron gain enthalpy.(ii) Low bond dissociation energy of X–X bond.

1. Oxidising nature : – –2

1X + e X

2⎯⎯→

Standard reduction potentials for halogens :

Halogen F2 Cl2 Br2 I2

Eº volt 2.87 1.40 1.05 0.62

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Due to strong tendency to accept electrons and undergo reduction, they act as strong oxidisingagents. Oxidising power decreases from F to Cl due to increase in atomic size.A halogen can displace other halogen below it in the group.

– –2 2F + 2X 2F + X⎯⎯→ [X = Cl, Br, I]

– –2 2Br + 2I 2Br + I⎯⎯→

Fluorine is strongest oxidising agent, due to small size, low bond enthalpy of F–F bondand high energy of hydration of F– ion. Metal fluorides in the highest oxidation state ofmetal are stabilised.

2. All halogens react with metals and non metals to form halides. Reactivity decreases downthe group.(a) The ionic nature of M–X bond decreases from F to I.(b) Halides of metals with low ionisation enthalpy are ionic and possess low melting and

boiling point. Halides of non metals are covalent and possess high melting and boilingpoints.

(c) In metal halides with more than ONE oxidation state of metal, halide with higher oxidationstate is more covalent than halide with lower oxidation state.PbCl4 > PbCl2, SbCl5 > SbCl3 etc.

(d) Most metals show highest oxidation state in fluorides.(e) Strength of bond M – X, is MF > M – Cl > M – Br > M – I.

3. Reactivity towards hydrogen. 2 2H + X 2HX⎯⎯→

Reactivity varies as F2 > Cl2 > Br2 > I2.

Properties of hydrogen halidesProperty HF HCl HBr HI

Boiling Point (K) 293 189 206 238

Melting point (K) 190 159 185 222

Bond length (H–X) pm 91.7 127.4 141.4 160.9

Diss. ΔH kJmol–1 574 431.6 362.5 294.6

(Bond dissociation enthalpy)

Thermal stability of H – X decreases from HF to HI due to increase in atomic size downthe group. So,Reducing property increases from HF to HI. H – F doesn’t show reducing property dueto association of HF molecules by hydrogen bonding.

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+ – + – + –H – F ........ H – F ........ H – F

δ δ δ δ δ δ

It is due to hydrogen bonding that HF is liquid and other halogen acids are gases.4. Reactivity towards oxygen : Halogens form different oxides. Many are unstable.

Halogen Oxide Stability and use

F OF2 OF2 is stable and both are used as fluorinting agents.

O2F2 e.g. Oxidised6Pu PuF⎯⎯⎯⎯→ from used nuclear fuel.

Cl Cl2O, ClO2 Highly explosive, ClO2 is used as bleaching agent.Cl2O6, Cl2O7 In paper industry and paper purification.

Br Br2O, BrO2, BrO3 Unstable used as oxidising agents.

I I2O4, I2O5, I2O7 I2O5 used as oxidising agentto estimate amount of CO.

5. Reaction with metals : Halogens react with metals to form halides. Ionic character variesas MFn > MCln > MBrn > MIn.

6. Reaction with water : Order of reactivity decreases from F to I.

2 2 22F + 2H O 4HF(aq) + O (g)⎯⎯→

2 2X (g) + H O( ) HX(aq) + HOX(aq)l ⎯⎯→ when X = Cl, Br

Non spontaneous reaction – +2 2 24I (aq) + 4H (aq) + O 2I (s) + 2H O(l)⎯⎯→

7. Reaction with hydrocarbons : Hydrogen of hydrocarbons are substituted by halogens.Fluorine decomposes hydrocarbons. Reactivity decreases down the group.

4 2CH + 2F C + 4HF⎯⎯→

4 2 4CH + 4Cl CCl + 4HCl⎯⎯→

8. Reaction with alkali : Fluorine reacts vigorously.

2 2 24NaOH + 2F 4NaF + O + 2H O↑⎯⎯→ other halogen react with cold NaOH

2 22NaOH + Cl NaCl + NaOCl + H Ocold and dil. hypochlorite

⎯⎯→

2 3 26NaOH + 3Cl 5NaCl + NaClO + H Ohot and Conc. sodium chlorate

⎯⎯→

9. Bleaching action : Only chlorine is a good bleaching agent due to formation of nascent oxygen.

2 2Cl + H O HCl + HOClhypochlorous acid

⎯⎯→

Vegetable Colouring matter + [O] colourless matter⎯⎯→

HOCl HCl + [O]⎯⎯→

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10. Reaction of halogen with halogens.They form inter halogen compoundsXX′, XX′3, XX′5, XX′7X = larger halogen atom, less electronegativeX ′ = smaller halogen atom, more electronegative.

(d) Anomalous behaviour of Fluorine :Fluorine behaves differently than other halogens due to :

1. small atomic size2. highest electronegativity3. low F–F bond dissociation energy4. non availability of d orbitals in valence shell.

No Property Fluorine Other halogens Reason1. Reactivity Very high Moderate Low F–F bond

dissociation energy.

2. Hydrogen bonding Exhibits Not possible Small atomic sizeand highestelectronegativity.

3. Behaviour of HF

a) State HF is liquid OtherHX are Due togases hydrogen Bonding

b) Acidity F is weak acid other HXare strong

4. Oxidation state –1 –1 and +1, +3, Highest+5, +7 electronegativity

and nonavailability of dorbitals.

