THE ELECTROMAGNETIC SPECTRUM · Web viewthe same volume of sodium hydroxide will be required to...

Unit 1

THE ELECTROMAGNETIC SPECTRUM

The electromagnetic spectrum is the full range of electromagnetic radiation which travels

through space with a constant velocity of 3 x 108 ms-1. This radiation can be described in terms

of wave motion.

QUANTITY SYMBOL UNITS DESCRIPTION

Wavelength

Velocity

Frequency

Wavenumber

The wavenumber is a useful measure of frequency. This arises from the following relationship :

velocity = frequency x wavelength

c = f x

and soc

f =

Now try exercise 1.1

Unit 1

THE POSITION OF THE VISIBLE SPECTRUMThe full electromagnetic spectrum extends from radio and T.V. waves some of which are over

10 km (104 m) in length to gamma rays with a wavelength of a millionth of a millionth of a

centimetre (10-14 m).

The visible spectrum lies between 400 nm (400 x 10-9 m) which is violet and 700 nm (700 x 10-9

m) which is red. The red end of the visible is the long wavelength, low energy part of the visible

spectrum and the violet end of the visible spectrum the short wavelength, high frequency end of

the spectrum.

f/Hz

Unit 1

QUANTUM THEORY OF RADIATIONThe wave theory did not adequately explain all of the phenomena associated with

electromagnetic radiation and in 1905 Einstein proposed that electromagnetic radiation could in

some respects be regarded as small packets of energy (quanta) called photons, the energy of

these photons being proportional to frequency.

Low frequency = Low energy and High frequency = High energy.

The energy of any photon is given by the expression :

E = hf

where E = energy of a photon (or quantum) expressed in joules (J)

h = Planck’s constant, 6.63 x 10-34 Joule seconds (Js)

f = frequency of the radiation in Hertz (Hz)

Thus for 1 mole of photons

E = Lhf

where L = Avogadro’s constant, 6.02 x 1023 (mol-1).

but f = c

which gives E = Lhc for 1 mole of photons of wavelength

CALCULATING THE ENERGY ASSOCIATED WITH ONE MOLE OF PHOTONSFor example, calculate the energy associated with one mole of photons of wavenumber

2000 cm-1.

since = 1 wavenumber = 1 wavelength

= 1 = 5 x 10-4 cm or 5 x 10-6 m 2000

Using the relationship

E = Lhc = 6.02 X 1023 X 6.63 X 10-34 X 3 X 108 x 10-3 (conversion factor for kJ) 5 X 10-6

= 23.947 kJ.mol-1


Unit 1 ENERGY LEVELS IN ATOMS

Radiation can be absorbed or emitted by a species. When radiation is absorbed, the total

energy of the species is increased and when it is emitted there is a corresponding decrease in

the total energy of the species.

From Quantum theory of radiation, a species can only absorb or emit energy in certain fixed

amounts and so the internal energy of an atom must be restricted to certain values. These

energy values are called energy levels.

Unit 1

SPECTRAContinuous SpectrumWhen a beam of pure white light is passed through a prism a continuous spectrum is seen (all

the colours of the rainbow).

Emission SpectrumAtomic emission spectra are produced when atoms of elements, usually in their gaseous state,

are excited by heat or electrical discharge so that they emit radiation. The radiation emitted is

passed through a prism and the spectrum obtained is a series of sharp coloured lines on a

black background.

Absorption SpectrumIf a beam of pure white light is passed through a sample to be analysed e.g. a gaseous sample

of an element, the radiation that emerges has certain wavelengths (colours) missing. This

shows up as dark lines on a continuous spectrum. This provides a pattern that can be used in

identification of an unknown sample.

Both emission and absorption spectroscopy can be used to determine whether a certain

species is present in a sample and how much of it is present, since the intensity of transmitted

or absorbed radiation can be measured.

Unit 1

ATOMIC EMISSION SPECTRUM OF HYDROGEN

Examination of the atomic emission spectrum of hydrogen shows that this consists of a number

of lines (a line spectrum) of very precise frequency, corresponding to precise

amounts of energy.

Using the relationship

E = Lhc

Calculate the energy of the red line in the spectrum.

The lines in the spectrum are produced when electrons, which have been promoted to a higher

energy level either by heat or by electrical discharge, fall back down to lower energy levels. In

falling back down to lower energy levels from the excited state the atom emits radiation of a

specific wavelength or frequency from which the difference in energy between the levels can be

calculated.

The lines in the visible part of the spectrum (the Balmer series) are produced by electrons falling

back down to the second energy level from higher levels within the atom. The lines in the

ultraviolet part of the spectrum (the Lyman series) are produced by electrons falling back down

to the first energy level (ground state) and the lines in the infra red part of the spectrum (the

Paschen series) are produced by electrons falling back down to the third & fourth energy

levels.

e.g.

Unit 1 HYDROGEN SPECTRUMThe hydrogen spectrum shown below consists of the three series of lines, the Lyman series in

the Ultraviolet (higher energy) region of the spectrum, the Balmer series in the visible region

and the Paschen series in the Infra Red (lower energy) region.

POINTS TO NOTE1. The spectrum appears as lines at very precise frequencies which indicates that the energy

levels are themselves fixed – the electrons have fixed energies – the energy of the electrons

are said to be quantised.

2. The fact that the lines in all the series (UV, visible & IR) all converge towards the violet (high

energy) end of the spectrum shows that the energy levels come closer together until they

converge.

3. By determining the wavelength at which the lines converge in the ultraviolet series of lines in

a hydrogen spectrum we can determine the amount of energy given out when an electron

falls from the outermost energy level to the first energy level (ground state). This is,

therefore, also the amount of energy that has to be supplied to remove an electron from an

isolated, gaseous atom – the (1st) ionisation energy of hydrogen.

Calculate the ionisation energy for hydrogen if the Lyman series converges at 91.2 nm.

Unit 1

ELECTRONS and QUANTUM NUMBERSThe energy possessed by an electron in an atom can be defined in terms of 4 quantum

numbers.

The principle quantum number, symbol n, can be any positive integral number i.e.

1,2,3,4…..n. The principle quantum number tells us which shell or energy level the electron is

in.

Closer examination of emission spectra under high resolution shows that the lines are often not

single lines but doublets, triplets etc. This suggests that the electron shells are further sub-

divided into sub-shells. These are defined by a further quantum number, symbol l.The values of l are related to those of the principal quantum number n. For any given value of

n, l may take the value of 0, 1, 2…..(n-1). To avoid confusion the values of l corresponding to

0, 1, 2 and 3 are given the letters s, p, d and f.

The following picture of the electronic energy levels can be constructed.

Value of n Value of l Energy level12

3

4

This gives a picture of the atom in which electrons are in sub shells defined by the quantum

number l, within shells defined by the quantum number n.

Unit 1

ATOMIC ORBITALSHeisenberg’s uncertainty principle states that it is impossible to know both the momentum

and position of an electron precisely enough to draw a picture of the path of an electron in a

particular energy level. An experiment, no matter how well designed, to measure the location or

momentum of an electron will, by the measuring process, change either its momentum or its

location.

Although it is not possible to define a point in space where the electron is to be found, it is

possible to calculate the probability of finding an electron within a certain volume of space. The

regions where there is a high probability (usually greater than 90% chance) of finding an

electron are called atomic orbitals.

When we draw an orbital we are defining the volume of space where there is this high

probability of finding an electron. The overall size of the atomic orbital is governed by the

principal quantum number, n.

1s 2s 3s

The shape of the orbital is governed by the value of the 2nd quantum number, l. All s orbitals are

spherical in shape – the diameter increasing as the value of n increases. Outside the boundary

represented by the surface of the sphere, the probability of finding an electron is low (although it

can never be zero).

Unit 1

For p, d and f orbitals, it is necessary to define a third quantum number, m, called the magnetic

quantum number. For any given value of l, m can have any integral value between – l and + l. This quantum number indicates the orientation of the orbital in space. Where n=1 (and

therefore l=0) a value of m is not necessary as a sphere can have only one orientation in space.Complete the table.

Value of n Value of l l = 0……(n-1)

Value of m m = - l…0…+ l

1

2

3

4

P-ORBITALSEach of the p orbitals, unlike the s orbitals, are not spherical in shape but are dumb-bell shaped

and they lie along the x, y and z axes as shown below.

For the p orbitals (l = 1) there are three possible values for m, namely –1, 0, +1. This gives rise

to three p-orbitals which have equal energy in an isolated atom. These orbitals of equal energy

are said to be degenerate. The three p-orbitals are arranged in space along the three mutually

Unit 1 perpendicular axes x, y and z. The value of m governs the orientation in space, it has no effect

on the energy of the orbital.

D-ORBITALSWith the d-orbitals (l = 2) there are 5 possible values of m (-2, -1, 0, +1, +2) and so there are 5

d-orbitals. Like the p-orbitals, the d-orbitals are all of equal energy (they are degenerate) in an

isolated atom.

The three d orbitals dxy, dxz and dyz lie between the three principal axes and the two orbitals dz2

and dx2

-y2 lie along the axes.

ELECTRONIC CONFIGURATIONSUsing evidence from emission spectra, as well as information from mathematical models that

describe the energies and the likely locations of electrons in atoms, the number and nature of

the various electron energy levels has been established. However before doing this it is

necessary to introduce a fourth quantum number, s, which is called the spin quantum number.

The spin quantum number can have only 2 values, + ½ and – ½ .

Thus given values for the 4 quantum numbers, n, l, m and s we can define any single electron

in an atom in terms of its energy and likely location.

SUMMARY OF 4 QUANTUM NUMBERS

Quantum Number Name Symbol Meaning Value

1st Principal n Energy level 1, 2, 3…..

2nd Angular momentum l Shape (s, p, d, f) 0, 1, 2…..(n-1)

3rd Magnetic m Orientation in space - l…….0……+ l

Unit 1 4th Spin s Electron spin - ½ , + ½

PAULI EXCLUSION PRINCIPLEThe Pauli exclusion principle states that no two electrons in the same atom can have the

same set of all 4 quantum numbers. This is important in two ways:

1. The total number of electrons which may occupy any one orbital is two.

2. If there are two electrons in an orbital then the spins must be opposed.

SHOWING ELECTRON ARRANGEMENTSThere are a number of ways in which we can express the various orbitals in atoms. Using the

hydrogen atom we can see that the single electron will occupy the lowest energy level. This

can be represented as H = 1s1.

Another way is by using the notation where an orbital is represented by a box and each electron

by an arrow. We can represent the electron configuration of helium as He = 1s2 or as ;

He s = + ½ s = - ½

1s2

The significance of one arrow pointing up and the other down is to show the opposed spins of

the electrons.

THE AUFBAU PRINCIPLEBefore we can write the configuration of a multi-electron atom it is necessary to know the order

in which the various orbitals in an atom are filled. The Aufbau principle states that the lowest

energy levels are filled first and thus, provided the relative energies are known, the electron

configuration can be deduced.

Spectroscopic data gives the following arrangement of the energies of the orbitals.

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 7s2 5f14 6d10 7p6

The filling order can be remedied by using the diagram shown below.

Unit 1

HUND’S RULE OF MAXIMUM MULTIPLICITYWhen the situation is reached that there is more than one degenerate orbital available for the

electrons, it is necessary to apply Hund’s rule of maximum multiplicity which states that

“when two electrons occupy degenerate orbitals they do so in such a way as to maximise the

number of parallel spins” – i.e. electrons do not pair together until they have to.

Thus for a nitrogen atom containing a total of 7 electrons, the electron configuration can be

written in two ways;

(i) N = 1s2 2s2 2p3

or

(ii) N =

The example below is NOT correct

In the same two ways as above show the electron configurations of the following atoms.

1) Sodium

2) Calcium

Unit 1

It is often acceptable to write the electronic configurations of other species in a shortened

version, taking account of the electronic configuration of the preceding noble gas. e.g. the

electronic configuration for sodium can be written as [Ne] 3s1, where [Ne] represents 1s2 2s2

2p6, and that of calcium can be written as [Ar] 4s2 where [Ar] represents 1s2 2s2 2p6 3s2 3p6.


THE PERIODIC CLASSIFICATION OF THE ELEMENTSThe electron arrangements of the first 36 elements are shown below;

Element Electron arrangement Element Electron arrangementH 1s1 K 1s22s22p63s23p64s1

He 1s2 Ca 1s22s22p63s23p64s2

Li 1s22s1 Sc 1s22s22p63s23p64s23d1

Be 1s22s2 Ti 1s22s22p63s23p64s23d2

B 1s22s22p1 V 1s22s22p63s23p64s23d3

C 1s22s22p2 Cr 1s22s22p63s23p64s13d5

N 1s22s22p3 Mn 1s22s22p63s23p64s23d5

O 1s22s22p4 Fe 1s22s22p63s23p64s23d6

F 1s22s22p5 Co 1s22s22p63s23p64s23d7

Ne 1s22s22p6 Ni 1s22s22p63s23p64s23d8

Na 1s22s22p63s1 Cu 1s22s22p63s23p64s13d10

Mg 1s22s22p63s2 Zn 1s22s22p63s23p64s23d10

Al 1s22s22p63s23p1 Ga 1s22s22p63s23p64s23d104p1

Si 1s22s22p63s23p2 Ge 1s22s22p63s23p64s23d104p2

P 1s22s22p63s23p3 As 1s22s22p63s23p64s23d104p3

S 1s22s22p63s23p4 Se 1s22s22p63s23p64s23d104p4

Cl 1s22s22p63s23p5 Br 1s22s22p63s23p64s23d104p5

Ar 1s22s22p63s23p6 Kr 1s22s22p63s23p64s23d104p6

The present day Periodic Table is based on 8 vertical groups in which atoms have the same

number of electrons in the outer shell and on 7 horizontal periods. The start of a new row

(period) represents the start of filling a new energy level.

