AS Revision PowerPoint

Atomic Structure Amount of Substance Bonding Periodicity Introduction to Organic Chemistry Alkanes

Fundamental ParticlesProtons, Neutrons and ElectronsMass Number and IsotopesElectron Arrangement

Atoms are the fundamental building blocks of matter. However, even atoms are constructed of smaller, sub-atomic particles.

The fundamental sub-atomic particles are protons, neutrons and electrons. The protons and neutrons are held together by strong nuclear binding forces in the nucleus of the atom. The electrons may be considered to be tiny particles that exist in regions of space known as orbitals around the atom.

Mass Charge

Proton 1 +1

Neutron 1 0

Electron 1/1836 -1

Isotopes are atoms of a specific element that have a definite number of neutrons and consequently a different mass. In effect all atoms are isotopes of one element or another.

Most elements have several isotopes, some of which are stable, and others that spontaneously break apart releasing radioactivity.

The percentage of an isotope occurring in a natural sample of an element is called its natural abundance. The natural abundance of each isotope on the earth are usually very similar wherever the sample is obtained.

The 'relative abundance' of an isotope means the percentage of that particular isotope that occurs in nature.

Most elements are made up of a mixture of isotopes.

The mass spectrometer is an instrument used for analysing samples of elements and compounds. It consists of six basic stages.

The sample is injected into the instrument. It may need vaporising by heating in which case the initial stage also has an oven. The sample in gaseous state then passes into an ionising beam of electrons which knock electrons off the sample creating positive ions.

These positive ions are then accelerated by some electrostatically charged plates into magnetic field, which then deflects the particles according to their mass/charge ratio.

The deflected ions then arrive at the detector. The five stages are then:

◦ Vaporisation◦ Ionisation◦ Acceleration◦ Deflection◦ Detection◦ (VIADD)

The mass spectrometer is a very sensitive instrument and can detect particles with very small differences in relative mass. Isotopes have mass numbers that differ by at least one atomic mass unit and so are easily differentiated. The intensity of the signal in a spectrum is directly proportional to the amount of that species in the sample. Hence, the spectrum tells us both which isotopes are present and their relative proportions.

This is the arrangement of electrons around the atom. They are arranged in electron shells and within each shell there are sets of sub-shells. The sub-shells themselves are made up of orbitals each of which can hold a maximum of two electrons, each with opposite spin.

The order in which the electrons fill up the shells is also called the Aufbau principle.

The first ionisation energy is defined as the energy required to remove one mole of electrons from one mole of gaseous atoms to provide one mole of gaseous single charged ions.◦ Na(g)  Na+(g) + e¯

Subsequent ionisation energies are defined in a similar way only by removing electrons from already charged ions.

The second ionisation energy◦ Na+(g)  Na2+(g) + e¯

If the energy required to achieve these ionisations is plotted on a graph against the ionisation number, the 'jumps' in the required energy clearly show the main and sub energy levels.

Relative Atomic and Molecular MassThe Mole and Avogadro’s ConstantThe Ideal Gas EquationEmpirical and Molecular FormulaBalanced Equations and Calculations

The atomic mass number is represented by the symbol (letter) 'A'.

The mass number gives the integral number of nucleons, protons and neutrons found in the nucleus of an atom.

The relative mass is a value that is not necessarily integral that compares a mass to the mass of a carbon isotope, assigned a value of exactly 12 units.

The atomic number is represented by the symbol (letter) 'Z'. It shows us the number of protons in an atom (and the number of electrons) in a neutral atom.

Avogadro's number, or constant, is the number to which the mass of an atom must be multiplied to give a mass in grams numerically equal to its relative atomic mass.

Avogadro's constant (L) = 6.02 x 1023

Carbon has a relative mass of 12 therefore 6.02 x 1023 carbon atoms have a mass of 12g

Magnesium has a relative atomic mass of 24 therefore 6.02 x 1023 magnesium atoms have a mass of 24g

This gives rise to two important definitions◦ The amount of any substance containing an Avogadro number

of particles of that substance is called a mole.◦ 1 mole of any substance has a mass equal to its relative mass

expressed in grams

Example 1 mole of Mg contains 6.02 x 1023 Mg atoms 1 mole of Mg = 24g 12g of Mg = 0.5 moles of Mg 0.5 x 6.02 x 1023 = 3.01x 1023 magnesium atoms The relationship between moles, mass and number of

particles can be expressed by simple formulae:

no. of particles = moles x Avogadro's constant

mass in grams = moles x relative mass

These formulae can be used to find any quantity when the other two quantities are known.

The equation of state refers to a fixed mass of gas. From Avogadro's law we know that the same volume of all gases contain the same number of moles and from this, it follows that the volume is proportional to the number of moles.

Volume ∝ number of moles (n) These two equations can be combined to obtain an expression involving all the

quantities:

PV = nRT where:

◦ P = pressure (Pa)◦ V = volume (m3)

◦ n = number of moles of gas◦ R = Universal Gas constant = 8.314 J K-1 mol-1◦ T = temperature (K)

It is often more convenient to express the pressure in kPa and the volume in litres (dm3).

Example: Calculate the number of moles of gas present in 2.6 dm3 at a pressure of 1.01 x 105 Pa and 300 K.