5. Electron gain enthalpy Smaller than Cl Repulsionbetweeenelectrons due tosmall atomic size.

6. Nature of compounds Ionic Covalent Highestelectronegativity

7. Poly halide ion –3X Not possible Formed Non availability

of d-orbitals.

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(e) Compounds of halogen — Chlorine1. Occurrence : Origin of name: [Chloras – Greek meaning greenish yellow]

Minerals containing chlorine.a) Rock salt NaClb) Carnalite KCl.MgCl2.6H2Oc) Sylvine KCld) Horn Silver AgCle) Sea water contains 2.8% chlorine

2. Preparation of chlorine :(a) By oxidation of HCl :

(i) From MnO2;

2 2 2 2MnO + 4HCl MnCl + Cl + 2H O⎯⎯→Δ

(ii) From Lead dioxide or red Lead;

2 2 2 2PbO + 4HCl PbCl + Cl + 2H O⎯⎯→

3 4 2 2 2Pb O + 8HCl 3PbCl + Cl + 4H O⎯⎯→

(iii) From KMnO4 - Potassium Permaganate;

4 2 2 22KMnO + 16HCl 2KCl + 2MnCl + 8H O + 5Cl⎯⎯→

(iv) From K2Cr2O7 - Potassium dichromate;

2 2 7 3 2 2K Cr O + 14HCl 2KCl + 2CrCl + 3Cl + 7H O⎯⎯→

(b) From bleaching powder and mineral acid :

2 2 2 2CaOCl + 2HCl CaCl + Cl + H O⎯⎯→

2 2 4 4 2 2CaOCl + H SO CaSO + Cl + H O⎯⎯→

(c) From NaCl + MnO2 and conc. H2SO4

2 2 4 4 4 2 22NaCl + MnO + 3H SO 2NaHSO + MnSO + Cl + 2H O⎯⎯→

3. Manufacture of chlorine :

(a) Deacon Process : CuCl2

2 2 2723K4HCl+O 2Cl + 2H O

(gas)

⎯⎯⎯→

(b) Electrolytic Process : electrolysis

(aq) 22NaCl 2Na + Cl

Brine solution cathode Anode

⎯⎯⎯⎯⎯→

Anode and cathode are separated so that chlorine at anode doesn’t come in contactwith NaOH or H2 at Cathode.

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4. Physical Properties of Chlorine :Chlorine is : (1) It is greenish yellow gas with Pungent smell.

(2) It can be liquified easily into greenish yellow liquid with b.p 239K.(3) It is heavier than air.(4) It is poisonous and fatal if inhaled in large quantities.(5) Chlorine soluble in water forming chlorine water.(6) On cooling the aqueous solution on cooling forms yellow crystals of

Cl2.8H2O.

5. Chemical properties of chlorine :(a) Combustibility : It is non combustible and does not support burning.(b) Metals burn in chlorine forming respective chlorides

22Na + Cl 2NaCl⎯⎯→

2 2Cu + Cl CuCl⎯⎯→

2 32Al + 3Cl 2AlCl⎯⎯→

(c) Non metals form respective chlorides

2 32As + 3Cl 2AsCl⎯⎯→

4 2 3P +6Cl 4PCl⎯⎯→

(d) Water reacts to form mixture hypochlorous acid and HCl.

2 2Cl + H O HCl + HOCl⎯⎯→

22HOCl 2HCl + Ounstable

⎯⎯→ ↑

(e) Hydrogen reacts with high affinity

2 2H Cl 2HCl+ ⎯⎯→

Reacts with hydrogen from hydrocarbons

10 16 2C H + 8Cl 16HCl + 10C⎯⎯→

Turpentine Oil.(f) Ammonia reacts vigorously.

3 2 4 28NH + 3Cl 6NH Cl + Nexcess

⎯⎯→

If Cl2 is in excess 2 3 33Cl + NH 3HCl + NCl⎯⎯→

(g) CS2

2 2 4 2 2CS + 3Cl CCl + S Cl⎯⎯→

2 2 2 42S Cl + CS CCl + 6S⎯⎯→

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(h) Bleaching action : It has a powerful bleaching action due to liberation of nascentoxygen.

[ ]2 2Cl + H O 2HCl + O⎯⎯→

Coloured vegetable matter + [O] ⎯⎯→ Colourless matter.(i) Disinfectant action : Due to liberation of nascent oxygen, it is used to kill

microorganisms in water.(j) Oxidising action :

(i) Fe2+ salts to Fe+3

4 2 4 2 2 4 32FeSO + H SO + Cl Fe (SO ) + 2HCl⎯⎯→

(ii) Sulphites to sulphates

2 2 3 2 2 4Cl + Na SO + H O Na SO + 2HCl⎯⎯→

(iii) H2S to S

2 2Cl + H S 2HCl + S⎯⎯→

(iv) SO2 to H2SO4

2 2 2 2 4Cl + SO + 2H O 2HCl + H SO⎯⎯→

(k) Alkali :(i) Dilute and cold alkali - it forms Hypoctitorite and Chloride

2 2Cl + 2NaOH NaCl + NaOCl + H O⎯⎯→

( ) ( )2 22 2 22Cl + 2Ca OH CaCl + Ca OCl + 2H OCalcium Hypo Chloride

⎯⎯→

(ii) Conc. and hot alkali forms chlorate and chloride.