The present day Periodic Table has the form shown below;

Unit 1


TRENDS IN THE FIRST IONISATION ENERGY VALUESThe graph below shows the trend in the first ionisation energy values for the first 30 elements in

the Periodic Table.

TRENDS IN IONISATION ENERGY VALUES1. The noble gases have the _____________ first ionisation energy values as they have a

complete set of _______ orbitals in their outer energy level. Having a completely filled

set of _________ orbitals is a particularly stable electron arrangement.

Unit 1 2. There is a ______________ in the first ionisation energy from beryllium to boron.

Beryllium has a relatively high value as the outer electrons occupy a completely filled 2s

orbital – the completely filled orbital being a very stable electron arrangement. The

electron to be removed from the boron atom comes from the 2p orbital which contains

only one electron.

3. There is a ____________in first ionisation energy from nitrogen to oxygen. Nitrogen

has an electron arrangement of 1s2 2s2 2p3, which gives it a half-filled outer p orbital.

Like filled orbitals, half-filled orbitals are particularly stable arrangements.

4. The transition metals (scandium through to zinc) show fairly similar values for the first

ionisation energy. This is because it is not the 3d electron that is lost first but the 4s

electrons. The 3d electrons provide extra shielding effect and so partially cancel out the

extra attraction of the nucleus as we go across the period. Zinc has a higher ________

ionisation energy value due to its filled 3d and 4s orbitals.

Conclusion : the more stable the electronic configuration the higher the ionisation energy, and

vice versa.

SECOND AND SUBSEQUENT IONISATION ENERGIESFor sodium (1s2 2s2 2p6 3s1) 502 kJ of energy is required to remove the first electron from each

of one mole of sodium atoms in the gaseous state – this value is equal to the first ionisation

energy and is represented as;

Na(g) Na+(g) + e- H = (+) 502 kJ.mol-1

It is possible to remove successive electrons from sodium atoms, e.g. the second ionisation

energy is the enthalpy change associated with

Na+(g) Na2+

(g) + e- H = (+) 4560 kJ.mol-1

The second ionisation energy is much greater than the first because having lost the first

electron, each sodium atom has gained the stability of a filled outer orbital (1s2 2s2 2p6) like the

noble gases.

This is the same for other elements, e.g. magnesium (1s2 2s2 2p6 3s2)

Mg(g) Mg+(g) + e- H = (+) 744 kJ.mol-1

Mg+(g) Mg2+

(g) + e- H = (+) 1460 kJ.mol-1

Mg2+ (g) Mg3+(g) + e- H = (+) 7750 kJ.mol-1

Unit 1

The third equation represents the third ionisation energy of magnesium.

MORE ON ION FORMATIONThe overlapping of energy levels leads to the 4s orbital filling before the 3d (an electron in a 4s

orbital has a lower energy than an electron in a 3d orbital), when the atom ionises it is the 4s

electrons that are lost first. It appears that the presence of the 3d electrons increases the

energy of the 4s electrons so that they will be lost first on ionisation.

Use this information to write the electron arrangement of;

1. (i) A vanadium atom

(ii) A vanadium(II) ion

2. (i) An iron atom

(ii) An iron(III) ion

SPECTROSCOPYThe light emitted by stars contains all the frequencies between the ultraviolet and infra red parts

of the spectrum. Our own star, the Sun, is fairly cool with a surface temperature of about 6000K

and emits mainly visible light (400nm to 700nm). Some stars are cooler than ours and emit

mainly infra red radiation whereas some are very hot with surface temperatures in excess of

40,000K. Very hot stars emit mainly ultraviolet radiation (approximately 200nm to 400nm).

Any atoms in the outer regions of a star will absorb some of the light emitted from the surface of

the star and so the light which reaches us will have certain frequencies missing – frequencies

which have been ____________ by the atoms. Different atoms absorb ____________

frequencies of light.

When the light from the star is analysed in a spectrometer black absorption bands appear on a

bright background of light emitted from the star. The ____________ bands correspond to the

frequencies which are missing – the ones which are absorbed.

Unit 1

Unit 1

ABSORPTION SPECTRAShown below are the visible absorption spectra for two stars Centauri and our nearest star –

the Sun.

The dwarf star, Sirius, shows absorption bands for the elements hydrogen, helium, sodium,

magnesium and iron. Show the absorption bands for these elements on the diagram above.


COLOUR IN CHEMICALSMost of the colour we see around us is due to chemicals which absorb light from the visible

spectrum. If some of the electrons in a chemical absorb energy from the red/orange part of the

spectrum the chemical will appear to be blue/green in colour since this is the part of the

spectrum that is reflected.

Unit 1

EMISSION SPECTRUMWhen atoms, molecules and ions around stars absorb electromagnetic radiation, electrons in

the atoms are promoted from their usual energy level to higher energy levels called ‘excited

states’.

The electromagnetic radiation emitted as the electrons fall back to their original energy level

gives an emission spectrum.

Different elements produce different emission spectra – bright lines at certain frequencies on a

dark background.

Use page 15 of the data book to complete the table.

Element Wavelength / nm Colour of line

Hydrogen 656 red

486

Helium 706 red

Sodium

Unit 1

FLAME COLOURSAtoms of certain elements, when excited in a flame, emit light in the visible part of the

electromagnetic spectrum producing characteristic colours.

When a substance containing atoms of these metals is placed in a flame, energy is absorbed

from the flame in order to promote electrons in the atoms to higher energy levels. When the

electrons fall back down to their original energy level they emit electromagnetic radiation. In

some elements this emitted radiation is of the correct wavelength to be part of the visible

spectrum (400 – 700 nm) and so gives the flame a characteristic colour.

Flame TestsWhat to do :1. Clean the flame test wire by dipping it in a little

concentrated hydrochloric acid in a small beaker.

2. Dip the flame test wire into one of the substances

provided in the test tubes given.

3. Observe closely the colour of the flame produced

- the colour may only appear very briefly so watch

carefully.

4. Record your results in the table below.

Compound Metal present Flame colour Compound Metal present Flame colour

sodium chloride sodium bromide

lithium chloride lithium bromide

copper chloride copper sulphate

barium chloride barium sulphate

potassium chloride potassium iodide

Conclusion : It is the __________ atoms in the compound which are responsible for the

characteristic flame colour.The following metals produce the characteristic colours shown.

Metal Colour of flame Metal Colour of flame

Unit 1

COLORIMETERA colorimeter is an instrument that can be used to determine the concentration of a species by

determining the amount of light energy absorbed, or transmitted.

A schematic diagram of what is in a colorimeter is shown below.

Firstly we determine which frequency of light is absorbed most by the coloured sample, e.g.

potassium permanganate crystals are purple as they absorb green light and transmit the red

and blue light which together make the sample appear purple. In this case a green filter would

be used.

The higher the concentration of the coloured sample the lower the % of light which would be

transmitted through the sample. By determining the amount of light transmitted by solutions of

known concentration we can construct a calibration graph by plotting transmittance against

concentration. A sample of unknown concentration can then be inserted into the colorimeter

and from its % transmittance we can determine its concentration using the calibration graph.

Practical – Colorimetric Determination of Manganese in Steel.

Unit 1

WHAT IS A CHEMICAL BOND?One possible way of looking at a chemical bond is to consider a bond as being an electron (or

electrons) positioned between two (or more) nuclei so holding them together.

e- + nucleus+ +

e- e- shared electrons

In this model of a bond the positively charged nuclei are bonded together by the electrostatic

attraction of the nuclei to the negatively charged electron(s) between the two nuclei.

As the two hydrogen atoms approach one another to form a hydrogen molecule, the nucleus of

each atom attracts the other atom’s electron and a drop in potential energy occurs;

A atoms are infinitely apart

B to D atoms approach each other

B

C

D atoms are now a bond length apart (r)E atoms too close, nuclei repel and the potential energy increases rapidly

The energy required to break a mole of these bonds is the bond enthalpy and is shown as H

on the diagram.

Unit 1

BOND TYPESIn previous years we have established that there appear to be two kinds of bonding situation in

compounds. In the first type electrical conduction when molten or in solution is a characteristic

feature. The other type of compound shows little tendency to conduct (although some do

conduct when dissolved in water). The second group of compounds tend to have low melting

and boiling points. Until now it has been convenient to classify these compounds into two main

groups – ionic and covalent respectively.

However by looking at bonding as being electrons attracting nuclei to each other we can draw

together many aspects of bonding into a more unified picture. Using this model of bonding it is

possible to conceive of 4 situations;

1. If both nuclei have a strong, but equal attraction for electrons then the electrons will be

tightly held between the nuclei. The electrons will not spend their time nearer one nucleus

than the other. This gives equal sharing of electrons – a pure covalent bond. The pure

covalent bond only occurs when the atoms sharing the electrons have the same

electronegativity value – most often this is where the two atoms sharing electrons are

identical e.g. in diatomic elements like chlorine (Cl2).

2. If both nuclei have strong but unequal attraction for electrons then the electrons will spend

their time nearer one nucleus than the other. The electrons will be drawn towards the atom

with the higher electronegativity value. This unequal sharing of electrons produces a

molecule with a dipole (a separation of charge). This type of bond is called a polar covalent bond e.g. hydrogen chloride;

3. If one nucleus has a weak attraction for electrons and the other has a strong attraction for

electrons then the electrons may be totally transferred from one atom to the other. When this

occurs an ionic bond is formed. In ionic bonding a positive ion (a cation) and a negative ion

(an anion) are formed e.g. in sodium chloride, Na+ and Cl- ions are formed.

4. If both nuclei have a weak attraction for electrons the electrons are relatively free to move

around leaving the nuclei ‘floating’ in a ‘sea of electrons’ – the metallic bond is formed. The

mobile electrons bind the nuclei together.


Unit 1

RESONANCE STRUCTURESIt sometimes happens that no reasonable picture can be drawn for a molecule which

satisfactorily accounts for the observed properties.

For example, in the case of sulphur dioxide we know that the molecule is polar and therefore

non-symmetrical – a bent molecule. Sulphur has 6 electrons in its outer shell (valence

electrons) as does oxygen. Therefore in sulphur dioxide there are 18 valence electrons. To

agree with the octet rule these can be arranged in several ways;

Neither (1) or (2) agree with the fact that the molecule does not have one long sulphur to

oxygen bond and one short one – both are the same length.

(3) does not obey the octet rule and so is usually not considered likely.

(4) contains unpaired electrons which also makes it very unlikely. In such a situation, where

none of the proposed bonding models fits the observed facts, the actual electron distribution in

the molecule is said to be a resonance hybrid of the contributing formulae – in this case it is

thought that the electron structure resonates (moves between) structures (1) and (2).

Other substances where this occurs include ozone and the carbonate ion.

In ozone the resonance structures are;

In the carbonate ion the resonance structures are;

Unit 1

SHAPES OF MOLECULES AND POLYATOMIC IONS

Because we are dealing with individual molecules, the structure can be described in terms of

the shapes of the individual molecules. The shape of a molecule can be determined by

considering the number of electron pairs round the central atom in the bonding shell(s).

The structure of such compounds can be determined by taking all the electrons in the outer

shell of the central atom and adding to these 1 electron from each of the surrounding atoms.

From this the number of electron pairs can be determined. Knowing the number of electron

pairs it is an easy matter to determine the shape of the molecule if we assume that pairs of

electrons will repel each other and so try to take up positions as far away from each other as

possible.

Note : Electron pair repulsion decreases in the order;

non-bonding pair : non-bonding pair > non-bonding pair : bonding pair > bonding pair : bonding pair

This is the reason why the bond angles in ammonia and water are not the 109.5 we would

expect of 4 pairs of electrons.

For example, consider the molecule iodine pentafluoride (IF5).

The central iodine atom has 7 electrons in its bonding shell and each of the 5 fluorine atoms

contribute 1 electron. This gives a total of 12 electrons – 5 pairs of bonding electrons and one

pair of non-bonding electrons, 6 pairs in total. These 6 pairs arrange themselves in an

Unit 1

octahedral shape, the non-bonding pair is not included in the final description of the shape of

the molecule. So this gives a square pyramid structure.

EXAMPLES FOR PRACTICEIn each example below draw and name the structures formed.

(a) Phosphorus pentachloride (PCl5) (b) Iodine pentabromide (IBr5)

(c) The ion ICl4- (d) Sulphur hexafluoride (SF6)Note : the negative charge on the ion is counted as being on thecentral ion and so we take the number of electrons on the centraliodine as being 8

(e) BF3 (f) Ammonium ion (NH4+)

(g) BF4- ion (h) H3O+ ion


Unit 1

TRANSITION METAL CHEMISTRYThe group of metals that lie between group 2 and group 3 in the Periodic Table are referred to

as the transition metals. A transition metal can be defined as a metal with an incomplete d subshell in at least one of their ions. It is this feature which gives them their distinctive

properties. However, as scandium exists in all of its compounds as the 3+ ion which contains

no d-electrons, this element is sometimes excluded as a transition metal. In this unit we will

look at the broad characteristics associated with the elements from scandium to zinc.

Within this block of elements there are many metals which have familiar commercial

applications;

copper electrical wiring, pipes, coinage

silver, gold coinage, jewellery

iron cars, bridges, buildings

Many have important applications in catalysis, e.g. platinum, palladium, rhodium, iron, nickel.

Others have biological significance, the most familiar being iron in haemoglobin but cobalt,

copper and zinc are also important.