PV = nRT 2.6 dm3 = 0.0026 m3

0.0026 x 1.01 x 105 = n x 8.314 x 300 n = 0.0026 x 1.01 x 105 / 8.314 x 300 n = 0.105 moles

1 litre = 1 dm3 = 1000cm3

The empirical formula is the simplest ratio of atoms within a chemical compound.

Example: Find the empirical formula of the compound which has the following percentage by mass composition: Carbon 12.12%; Oxygen 16.16%; Chlorine 71.17%.

Steps Carbon Oxygen Chlorine

Percentage 12.12 16.16 71.17

Ar 12 16 35.5

%/Ar = X 1.01 1.01 2

X/(smallest X value)

1 1 2

Formula C O Cl₂

The molecular formula is the number and type of atoms that are present in a single molecule of a substance.

Determine the molecular mass in grams. Divide the molecular mass of the compound by the

molecular mass of the empirical formula. Round the quotient to the closest integer. Multiply the rounded number by all the subscripts, using

the product as the new subscripts. Example: A compound has an empirical formula of CH₂ and a

molecular mass of 42. Determine its molecular formula.◦ 12 + (2x1) = 14◦ 42 ÷ 14 = 3◦ CH₂ x 3 = C₃H₆

To balance a chemical equation it is important to remember that the formula of the reactants and products cannot be changed and that coefficients may only be placed before the formulae, multiplying them by whole numbers.

These equations are constructed by writing the formula of each of the compounds in the reaction, and then by counting up the number of atoms on each side to make sure they are equal. If they are not equal, balancing numbers (coefficients) are added in front of each chemical formula (where needed), so that the numbers of each type of particle on each side of the equation are the same.

Step 1 - write the chemical equation◦ ammonia + oxygen nitrogen monoxide + water

Step 2 - write the formula of each of the reaction components◦ NH3 + O2  NO + H2O

Step 3 - add coefficients in front of the formulae to balance the equation◦ 4NH3 + 5O2  4NO + 6H2O

Avogadro suggested that "equal volumes of all gases contain equal number of molecules (the gases being measured at the same temperature and pressure).

This means that in the reaction: N2 + 3H2  2NH3 For a given volume of nitrogen, three times the volume of

hydrogen is needed for complete reaction. Example: Find the volume of hydrogen required to react

completely with 200 cm3 of nitrogen according to the equation: N2 + 3H2  2NH3 The equation tells us that 1 volume of nitrogen reacts

completely with 3 volumes of hydrogen: Therefore volume of hydrogen = 3 x volume of nitrogen Therefore volume of hydrogen = 3 x 200 cm3 = 600 cm3

The concentration of a solution is the quantity of solute that it contains per unit volume.

This may be given in grams per 100cm3 or grams per litre, but it is usually given in terms of molarity as this gives a direct measure of the number of solute particles contained by the solution.

Molarity = number of moles of solute per litre of solution Molarity = moles/litres The molarity is denoted by the capital letter M, or the units mol

dm¯³ Example: Calculate the molarity of a solution containing 0.15

moles of potassium nitrate in 100cm3 of solution.

Molarity = moles/litres 100cm3 = 0.1dm³ Molarity = 0.15/0.1 = 1.5 M

A titration of a 25ml sample of a hydrochloric acid solution of unknown molarity reaches the equivalence point when 38.28ml of 0.4370M NaOH solution has been added.

What is the molarity of the HCl solution?

HCl(aq) + NaOH(aq) NaCl(aq) + H₂O(l)

Use the volume and molarity of the calculate the molarity of the NaOH to calculate the number of HCl solution? moles of NaOH that reacted. Use the mole ratio between base and acid to determine the moles of HCl that reacted. Use the volume of the acid to calculate molarity.

=0.6691M

This term refers to the efficiency of a chemical process in terms of the atoms that are lost as by-products to the intended product.

The yield of a reaction is the actual mass of product obtained. The percentage yield can be calculated:

Ionic, Covalent and Metallic BondingBond PolarityForces Acting Between MoleculesStates of MatterShapes of Simple Molecules and Ions

Ions are charged atoms or groups of atoms. Atoms form ions to gain a noble gas configuration and become more stable. Atoms form ions by losing or gaining electrons.

Positive ions are formed when atoms lose electrons. Negative ions are formed when atoms gain electrons

Ionic bonds never exist in isolation, instead they form part of a giant ionic lattice in which each positive ion attracted to negative ions which surround the positive ion in a regular arrangement. The positive ions are electrostatically attracted to the negative ions.

They have high melting and boiling points; the attraction between the positive and negative ions is very strong and thus a great deal of energy is needed to break the bonds.

They are solid at room temperature (due to high melting and boiling points) They are brittle; when a force is applied it knocks the regular pattern and

positive ions end up next to positive which cause repulsion. Soluble in water Conduct electricity when molten or in aqueous solution, but not when solid. The

ions are fixed in position in the solid, but in the liquid state, or in solution, they are free to move and carry the current.

In a covalent bond, two atoms share a pair of electrons. The electron pair creates a ‘bond’ between the two atoms because it attracts the nucleus of each atom and therefore resists the separation of the two atoms.

Simple molecular: low melting and boiling points, do not conduct electricity

Giant molecular: high melting and boiling points, do not conduct electricity (except carbon)

A co-ordinate (dative) bond is formed when both electrons in the shared pair are from one atom (from the lone pair of electrons on the donor atom)

Metallic bonds in solids do not exist in isolation. They make up a giant metallic lattice which comprises of closely packed metal ions surrounded by delocalised electrons which are free to move through the lattice.