2 3 23Cl + 6NaOH 5NaCl + NaClO H O⎯⎯→

(iii) Dry slaked lime forms bleaching powder.

( )22 2 2Cl + Ca OH CaOCl + H OCalcium oxychloride

⎯⎯→

(l) Bromine and Iodine are displaced from salt solution :

2 22KBr + Cl 2KCl + Br⎯⎯→

2 22KI + Cl 2KI + I⎯⎯→

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(m) Hydrocarbons :(i) Saturated hydrocarbons undergo substitution.

2 6 2 2 5C H + Cl C H Cl + HClethane ethylchloride

⎯⎯→

(ii) Unsaturated hydrocarbons form addition products.

2 2 2 2 2H C=CH + Cl H C – CH| |ethylene 1, 2-dichloroethaneCl Cl

⎯⎯→

(n) Formation of addition products in sunlight.

2 2 2 2SO + Cl SO Cl⎯⎯→ Sulphuryl chloride

2 2CO + Cl COCl⎯⎯→ Carbonyl chloride (phosgene)

6. Uses of chlorine :(a) As bleaching agent in paper and textile industry.(b) Purification of water by muncipality.(c) For extraction of gold and platinum(d) For manufacture of bleaching powder.(e) Chlorine is used in manufacture of dyestuffs and explosives.(f) For manufacture of Freon CCl2 F2 a refrigerant for manufacture of Cl – C2H4 –

S – C2H4Cl – mustard gas.(g) For manufacture of COCl2 Phosgene.(h) For manufacture of CCl3NO2 tear gas used in warfare.(i) For manufacture of HCl, CCl4, Perchlorates.(j) For manufacture of Synthetic plastic PVC.(k) For manufacture of insecticide DDT, BHC etc.

(f) Hydrogen chloride :1. Method of preparation :

420k2 4 4NaCl + H SO NaHSO + HCl

Conc.⎯⎯⎯→

4 2 4NaHSO + NaCl Na SO + HCl⎯⎯→

conc H SO2 4HCl ⎯⎯⎯⎯⎯→ dry gas collected by upward displacement of air.

2. Properties of HCl :(a) It is colourless gas with pungent odour

Liquification freezingHCl HCl HCl

(gas) (liquid) (Solid)b.p.189k m.p.159k

⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯→

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(b) It is highly water soluble forming strongly acidic solution

+ – 7(g) 2 (l) 3 (aq) aHCl + H O H O + Cl k =10⎯⎯→

Aqueous solution of HCl is Hydrochloric acid. conc HCl is 37% or 12 NAcid fumes in moist air forming fuming Hydrochloric acid.

(c) Sodium burns in HCl gas with yellow flame.

22Na + 2HCl 2NaCl + H⎯⎯→

(d) Ammonia reacts with HCl to form NH4Cl.

3 4NH + HCl NH Cl⎯⎯→

(e) Aquaregia is a mixture of 3 parts conc. HCl + 1 Part conc. HNO3. It can dissolvenoble metals.

. (f) Salts of weaker acids, sulphites, carbonates and hydrogen carbonates are decomposed.

3 2 2NaHSO + HCl NaCl + H O + SO⎯⎯→

3 2 2NaHCO + HCl NaCl + H O + CO⎯⎯→

2 3 2 2Na CO + 2HCl 2NaCl + H O + CO⎯⎯→

(g) Basic oxides and alkalies dissove in HCl.

2 2CaO + 2HCl CaCl + H O⎯⎯→

( ) 2 22Mg + 2HCl MgCl + 2H OOH ⎯⎯→

2NaOH + HCl NaCl + H O⎯⎯→

3. Uses of hydrochloric acid :(a) Manufacture of dye used as laboratory reagent.(b) Manufacture of chlorine and NH4Cl.(c) Manufacture of Glucose from Corn starch.(d) It is used for extraction of glue from bones, in medicines and in galvanising.(e) For preparation of H2 gas from metals.

2 2Zn + 2HCl ZnCl + H⎯⎯→

Powdered Iron :

2 2Fe + 2HCl FeCl + H⎯⎯→

H2 prevents oxidation of +2 +3e eF F⎯⎯→

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(g) Interhalogen compounds :1. Introduction : Interhalogen compounds are formed between different halogens.

Type Example

XX′ ClF, BrF, BrCl, ICl

XX′3 ClF3, BrF3, IF3, ICl3(unstable)

XX′5 IF5, BrF5, ClF5

XX′7 IF7

X is larger atom than X′.X is more electropositive than X′X is below X′ in the group.Number of X′ attached to X increase, when difference in atomic radii (X and X′) increases.eg. IF7

2. Preparation : By reaction between halogens under certain condition.(a) ClF [Chlorine monofluoride] is colourless gas.

437K2 2Cl + F 2ClF⎯⎯⎯→

(b) ClF3 [chlorine trifluoride] is colourless gas.373K

2 2 3Cl 3F 2ClF+ ⎯⎯⎯→

(c) ICl3 [Iodine trichloride] is yellow powder.

2 2 3I + 3Cl 2ICl⎯⎯→

(d) BrF3[Bromine trifluoride] is yellow green liquid.

2 2 3Br + 3F 2BrF⎯⎯→

(diluted with water)(e) ICl [Iodine monochloride] α − form is ruby red β − form is red brown.

2 2I + Cl 2ICl⎯⎯→

(f) BrF5 [Bromine pentafluoride] colourless liquid.