Unit 1

ELECTRONIC CONFIGURATIONAcross the first transition series (Sc to Zn) the 3d orbitals are being filled, the 4s orbital having

already been filled (according to the Aufbau Principle). The electronic configurations of the

elements of the first transition series are shown below;

Element Atomic Number Electron Arrangement

Scandium 21 1s22s22p63s23p64s23d1

Titanium 22 1s22s22p63s23p64s23d2

Vanadium 23 1s22s22p63s23p64s23d3

Chromium 24 1s22s22p63s23p64s13d5

Manganese 25 1s22s22p63s23p64s23d5

Iron 26 1s22s22p63s23p64s23d6

Cobalt 27 1s22s22p63s23p64s23d7

Nickel 28 1s22s22p63s23p64s23d8

Copper 29 1s22s22p63s23p64s13d10

Zinc 30 1s22s22p63s23p64s23d10

Note:1. When an atom forms an ion it is the 4s electrons that are lost before the 3d electrons.

2. For chromium and copper there is a slight difference from the arrangement of electrons

predicted from application of the rules given in the section on atomic structure. Such

deviations occur when the d-orbitals are either nearly half-filled or nearly filled.

Since a half-filled, or completely filled, d-orbital gives rise to increased stability then one

of the 4s electrons is promoted to produce a filled (or half-filled) orbital in these atoms.

This is better seen using the orbital box notation.

Unit 1

ELECTRONIC CONFIGURATIONS OF THE 1 ST SERIES TRANSITION METALS

Spectroscopic Orbital box notation

Element notation 3d 4s

Sc [Ar] 4s2 3d1

Ti [Ar] 4s2 3d2

V [Ar] 4s2 3d3

Cr [Ar] 4s1 3d5

Mn [Ar] 4s2 3d5

Fe [Ar] 4s2 3d6

Co [Ar] 4s2 3d7

Ni [Ar] 4s2 3d8

Cu [Ar] 4s1 3d10

Zn [Ar] 4s2 3d10

When transition metals form positive ions it is the 4s electrons which are lost first, rather than

the d electrons e.g. the electronic configuration for the nickel atom is [Ar] 4s2 3d8, and the

electronic configuration for the nickel (II) ion is [Ar] 3d8.

Variable valency is a familiar feature of transition metal chemistry. Variable valency is possible

because transition metals can lose the 4s electrons and some, or all, of the 3d electrons to form

positive ions. In forming positive ions, the electrons being lost come from the subshell with the

highest energy. N.B. the 4s electrons are lost before the 3d.

Unit 1

TYPICAL PROPERTIES OF TRANSITION METALSThe typical properties of transition metals are;

1. They are metallic so conduct both heat and electricity

2. They show marked catalytic ability both as the metal and in compounds

3. They exhibit variable valency (exception : zinc – always has a valency of 2)

4. They form many complex salts

5. They form coloured ions (exception – zinc)

VARIABLE OXIDATION STATEIn simple ionic compounds the oxidation state of an element is equal to the charge on the ion in

a compound. e.g. in iron(III) chloride the oxidation state of the iron(III) is +3 (note that the

charge on the ion is written 3+, whereas the oxidation number is written as +3) and the

oxidation state of the chloride is –1. The two terms oxidation state and oxidation number are

usually interchangeable and so an element is said to be in a particular oxidation state when it

has a specific oxidation number.

In less obvious cases, and particularly in polyatomic species, the oxidation state of an atom can

be determined by following a set of rules.

1. All free (uncombined) elements are given the oxidation number zero. e.g. metallic

magnesium has an oxidation number of zero, as does Cl in chlorine gas, Cl2.

2. For monatomic ions, the oxidation number is the same as the charge on the ion (see

examples above).

3. In most compounds the oxidation number for hydrogen is +1 (exception : metallic

hydrides where it is –1).

4. In most compounds the oxidation number for oxygen is –2 (exception : peroxides e.g.

hydrogen peroxide, where it is –1).

5. In all its compounds fluorine has an oxidation number of –1.

6. In polyatomic ions, the sum of all the oxidation numbers for all the atoms is equal to the

overall charge on the ion. For neutral compounds the sum is equal to zero.

Unit 1

e.g. to work out the oxidation number for sulphur in sulphuric acid;

In sulphuric acid the overall oxidation number is zero, the 4 oxygen atoms give a total oxidation

number of –8, the hydrogen atoms give a total oxidation number of +2, so the sulphur atom

must have an oxidation number of +6 to leave the overall charge of H2SO4 neutral.

Now try exercise 1.8.

Oxidation numbers can be used to determine whether a redox reaction has taken place. An

increase in oxidation number means that oxidation of the species has occurred, a decrease in

oxidation number means that reduction has occurred.

e.g.

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

The oxidation number of manganese in MnO4- is +7, and in Mn2+ it is +2, showing a decrease in

oxidation number and therefore reduction has occurred (the ion-electron equation shows a gain

of electrons).

It can be seen that manganese in oxidation state +7 has been reduced and so MnO4- is acting

as an oxidising agent when it reacts in this manner. It is generally true that compounds

containing metals in high oxidation states tend to be oxidising agents whereas compounds with

metals in low oxidation states are often reducing agents.


Unit 1

Transition metals exhibit variable oxidation states of differing stability. A common oxidation

state of most of these elements is +2 when the atom has lost its 4s electrons. However

because the 3d subshells have energy levels very close to that of the 4s subshell, it is fairly

easy for the 3d electrons to be also lost to form other oxidation states. The different ions in

different oxidation states have different stabilities, as in the case of Fe2+ and Fe3+, using orbital

box notation we can see which of the two ions is more stable;

Spectroscopic Orbital box notation

Element notation 3d 4s

Fe2+ [Ar] 3d6

Fe3+ [Ar] 3d5

Changing from one oxidation state to another is an important aspect of transition metal

chemistry, often characterised by a distinct colour change, as shown in the table below.

Ion Oxidation state of transition metal ColourVO3

- +5 Yellow

VO2+ +4 Blue

V3+ +3 Green

V2+ +2 Violet

The common oxidation states for the elements in the first transition series are shown below;

7+6+ 6+ 6+

5+ 5+ 5+ 5+

4+ 4+ 4+ 4+ 4+ 4+ 4+

3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+

2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+1+ 1+ 1+ 1+ 1+ 1+ 1+

Sc Ti V Cr Mn Fe Co Ni Cu Zn

The more common oxidation states are shown in bold.

Unit 1

Oxidation states greater than +4 are observed but never in simple ions. In high oxidation states,

the central ion (e.g. vanadium) is usually bonded to an electronegative element like oxygen.

Formula of ion Oxidation stateV2+ +2

V3+ +3

VO2+ +4

VO3- +5

TRANSITION METAL COMPLEXESAn important feature of the transition metals is their ability to form complex ions and molecules,

often called coordination compounds. A complex consists of a central metal ion surrounded by

ligands. A ligand is a molecule or negative ion with at least one lone (non-bonding) pair of

electrons available for bonding with the metal ion. The electron orbitals of the ligand have an

effect on the electron distribution in the central ion and this in turn results in significant effects in

physical properties of the complexes e.g. colour.

Typical ligands involved in the formation of complexes are negative ions such as;

CN-

F-, Cl-, Br-, I-

NO2-

OH-

and molecules such as;

H2O

NH3

In each case the ligand uses a pair of electrons to form a dative covalent bond (coordinate

bond) with the metal ion. Ligands such as CN-, H2O, NH3 and other molecules or negative ions

donating one electron pair to the metal ion are said to be monodentate (Latin : literally ‘one

toothed’). Those that donate more than one electron pair are said to be polydentate. Ligands

are also said to be chelating (Greek : literally ‘a claw’).

Unit 1 Common bidentate ligands are;

Ethandioate (oxalate) Ethylenediamine (en)

EDTA (ethylenediaminetetraacetic acid) contains the EDTA4- ion which is a hexadentate ligand

which forms very stable complexes with metal ions;

EDTA4- Nickel EDTA complex

HAEMOGLOBINIron forms the red complex, haemoglobin, responsible for the transport of oxygen in the blood.

The porphyrin ring (a tetradentate ligand) attaches itself to a central iron(III) ion via its 4 nitrogen

atoms. In haemoglobin the nitrogen atoms in the porphyrin ring occupy four ligand sites. The

other two sites are occupied by the protein, globin, and a molecule of oxygen.

SHAPES OF COMPLEXES

Unit 1 The number of bonds from the central metal ion to the ligand(s) is known as the coordination

number of the central ion. The same term was used in a similar way in describing the

arrangement of ions in a crystal lattice earlier in this topic. The coordination number will

determine the shape of the complex ion.

With coordination number of 6 a complex will have an octahedral shape;

With a coordination number of 4 the complex could have a tetrahedral shape;

or a square planar shape, see table below;

Coordination number Shape

4 square planar

4 tetrahedral

6 octahedral

Unit 1

NAMING COMPLEXESComplex ions and complexes are written and named according to IUPAC rules.

1. The formula of the complex ion should be enclosed within square brackets. The metal

symbol is written first, then the negative ligands followed by the neutral ligands, e.g.

[Fe(OH)2(H2O)4]+.

2. The name of the complex ion or molecule consists of two parts written as one word. The

ligands are named first and the central metal ion second.

3. The ligand name is preceded by a prefix showing the number of ligands (di, tri, etc). If

the ligand is a negative ion the name of which ends in -ide, the ending changes to ‘o’,

e.g. Chloride, Cl- becomes chloro, cyanide, CN- becomes cyano. If the ligand is a

negative ion which ends in -ite, the final ‘e’ changes to ‘o’, e.g. nitrite, NO2-, changes to nitrito.

Some common ligand names are given in the table below.

chloride, Cl- chloro

oxide, O2- oxo

cyanide, CN- cyano

oxalate, C2O42- oxalato

ammonia, NH3 ammine

water, H2O aqua

carbon monoxide, CO carbonyl

4. The metal name is followed by its oxidation state in Roman numerals in brackets. If the complex is an anion

(negative ion), ‘ate’ is added to the name of the metal, sometimes the name ending of the metal is dropped.

Sometimes the Latin name is used, e.g. ferrate not ironate, cuprate not copperate. Some common negative ion names

are shown in the table below.

Vanadium Vanadate

Chromium Chromate

Manganese Manganate

Iron Ferrate

Cobalt Cobaltate

Nickel Nickelate

Copper Cuprate

Tin Stannate

Lead Plumbate

Unit 1 For example, naming K3[Fe(CN)6]

1. Since there are three potassium ions (each 1+), overall charge on the complex ion must

be 3-

2. There are 6 cyanide ions surrounding the central metal ion , each with a charge of 1- so;

(Oxidation number of Fe) + (6 x -1) = -3

Oxidation number of Fe = -3 + 6

Oxidation number of Fe = +33. Six cyanide ions gives ‘hexacyano-‘4. Iron is the central metal ion. Since the complex ion is a negative ion (anion), this gives

‘ferrate(III)’The positive ion name precedes the negative ion name, giving the name;


Practical - Preparation of Potassium Trioxolatoferrate(III)

Unit 1

COLOUR OF TRANSITION METAL IONS AND COMPLEX IONSWhite light consists of a complete spectrum of wavelengths ranging from 400nm to 700nm (all

the colours of the rainbow from violet to red). A compound appears coloured when light from

some region of this spectrum is absorbed, the colour seen is the complementary colour to that

absorbed, i.e. it is a combination of the colours not absorbed. If no light is absorbed then the

compound appears colourless or white, if all the visible light is absorbed the compound is black.

To understand this outcome it has to be appreciated that white light is a combination of the

three primary colours red, blue and green.

red greenyellow

whitemagenta cyan

blue

If red light is absorbed, the colours transmitted are blue and green, which is seen as cyan.

If blue light is absorbed, the colours transmitted are red and green, the colour which is seen is

yellow.

If green light is absorbed, the colours transmitted are blue and red, which is seen as magenta.

The reason that compounds of transition metals absorb white light is due to the loss of

degeneracy of the d orbitals in these compounds. In the free ion, e.g. Ti3+ (1s22s22p63s23p63d1),

the 5 d orbitals (dxy, dyz, dxz, dz2, dx

2-y

2) are degenerate;

Unit 1

However in a complex ion such as [Ti(H2O)6]3+ the metal ion is no longer isolated but surrounded

by 6 water ligands. The complex has an octahedral shape and the water molecules can be

considered to be approaching the central Ti3+ ion along the x-, y- and z- axes.

Due to electrostatic repulsion, the orbitals which lie on these axes (dz2, dx

2-y

2) will be raised to a

higher energy than those orbitals which lie between the axes (dxy, dyz, dxz). Consequently the

degeneracy of the d orbitals is split as shown below;

Unit 1

The difference in energy between the two sets of orbitals, , is called the Crystal Field Strength.

The single d electron in the [Ti(H2O)6]3+ will occupy one of the lower energy orbitals. If radiation

of the correct energy is absorbed the electron can be promoted to one of the higher energy

orbitals. This is normally called a d→d transition. In this example the difference in energy, ,

corresponds to light in the visible region of the spectrum, in fact light which has a = 500 nm

(green light) is absorbed. The transmitted light will be blue and red, giving this solution a

magenta (purple) colour.

The d orbitals are split differently in complexes of different shapes. The amount of splitting, ,

of the d orbitals depends on the ligand. The ability of ligands to cause the splitting of the d

orbitals is given by the spectrochemical series;

small orbital splitting large orbital splittingI- < Br- < Cl- < F- < OH- < H2O < NH3 < CN- < CO

increasing

The colour of the complex can be found by subtracting the colour of the light absorbed from

white light and adding together the remaining colours of transmitted light.