They have high melting and boiling points (as a large amount of energy is needed to remove a metal atom from the attraction of the delocalised electrons)

They can conduct electricity (the sea of delocalised electrons can carry an electrical charge)

They are malleable and ductile (because the forces in a metallic lattice are strong, but not directed at one particular atom, the layers of atoms can slide over each other – known as slip)

Metal lustre (the delocalised electrons also cause the meta to reflect light)

Electronegativity is the power of an atom to withdraw electron density from a covalent bond.

It can be calculated using at various methods and is usually given a value between 0.7 and 0.4

Small atoms with a large numbers of protons in the nucleus attract electron density more strongly. Thus electronegativity increases from left to right across a period of the Periodic Table and from the bottom to the top of a group.

When a covalent bond exists between atoms of different electronegativity, the shared pair of electrons is displaced towards the more electronegative atom.

The displacement of electron density makes the less electronegative atom slightly electron deficient (δ⁺) and the more electronegative atom has a slight excess of electron density (δ⁻)

This charge separation creates an electric ‘dipole’ and the molecule is said to be polar.

The polarity of a bond can be measured in a unit called the debye. Its size depends on the difference in electronegativity between the elements involved.

Covalent bonds are attracted to each other by intermolecular forces. They are much weaker than intramolecular forces.

The three types of intermolecular force are van der Waal’s (temporary induced dipole-dipole), permanent dipole-dipole and hydrogen bonding.

Permanent dipole-dipole interactions occur between molecules that have permanent dipoles.

The permanent dipole consists of regions of partial positive charge and regions of partial negative charge within the same molecule.

By convention we use the small Greek letter delta, ∂, to represent a partial charge. Hence, ∂⁺ means a partial positive charge and ∂⁻ means a partial negative charge.

Van der Waals' forces of attraction are temporary, induced dipole-dipole attractions

At any moment in time, the electron distribution in a non-polar covalent molecule may be unequal due to the fluctuating movement of electrons. This leads to a temporary dipole which induces an opposite dipole on an adjacent. The second molecule is thus attracted to the first one

The size or magnitude of van der Waal’s forces increases with the size of molecules and also depends upon their shape. Branched chain hydrocarbons have weaker intermolecular forces than straight chain molecules because they are less polarisable.

Hydrogen bonding is the strongest type of intermolecular forces in neutral molecules. It is a special case of dipole-dipole force which exists between a lone pair of electrons on a N, O or F atom and a hydrogen atom that has a strong partial charge (∂⁺) because it is attached to a highly electronegative atom e.g. N, O or F

The electronegative atom pulls electrons away from hydrogen atom so that, on the opposite side to the bond, the hydrogen atom appears almost like an unshielded proton

State Particles/Motion Forces

SolidFixed in position

Vibrate

Ionic attractions; covalent bonds; metallic bonds;

permanent dipole-dipole forces; all van der Waal’s

forces

LiquidResembles a disorded solid

Some movementSeparation is 10% more than solids

Similar to a solid

Gas Rapid, random motion

Intermolecular forces can be ignored unless the

particles are close together at high pressure

or temperature

SOLID LIQUID GASMELTING

SOLIDIFYING

EVAPORATING

CONDENSING

SUBLIMATION

SOLID LIQUID Heat energy is required to reduce forces which hold the particles together. This heat energy is called the enthalpy of fusion.

LIQUID GAS More energy is required than to change from solid to liquid. The energy is used to overcome the forces which hold the particles together so that the particles can be completely separated. This heat energy is called the enthalpy of vaporisation.

A crystal is a solid with a regular shape, containing particles arranged in a regular structure. Crystals can be classified according to the type of bonding between the particles.

Crystal TypeMelting and

Boiling Points

Electrical ConductivitySolubility in WaterSolid Molten

Ionic High None GoodVariable but often good

Macromolecular Very HighNone – except

graphiteNone Insoluble

Molecular Low None None Variable

Metallic Usually High Good Good Insoluble

1. The number of outer shell electrons originally in the centre atom

2. The number of additional shared electrons in covalent or dative bonds

3. The loss or gain of additional electrons if the species is a positive or negative ion

The final shape of a molecule is also altered if some of the electron pairs are lone (non-bonding) pairs

Lone pairs are more compact than bonding pairs so they repel more strongly, leading to bond angles between bonding pairs which are smaller than those found in totally symmetrical shapes

Lone pair-lone pair repulsions are stronger than lone pair-bonding pair repulsions. These principles can be used to predict different shapes.

Lone Pair – Lone PairLone Pair – Bonding Pair

Bonding Pair – Bonding PairIncreasin

g Repulsio

n

Classification of Elements in the s, p and d blocksProperties of the Elements ofPeriod 3 and their PeriodicTrends

In these blocks, the elements have their highest energy electrons in s, p, d or f electronic sub-levels.

The atomic radii decrease because, across the period, the nuclear charge increases and the outer electrons are attracted more strongly. Thus, they are drawn closer to the nucleus, without any additional shielding of the nuclear charge by the added electrons.

The atomic radius of these elements is measured in suitable compounds of the elements is measured in suitable compounds of the element. Argon forms no compounds and thus is not possible to measure its atomic radius in a comparable way to the other elements.