2 2 5Br + 5F 2BrFexcess

⎯⎯→

3. Characteristics of Interhalogen compounds :(a) Covalent due to small difference in electronegativities.(b) Volatile but not explosive.(c) Diamagnetic.(d) Due to weaker X–X′ bond than X–X or X′–X′, they are more reactive than halogen(e) Hydrolysed.

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+ –2 3ICl + 2H O H O + Cl + HOI⎯⎯→

3 2 32ICl + 3H O ICl + HIO + 5HCl⎯⎯→

5 2 2 52IF + 5H O 10HF + I O⎯⎯→

(f) Strong oxidising agentsSummary of structure of Interhalogens

Formula Hybridisation of X Geometry

X – X nilX – X′3 sp3d trigonal pyramidal or ‘T’ Shaped.X – X′5 sp3d2 square pyramidalIF7 sp3d3 pentagonal bipyramidal

4. Structure of Interhalogens :(a) XX′ [e.g. ClF] structure is simple, formed by covalent bond is p-p overlap

(b) XX′3 [e.g. ClF3]

(c) XX′5[e.g. ClF5]

Square pyramidal Structure of XX’5 (XF5)

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(d) XX′7[e.g. ClF7]

5. Uses of Interhalogens :(a) ICl is used for estimation of Iodine number of fats and Oils.(b) Preparation of polyhalides.(c) As non aqueous solvents.(d) ClF3, BrF3 used as fluorinating agents.(e) ClF3, BrF3 used as oxidisers in propellants.(f) ClF3, BrF3 used for production of UF6.

(h) Oxyacids and oxoacids of halogens :1. Introductiion

Oxyacids of Hypohalous acid Halous acid Halicacid Per halic acid halogens

Fluorine HOF — — —Hypofluorous acid

Chlorine HOCl HOClO HOClO2 HOClO3

Hypochlorus acid Chlorous acid Chloric acid Per chloric acid

Bromine HOBr — HOBrO2 HOBrO3

Hypobromous acid Bromic acid Per bromic acid

Iodine HOI — HOIO2 HOIO3

Hypoiodous acid Iodic acid Per iodic acid

Fluorine forms only HOF due to it’s small atomic size and high electronegativity.

495


H

O

ClHOCl

(Oxidation state +1)

H

O

Cl

O

HOClO(Oxidation state +3)

H

O

Cl

OO


2

H

O

Cl

OO O


3

2. Structures of oxyacids of chlorine :

Acidic strength

2 3HOCl < HOClO < HOClO < HOClO

As thermal stability increases with increase in oxidation state of halogen.

( ) ( ) ( ) ( )4 3 2HClO HClO HClO HClO

7 5 3 1> > >

+ + + +

Oxidising power decreases with increase in oxidation number.HClO4 < HClO3 < HClO2 < HClO


*1. Explain, fluorine reacts with sulphur to form SF6 whereas iodine does not react.Ans. (a) As F is more electronegative than I, F can oxidise S to +6 oxidation state and I cannot.

(b) Also as atomic size of F is less than that of I ; six Fluorine atoms can be accommodatedin the molecule SF6. With larger Iodine atoms, molecule SF6 is not stable.

*2. Why does fluorine shows anomalous behaviour?Ans. Refer 7.3 (d)

*3. Sea is the greatest source of halogen. Explain.Ans. Halide ions are part of most of the salts found in sea like Cl–, Br–, I– of Na, K, Ca and Mg.

*4. Fluorine exhibits only -1 oxidation state whereas other halogens exhibit +1,+3, +5 and +7 states.Explain.

Ans. Due to non-availability of d orbitals in valance shell for excitation of electrons and highestelectronegativity.

*5. Give the reason for bleaching action of chlorine and explain bleaching action of Cl2.Ans. It has a powerful bleaching action due to liberation of nascent oxygen.

[ ]2 2Cl + H O 2HCl + O⎯⎯→

Coloured Vegetable matter + [O] Colourless matter.

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*6. What happens when ;(a) Thin copper leaves are thrown in a jar of chlorine.

Ans. 2 2Cu + Cl CuCl⎯⎯→

(b) Chlorine is treated with excess ammonia.

Ans. 3 2 4 2 2 3 38NH + 3Cl 6NH Cl + N or 3Cl + NH 3HCl + NCl(excess) (excess)

⎯⎯→ ⎯⎯→

(c) Chlorine is passed over dry slaked lime.

Ans. 2 2 2 2Cl + Ca(OH) Ca(OCl) + H O⎯⎯→

*7. What is the action of Cl2On?(a) Cold and dil NaOH.

Ans. 2 2Cl + 2NaOH NaCl + NaOCl + H Ohypochlorite

⎯⎯→

(b) Hot and conc. NaOH.

Ans. 2 3 23Cl + 6NaOH 5NaCl + NaClO + 3H Osodium chlorate

⎯⎯→

(c) P4 molecule.

Ans. 4 2 3P + 6Cl 4PClphosphorustrichloride

⎯⎯→ and 4 2 5P + 5Cl 5PCl⎯⎯→

(d) Iron (II).

Ans. 4 2 4 2 2 4 32FeSO + H SO + Cl Fe (SO ) + 2HCl (Oxidation reaction)⎯⎯→

*8. Write the formulae of tear gas, phosgene and mustard gas.Ans. Tear gas (CCl3.NO2), Phosgene (COCl2) and mustard gas (Cl.C2H4–S–C2H4.Cl).