Unit 1

UV and VISIBLE ABSORPTION SPECTROSCOPYThe effects of these d→d transitions can be studied using ultra-violet and visible absorption

spectroscopy. The absorption of uv or visible radiation corresponds to outer electrons

becoming excited. When an atom, molecule or ion absorbs energy, electrons are promoted

from their ground state to an excited state. The energy difference between these states

corresponds to the ultra-violet and visible regions of the spectrum. The wavelength ranges are

approximately 200-400 nm for ultraviolet and 400-700 nm for visible. Ultraviolet radiation is of

shorter wavelength and therefore higher energy than visible radiation. So the difference

between the energy levels is greater when uv radiation is absorbed. Transition metal

compounds are often coloured because they absorb in the visible region. If a compound

absorbs solely in the uv it will appear colourless.

A schematic diagram of a uv/vis spectrometer is shown in the diagram;

Samples are in solution and placed in a cell (a small cuvette). Another identical cell containing

the pure solvent used for the solution is also placed in the machine. Radiation across the whole

range is scanned continuously through both sample and pure solvent. The spectrometer

compares the two beams, the difference is the light absorbed by the compound in the sample.

This data is produced as a chart of wavelength against absorbance.

Unit 1

E.g. for Ti3+(aq) visible spectrum

In the case of visible light being absorbed, the colour transmitted is white light minus the

absorbed light and the complimentary colour is observed. In this example, from the position of

the main peak, it can be seen that mainly green light is absorbed, hence the purple colour of the

solution (red and blue are transmitted).

Watch the Royal Society of Chemistry video on ‘UV/vis Spectrophotometry’

A colorimeter is a simpler device which can also be used in analysis. In this, a light source used

with suitable filters provides light with a narrow band of wavelengths. This is passed through

the sample in a cell as before and the absorbance measured.

The absorbance is directly proportional to the concentration of the absorbing species. By

choosing a range of known different concentrations a calibration curve be drawn and the

concentrations of unknown solutions can then be calculated from the calibration curve.


Unit 1

CATALYSTSTransition metals and their compounds are important catalysts in biological and industrial

chemical reactions. Elements such as iron, copper, manganese, cobalt, nickel and chromium

are essential for the effective catalytic activity of certain enzymes. Some examples of transition

metals and their compounds used as catalysts in industrial reactions are shown in the table;

Process CatalystHaber Iron

Contact Vanadium(V) oxide

Ostwald Platinum

Catalytic converters in car exhausts Platinum, palladium and rhodium

Preparation of methanol Copper

Preparation of margarine Nickel

Polymerisation of alkenes Titanium compounds

It is thought that the presence of unpaired d electrons or unfilled d orbitals allows intermediate

complexes to form, providing reaction pathways of lower energy compared to the uncatalysed

reaction. Many transition metals act as catalysts because of their ability to exist in a variety of

different oxidation states, the transition metal reverts to its original oxidation state once the

reaction is complete.

A good example is the homogenous catalysis by cobalt(II) chloride of the reaction between

hydrogen peroxide and potassium sodium tartrate (Rochelle salt). In the reaction the cobalt

changes oxidation state from +2 (pink) to +3 (green) and then back to +2 (pink);

Co2+ → Co3+ → Co2+

pink green pink

Unit 1

REACTIONS AT EQUILIBRIUMMany reactions never go to completion but are in a state of equilibrium, with reactants and products always present. A chemical reaction is said to be in equilibrium when the concentration of the reactants and products remains constant. For an equilibrium to be established the reaction must take place in a closed system. A closed system is one which allows energy to be transferred to or from the surroundings but not the reactants or products, which remain trapped in the reaction vessel, reactants forming products and products reforming reactants.

For example;

Ethanoic acid, a weak acid, shows little tendency to dissociate;

CH3COOH(aq) CH3COO-(aq) + H+

(aq)

This is a closed system as all the species are in solution and cannot ‘escape’.

However adding two solutions together does not always form a closed system. Sodium carbonate solution reacts with dilute acid to form a salt, water and carbon dioxide gas, which escapes therefore this is not a closed system and equilibrium is never established. The reaction goes to completion;

Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)

When a system is in equilibrium and the concentration of reactants and products is constant, the forward and reverse reactions do not stop. The rate of the forward reaction equals the rate of the reverse reaction.

It is important to realise that when equilibrium is reached the equilibrium will very rarely contain 50% reactants and 50% products.


Unit 1

The Equilibrium Constant (K)Every equilibrium is described by an equilibrium constant (K), which is normally measured in terms of the concentrations of the reactants and products at equilibrium but can in cases of gaseous reactions be measured in terms of pressure (more strictly by the partial pressures of the gases involved, see later).

For the general reaction;

aA + bB cC + dD

K is given by the equation;

K = [products] = [C]c.[D]d

[reactants] [A]a.[B]b

Where [A], [B], [C] and [D] are the equilibrium concentrations of A, B, C and D respectively anda, b, c and d are the stoichiometric coefficients (no. of moles) in the balanced equation.

For example, for the reaction between iron(III) ions and cyanide :

Fe3+(aq) + 6CN-

(aq) [Fe(CN)6]3-(aq)

K = [ [Fe(CN)6]3- ]

[Fe3+].[CN-]6

The same equilibrium position is always reached, whether starting with the reactants or with the products at a given temperature (see Higher unit 3 part 1 appendix 3.7). This means that K is independent of concentrations or pressures of the species in the equilibrium, but is dependant on the temperature.

Since the equilibrium constant is the ratio of the concentration of products divided by the concentration of reactants, its actual value gives an indication of the extent a reaction has progressed i.e. the position of the equilibrium. The greater the value of K, the greater the concentration of products compared to reactants and therefore the further a reaction has gone to the products (remembering that a reaction which has reached equilibrium has gone tocompletion).

ExamplesFor the explosive reaction;

H2 + F2 2HF K = 1 x 1047 reaction goes almost to

completion

Whereas the dissociation;

Unit 1 Cl2 2Cl K = 1 x 10-38 reaction hardly occurs

The value of K gives an indication of how far the equilibrium lies to one side of a reaction or theother.Generally;

Value of K Extent of reaction< 10-3 effectively no reaction

10-3 to 103 significant quantities of reactants and products at equilibrium

> 103 reaction is effectively complete

Note: When K = 1, [products] = [reactants]

K gives no indication of the rate at which equilibrium is established.

Pure solids and solventsWhen a pure solid is present in an equilibrium reaction or a liquid is used as a solvent the concentration, at a given temperature, does not vary to a measurable extent (treated as constant) and it is given the concentration value of 1 in the equilibrium equation.

Unit 1

Calculations using Equilibrium ConstantsIf all the equilibrium concentrations for the reactants and products are known, calculation of the equilibrium constant is simply a matter of substitution into an appropriate equilibrium expression.

WORKED EXAMPLE – equilibrium constant calculations

An equilibrium mixture of gaseous O2, NO and NO2 at 500K contains 1.0 x 10-3 mol.l-1 O2, 1.9 x 10-3 mol.l-1 NO and 5.0 x 10-2 mol.l-1 NO2.

Calculate the value of K at 500K.

O2 + 2NO 2NO2

K = [NO2]2 [NO]2.[O2]

= (5.0 x 10-2)2

(1.9 x 10-3)2.(1.0 x 10-3)

= 6.9 x 105

Unit 1

WORKED EXAMPLE – equilibrium constant calculations

The equilibrium constant K for the reaction;

H2 + I2 2HI

is 57.0 at 700K.

If 0.10 mol of H2 and 0.10 mol of I2 react in a 1.0 litre vessel at 700K, what are the concentrations of H2, I2 and HI at equilibrium?

H2 + I2 2HI

Initial concentration : 0.10 0.10 0

let x moles of H2 react so;

change in moles - x - x + 2 x

equilibrium concentration 0.10 – x 0.10 – x + 2 x

substitute these values into the equilibrium expression;

K = [HI]2 [H2].[I2]

57.0 = (2 x)2

(0.10 – x).(0.10 – x)

57.0 = (2x)2

(0.10 – x)2

take the positivesquare root of each side 7.55 = 2 x

0.10 – x

7.55(0.10 – x) = 2 x

0.755 – 7.55 x = 2 x

0.755 = 2 x + 7.55 x

0.755 = 9.55 x

x = 0.755 9.55

x = 0.0791

so at equilibrium the concentrations are;

[H2] = [I2] = 0.10 – 0.0791 = 0.021 mol.l-1

and [HI] = 2 x 0.0791 = 0.158 mol.l-1

Unit 1

EXAMPLES – equilibrium constant calculations1. For the reaction;

H2 + I2 2HI

The equilibrium constant at 700K is 57.0. If at equilibrium, the concentration of H2 equals the concentration of I2 at 0.08 mol.l-1, what is the concentration of HI?

2. Calculate the equilibrium concentration of PCl5 at 300K when the equilibrium concentrations of PCl3 and Cl2 are 0.2 mol.l-1. The equilibrium constant for the reaction;

PCl5(g) PCl3(g) + Cl2(g)

at 300K is 11.5. Calculate your answer in mol.l-1 and give your answer to 2 significant figures.

Unit 1

Factors affecting the composition of an equilibrium mixture

Le Chatelier’s PrincipleLe Chatelier’s principle states that any change in a system at equilibrium results in a shift of the equilibrium in the direction which minimises the change.

ConcentrationAt a given temperature changes in concentration may alter the position of equilibrium but these changes do not alter the value of K. Following addition of a reactant, or product, the forward and reverse reactions will continue until equilibrium is re-established.

For example in the reaction;

CO(g) + H2O(g) CO2(g) + H2(g)

the equilibrium constant K = 4.31 at 800K.

Starting with 0.2 mol.l-1 each of CO and H2O you can calculate that the equilibrium mixture will contain;

[CO] = [H2O] = 0.065 mol.l-1

and [CO2] = [H2] = 0.135 mol.l-1

If an additional 0.2 mol.l-1 of steam is added, the mixture will no longer be at equilibrium since;

[CO2].[H2] = 1.06 ≠ K[CO].[H2O]

In order to restore the equilibrium ratios, some of the added steam will react with CO to form more products (CO2) and H2) until the ratios again are correct for the equilibrium situation.

Using the method for calculating equilibrium concentrations described on page 19, this will occur when :

[H2O] = 0.233 mol.l-1

[CO] = 0.028 mol.l-1

[CO2] = [H2] = 0.167 mol.l-1

Checking with the equilibrium expression, these values give K ≈ 4.31.

Pressure (only applies to gases)At a given temperature changes in pressure may alter the position of equilibrium but again, these changes do not alter the value of K.

Le Chatelier’s principle states that when a change is applied to a reaction mixture at equilibrium, reaction will occur in a direction which minimises the change.

Unit 1

If the pressure is increased by decreasing the volume of a closed system, the equilibrium will move to reduce this increased pressure. This is achieved by shifting to the side with the least number of gas molecules.

Consider the equilibrium;

N2(g) + 3H2(g) 2NH3(g)

An increase in total pressure on this system will cause the equilibrium to move to favour the side with the least pressure (least gas molecules, the products side) i.e. the equilibrium will move from left to right.This can be seen from the experimental results;

Pressure (atm) % of ammonia at equilibrium1.0 0.255.0 9.3100 16.7200 25.3

In a system such as;

C(s) + O2(g) CO2(g)

A change in pressure will have no effect on the equilibrium position, since there are equal nos.of moles of gas molecules on each side.

TemperatureLe Chatelier’s principle states that when the temperature of a system is raised, the system will move in a direction to reduce the temperature, i.e. the endothermic reaction will be favoured.

Consider the equilibrium;

N2O4(g) 2NO2(g) ∆H = +veyellow brown

The reaction between concentrated nitric acid and copper turnings produces this mixture of N2O4 and NO2. 3 test tubes are filled with this gas mixture, one is placed in ice water (~ 0ºC) one is placed in hot water (~ 80ºC) while the third is kept, as a control, at room temperature(~ 20ºC);

Unit 1

From the colour changes which take place it is evident that the relative concentrations of N2O4 and NO2 have changed with temperature. This means that K is affected by changes in temperature and that equilibria are temperature dependant. Most values of K are quoted at a specific temperature.

Results for K for the Haber process;

N2(g) + 3H2(g) 2NH3(g) ∆H = -92 kj.mol-1

carried out at different temperatures, show clearly the effect of temperature on K.;

Temperature (K) Equilibrium Constant (K)

400 4.39 x 104

500 4.03

600 3.00 x 10-2

Since K is so large at 400K why is the Haber process not carried out at this temperature?

Effect of a catalystCatalysts have no effect on the position of equilibrium. Catalysts affect the rate of a reaction, both forward and reverse reactions, and will therefore speed up the rate of attainment of equilibrium but will not alter the composition of the equilibrium mixture.

UnitsThere is no need to work out the units for equilibrium constants, for the purposes of this course you may consider equilibrium constants to have no units.

Summary of factors affecting KK is unaffected by changes in;

concentration

pressure

K is affected by changes in temperature, an increase in temperature will shift the equilibrium in the direction of the endothermic reaction.


Unit 1

The pH Scale

Water

Water has the following EQUILIBRIUM:-

H2O (l) H+(aq) + OH-(aq)

Water CONDUCTS ELECTRICITY due to water containing LOW CONCENTRATIONS of hydrogen ( ) and hydroxide ( ) ions in solution.

Water DISSOCIATES or IONISES into hydrogen ( ) and hydroxide ( ) ions. The concentrations of the two ions are EQUAL, thus water is pH NEUTRAL.

Equilibrium lies to the LEFT. This means:-

[H2O] = very high

[H+] = very low[OH-] = very low

(The ratio of water molecules to hydrogen ions is 86,000,000:1)

A very important equation called the IONIC PRODUCT OF WATER can be derived.