A general increase because nuclear charge increases Mg-Al: removing electrons from 3s-orbital to 3p-orbital P-S: pairs in orbitals in sub-levels

The electronegativity of an element is the power of an atom to attract electrons in a covalent bond. Electronegativity increases across a period because the number of protons increases and thus the attraction between the nucleus and a pair of electrons in a covalent bond increases.

The elements Na-Al are good electrical conductors because they have metallic bonding, with delocalised electrons which are free to move through the whole structure. The elements Si-Cl have covalent bonding with electrons which are not free to move from one atom to another. Their conductivities are almost zero.

The variation in the melting and boiling points is related to the bond strengths and structures of the elements.

Argon has very low melting and boiling points because it exists as individual atoms. The atoms are not very polarisable and thus van der Waal’s forces are very weak.

Na-Al are metallic elements. The melting and boiling points increase because from Na-Al the atoms are smaller and have an increasing nuclear charge. Thus, the strength of the metal-metal bonds increase.

Si has high melting and boiling points because it is macromolecular with a diamond structure and strong covalent bonds link all the atoms in three dimensions. A large amount of energy is needed to break these bonds.

P, S and Cl are all molecular structures. The melting points are determined by the strength of the van der Waal’s forces between molecules which, in turn, are determined by the size of the molecules (P₄ S₈ Cl₂). Each of these elements has a low melting and boiling point because the van der Waal’s forces are weak and easily broken.

NomenclatureIsomerism

When naming an alkene derivative, look for the longest carbon chain in the skeleton. This determines the stem of the name.

Many compounds have a branched carbon skeleton, in which case the name of the side chain also depends on the number of carbon atoms in them e.g. a single carbon branch is CH₃

The position of the branch also has to be made clear which is done by numbering the carbon atoms in the skeleton. The numbers used are kept as low as possible.

1. Use the name to determine the number of carbon atoms in the longest chain

2. Draw this carbon skeleton and number the carbon atoms

3. Add any functional groups in the correct positions

4. Add hydrogen atoms until each carbon atom has 4 bonds

Isomerism occurs where molecules with the same molecular formula have their atoms arranged in different ways. Isomerism is divided into main types:◦ Structural Isomerism◦ Stereoisomerism

This occurs when there are two or more ways of arranging the carbon skeleton of a molecule.

e.g. C₄H₁₀ can be butane or 2-methylpropane

e.g. C₅H₁₂ can be pentane, 2-methylbutane or 2,2-dimethylpropane

These isomers have similar chemical properties but slightly different physical properties.

These isomers have the same carbon skeleton and the same functional groups but the functional group is joined in a different place.

e.g. 1-bromopropane and 2-bromopropane e.g. but-1-ene and but-2-ene These isomers have similar chemistry

because they have the same functional group but the different positions can cause different properties

These isomers have different functional groups so different chemical and physical properties.

e.g. aldehydes and ketones: C₃H₆O can be propanal or propanone

e.g. carboxylic acids and esters: C₃H₆O₂ can be propanoic acid, methyl ethanoate or ethyl methanoate

e.g. alcohols and ethers: C₃H₆O can be ethanol or methoxymethane

Stereoisomerism have the same structural formula but the bonds are arranged differently in space.

Due to the π electron clouds present above and below the plane of the bond, carbon-carbon double bonds cannot rotate. If an alkene has two different groups on each end of the double bond (E/Z) isomerism occurs.

If the groups (often identical) are on the same side, then they show Z-isomerism. If they are on different sides, then they show E-isomerism.

One way to remember this is that Enemies are on different sides, and people must be on Zee-Zame-Zide (with a really fake German accent)

Z-isomers usually have higher boiling points, as they will have some polarity, whereas E-isomers are less polar. For example, Z-but-2-ene has a boiling point of 4˚C and E-but-2-ene is 1˚C.

E-isomers however have higher melting points because they pack together more closely. For example, E-but-2-ene’s melting point is -106˚C while Z-but-2-ene’s is -139˚C

NB: molecules with two identical groups on the same carbon atom of the double bond do not show geometric isomerism.

Fractional Distillation of Crude OilCracking of AlkanesCombustion of Alkanes

Crude oil is a mixture of hydrocarbons, mainly alkanes, formed by the decay of remains of sea creatures and plants over millions of years.

Crude oil is a complex mixture of hydrocarbons which has no use in its raw form. To provide useful products, the components must be partly separated and, if necessary, modified.

The complex mixture of hydrocarbons in crude oil is separated into less complicated mixtures or fractions by fractional distillation (primary distillation). The crude oil is heated and the vapour/liquid mixture is passed into a tower.

The top of the tower is cooler than the bottom i.e. there is a mixture into fractions depending on those boiling points of the hydrocarbons present. Only those with low boiling points (more volatile) reach the top; others condense in trays at different levels up the tower, and are drawn off.

The residue from the primary distillation still contains useful materials such as lubricating oil and waxes which boil above 350˚C at atmospheric pressure. At such high temperatures some of the components of the residue decompose. To avoid this, the residue is further distilled under reduced pressure (vacuum distillation). The reduced pressure lowers the boiling points, so they can be distilled without decomposing.