*9. Explain, ICl is more reactive than I2.Ans. I–Cl bond is weaker than I – I bond. Due to lower bond dissociation enthalpy. Hence,

I – Cl is more reactive.

*10. Bromine has highest negative electron gain enthalpy as compared to other elements of 4th

period. Explain.Ans. (i) Atomic size decreases from left to right across the period.

So Br has smallest atomic size in the period and highest electronegativity and maximumattraction for electron.

(ii)– –

2 5 2 6Br + e Br

ns, np ns, np

⎯⎯→ Δeg H = –324 kJ mol–1.

By gaining an electron Br acquires stable inert gas configuration.Hence, maximum amount of energy is released.

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*11. Write formulae and names of fluorine minerals.

Ans. Fluorine occurs as (a) fluorspar CaF2 (b) Cryolite Na3AlF6 (c) Fluoropatite 3Ca3(PO4)2.CaF2.

*12. What are the trends in ionization enthalpy and electronegativity of group 17 elements?

Ans. (a) Ionisation enthalpy : They show least tendency to lose electron due to small size. So ionisation

enthalpy is very high. It decreases down the group as atomic size increases. I forms I+ ion.

(b) Electronegativity : Halogen posses highest electronegativity in the respective period due

to smallest size. Electronegativity decreases down the group with increase in atomic size.

Fluorine is most electronegative.

*13. Explain reactivity of halogen’s towards hydrogen.

Ans. Refer 7.3 (c) 3

*14. Fluorine shows only -1 oxidation state whereas other halogens show along with -1 also +1,+3,+5

and +7. Explain.

Ans. Oxidation state : Fluorine exhibit only in –1 oxidation state in all its compounds. Other elements

of the group exhibit +1, +3, +5 and +7 oxidation states in addition to the –1 oxidation state.

The existence of only –1 oxidation state of fluorine is due to its highest electronegativity and

absence of the vacant d-orbitals in its valence shell.

*15. How is chlorine prepared?

Ans. Refer 7.3 (e) 2

*16. What are interhalogen compounds? Give examples.

Ans. Refer 7.3 (g)

*17. Draw the structures of ClF3 and IF7.

Ans. Refer 7.3 (g) 4 for structure of ClF3 and IF7

*18. Which are the different oxyacids of halogens?

Ans. Refer 7.3 (h) 1 table

*19. Write the uses of fluorine.

Ans. Fluorine is used in :

(a) Preparation of Freon CCl2F2

(b) Preparation of Teflon

(c) Preparation of SF6 used in vulcanisation of rubber.

(d) Preparation of 2356UF to separate uranium isotopes.

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• Theoretical MCQs :*1. The shape of IF7 molecule is .......

a) Tetrahedral b) Octahedralc) Trigonal bipyramidal d) Pentagonal bipyramidal

*2. The strongest oxidizing agent is .......a) HOCl b) HClO4 c) HClO3 d) HClO2

*3. Iodine can exist in the oxidation states .......a) +1, +3, +5 b) –1, +1, +3 c) +3, +5, +7 d) –1, +1, +3, +5, +7

*4. Fluorine does not show positive oxidation state due to absence of .......a) d-orbitals b) s-orbitals c) p-orbitals d) f-orbitals

*5. Which of the following halogen is radioactive in nature?a) Chlorine b) Bromine c) Iodine d) Astatine

*6. Cl2 is liberated from the cold hydrochloric acid by the action of .......a) KMnO4 b) K2MnO4 c) KMnO8 d) K2CrO4

*7. Most powerful oxidizing agent is .......a) Fluorine b) Chlorine c) Iodine d) Bromine

*8. Which one is the strongest reducing agent?a) HF b) HCl c) HBr d) HI

*9. Fluorine can exist in the oxidation state ........a) –1 only b) –1 and +1 c) –1, +1, +3 only d) –1. +1, +3, +7

10. Which of the following orders of electron affinities of halogens is correct?a) F < Cl < Br b) F > Cl > Br c) F > Cl < Br d) F < Cl > Br

11. Which of the following halogens does not exhibit positive oxidation state?a) F b) Cl c) Br d) I

12. Which of the following orders of melting point of hydrides of halogens is correct?a) HF > HCl > HBr b) HF < HCl < HBrc) HF > HCl < HBr d) HF < HCl > HBr

13. Which of the following does not contain fluroine?a) Chile salt petre b) Cryolite c) Fluoroapatite d) Fluorspar

14. Which of the following reactions is quite violent in nature?a) Between hydrogen and fluorine b) Between hydrogen and chlorinec) Between hydrogen and bromine d) Between hydrogen and iodine

15. The strength of halogen acids in water follows the order .........a) HF > HCl > HBr b) HF < HCl < HBrc) HF > HCl < HBr d) HF < HCl > HBr

16. Which of the following statements is not correct?a) Halides formed with nonmetals are covalent in nature.b) The reactivity of halogens decreases with increasing atomic number

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c) Hydrogen halides are ionic molecules in the gaseous phase.d) Metals with low ionization energies form ionic halides while those of high ionization energies

form covalent molecules.17. Which of the following statements is not correct?

a) Binary compounds of fluorine and oxygen are known as oxygen fluorides and notfluorine oxides.

b) For a metal exhibiting more than one exidation state, the halides in the loweroxidation state is more covalent than the one in the higher state.

c) Halogens do not occur freely in nature.d) Most of binary compounds between oxygen and halogens are unstable.