BUT EQUAL

[H+] X [OH-] = 1 x 10 –14mol2l-2

[H+] = 10 –14 / [OH-]

It’s a CONSTANT VALUE.

Unit 1

CONCENTRATION OF IONS

We can work out the pH of an ACIDIC or ALKALINE solution if we know the concentration of H+ (aq) ions.

(In chemistry p = -log).

Be aware that the pH scale actually STARTS BELOW 0 and GOES ABOVE 14.

Examples

Calculate the pH of HCl, concentration 2.5mol l-1.

[H+] = same as the acid concentration = 2.5

pH = -log [2.5] = -0.397

pH = - log [H+]

[H+] = 10-pH

Unit 1

Calculate the concentration of hydroxide ions if the hydrogen ion concentration is given as 1 x 10-3 mol l-1.

[H+] = 1 x 10-3 mol l-1

[H+] x [OH-] = 1 x 10-14 then

[OH-] = 1 x 10 -14 = 1 x 10 -11 mol l -1 1 x 10-3

( The two powers are taken away from each other ).

More on the Ionic Product of Water

Kw = Ionic Product of Water

= [H+] X [OH-] = 1 x 10 –14mol2l-2

This means:-

[H+] = 1 x 10-7 mol l-1

[OH-] = 1 x 10-7 mol l-1

Kw is a constant value and can be used to work out unknown hydrogen and hydroxide values for different solutions other than water.

[H+] X [OH-] = 1 x 10 –14mol2l-2

[H+] = 1 x 10 –14 mol 2 l -2 [OH-]

[OH-] = 1 x 10 –14 mol 2 l -2 [H+]

FOR PURE WATER ONLY

Unit 1

More Examples

Calculate the concentration of H+ ions if the OH- ion concentration is 1 x 10-9mol l-1.

Calculate the pH of a solution if the OH- ion concentration is 1 x 10-8 mol l-1.

Unit 1

ACID/BASE EQUILIBRIAHistorically acids were defined as producing hydrogen ions in solution and bases as producing hydroxide ions in solution. The two ions combine in the neutralisation equation;

H+(aq) + OH-

(aq) → H2O(l)

As research continued in the early twentieth century these definitions had to be refined. For example, pure hydrogen chloride gas contains no H+ ions, and ammonia, which has no OH- ions, can neutralise an acid. It was also discovered that the H+ ion could not exist in solution, it is so small (basically just a proton) but creates a huge electric field relative to its size. In water it attracts a lone pair of electrons to form H3O+, the hydronium ion (also called the oxonium ion or the hydroxonium ion) (see Unit 1, page 24, NH4

+ ion, Unit 3 page 10, where H3O+ acts as an electrophile).

H+(aq) + H2O(l) → H3O+

(l)

The shorthand H+(aq) is always used in stoichiometric and equilibrium equations, although this is

not strictly accurate.

In 1923 Bronsted and Lowry defined acids and bases as;

An acid is any substance capable of donating a proton.A base is any substance capable of accepting a proton.

When an acid donates a proton the species left is called the conjugate base of that acid.For every acid there is a conjugate base formed by the loss of a proton.

When a base accepts a proton the species formed is called the conjugate acid of that base.For every base there is a conjugate acid formed by the gain of a proton.

For example dissolving hydrogen chloride in water;

HCl + H2O Cl- + H3O+ acid base conjugate base conjugate acid

Complete the following table;

acid + base conjugate base + conjugate acid

CH3COOH + H2O +

H2O + NH3 OH- + NH4+

H2O + CH3COO- +

These examples illustrate the amphoteric nature of water, since it can act both as a proton acceptor (base) and a proton donor (acid).

Unit 1

EXAMPLES – acid/base equilibria

Label the acid, base, conjugate acid and conjugate base in the above reaction.

Unit 1

The Dissociation of Water

When water is purified by repeated distillation its conductivity falls to a constant, but definite value. Even the purest water has a tiny conductivity indicating that even the purest water contains a small number of _________.

This is evidence that water forms ions as a result of its own dissociation;

H2O(l) + H2O(l) OH-(aq) + H3O+

(aq) acid base conjugate conjugate

base acid

Again it can be seen that water is amphoteric. In the forward reaction the water molecules act both as a proton donor (acid) and proton acceptor (base).

The concentration of ions is very small and the equilibrium for this reaction lies very far to the left.The equilibrium constant is K = [H3O+].[OH-]

[H2O]2

Since water is a liquid, [H2O] = 1 (see page 17), giving;

K = [H3O+].[OH-]

This particular equilibrium constant is know as the ionic product (Kw) for water and has the value of 1.0 x 10-14 at 25 ºC.

The water equilibrium is more usually written as;

H2O(l) H+(aq) + OH-

(aq) Kw = [H+].[OH-]

In pure water, for every molecule which ionises, one H+ and one OH- ion are produced, hence (in mol.l-1);

[H+] = [OH-]

On average there is one H+ ion and one OH- ion for every 555 million H2O molecules!

Kw is always quoted as 1.0 x 10-14 at 25 ºC since the reaction is endothermic,

H2O(l) H+(aq) + OH-

(aq) ∆H +ve

An increase in temperature moves the equilibrium to the _____________ and a decrease moves it to the _______________ see the table below.

Unit 1

Temperature (ºC) Kw (x 10-14)

18 0.6

25 1.0

40 2.9

75 16.9

The ionic product of water _____________ as the temperature increases.

The pH scalepH = - (the power to which [H+] is raised)

which allowed the calculation of (whole number) pHs.

If an acid is strong (completely dissociated) and monoprotic (e.g. Hydrochloric acid, HCl), the molar concentration of the acid will equal [H+] and so the pH can be calculated. Similarly, for a strong alkali, the molar concentration can be used to calculate [OH-]. Using Kw, [H+] can becalculated and hence the pH.

WORKED EXAMPLES – revision of pH calculations1. Calculate the pH of a 0.01 mol.l-1 solution of HCl.

[H+] = 0.01 mol.l-1

[H+] = 1 x 10-2 mol.l-1

pH = 2

2. Calculate the concentration of OH- ions in the above solution.

Kw = [H+].[OH-] = 1.0 x 10-14

[OH-] = 1.0 x 10-14

[H+]

[OH-] = 1.0 x 10-14

1.0 x 10-2

= 1.0 x 10-12 mol.l-1

Unit 1

Calculating pHIn fact the pH of any aqueous solution of a strong acid or alkali can be calculated using theexpression;

pH = - log[H+]

Remember the [H+] in a solution of pH 2 is 10 times that of a solution of pH 3.

WORKED EXAMPLES – pH calculations1. Calculate the pH of a solution with [H+] = 2.5 x 10-3 mol.l-1

pH = - log 2.5 x 10-3

= -(-2.6) = 2.6

2. Calculate the concentration of H+ ions and OH- ions in a solution of pH 3.6

pH = - log [H+]

3.6 = - log [H+]

log [H+] = -3.6

[H+] = antilog (-3.6)

= 10-3.6

= 0.000251

= 2.51 x 10-4 mol.l-1

Kw = [H+].[OH-] = 1 x 10-14

[OH-] = 1 x 10-14

[H+]

= 1 x 10-14

2.51 x 10-4

= 3.98 x 10-11 mol.l-1

This expression also allows calculation of [H+] and [OH-] of all aqueous solutions.


Unit 1

Concentrated and Dilute Acids

Concentrated Acids

An acidic solution which has a LARGE MASS of acid dissolved in a litre of water.

Dilute Acids

An acidic solution which has a SMALL MASS of acid dissolved in a litre of water.

Units of concentration = __________________________

The higher the number, the more concentrated an acid (or any solution is).

Eg, a 6 mol l-1 solution contains more chemical dissolved in one litre than a 0.6 mol l-1 solution.

Remember, to reduce a concentration, the solution can be diluted with water.

The more concentrated a solution is, the better it CONDUCTS since THE MORE IONS IT CONTAINS.

STRONG and WEAK ACIDS

Experiment Ethanoic acid Hydrochloric acidpHConductivityVolume of 0.2 mol l-1

NaOH required to neutralise 25ml of acid.Rate of reaction with Magnesium

Hydrochloric, nitric and sulphuric are STRONG ACIDS.

Phosphoric, carbonic (found in fizzy drinks), sulphurous and all the alkanoic acids (methanoic acid, ethanoic acid, propanoic acid) are WEAK ACIDS.

Concentration of the acids must be kept the same to allow for a fair comparison.

Unit 1

Results

Ethanoic Acid

This is a weak acid and doesn’t dissociate fully in water (5%). It mainly exists as molecules.

This means that there is a LOW CONCENTRATION of hydrogen ions, causing a LESS ACIDIC pH.

Less ions in solution results in a LOWER CONDUCTIVITY since there are LESS IONS to carry the charge.

There is a slower reaction with magnesium because there are less hydrogen ions present in the acid to react with the magnesium.

FINALLY, the same volume of alkali will be needed to neutralise as a strong acid. This is because of the equilibrium system.

CH3COOH (l) + H2O (l) CH3COO- (aq) + H+ (aq)

The OH- (aq) ions react with the H+ (aq) ions removing it from the equilibrium. This unbalances the equilibrium and to re-establish it again and produce more H+, the equilibrium gets pushed to the right hand side causing all the ethanoic acid molecules to dissociate over time.

Hydrochloric Acid

This is a STRONG ACID. And fully dissociates 100% in water, forming ions.

HCl(g) + H2O (l) H+ (aq) + Cl- (aq) MOLECULES IONS

Since a high concentration of hydrogen ions are present, this results in a LOW pH value, which means the solution is very acidic.

Since there is a high concentration of both the hydrogen and chloride ions, this means that the solution will conduct better.

FINALLY, the hydrogen ion concentration is so high this means that the

reaction with magnesium is fast

Unit 1

BUT the same volume of sodium hydroxide will be required to neutralise as with

the weak acid. This is because all the acid molecules are in the form of ions, whereas for a weak acid of the same concentration this happens over time.

STRONG and WEAK BASES

Experiment Sodium hydroxide solution

Ammonia Solution

pH

Conductivity

Volume of 0.2M acid required to neutralise 25ml of alkali

STRONG BASES include SODIUM HYDROXIDE, CALCIUM HYDROXIDE, LITHIUM HYDROXIDE etc etc.

AMMONIA as a WEAK BASE.

Concentrations of the bases are kept the same in the experiment to allow for a FAIR COMPARISON to be made.

Unit 1

RESULTS

Ammonia Solution

NH3(g) + H2O(l) NH+4(aq) + OH-(aq)

This is a weak base and does not fully dissociate in water (~5%).

This means there is a LOW CONCENTRATION of hydroxide ions, causing the pH of the solution to be less alkaline.

Less ions in solution results in a LOWER CONDUCTIVITY since there are LESS IONS to carry the charge.

FINALLY, the same volume of acid will be needed to neutralise (compared to a strong base). This is because:-

NH3(g) + H2O(l) NH+4(aq) + OH-(aq)

H+

The hydrogen ions react with the hydroxide ions removing it from the equilibrium. This unbalances the equilibrium again and to re-establish equilibrium again it gets pushed to the right hand side causing all the ammonia to dissociate over time into NH4

+ .

Sodium Hydroxide

This is a strong base since it is fully DISSOCIATED when added to water. The arrow means the reaction completely turns to the products on the right hand side (100% conversion).

NaOH(s) + H2O(l) Na+(aq) + OH-(aq)

Since a high concentration of sodium and hydroxide ions are present, this means the solution will conduct better.

Since there is a high concentration of hydroxide ions (OH-(aq)) this means that the solution will have a high pH value (more alkaline).

Finally, the hydroxide ion concentration is high, so it will still need the same volume of acid to neutralise it, as a weak acid does.

Unit 1

SHIFTING THE EQUILIBRIUM

Changing the concentration of one of the PRODUCTS in an EQUILIBRIUM REACTION can shift the position of EQUILIBRIUM.

This is true for WEAK ACIDS and BASES.

1. Ethanoic Acid (equilibrium in water) – WEAK ACID

CH3COOH (l) + H2O (l) CH3COO- (aq) + H+ (aq)

Molecules Ethanoate ions Hydrogen ions

If the salt SODIUM ETHANOATE (this is the salt of a strong base and a weak acid - CH3COO-Na+) is added to the ETHANOIC ACID EQUILIBRIUM ABOVE then the following will happen:-

The CH3COO-(aq) concentration increases on the right hand side of the equilibrium.

Equilibrium moves to the LEFT to reduce this.

Due to this the pH will become LESS ACIDIC as there are LESS H+(aq) ions on the LEFT of the equilibrium compared to the RIGHT.

2.Ammonia (equilibrium in water) – WEAK BASE

NH3 (g) + H2O (l) NH4+ (aq) + OH- (aq)

Molecules Ammonium ions Hydroxide ions

Unit 1

If the salt AMMONIUM CHLORIDE (this is the salt of a strong acid and a weak base – NH4

+Cl-) is added to the AMMONIUM SOLUTION EQUILIBRIUM ABOVE then the following will happen:-

The NH4+(aq) concentration increases on the right hand side of the

equilibrium.

Equilibrium moves to the LEFT to reduce this.

Due to this the pH will become LESS ALKALINE as there are LESS OH-

(aq) ions on the LEFT of the equilibrium compared to the RIGHT.

Now Try Exercise 1.15

Unit 1

The pH OF SALT SOLUTIONS

REMEMBER:-

All salts are STRONG ELECTROLYTES (solutions which conduct electricity). This means that they completely IONISE.

Ions of a weak acid/ base set up an equilibrium when dissolved in water.

Salts formed in NEUTRALISATION REACTIONS can be formed from 4 different combinations of ACID and BASE.