Thus, fractional distillation is a physical process

Name of Fraction

Boiling Range (˚C)

UsesLength of

Carbon Chain (approx)

LPG <25 Camping gas 1-4

Petrol 40-100 Fuel for cars 4-12

Naphtha 100-150 Petro chemicals 7-14

Paraffin 150-250 Aviation fuels 11-15

Diesel 220-350Central heating

fuel15-19

Lubricating Oil 350< Lubricating oil 20-30

Fuel Oil 400< Fuel for ships 30-40

Wax, Grease 400< Candles 40-50

Bitumen 400< Roofing 50<

At high temperatures, the atoms in an alkane molecule vibrate rapidly. If this temperature is high enough, the vibration becomes sufficiently vigorous for chemical bonds to break. This breakage of the C-C bonds in alkanes leads to the formation of smaller hydrocarbon fragments and is called cracking. The temperature required for cracking can be reduced by using a solid catalyst.

Thermal cracking results in the production of a high proportion of alkenes. The energy required for bond breaking is provided by heat, with temperatures used in the range 400-900˚C and pressure of up to 700kPa

At the lower end of the temperature range, carbon chains break preferentially part way along the carbon chain of the molecule, leading to a greater proportion of low Mr alkenes

The process is initiated by homolytic fission of C-C bonds forming two alkyl radicals. Each radical can then abstract a hydrogen atom from an alkane molecule to produce a different alkyl radical and a shorter alkane.

Alternatively, the alkyl radical can undergo further bond breaking to give an alkene and a shorter alkyl radical.

Dehydrogenation, isomerism and cyclisation can also occur. The mixtures of products are separated by fractional distillation.

Catalytic cracking uses zeolite catalysts, a slight excess of pressure and a temperature of about 450˚C. Large alkanes are converted mainly into branched alkanes, cycloalkanes and aromatic hydrocarbons.

The proportion of alkenes is small so catalytic cracking is primarily used to produce motor fuels. Branched chain alkanes burn more smoothly than unbranched chains. In an engine, due to the pressures involved, the fuel-air mixture may ignite before the ‘spark’ is produced, causing ‘knocking’. Using branched-chain alkanes prevents this problem.

The mechanism for catalytic cracking involved the formation of carbocations wit the catalyst acting as a Lewis acid.

Benzene (and its methyl and dimethyl derivatives) arise through cyclisation and dehydrogenation.

Thermal Catalytic

ConditionsHigh temperature: 400-900˚C

High pressure: 700kPaShort residence time

Lower temperature: 450˚CZeolite catalyst

MechanismHomolytic fission forming free

radicalsHeterolytic fission forming

carbocations

ProductsLow Mr alkenesShorter alkanesSome hydrogen

Branched chain alkanesCycloalkanes

Aromatics

Product UsesChemical industry (including

polymers)Motor fuel

Other ProductsBranched chain alkanes

CycloalkanesAromatics

Low Mr alkenesShorter alkanesSome hydrogen

Complete Combustion In the presence of a plentiful supply of oxygen,

complete combustion of alkanes occurs exothermically to produce carbon dioxide and water.

As the number of carbon atoms increases, more oxygen is required per mole of hydrocarbon for complete combustion, and more energy is released. Sulphur-containing impurities in the hydrocarbons produce oxides of sulphur e.g. sulphur di- and trioxide which are toxic and dissolve in water to form acid rain.

Incomplete Combustion If insufficient oxygen is available,

incomplete combustion occurs and water forms with either carbon monoxide or carbon.

Carbon monoxide is also formed by the incomplete combustion of petrol vapour (C₈H₁₈) in a car engine.

Such engines produce other pollutants, notably oxides of nitrogen, unburned hydrocarbons and, when leaded petrol is used, compounds of lead. Cars with diesel engine produce carbon, not carbon monoxide.

Oxides of nitrogen are also formed when the air/petrol mixture is sparked and explodes. The temperature can reach values of 2500˚C and this provides sufficient activation energy for nitrogen (in air) to react with oxygen (in air), forming nitrogen monoxide.

On cooling, nitrogen monoxide easily reacts with more oxygen to form nitrogen dioxide.

Nitrogen dioxide reacts with water and more oxygen to produce nitric acid, which can lead to acid rain

Nitrogen dioxide also reacts with oxygen or hydrocarbons, in the presence of sunlight, to form an irritating photochemical smog.

Catalytic converters help to remove harmful gases from car exhausts. The converter contains a honeycomb of ceramic material on which metals e.g. platinum, palladium and rhodium are spread in a thin layer. The metals catalyse reactions between the pollutants, and help to remove up to 90% of the harmful gases.

2CO + 2NO 2CO₂ + N₂ C₈H₁₈ + 25NO 8CO₂ + 12.5N₂ + 9H₂O Thus the pollutant gases (CO,NO) and

hydrocarbons are replaced by the harmless products (O₂ , N₂ , H₂O)

Energetics Kinetics Equilibria Redox Reactions Group 7: the Halogens Group 2: the Alkaline Earth Metals Extraction of Metals Haloalkanes Alkenes Alcohols Analytical Techniques

Enthalpy ChangeCalorimetryHess’s LawBond Enthalpies

If a reaction produces heat (i.e. increases the temperature of the surroundings) it is exothermic. If the temperature of the reaction mixture decreases (i.e. heat is absorbed) then the reaction is endothermic.