18. Which of the following halogen oxides are ionic in nature?a) Cl2O b) Br2O c) BrO2 d) I2O4

19. Which of the following interhalogens does not exist?a) ClF b) ClF2 c) ClF3 d) ClF5

20. Which of the following represents hypohalous acid?a) HOX b) HXO2 c) HXO3 d) HXO4

21. Which of the following represents halous acid?a) HOX b) HXO2 c) HXO3 d) HXO4

22. Which of the following represents halic acid?a) HOX b) HXO2 c) HXO3 d) HXO4

23. Which of the following represents perhalic acid?a) HOX b) HXO2 c) HXO3 d) HXO4

24. Which of the following exhibits +7 oxidation state in interhalogen compounds?a) F b) Cl c) Br d) I

25. Iodine in IF7 ivolves .......a) sp3d3 hybridization b) s2p3d2 hybridizationc) p3d4 hybridization d) p2d5 hybridization

26. The anhydride of HClO4 is .......a) ClO4

– b) Cl2O7 c) ClO2 d) ClO3

27. Hypobromous acid is .......a) HOBr b) HOBrO c) HOBrO2 d) HOBrO3

28. Hydrofluoric acid .......a) is a weak acid b) is a strong acidc) does not exist d) is a good oxidizing agent

29. Fluorine is highly reactive because .......a) F–F bond energy is high b) F–F bond energy is lowc) It is gaseous at room temperature d) F has smaller size

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30. Which of the following molecules has minimum dipole moment?a) HF b) HCl c) HBr d) HI

31. The order of strength of oxyacids of chlorine is .......a) HClO < HClO2 < HClO3 < HClO4 b) HClO > HClO2 > HClO3 > HClO4

c) HClO > HClO2 > HClO3 < HClO4 d) HClO < HClO2 < HClO3 < HClO4

32. The gaseous fluorine is pale yellow in colour. The light absorbed by fluorine in the visible regionwould be .......a) violet light b) red light c) green light d) orange light

33. The gaseous iodine molecules absorb yellow light from the visible region. Its colour would be .......a) violet b) red c) green d) yellow

34. Which of the following statements regarding halogens is not true?a) Ionization energy decreases with increases in atomic numberb) Electronegativity decreases with increase in atomic numberc) Electron affinity decreases with increase in atomic numberd) Enthalpy of fusion increases with increase in atomic number

35. Which of the following statements regarding halogens is not true?a) Halogens act as strong oxidizing agentsb) Fluorine oxidizes water to oxygen and ozonec) Chlorine oxidizes water to oxygend) Bromine disproportionates in water producing the reaction

2 2Br + H O HBr + HOBr⎯⎯→

36. Which of the following reactions would not proceed to right hand side?

a) –2F + Cl ⎯⎯→ b) –

2Cl + Br ⎯⎯→ c) –2Br + I ⎯⎯→ d) –

2I + Cl ⎯⎯→

37. Which of the following orders regarding the acid character of hydrogen halides in the aqueousmedium is correct?a) HCl < HBr < HI b) HCl < HBr < HI c) HCl > HBr > HI d) HCl > HBr < HI

38. Which of the halogens is found in Teflon?a) Fluorine b) Chlorine c) Bromine d) Iodine

39. Which of the halogens is found in DDT?a) Fluorine b) Chlorine c) Bromine d) Iodine

40. When chlorine reacts with cold and dilute solution of sodium hydroxide the products obtainedare .......a) Cl– + ClO– b) Cl– + ClO2

– c) Cl– + ClO3– d) Cl– + ClO4

–

41. The structure of interhalogen AX3 is .......a) triangular planar b) pyramidalc) T-shaped with two lone pairs of electrons at the equilateral positionsd) tetrahedral with a single electron

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42. Which of the following is anhydride of hypochlorous acid?a) Cl2O b) ClO2 c) Cl2O6 d) Cl2O7

43. Which of the following is anhydride of perchloric acid?a) Cl2O b) ClO2 c) Cl2O6 d) Cl2O7

7.4 GROUP 18 ELEMENTS (NOBLE GASES)


(a) Introduction and Electronic configuration :

Electronic configuration of noble gasesSince noble gases have a very stable configuration of completely filled orbitals i.e. 1s2

or ns2np6, they do not show any tendency to take part in any chemical reaction.So they occur in free state in atmosphere to 1% of volume of air.Ar is formed by electron capture of K (potassium).Kr, Xe are formed by fractional distillation of liquid air.Certain spring water contain He, Ar, Xe.

22686 Rn is obtained by radioactive decay of Ra.

226 222 488 86 2Ra Rn + He⎯⎯→

22286 Rn is longest living isotope of radon with life 3.8 days.

Element

Helium

Neon

Argon

Krypton

Xenon

Radon

Symbol

He

Ne

Ar

Kr

Xe

Rn

Atomic No.