1. Salts of strong acids and strong bases.

2. Salts of weak acids and weak bases.

1. Salts of strong bases and weak acids.

2. Salts of strong acids and weak bases.

We will look at the pH of each of these salts in turn when they are dissolved in water.

Salts of strong acids and strong bases (neutral salts)

Examples – sodium chloride, ,

Made from __________________________________________.

NaCl (s) + H2O(l) Na+(aq) + Cl-(aq)

All these salts ionise 100% in water to form NEUTRAL SALTS of pH7. This is because

there is no equilibrium set up in the water when they dissolve.

Unit 1

Salts of weak acids and weak bases (neutral salts)

Examples – ammonium ethanoate

Made from __________________________________________.

All these salts dissolve in water to form salts with pH7.

Salts of strong bases and weak acids (alkaline salts)

Examples – potassium carbonate, sodium ethanoate

Sodium ethanoate (salt)

CH3COONa (s) + H2O (l) CH3COO- (aq) + Na+ (aq) SALT ethanoate sodium ion

ion

This equation shows that the salt dissociates 100% when added to water. There is no equilibrium arrow present here.

In water an EQUILIBRIUM is present:-

Unit 1

CONSEQUENCE:-

The hydrogen ions (H+(aq)) from the water react with the ethanoate ions (CH3COO- (aq) from the salt to form:-

CH3COO- (aq) + H+ (aq) CH3COOH (aq)

From salt from water ethanoic acid (undissociated form)(more stable form)

However, the hydrogen ions have been removed from the water equilibrium leaving an excess of hydroxide ions in the water equilibrium. This will also push the water equilibrium to the RHS making even more OH-(aq) ions compared to H+ (aq) ions.

More hydroxide ions compared to hydrogen ions means ____________________________________________________________________________

____________________________________________________________________________

Salts of strong acids and weak bases (acidic salts)

Examples – ammonium nitrate, ammonium chloride

Ammonium chloride (salt)

NH4Cl (s) + H2O (l) NH4+ (aq) + Cl- (aq)

This equation shows that the salt dissociates 100% when added to water. There is no equilibrium arrow present here.

In water an EQUILIBRIUM is present:-

Unit 1

CONSEQUENCE:-

The hydroxide ions (OH-(aq)) from the water react with the ammonium ions (NH4+ (aq))

from the salt to form:-

NH4+ (aq) + OH- (aq) NH3(aq) + H2O(l)

From salt from water ammonia (undissociated form)(more stable form)

However, the hydroxide ions have been removed from the water equilibrium leaving an excess of hydrogen ions in the water equilibrium. This also causes the equilibrium to be pushed to the RHS making even more hydrogen ions along with some hydroxide ions.

More hydrogen ions compared to hydroxide ions means ____________________________________________________________________________

____________________________________________________________________________

Soaps

Soaps are salts of weak acids and strong bases.


Unit 1

Dissociation of AcidsThe dissociation of any acid, HA, in aqueous solution can be represented by the equation;

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

acid base conjugate conjugate acid base

Since water is a liquid, [H2O] = 1, giving an equilibrium constant;

Ka = [H3O+].[A-] [HA]

Ka is known as the acid dissociation constant, and is a measure of the strength of the acid.

If the acid is strong the equilibrium will lie far to the right with effectively complete dissociation so that Ka is very large. However some acids do not dissociate well in aqueous solution and these are called weak acids. The smaller the value of Ka the weaker the acid.

Page 12 of the data book lists the dissociation constants of some acidic species. The dissociation constant can also be expressed as pKa where;

pKa = - log Ka

Both Ka and pKa can be used as a measure of the strength of an acid. It can be seen from page 12 of the data book that the values of Ka are spread over a very wide range (10-11 to 10-2), pKa condenses the range of values and is often more convenient to use.Complete the following table;

Acid Ka pKa

Methanoic 1.8 x 10-4 3.75

Ethanoic 1.7 x 10-5

Benzoic 6.3 x 10-5

Hydrocyanic 6.2 x 10-10

Calculation of pH of a Weak Monoprotic AcidA monoprotic acid contains a single hydrogen atom per molecule that is susceptible to dissociation in aqueous solution. Example; CH3COOH. (Strong monoprotic acid examples; HCl,HNO3)

The pH of a weak acid depends on both the concentration and the extent of the dissociation of the acid. The pH cannot be calculated from the molar concentration since only a small proportion of the molecules are dissociated into ions. However the pH can be calculated if theKa value for the acid is known, the required equation is;

pH = ½ p Ka – ½ log c

Unit 1

where c is the molar concentration of the weak acid.

Note: this equation can only be used for weak acids. It becomes inaccurate for strong acids

WORKED EXAMPLES – further pH calculations

1. Calculate the pH of a solution of 0.1 mol.l-1 ethanoic acid.

From page 13 of the data book, pKa = 4.76;

pH = ½ pKa – ½ log c

= ½ (4.76) – ½ (log 0.1)

= 2.38 – (-0.5)

= 2.88

≈ 2.9

2. A solution of benzoic acid has a pH of 3.0. What is the molar concentration?

From page 13 of the data book, pKa = 4.20;

pH = ½ p Ka – ½ log c

3.0 = ½ (4.20) – ½ (log c)

½ log c = ½ (4.20) – 3.0

log c = -1.8

c ≈ 0.016 mol.l-1

Dissociation of Bases

In the presence of a more powerful base, B, water molecules behave as an acid by donating protons to the base. The dissociation of any base, B, in aqueous solution can be represented by the equation;

B(aq) + H2O(l) BH+(aq) + OH-

(aq)

base acid conjugate conjugate acid base

If the base is strong the equilibrium will lie far to the right. For a weak base the equilibrium lies far to the left. Since water is a liquid, [H2O] = 1 (see page 17), the base dissociation constant Kb, is given by;

Kb = [BH+].[OH-] [B]

Unit 1

For simplicity, the dissociation constant of any base B is defined by the dissociation constant Ka of the conjugate acid, BH+;

BH+(aq) + H2O(l) B(aq) + H3O+

(aq)

Again [H2O] = 1 (see page 13), giving an equilibrium constant;

Ka = [B].[H3O+] [BH+]

Ammonia can be used as an example;

NH3(aq) + H2O(l) NH4+

(aq) + OH-(aq)

base acid conjugate conjugate acid base

dissociation of the conjugate acid NH4+ is;

NH4+

(aq) + H2O(l) NH3(aq) + H3O+(aq)

giving an equilibrium constant;

Ka = [NH3].[H3O+] [NH4

+]

The Ka value of the conjugate acid can be used to give information about the strength of the base using the expression;

Ka x Kb = 1.0 x 10-14

Therefore as Ka increases the base gets weaker, e.g. methylamine, Ka = 2.2 x 10-11 and ammonia, Ka = 5.8 x 10-10.


Unit 1

IndicatorsIndicators are usually weak acids which change colour depending on the pH of the solution. They are dyes whose colours are sensitive to pH, and are used to determine the end-point of an acid-alkali titration. A suitable indicator must be chosen for any given reaction. HIn can be used as a general formula for an indicator and its dissociation can be represented as;

HIn(aq) + H2O(l) H3O+(aq) + In-

(aq)

For a good indicator, the undissociated acid, HIn, will have a distinctly different colour from its conjugate base, In-.

e.g. For the indicator litmus, HIn is red and In- is blue.

Unit 1

The acid dissociation constant for an indicator HIn is given the symbol KIn and is represented

by;

KIn = [H3O+].[In-] [HIn]

Rearranging this expression gives;

[In-] = KIn

[HIn] [H3O+]

This shows that the relative concentrations of the acid, [HIn], and its conjugate base, [In-], is a function of the hydronium ion concentration [H3O+], i.e. the pH of the solution. The overall colour of the indicator in any given solution depends on the relative concentrations of the two coloured forms, which in turn is determined by the pH of the solution. If the [H3O+] is large (on adding acid) the colour will be that of the undissociated acid, [HIn]. If the [H3O+] is small (on adding alkali) the colour will be that of In-, see table below.

For the general indicator equilibrium;

HIn(aq) + H2O(l) H3O+(aq) + In-

(aq)

In neutral solution the colour of the indicator would be seen as a mixture of the two colours of HIn and In-, the colour of this mixture would be dictated by their relative concentrations.

Addition of indicator to an acidic solution can be thought of as adding H+ ions (H3O+ ions) to the above equilibrium. Similarly adding indicator to an alkaline solution can be thought of as adding OH- ions to the indicator equilibrium. The effect in each case would be;

Ions added Effect Equilibrium shifts Colour seenH+ adds H3O+ to the left HIn

OH- removes H3O+ ions to the right In-

Examples of indicator colours at different pHs;

indicator colour seen in

Acid [HIn] Neutral Alkali [In-]

Phenolphthalein Colourless Colourless Pink

Bromothymol blue Yellow Green Blue

Unit 1

When does an indicator change colour?From the relationship;

KIn = [H3O+].[In-] [HIn]

In the middle of the indicator’s colour change (i.e. the theoretical point at which the colour change occurs) the two forms of the indicator, HIn and In- are present in equal amounts and so [HIn] = [In-], the expression above then simplifies to;

KIn = [H3O+]

Or, since pKIn = - log KIn = -log[H3O+] (= -log[H+]) = pH (see page 33);

pKIn = pH

***Only true at the equivalence point***NOTE : equivalence points change according to the strengths of acid andalkali used in the titration (marked ‘x’ on diagrams on pages 78 and 79).

For the two forms of the indicator to be distinguishable by the naked eye, the ratio of HIn : In- must be around 10:1.

For example if HIn is yellow and In- is blue, then must be at least as small as for the

solution to appear yellow and at least as large as for it to appear blue. Therefore, thecolour changes are only distinguishable when the concentrations differ by a factor of 10. Thisresults in the indicator changing colour over a range of pHs.

The pH range over which a colour change can be seen is therefore established using the expression;

pH = (pKIn) +1 (remember : log 10 = 1)

This means that a particular indicator could change colour between pH 4.0 – 5.0, in practice most indicator changes are given a slightly wider range over which the colour changes. Methyl orange changes colour between pH 3.0 – 4.4.

Choosing the indicatorThe appropriate indicator for any titration must be chosen such that the colour change occurs over the pH range when the pH is changing very rapidly. This means that the indicator must change colour during the addition of, say, half a drop of reagent (this gives a very accurate end-point).

When an acid is gradually neutralised by a base, the change in pH can be monitored by using a pH meter. The results can be used to produce a titration curve (see pages 78 and 79) from which the equivalence point (marked ‘x’ on diagrams on page 78 and 79) can be identified. The indicator for a reaction can therefore be chosen by reference to these titration curves.

Unit 1

Strong acid / strong baseConsider the titration of a strong acid with a strong alkali. The pH of the salt formed is 7, therefore full neutralisation occurs at pH 7. As both the acid and the alkali are strong the concentration of ions in solution is high and reaction occurs rapidly as the end point is approached.

This results in a very fast change in pH (~ vertical part of the curve). A suitable indicator for astrong acid / strong base reaction will change colour within the pH range 3-10.

Strong acid / weak baseThe pH of the salt formed is <7 which means neutralisation will be achieved at less than pH 7.

Still rapid change of pH close to neutralisation. A suitable indicator for a strong acid / weak base reaction will change colour within the pH range 3-7.

Unit 1

Weak acid / strong baseThe pH of the salt formed is >7 which means neutralisation will be achieved at a pH greater than 7.

Still a rapid change of pH close to neutralisation. A suitable indicator for a weak acid / strong base reaction will change colour within the pH range 7-10.

Weak acid / weak baseThe pH of the salt formed is variable and depends on the combination of weak acid and base used.

There is no suitable indicator for this type of reaction since the pH does not change rapidly enough at the end point (due to the low concentration of ions) which means you have a gradual colour change with the indicator which would not give an accurate result.

Weak acid / weak base titrations are not used for volumetric analysis as no clear end point can be identified.

Unit 1

From these titration curves it can be seen that the middle of the indicator’s colour change range needs to correspond to the equivalence point of the titration. Since the equivalence points change according to the strengths of acid and alkali used in the titration , the appropriate indicator is chosen using the relationship;

pH = pKIn + log

This shows that the pH of a solution is determined by the pKIn of the indicator and the ratio of [In-] : [HIn]. Since these are different colours, the ratio of [In-] : [HIn] determines the overall colour of the solution. For a given indicator, the overall colour of the solution is dependant on the pH of the solution.

WORKED EXAMPLES – indicator calculations

1. Bromothymol blue has a pKIn value of 7.0.

What value will KIn have?

pKIn = - log KIn

so KIn = inverse log (- pKIn)

KIn = inverse log(- 7.0)

= 1 x 10-7.0

= 0.0000001

2. Calculate the ratio of [In-] : [HIn] for Bromothymol blue at pH 3.0.

pH = pKIn + log

log = pH - pKIn

log = 3.0 - 7.0

= inverse log (- 4.0)

= 1 x 10-4.0

= 0.0001

Common Indicators – see data book page 20

indicator pH range of colour change colour of[HIn] [In-]

methyl orange 3.2 – 4.4 orange yellow

methyl red 4.8 – 6.3 red yellow

bromothymol blue 6.0 – 7.6 yellow blue

phenolphthalein 8.2 – 10 colourless red


Unit 1

Buffer solutions

A buffer solution is one in which the pH remains approximately constant when small amounts of acid or alkali are added.

There are two types of buffer solution;

an acid buffer consists of a solution of a weak acid and one of its salts

a basic buffer consists of a solution of a weak base and one of its salts

If a buffer is to stabilise pH, it must be able to absorb extra acid or alkali if these are encountered.

Both types of buffer solution work in the same way.