Exothermic:  a reaction which produces heat (ΔH has a negative value by convention, -ve)

Endothermic:  a reaction which absorbs heat (ΔH has a positive value by convention, +ve)

Enthalpy of reaction: The change in internal (chemical) energy (H) in a reaction = ΔH

Standard state: Pressure = 100 kPa, Temperature = 298 K

The standard state of an element or compound is the form in which it exists under standard conditions (not to be confused with STP)

 Standard enthalpies of combustion (ΔHc ) The standard enthalpy of combustion is the energy released when 1

mole of a substance is burnt in excess air or oxygen, all quantities being measured under standard conditions.

 Standard enthalpies of formation (ΔHf ) The enthalpy change when 1 mole of a substance is made from its

elements in their standard states. There MUST be 1 mole of the substance formed AND all of the elements forming it MUST be in their standard states.

Change in energy = mass x specific heat capacity x change in temperature  

q = m x c x ΔT Experimental A known mass of solution should be placed in a container, as insulated

as possible, to prevent as much heat as possible from escaping. The temperature is measured continuously, the value used in the equation is the maximum change in temperature from the initial reading.

The result will be a change in temperature. This can be converted into a change in heat (or energy) by using the above equation q = m x c x ΔT.

Δ-H may then be calculated for the amount of reactants present, and then this can be used to calculate for a given number of mols.

Hess' Law states that the total enthalpy change between given reactants and products is that same regardless of any intermediate steps (or the reaction pathway).

This basically means that energy changes can be calculated from other energy changes which may be found experimentally. It is the fundamental basis of nearly all the thermodynamic data that we have available today.

It is not possible in the majority of cases to calculate the formation enthalpy of a substance, whereas it is possible to find the combustion enthalpy by experimentation. In many cases historically it was possible to calculate the enthalpy of formation from the combustion enthalpies using Hess' law.

Bond enthalpy (mean bond energy) is the average value of a particular type of bond which has been measured over a range of molecules.

Example: CH4 has four C-H bonds, and so will have four different bond

dissociation enthalpies corresponding to the following bonds breaking:◦ CH4  CH3 + H◦ CH3  CH2 + H◦ CH2  CH + H◦ CH  C + H

The C-H bond enthalpy is the average value of the four bond dissociation enthalpies.

Bond energies (enthalpies) can be used to calculate unknown enthalpy changes in reactions where only a few bonds are being formed/broken.

Bonds broken (left hand side) - bonds formed (right hand side) = enthalpy change for the reaction.

Collision TheoryMaxwell–Boltzmann DistributionEffect of Temperature on Reaction RateEffect of ConcentrationCatalysts

Collision theory - reactions take place as a result of particles (atoms or molecules) colliding and then undergoing a reaction. Not all collisions cause reaction, however, even in a system where the reaction is spontaneous.

The particles must have sufficient kinetic energy, and the correct orientation with respect to each other for them to react. This energy is needed to cause bonds to break, which is essential if reaction is to take place.

Activation Energy: This is the minimum energy that particles colliding must have in order to produce successful reaction. It is given the symbol Ea (Energy of Activation). 

The Maxwell Boltzmann distribution is a statistically derived function showing how the energy of particles is distributed over all of the particles with some particles having no energy to particles with a great deal of energy. The majority of particles have energies intermediate between these two situations.

Increasing the temperature of a substance increases the average speed (energy) of the particles and consequently the number of particles colliding with sufficient energy (Ea) to react. At higher temperatures there are more successful collisions and therefore a faster reaction.

Increasing temperature has two effects on a reaction:◦ There are more collisions because the particles are moving faster on average◦ The collisions have more energy

Increasing the concentration of a reactant increases the number of particles in a given volume and increases the chance of successful collisions and in turn increases the rate of reaction.

A catalyst is a substance which alters the rate of reaction without being chemically changed itself

The catalyst provides an alternative reaction route of lower activation energy than the uncatalysed one.

The catalyst has no effect on the overall enthalpy change for the reaction.

The Dynamic Nature of EquilibriaQualitative Effects of Changes of Pressure, Temperature and Concentration on a System in EquilibriumImportance of Equilibria in Industrial Processes

If a body of liquid is placed in a closed container, the higher energy particles will continue to leave but now they will be held in the same place (the air space above the liquid surface). Occasionally, due to collisions, some of these vapour particles will lose energy and rejoin the body of liquid. As more particles gather in the air space there will be more possibility of particles losing energy and returning to the liquid.

We now have two processes happening. Liquid particles turning to vapour and vapour particles turning to liquid. Eventually the rate at which the particles move from the liquid to the vapour phase will equal the rate at which the particles move from the vapour phase to the liquid. When this situation is arrived at the concentration of vapour particles in the air space will be constant even though there is movement in both directions. We call this Dynamic Equilibrium

In reactions where gases are produced (or there are more mols of gas on the left), and increase in pressure will force the reaction to move to the left (in reverse). If pressure is decreased, the reaction will proceed forward to increase pressure. If there are more mols of gas on the left of the equation, this is all reversed.

Pressure change only affects gaseous equilibria and only then when there are an unequal number of moles of gas on either side of the equilibrium.

When the pressure is increased on a gaseous mixture at equilibrium the equilibrium will respond so as to release the applied pressure. In other words the equilibrium will move towards the side of the fewer number of moles.

It should be remembered that the pressure of a gas is directly proportional to the number of moles of gas and that this does not depend on the nature of the gas in question - all gases behave in the same way in terms of their physical properties.