2

10

18

36

54

86


1s2

1s2 2s2 2p6

Or [He] 2s2 2p6

1s2 2s2 2p6 3s2 3p6

Or [Ne] 3s2 3p6

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6

Or [Ar] 3d10 4s2 4p6

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6

Or [Kr] 4d10 5s2 5p6

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6

5d10 6s2 6p6

Or [Xe] 4f14 5d10 6s2 6p6

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(b) Trends in physical properties :

1. Physical state : All group 17 elements are colourless, tasteless, odourless, monoatomic

gases. (Cp/Cv ≅ 1.66 as for monoatomic gas)

2. Atomic radius : It increases down the group due to introduction of new shells.

Although atomic radius decreases across the period, atomic radii of inert gas elements

is larger than the preceding halogen. This is due to repulsion between electrons in completely

filled valence shell.

3. Ionisation enthalpy : They possess highest ionisation enthalpy due to stable electronic

configuration. However, it’s value dereases down the group due to increase in atomic size.

4. Electron gain enthalpy : Again due to very stable configuration they do not show any

tendency to accept electrons. So they have (+) value of electron gain enthalpy.

5. Melting and Boiling point : Values of m.p and b.p are small as compared to substances

with comparable molecular masses. This is due to weak interatomic forces. Atomic size

increases down the group, causing polarisation of atom. This increases van der Waal’s

attractive forces. So m.p, b.p increases down the group.

6. Liquefaction tendency : Inert gases have little tendency for liquefaction. However, this

tendency increases down the group with increase in the van der waal’s forces with increase

in atomic size.

All inert gases except Rn solidify at 280 K.

He solidifies at 1K under 25 atm pressure. So it is used as cooling agent.

Property

Atomic number

Atomic radii (pm)

Ionisation enthalpykJmol–1

Electron gainenthalpy kJmol–1

Density at STP(gm/cm–3)

Melting point (K)

Boiling point (K)

Ne

10

160

2080

116

9.0 × 10–4

24.0

27.0

Ar

18

192

1520

96

1.8 × 10–3

84

87.0

Kr

36

197

1350

96

3.7 × 10–3

116.0

121

Xe

54

217

1170

77

5.9 × 10–3

161.0

165.0

Rn

86

–

1037

68

9.7 × 10–3

202

211

He

2

99

2372

48

1.8 × 10–4

3

4.0

Elements

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7. Diffuse : All noble gases diffuse easily through glass, rubber or plastic.

8. Solubility in water : Solubility is low. But it increases down the group with increase in

atomic size. As atomic size increases polarisation of inert gas molecules by water increases

and solubility increases.

(b) Trends in chemical properties of inert gases. Due to very stable 1s2 or ns2, np6

configuration.

(i) Positive electron gain enthalpy is very large.

(ii) Ionisation enthalpy is very large.

So they do not show any tendency for chemical reaction.

Bartlett reaction was carried out in 1962.+ –

6 6Xe + PtF Xe Pt F⎯⎯→

[Xenon hexafluo platinate (v) is orange yellow crystalline solid].

Later some compounds of rare gases were prepared. [KrF2] [RnF2]

(c) Uses of noble gases :1. Helium gas :

(a) For filling balloons and airships being lighter and non combustible.

(b) For producing inert atmosphere during welding.

(c) For producing low temperature.

(d) As mixture of He + O2 is used by deep sea divers, because He is less soluble in

blood under pressure than N2. Used for treating Asthama.

(e) In NMR spectrometers and MRI – magnetic resonance imaging diagnostic technique.

(f) Nuclear reactors

(g) Cooling agent

2. Neon gas :

(a) In Neon lights, due to the glow produced on passing electric discharge.

(b) Neon light has good penetrating power in mist and fog.

(c) Television sets.

(d) Voltage stabilisers.

(e) Neon bulbs are used for botanical garden.

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3. Argon gas :

(a) For filling discharge lamps.

(b) For providing inert atmosphere for welding.

(c) For glass chromatography.

(d) Mixed with Neon, in Neon lights.

4. Krypton gas :

(a) For filling discharge tubes.

(b) For miner cap lamps.

(c) For measuring thickness of papers.

(d) For signal lights for airport runway.

(e) Krypton lamps are used in cosmic ray instrument.

5. Xenon gas :

(a) Xe/Kr mixture is used as flash bulbs in high speed photography.

(b) For detection of mesons and gamma photons.

6. Radon :

(a) In radiotherapy for cancer treatment.

(b) Photography of interior of opaque materials like steel castings.

(c) Radioactivity research.


• Answer in short :

*1. Name the elements of group 18.

Ans. Group 18 elements are noble gases and they are helium, neon, argon, krypton, xenon and radon.

*2. Write the uses of inert gases.

Ans. Refer 7.4 (c)



*1. Name the noble gas not present in the air .......

a) Radon b) Argon c) Krypton d) Helium

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*2. Which is the most abundant noble gas?

a) Argon b) Helium c) Neon d) Krypton

*3. Noble gas used in the miner’s cap lamp is .......

a) krypton b) argon c) helium d) Radon

*4. A mixture used for respiration by the seadivers is .......

a) He + O2 b) Ne + O2 c) Ar + O2 d) Kr + O2

*5. Which one of the following does not exist?

a) HeF4 b) XeF4 c) CF4 d) SF6

6. The number of elements in Group 18 is .......

a) 4 b) 5 c) 6 d) 7

7. Which of the following inert gases has the largest abundance (by volume) in air?

a) Helium b) Neon c) Argon d) Krypton

8. The formula of freon – 12 is .......

a) CFCl3 b) CF2Cl2 c) CF3Cl d) CF4

9. The first noble gas compound ever synthesized is .......

a) XePtF6 b) Xe(RuF6)2 c) XeOF4 d) XeOF2

HOURS BEFORE EXAMFirst elements from group 15, 16, 17 show unusual behaviour due to

(i) Small size

(ii) Non-availability of d-orbitals in valence shell and formation of pπ – pπ multiple bonds.