Acid bufferIn an acid buffer solution, the weak acid supplies more hydrogen (hydronium) ions when existing ones are removed by a base being added, while the salt of the weak acid provides the conjugate base to react with the hydrogen (hydronium) ions when small amounts of acid are added. In both cases the pH hardly changes.

This is illustrated by an acid buffer of the weak acid HA and the sodium salt of that acid NaA. In solution the weak acid forms an equilibrium and the sodium salt fully ionises;


(aq) equilibrium


NaA(s) + (aq) Na+(aq) + A-

(aq) fully ionised conjugate base

NaA is fully ionised in solution providing large reserves of A- in the buffer solution, forcing the weak acid equilibrium to the left.

If acid is added to the mixture the A- ions will trap the extra hydronium ions and convert them to the weak acid HA. Since [A-] is high compared to [H3O+] in the original buffer solution its ability to remove H3O+ is substantial (but not infinite) and the pH of the solution is maintained.

In the same way, if alkali is added the addition of OH- does not alter the pH greatly since the OH- ions combine with the H3O+ of the weak acid. This shifts the equilibrium to the right which in turn replaces the removed H3O+.

A typical example of an acid buffer solution would be ethanoic acid and sodium ethanoate. The ethanoic acid is only partially dissociated, the sodium ethanoate salt completely dissociates and provides the conjugate base;

CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO -

(aq) equilibrium acid base conjugate conjugate

acid base

Na+CH3COO-(s) + (aq) Na+

(aq) + CH3COO -(aq) fully ionised

Unit 1 conjugate

base

The stable pH of the acid buffer solution is due to;

the weak acid which provides H3O+ to trap added OH-

the salt of the acid which provides A- to trap added H3O+

Basic bufferA basic buffer consists of a weak base and one of its salts. In a basic buffer solution the weak base removes added hydronium ions and the conjugate acid provided by the salt removes added hydroxide ions.

A typical example of a basic buffer solution would be ammonia, the weak base and ammonium chloride, the ammonium ion is the conjugate acid of the weak base;

NH3(aq) + H2O(l) NH4+

(aq) + OH-(aq) equilibrium

base acid conjugate conjugate acid base ions

NH4+Cl-(s) + (aq) NH4

+(aq) + Cl-(aq) fully ionised

conjugate acid

The buffer solution contains a large amount of NH4+, this pushes the equilibrium to the left

giving a large amount of NH3.

Addition of a small amount of alkali (OH-) will have little effect on the pH of the buffer as the OH- will react with the conjugate acid, NH4

+, to form ammonia and water, hence removing the added OH-.

Addition of a small amount of acid will have little effect on the pH of the buffer as the H3O+ ions will react with the OH-, this shifts the equilibrium to the right which in turn replaces the removed OH-.

The stable pH of the basic buffer solution is due to;

the weak base which provides OH- trap added H3O+

the salt of the base which provides NH4+ to trap added OH-

The pH and composition of buffer solutionsThe pH of buffer solutions depends upon two factors;

the acid dissociation constant

the relative proportions of salt and acid

The dissociation constant for a weak acid is given by the expression;


(aq)


Unit 1

giving an equilibrium constant;

Ka = [H3O+].[A-] [HA]

This can be rearranged to give;

[H3O+] = Ka. [HA][A-]

As the amount of dissociation of the weak acid, HA, is very limited we can make the assumption that [HA] is the same as the molar concentration of the weak acid put into the buffer solution. As the salt, e.g. NaA, completely dissociates, [A-] will be the same as the concentration of the salt.

The expression then becomes;

[H3O+] = Ka x [acid] [salt]

Taking the negative log of both sides gives;

pH = pKa – log

As adding water to a buffer will dilute both [acid] and [salt] equally, [H3O+] and therefore its pH will be unaffected.

If [acid] = [salt] when the buffer is made up, then pH = pKa, (or [H3O+] = Ka).

This equation allows calculation of the pH of an acid buffer from its composition and acid dissociation constant, or calculation of the composition from the other two values. Ka and pKa

values can be found on page 12 of the data booklet.

WORKED EXAMPLES – buffer calculations

1. Calculate the pH of a buffer solution made with 0.1 mol.l-1 ethanoic acid and 0.1 mol.l-1 sodium ethanoate. (from data booklet pKa = 4.76)

pH = pKa – log

pH = 4.76 – log

pH = 4.76 – log (1)

pH = 4.76

2. Calculate the pH of a buffer solution made with 0.1 mol.l-1 methanoic acid and 0.2 mol.l-1 sodium ethanoate. (from data booklet pKa = 3.75)

pH = pKa – log

pH = 3.75 – log

pH = 3.75 – log (0.5)

pH = 4.05

Unit 1

3. Calculate the concentrations of ethanoic acid and sodium ethanoate required to make a buffer solution with a pH of 5.3.

pH = pKa – log

5.3 = 4.76 – log

log = 4.76 – 5.3

log = = - 0.54

log = + 0.54

= 3.47

So the ratio of 3.47 to 1, salt to acid, is required, i.e. 3.47 moles sodium ethanoate mixedwith 1 litre of 1.0 mol.l-1 ethanoic acid.

or simply

log = - 0.54

= 0.29

So the ratio of 0.29 to 1, acid to salt is required, i.e. the concentration of ethanoic acid is 0.29 mol.l-1 and the concentration of sodium ethanoate is 1.0 mol.l-1


Buffer solutions are important in biological systems, especially those where enzymes work within narrow pH ranges, e.g. blood is a buffered solution of pH around 7.4 (CO2 / bicarbonate equilibrium is maintained by respiration and excretion of bicarbonate into the urine).

Phosphate solutions can act as buffers because the second and third dissociations of phosphoric acid are weak;

H3PO4 H+ + H2PO4-

H2PO4 H+ + HPO4

2-

HPO42- H+ + PO4

3-

Another useful buffer is potassium hydrogenphthalate, which is a weak acid and its salt all in one molecule;

Unit 1

THERMOCHEMISTRYThermochemistry is the study of the energy changes which occur during chemical reactions, and can be used to predict the feasibility of any reaction.

Energy is required to break bonds (endothermic) and is released when bonds are formed (exothermic). When a chemical reaction takes place these two processes occur and there is often an exchange of heat energy between the reaction and its surroundings, called the enthalpy change (∆H). Every substance has an enthalpy (a chemical potential energy or heat content) that cannot be measured. Only when the substances react can the enthalpy change be noted, the units are kilojoules per mole, kJ.mol-1.

The standard enthalpy change (∆Ho) is defined as;

the enthalpy change measured under standard conditions

standard conditions are one mole of a substance at one atmosphere pressure usually at 298K (25ºC)

the standard state of a substance is its most stable state under standard conditions.

Standard molar enthalpy of formation (∆Hof) is the enthalpy change when one mole of a

substance is formed from its elements in their standard states.

e.g.3C(s) + 4H2(g) + ½O2(g) C3H7OH(l) ∆H +/-ve

The standard molar enthalpy of formation of elements is defined as zero, giving a ‘base line’ from which enthalpy changes can be measured.

The enthalpy change for any reaction is defined as the sum of the enthalpy of the products minus the sum of the enthalpy of the reactants;

∆H = ∑ H(products) - ∑ H(reactants)

Exothermic reactions have a negative ∆H as energy is lost to the surroundings, while endothermic reactions have a positive ∆H as energy is taken in from the surroundings.

Unit 1

Hess’s Law (Revision from Higher)The first law of thermodynamics states that;

‘energy cannot be created or destroyed and it can only be converted from one form to another’

Hess, in 1840, applied this law to chemical reactions and Hess’s Law states (in short) that;

‘the enthalpy change associated with converting reactants into products is independent of the route taken’

e.g. Preparation of potassium chloride by two different routes

Route 1 : direct one step method. Solid potassium hydroxide is added to dilute hydrochloric acid to produce potassium chloride solution;

KOH(s) + HCl(aq) KCl(aq) + H2O(l) ∆H1

Route 2 : the indirect route. This involves 2 steps, the first step is to dissolve solid potassium hydroxide in water;

KOH(s) + (aq) KOH(aq) ∆H2

In the second step the potassium hydroxide solution is added to dilute hydrochloric acid to produce potassium chloride solution;

KOH(aq) + HCl(aq) KCl(aq) + H2O(l) ∆H3

Using Hess’s Law;

∆H1 = ∆H2 + ∆H3

This law allows the calculation of enthalpy changes for reactions that are difficult or impossible to determine by experiment, e.g. the reaction is too slow or too fast or by-products are formed. There are two general methods of carrying out calculations based on Hess’ Law;

The Pictorial Method

This method requires a thermochemical cycle to be set up to show the energy changes between reactants and products. Easy to use when bond enthalpies and ionic compounds are involved, but cumbersome for other calculations.

Unit 1

The Algebraic Method

Can be used for any calculations

WORKED EXAMPLES – Algebraic Method calculations

1. Calculate the standard enthalpy of formation of ethanoic acid given that its standard enthalpy of combustion is -876 kJ.mol-1 and that the standard enthalpies of formation of carbon dioxide and water are -394 and -286 kJ.mol-1 respectively.

Step 1 : write target equation

2C(s) + 2H2(g) + O2(g) CH3COOH(l)

Step 2 : write equations for all the information given ∆H o / kJ.mol -1

C(s) + O2(g) CO2(g) -394

H2(g) + ½O2(g) H2O(l) -286

CH3COOH(l) + 2O2(g) 2CO2(g) + 2H2O(l) -876

Step 3 : reverse, multiply or divide these equations and their enthalpy values

x 2 2C(s) + 2O2(g) 2CO2(g) -788

x 2 2H2(g) + O2(g) 2H2O(l) -572

reverse 2CO2(g) + 2H2O(l) CH3COOH(l) + 2O2(g) +876

Step 4 : combine these to obtain the target equation

2C(s) + 2H2(g) + O2(g) CH3COOH(l) -484

∆Hof = -484 kJ.mol-1

The standard enthalpy of a reaction can be calculated from tabulated standard molar enthalpies of formation using the relationship;

∆Ho = ∑ ∆Hof (products) - ∑ ∆Ho

f (reactants)

Unit 1

2. Calculate the standard enthalpy of reaction for the decomposition of copper(II) nitratefrom the given standard enthalpies of formation.

Step 1 : write the equation with ∆Hof values beneath each substance

Cu(NO3)2(s) CuO(s) + 2NO2(g) + ½O2(g)

∆Hfo/ kJ.mol-1 -307 -155 +34 0

Step 2 : multiply each ∆Hfo by its mole relationship

Cu(NO3)2(s) CuO(s) + 2NO2(g) + ½O2(g)

∆Hfo/ kJ.mol-1 -307 -155 +68 0

Step 3 : use the equation ∆Ho = ∑ ∆Hof (products) - ∑ ∆Ho

f (reactants)

∆Ho = ∑ ∆Hof (products) - ∑ ∆Ho

f (reactants)

= (-155 + 68) – (-307)

= -87 + 307

= +220

∆Ho = +220 kJ.mol-1


Unit 1

REACTION FEASIBILITY

Exothermic reactions have a negative value for the enthalpy change and release heat to the surroundings. In contrast endothermic reactions have a positive enthalpy change and heat is absorbed from the surroundings as the reaction takes place.

Almost all exothermic reactions are spontaneous at standard conditions, for example iron rusting;

2Fe(s) + 1½ O2(g) + 3H2O(l) → 2Fe(OH)3(s) ∆H = -791 kJ.mol-1

The endothermic reverse reaction, rusting reversing to give pure iron, with a positive enthalpy change, never occurs spontaneously at standard conditions.

However some endothermic changes do occur spontaneously at standard conditions. Two examples are;

Ice melting H2O(s) → H2O(l) ∆H = +6.01 kJ.mol-1

Water evaporating H2O(l) → H2O(g) ∆H = +44.1 kJ.mol-1

Note: these are not chemical reactions.

There are also reactions which are not spontaneous at room temperature but become spontaneous at higher temperatures. An example is the decomposition of limestone;

CaCO3(s) → CaO(s) + CO2(g) ∆H = +178.0 kJ.mol-1

which becomes spontaneous at temperatures above 1100K.

From this evidence, there is more to the question of spontaneity than simply a consideration of the enthalpy change and it would seem that temperature is an important factor. The other important factor is entropy (S).

Unit 1

Entropy

‘Everything in the Universe tends towards chaos’

In nature there is a tendency towards disorder. The entropy of a system is the degree of disorder of the system. The greater the disorder, the greater the entropy. Low entropy is associated with strongly ordered substances;

High entropy Low entropy

The second law of thermodynamics is the law of increasing disorder. In any spontaneous process, the overall degree of disorder (entropy) must always increase. A spontaneous process is one which is thermodynamically possible and this means that it can take place on itsown irrespective of how long this requires.

Entropy increases as temperature increases and changes in state, such as melting and evaporation, involve large changes in entropy.

Note : On graph a state change = change in gradient

Unit 1

When a substance changes from a solid to a liquid and then to a gas there is an increase in the amount of disorder with each change of state.

In the solid the atoms are held in a fixed position with restricted freedom of movement. In the liquid state the particles are free to move in any direction and in the gaseous state the particles are freely moving in any direction and will fill all the space available.

It is important to realise that the second law deals with an overall increase of disorder which means that it is essential to consider both the system (the reaction vessel) and the surroundings. A reaction may have an unfavourable local entropy change (system) but may have a sufficiently negative enthalpy change (exothermic) to cause an increase in the entropy of the surroundings. For example, when magnesium burns in oxygen to form magnesium oxide, there is a decrease in the entropy of the reactants in forming the product as a gaseous reactant is forming a solid product. However the heat energy released by this reaction produces an increase in entropy which exceeds the local decrease by increasing the disorder of the surrounding air molecules.