The effect of a change of temperature on a reaction will depend on whether the reaction is exothermic or endothermic.

When the temperature increases, Le Chatelier's principle says the reaction will proceed in such a way as to counteract this change, i.e. lower the temperature. Therefore, endothermic reactions will move forward, and exothermic reactions will move backwards (thus becoming endothermic). The reverse is true for a lowering of temperature.

If a reaction is exothermic in the forward direction in an equilibrium it will be endothermic in the reverse direction and vice-versa (law of conservation of energy)

Increasing the temperature of an equilibrium mixture will always favour the endothermic direction over the exothermic process. The rate of the endothermic process will increase more than the exothermic direction of change and the equilibrium will be re-established with new concentrations based on more of the endothermic product (i.e. the product of the endothermic direction of change)

To express this in non-scientific terms, it can be considered that the endothermic direction is absorbing heat therefore giving it more heat by increasing the temperature will favour it.

The reverse argument is true for the direction of exothermic change. The exothermic reaction is giving out heat and therefore applying more will hinder this process. Raising the temperature retards the reaction in the exothermic direction.

When the concentration of a product is increased, the reaction proceeds in reverse to decrease the concentration of the products. When the concentration of a reactant is increased, the reaction proceeds forward to decrease the concentration of reactants.

Some general rules:◦ If we add to the left hand side of an equilibrium we make more of the right hand side◦ If we add to the right hand side of an equilibrium we make more of the left hand side◦ If we remove from the left hand side the equilibrium makes more of the left hand side

component◦ If we remove from the right hand side the equilibrium makes more of the left hand side

component When we study rates of reaction one of the first conclusions drawn is that the

rate of a chemical reaction depends on the concentrations of the reactants. The greater the concentration the more collisions occur and the faster the rate of the reaction.

When we add more reactant, the forward rate will now be greater than the back rate. The reaction is now not at equilibrium and the forward reaction proceeds faster than the back reaction until the equilibrium conditions are re-established.

Similarly, removal of a component from one side will reduce the rate of its reaction and case the equilibrium to make more of it.

The Haber Process

N2(g) + 3H2(g)  2NH3(g)      ΔH = -92.4 kJ mol-1

There are more moles of gas on the left than the right, so a greater yield will be produced at high pressure.

The reaction is exothermic, therefore it will give a greater yield at low temperatures.

In practice, if low temperatures are used the time taken for the reaction to attain equilibrium becomes unfeasibly long. An intermediate temperature is chosen (450ºC) which allows the reaction to get to an equilibrium in a reasonable time and still has enough of the products in the equilibrium mixture.

A catalyst of finely divided iron is also used to help speed the reaction (finely divided to maximise the surface area).

To make the process more efficient the ammonia produced at equilibrium is removed by first cooling the mixture when the ammonia turns into a liquid which can be tapped off. The unreacted gases in the process are then mixed with fresh reactants and returned to the reaction chamber to re-establish the equilibrium again and the cycle is repeated continuously.

Manufacturing of Ethanol The reaction between ethene and steam in the presence

of a phosphoric acid catalyst at 300ºC and 60 atmospheres pressure is used in industry to manufacture ethanol.

C2H4(g) + H2O(g) C2H5OH(g)

This is a reversible reaction - alcohols can be dehydrated to alkenes - so to encourage the forward reaction high pressure is used.

You can see that in the equation there are two moles of gas on the left hand side and only one mole of gas on the right hand side. According to Le Chatelier the increased pressure pushes the reaction towards the side of fewer moles of gas, in this case the products.

Manufacturing of Methanol Synthesis gas, a mixture of carbon monoxide, carbon dioxide and

hydrogen, is first produced in a reformer. This is carried out by passing a mixture of a hydrocarbon feedstock and steam through a heated tubular reformer. The ratio of hydrogen and carbon oxides in the syngas may need to be adjusted by purging excess hydrogen or adding carbon dioxide.

H2O(g) + C(g)  CO(g) + H2(g) The syngas is cooled and then compressed before being fed to the

methanol converter. The methanol synthesis takes place in the presence of copper-based catalysts at 250-260ºC. The crude methanol is recovered and purified by distillation.

2H2(g) + CO(g)  CH3OH(g) The reaction is reversible. The reaction is exothermic, so the

temperature is kept to a minimum, high enough to ensure rapid reaction but low enough to ensure a high percentage of product at equilibrium. High pressure is also used to move the equilibrium to the side of fewer gaseous moles.

Oxidation and ReductionOxidation StatesRedox Equations

Oxidation occurs when the oxidation state of an element increases (becomes more positive)

Reduction occurs when the oxidation state of an element decreases (becomes more negative)

Oxidation states help to improve the definitions of oxidation and reduction.

Rules for Assigning Oxidation States1. The oxidation state of an atom in its element is always zero2. For a simple ion, the oxidation state is the same as the charge on the

ion; Na⁺, K⁺ are +1; Mg²⁺, Ca²⁺ are +2; F⁻ and Cl⁻ are -1; O²⁻ and S²⁻ are -2

3. The oxidation state of fluorine is always -1 because it is the most electronegative element

4. The oxidation state of oxygen is -2, except for peroxides and F₂O5. The oxidation state of chlorine is usually -1, except when combined

with F or O6. The oxidation state of hydrogen is +1, except when bonded to a metal

ion, when it is -17. The sum of all the oxidation states of the atoms or ions in a compound

is always zero8. The sum of the oxidation states in a polyatomic ion is always the

charge on the ion

Many compounds have traditional names e.g. water and ammonia which tell you nothing about the elements in it

As some elements can have different oxidation states in their compounds (e.g. iron oxide can be FeO or Fe₂O₃) using the oxidation state can help with naming them.