They exist in gaseous state

Their hydrides NH3, H2O, HF form hydrogen bonds and show anamolous properties.

Group 15 [Nitrogen Family] :Members : N, P, As, Sb, Bi electronic configuration ns2 np3

1. Transition is from non-metallic N to metallic Bi.

2. Except Bi, all show allotropy.

3. Oxidation states +3, +5.

Stability of +3 state down the group is due to ‘Inert pair effect’.

4. HNO3, NH3 are important compounds of nitrogen.

5. N forms many oxides from +1 to +5 oxidation state.

6. P forms different oxyacids in different oxidation states.

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Group 16 [Chalcogen] :

Members : O, S, Se, Te, Po electronic configuration ns2 np4.

1. Metallic trend increases down the group. Po is metal.

2. Oxygen shows oxidation state – 2.

exception +1 in peroxides + 2 in OF2.

other elements show oxidation states –2, +2, +4, +6.

3. Except H2O, all hydrides are strong reducing agents.

4. Dioxygen O2, Ozone O3 are allotropes of oxygen.

5. Oxides can be classified as acidic, basic, amphoteric or neutral.

6. Rhombic, monoclinic sulphur are important allotropes.

7. SO2, H2SO4 are important compounds of S.

H2SO4 is prepared in contact process by oxidation of S.

H O7 22 3 2 2 2 4S SO SO H S O H SO–

⎯⎯→ ⎯⎯→ ⎯⎯→ ⎯⎯⎯→

In Lead chamber process,

HNOO H O32 22 3 2 4S SO SO H SO⎯⎯⎯→ ⎯⎯⎯⎯→ ⎯⎯⎯→

8. Sulphur forms series of oxoacids with oxidation states ranging from –2, +2, +3, +4, +5,+6.

Group 17 [Halogens] :

Members : F, Cl, Br, I, At. electron configuration ns2 np5.

1. Highly electronegative and large negative value of electron gain enthalpy.

2. Fluorine has highest electronegativity and Cl with highest negative electron gain enthalpy.

3. F shows oxidation state –1 and other elements show –1, +1, +3, +5, +7.

4. They form Interhalogen compounds XX′, XX′3, XX′5, XX′7.

X is larger and more electropositive than X′.

5. They are very strong reducing agents.

6. They form oxyacids of four types.

HOX HOXO HOXO2 HOXO3

Hypohalous Halous Halic per Halic

oxidation state +1 +3 +5 +7

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Group 18 [Inert gases] :

Members : He, Ne, As, Kr, Xe, Rn electron configuration ns2 np6.

1. Very stable configuration, No tendency to lose or gain electrons. Large positive values

of Electron gain enthalpy and high Ionisation energy.

HIGHER ORDER THINKING SKILLS (HOTS)1. What is (i) Aqua fortis (ii) Aqua regia?

Ans. (i) 98% conc. HNO3 is aqua fortis (strong water)(ii) Aqua regia is mixture of conc HNO3 and conc. HCl in the ratio 1 : 3.

2. NO [Nitric oxide] is paramagnetic in gaseous state and diamagnetic in liquid and solid stateExplain.

Ans. In gaseous state. NO has an odd electron structure.

: N = O : : N = O :Ionic Covalent

: . :.–

So it is paramagnetic.In liquid and solid it dimerses. So it is diamagnetic.

N

N

O

O

3. What is ‘Chemiluminescence’ of phosphorous?Ans. Vapours of phosphorous undergo oxidation by atmospheric oxygen with faint, green glow.

It is called as ‘Chemiluminescence’

4 2 4 10P + 5O P O⎯⎯→ .

4. ‘Ionisation enthalpy increases across the period’. But ionisation enthalpy of group 16 elementsis lower than group 15 elements in the same period.

Ans. Configuration of group 15 elements is ns2 np3

Configuration of group 16 elements is ns2 np4

First Ionisation enthalpy for group 15 elements is very high because removal of an electrondisturbs stable half filled configuration np3.Group 16 elements attain stable ns2 np3 configuration by losing electron.So ionisation enthalpy is low.

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5. Give examples of compounds in which elements have following oxidation states.

O = –1, +2

N = –3, +4

S = +4, +6

Cl = +3, +5

Ans. Element Oxidation state Compound

O –1 BaO2

O +2 OF2

N –3 NH3

N +4 HNO3

S +4 SO2

S +6 SF6

Cl +3 ClF3

Cl +5 HClO3

6. Conc. H2SO4 is diluted by adding acid to water and not water to acid. Explain.

Ans. Conc. H2SO4 dissolves in water with great affinity and evolution of large amount of energy.

If water is added to acid it causes spurting due to steam formation.

So acid is added to water.

7. Colours of halogens vary from greenishyellow (chlorine) to violet (Iodine). Explain.

Ans. Electrons in the valence shell absorbs energy from visible light for excitation to higher energy

state.

Atomic size increases from F to I.

So in Iodine outer electrons are most loosely bound.

So Iodine molecules absorbs yellow light of lower energy and transmit violet light of higher

energy.

Fluorine absorbs violet light of higher energy and emits yellow light of lower energy.

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