Entropy change

Process System Surroundings OverallMixing / expansion + 0 +

Crystallisation - +

Polymerisation - +

Endothermic dissolving - +

Combustion of fuels +

Heat energy released by a reaction system (exothermic) into the surroundings increases the entropy of the surroundings, while heat absorbed by a reaction system (endothermic) from the surroundings decreases the entropy of the surroundings. The change in entropy of the surroundings which occurs as a result of such a reaction can be calculated from the temperature and the entropy change during the reaction.

The standard entropy of a substance (So) is the entropy for that substance in its standard state. These values can be used to calculate the standard enthalpy change for a reaction;

∆So = ∑ So (products) - ∑ So (reactants)

The unit of entropy change is Joules per degree Kelvin per mole (J.K-1.mol-1). Actual values of standard entropies for selected substances are given in page 17 of the data booklet.

Unit 1

EXAMPLECalculate the entropy change for the reaction below, given So (CaCO3) = 93J.K-1.mol-1;

CaCO3(g) CaO(s) + CO2(g)

Now Try Exercise 1.21

The third law of thermodynamics is that the entropy of a perfect (100% ordered) crystal at absolute zero (zero Kelvin) is zero. i.e. at absolute zero all motion of any kind in the crystal is stopped.

The total entropy change, including the surroundings, between reactants and products is zero for a system at equilibrium as the amount of order/disorder is unchanging once equilibrium has been established.

Gibbs free energyTo be able to determine if a reaction is thermodynamically feasible we need to use both the enthalpy (∆H) and the entropy (∆S) changes for the reaction.

If we are going to combine these two factors, as a way of determining if a reaction is feasible, we need to deal with their different units. We need to;

multiply ∆S by T (absolute temperature in K)and

divide ∆S by 103 (to convert joules to kilojoules)

Combining enthalpy and entropy gives;

∆H – T∆S

For a reaction to be favourable this will be negative, the units are kJ.mol-1.

The combination of enthalpy and entropy changes is known as the free energy change and is given the symbol ∆G;

∆G = ∆H – T∆S

Remember;

∆H – ve reaction is exothermic∆H + ve reaction is endothermic

Unit 1

For a reaction to proceed we need;

∆G ∆H T∆S Reaction- - + feasible

- + T∆S > ∆H feasible

+ + - not feasible

+ - T∆S < ∆H not feasible

Although a negative ∆G or ∆Go (see below) value allows us to predict that the reaction is feasible, i.e. the reaction will go in the direction of the products, it does not imply anything about the rate of the reaction. The activation energy for the reaction has to be overcome for the reaction to occur and this could be very high and so the reaction could be very slow.

For endothermic reactions (∆H +ve) the temperature at which the reaction becomes feasible is given by;

The sign of the free energy change obtained can be used to predict the possibility of the reaction proceeding. This can be summarised saying that a negative value for the change in free energy shows that the reaction is likely to happen spontaneously. On the other hand, a positive value for ∆G shows that the reaction is unlikely to happen unless external energy is available to do work on the system (e.g. the reaction is heated).

If the values for the enthalpy and entropy changes are known for a reaction, it is possible to calculate the value for the free energy change at any temperature. If the measurements are all made under standard conditions then the standard free energy change (∆Go) is given by;

∆Go = ∆Ho – T∆So

WORKED EXAMPLE – ∆G o calculations Calculate the value for ∆Go and thus predict whether or not the combustion of graphite is feasible. Enthalpy and entropy values can be found in the data booklet.

C(s) + O2(g) → CO2(g)

∆So = ∑ So (products) - ∑ So (reactants)

= 214 – (5.7 + 205)

= 3.3 J.K-1.mol-1

∆Go = ∆Ho - T∆So

= - 394 – (298 x (3.3 / 103) )

= - 394.9834

= - 394.98 kJ.mol-1 ∆Go is negative therefore the reaction is feasible.

Unit 1

In the same way that standard enthalpies of formation can be used to calculate enthalpy changes for a reaction (see page 48), the standard free energy change of a reaction can be calculated from the standard free energies of formation of reactants and products;

∆Go = ∑ Gfo (products) - ∑ Gf

o (reactants)

WORKED EXAMPLE – ∆G o calculations Calculate a value for ∆Go for the equilibrium reaction below given that the standard free energies of NO2 and N2O4 are +52 kJ.mol-1 and +98 kJ.mol-1 respectively.

2NO2 N2O4

∆Go = ∑ Gfo (products) - ∑ Gf

o (reactants)

= +98 – 2(+52)

= - 6 kJ.mol-1 ∆Go is negative therefore the reaction is feasible.

When ∆Go is negative the reaction is feasible and, at equilibrium, there will be more products than reactants.

When ∆Go is positive the reaction is not feasible and, at equilibrium, there will be more reactants than products.

It follows then, that when ∆Go = 0 the products and reactants will be present in equal concentrations (note: this very rarely occurs at equilibrium).

Unit 1

It is important to appreciate the difference between ∆Go and ∆G, the former applies only to the initial and final states of a reaction i.e. either pure reactants or pure products under standard conditions. As soon as the reaction starts ∆Go no longer applies and we must use ∆G.

As soon as the reaction starts we use ∆G because;

standard conditions no longer apply

there is an increased entropy due to the mixing of products and reactants

Equilibrium is reached when G(reactants) = G(products), i.e. ∆G = 0 (the minimum in the free energy curve).


Unit 1

KINETICSChemical kinetics is the study of the speed of chemical reactions. We already know that the rate of a chemical reaction depends on factors such as temperature, concentration, particle size and whether a catalyst is present or not. We also know that all reactions have an activation energy and that an activated complex is formed.

During the course of any chemical reaction, the concentrations of reactants decreases and the concentration of products increases. The rate of reaction can therefore be defined as the rate of reactant consumption or product formation.

For example, in the decomposition of methanoic acid;

HCOOH CO + H2O

the reaction rate can be determined either by the decrease in concentration of methanoic acid or the increase in concentration of carbon monoxide, with time (see diagrams next page).

From Higher (Unit 1, part 1) we know that;

Average Rate = change

time for the change

It is clear from the above graphs, as is the case in most reactions, the rate decreases with time.

Reaction Rates and ConcentrationIf we consider a very simple (single step) reaction;

A + B products

The rate of the forward reaction at any time depends on the concentrations of A and B and can be expressed as;

rate = k.[A].[B]

where k is the rate constant.In more general terms, for any simple reaction;

nA + mB products

giving the rate law equation;

rate = k.[A]n.[B]m

Unit 1

The quantities n and m are the orders of the reaction with respect to A and B respectively. The overall order of reaction is the sum of the powers of concentration terms which occur in the rate equation, in this example the overall order is n + m.

The units of k depend on the overall order of the reaction and can be worked out from the units of the other quantities in the rate equation (remember, here, rate = mol.l-1.s-1 always).

In simple reactions the rate law (or rate equation) can take one of the forms in the following table;

Rate Equation Order of reaction Units of k

rate k [A]0 0 mol.l-1.s-1

rate k [A]1 1 s-1

rate k [A]2 2 mol-1.l.s-1

rate k [A]1[B]1 2 mol-1.l.s-1

rate k [A]1[B]2 3 mol-2.l2.s-1

These values for the order of reaction refer to the number of particles involved in that singlestep which controls the overall reaction rate, the rate determining (or slowest) step.

The values of order of reaction do not necessarily refer to the number of particles in the stoichiometric equation.

Consider the reaction;

H2O2 + 2HI 2H2O + I2

The experimentally determined rate law is;

Rate = k [H2O2].[HI]

The overall order for the reaction is 2, whereas the stoichiometric equation has 3 reactant molecules. The order of reaction is entirely an experimental quantity that is determined solely by finding the rate equation which best fits the experimental data. It cannot be deduced from the balanced chemical equation.

Determining the Order of ReactionThe order of a reaction with respect to (w.r.t.) any one reactant is the power to which theconcentration of that reactant is raised in the rate equation. The overall order of a reaction is thesum of the powers to which the concentrations of the reactants are raised in the rate equation.

The rate constant of a reaction can be determined from a series of experiments where the initial concentrations of reactants are changed. The initial rate of each reaction is calculated.

Consider the following data for an imaginary reaction;

A + B + C D

Unit 1

Doubling [A] → doubles the rate (compare 1 & 2) therefore the reaction is 1st order w.r.t. A

Doubling [B] → no effect on the rate (compare 1 & 3) therefore the reaction is zero order w.r.t. B

Doubling [C] → quadruples the rate (compare 1 & 4) therefore the reaction is 2nd order w.r.t. C

So the overall order of reaction is : 1 + 0 + 2 = 3 (3rd order) and the rate equation is;

Rate = k [A].[C]2

Expt No. [A] [B]

(in mol.l-1)

[C] Initial Rate of D

Formed (mol.l-1.s-1)

1 1.0 1.0 1.0 20

2 2.0 1.0 1.0 40

3 1.0 2.0 1.0 20

4 1.0 1.0 2.0 80

Unit 1

The rate constant, k, for this reaction is obtained by substituting values from experiment 1 in the rate equation;

rate = k [A] [C]2

20 = k [1.0] [1.0]

k = 20 mol-2.l2.s-1

EXAMPLECalculate the rate constant for the following reaction;

2NO + O2 2NO2

Expt No. [NO] [O2] [NO2] formed

(in mol.l-1s-1)(in mol.l-1)

1 2.0 x 10-5 4.0 x 10-5 1.4 x 10-10

2 2.0 x 10-5 8.0 x 10-5 2.8 x 10-10

3 4.0 x 10-5 4.0 x 10-5 5.6 x 10-10

This data is consistent with the rate equation;

rate = k [NO]2 [O2] (reaction order = 3)

Substitute the values given in experiment 1 into this rate equation to determine the value of the rate constant (k), include the correct units in your answer. Check your answer by calculating in terms of the experimental values from the other two reactions (the value of k should be the same for all).

Unit 1

Kinetics and Reaction MechanismsAt Higher grade, simple collision theory was used to explain how reaction rates were affected by changes in concentration, temperature and particle size (surface area of solid reactants).

Consider the reaction between H2 and I2;

H2 + I2 2HI

Reaction can occur when a molecule of H2 collides with a molecule of I2 with sufficient energy and the correct orientation (collision geometry). A transition state forms, called an activated complex, which breaks down to give the products, or, if there is insufficient energy or the wrong collision geometry, the molecules simply bounce apart.

This is a simple, one step reaction. If the concentration of either reactant is doubled, the number of collisions will be doubled. In other words, the reaction will be first order with respect to both H2 and I2, i.e.;

Rate = k.[H2].[I2]

Experiments confirm this rate equation.

(Since we never know if a reaction is a simple one-step process we cannot use stoichiometry to work out the rate equation)

Most reactions are not so simple, many reactions proceed in a number of sequential steps.

Chemical kinetics can be thought of as an example of the ‘bottle-neck principle’ frequently observed in everyday life. Consider a production line in the bottling plant of a distillery.

Unit 1

There are 3 parts to the production line;

1. Filler 2 bottles filled per minute

2. Capper 120 bottles capped per minute

3. Labeller 60 bottles labelled per minute

30 seconds after the production line is switched on a bottle is passed to the capper, gets capped and then is passed to the labeller. It does not matter how fast the capper or labeller are the whole process can only go at the speed of the filler. The bottle filling is the rate determining step.

In any chemical reaction mechanism, one step will be significantly slower than the others and this step will determine the overall reaction rate, i.e. It will be the rate determining step. The kinetic information which we determine experimentally refers to this rate determining step. By reversing the argument, experimentally determined rate equations and orders of reaction can give us information about the way in which the reaction occurs.

Consider the following reactions;

1. H2O2 + 2HI 2H2O + I2

Experimentally the rate equation is;

Rate = k.[H2O2].[HI]

This tells us that the rate must be controlled by a step in which one molecule of hydrogen peroxide reacts with one molecule of hydrogen iodide. We can suggest that;

H2O2 + HI X (slow rate determining step)

X + HI products (fast step)

Where X is an intermediate formed in the reaction.

Note : kinetics alone can give no direct information about the nature of X or about the total number of steps. The generally accepted mechanism for this reaction is;

H2O2 + HI H2O + HOI (slow rate determining step)

HOI + HI H2O + I2 (fast step)

Where HOI is an intermediate formed with a transient lifetime.

Unit 1

2. Hydrolysis of halogenoalkanes

Example 1 : a primary halogenoalkane (See Unit 3, page 46)

CH3CH2Br + OH- CH3CH2OH + Br-

The experimentally determined rate equation is;

Rate = k.[CH3CH2Br].[OH-] (2nd order)

From the rate equation we can deduce that both of the reactants are involved in the rate determining step. The diagram below shows this ‘one on, one off’ mechanism, an Sn

2

mechanism (nucleophilic substitution, bi-molecular).

The hydroxide ion displaces the bromide ion by attack at the ‘back’ of the molecule.

Unit 1

Example 2 : a tertiary halogenoalkane

(CH3)3CBr + OH- (CH3)3COH + Br-

The observed rate law depends only on the concentration of the halogenoalkane;

Rate = k.[(CH3)3CBr] (1st order)

The rate determining step does not involve the hydroxide ion. This ‘one off, one on’ mechanism, an Sn

1 mechanism (nucleophilic substitution, uni-molecular) can be written as;

(CH3)3CBr X+ + Br- (slow rate determining step)

X+ + OH- (CH3)3COH (fast step)

The reaction is a two step process involving some intermediate X+, which by other means we can determine to be the carbocation (CH3)3C+;

Note : kinetics alone cannot establish an exact mechanism of a reaction. We can only suggest a possible mechanism which is consistent with the kinetics.


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