If a metal can only have one oxidation state in its compounds, the oxidation state is not put it i.e. NaCl is sodium chloride, not sodium (I) chloride

Example: HNO₃ Using the rules, H=+1 and O=-2, making the current

oxidation state -5. As there is no overall charge on the compound, N must

be +5 and therefore this is nitric (V) acid

Oxoanions are negative ions that contain oxygen. Oxoanions containing a metal need a systematic name to show the oxidation state of the metal.

Acids of oxoanions are named after their anions e.g. phosphoric acid, H₃PO₄

To name, first calculate the oxidation state of phosphorus in PO₄³⁻◦ Find out number of oxygen atoms (4)◦ Find out total oxidation number due to oxygen (-8)◦ Find out overall charge on the ion (-3)◦ Find the oxidation state of the central atom (+5)◦ Therefore the name of the species is Phosphoric (V) acid

The equation: Mg + 2HCl MgCl₂ + H₂ is a redox reaction

Oxidation state of H Oxidation state of Mg

Combined in HCl = +1 Uncombined metal = 0

Uncombined element = 0

Combined in MgCl₂=+2

The hydrogen ion is reduced by the

magnesium metal

The magnesium metal is oxidised by the

hydrochloric acid

The equation: MnO₂ + 4HCl MnCl₂ + Cl₂ + 2H ₂ O is a redox reaction

Oxidation state of Mn Oxidation state of Cl

Combined in MnO₂ = +4 Combined in HCl = -1

Combined in MnCl₂ = +2 Uncombined element=0

The manganese is reduced by the chloride

ions in HCl

The chloride is oxidised by manganese (VI) oxide

The equation: MgO + 2HCl MgCl₂ + H ₂ O is NOT a redox reaction

Oxidation state of Mg Oxidation state of Cl

Combined in MgO = +2 Combined in HCl = -1

Combined in MgCl₂ = +2 Combined in MgCl₂ =-1

The oxidation state does not change

The oxidation state does not change

1. Identify the oxidation states of the elements (only one will be changing in a half equation)

2. Put in the correct number of electrons on the appropriate side of the equation

3. If necessary, add water to balance oxygen atoms4. If necessary, add hydrogen ions to balance hydrogen

atoms5. Check that the equation balances for atoms and charge

1. Cr₂O₇²⁻ 2Cl³⁺; oxidation state of Cr (+6), O (-2), Cl (+3)2. Cr₂O₇²⁻ +6e⁻ 2Cl³⁺3. Cr₂O₇²⁻ +6e⁻ 2Cl³⁺ +7H₂O4. Cr₂O₇²⁻ +6e⁻ +14H⁺ 2Cl³⁺ +7H₂O

Trends in physical propertiesTrends in the oxidising abilities of the halogensTrends in the reducing abilities of the halide ionsIdentification of halide ions using silver nitrateUses of chlorine and chlorate(I)

ElementSymbo

lColour State

Atomic No.

Outer electro

ns

Atomic

Radius

(nm)

Radius f X⁻ ion

(nm)

Boiling Point (K)

Electronegativit

y

Electron Affinity

(kJmol⁻¹)

Fluorine FPale

yellowGas 9 5 0.071 0.133 85 4.0 -328

Chlorine ClLight green

Gas 17 5 0.099 0.181 239 3.5 -349

Bromine BrReddish-brown

Liquid 35 5 0.114 0.196 332 2.8 -324.6

Iodine I Violet Solid 53 5 0.133 0.220 457 2.5 -295.2

Trend - - - -

Astatine: Extremely rare (estimated that there is only 0.029g present in Earth’s crust), and radioactive, thus chemistry has not been studied.

The gaseous halogens vary in appearance (see table) and they all have a characteristic ‘swimming-bath’ smell

A number of the properties of fluorine are untypical. Many of these untypical properties stem from the fact that the F-F bond is unexpectedly weak, compared with the trend for the rest of the halogens. The small size of the F atom leads to repulsion between non-bonding electrons because they are so close together

Size of atoms: the atoms get bigger as we go down the group because each element has one extra filled main level of electrons compared with the one above it

Electronegativity: The shared electrons in the H-X bond get further away from the nucleus as the atoms get larger going down the group. This makes the shared electrons further from the halogens nucleus and increases are more important than the increasing nuclear charge, so the electronegativity decreases as we go down the group

Melting and boiling points: These increase as we go down the group. This is because larger atoms have more electrons and this makes the van der Waal’s forces between the molecules stronger. The lower the boiling point, the more volatile the element. So, chlorine, which is a gas at room temperature, is more volatile than iodine, which is a solid.

The Reactions of Cl₂ (aq) with Br¯ (aq) and I¯ (aq)

Halide Ion

Observations Conclusion Equation

Br¯ (aq) Yellow/BrownCl₂

displaced2Br¯ + Cl₂ 2Cl¯ +

Br₂

I¯ (aq)Brown Colour and/or

Black PrecipitateI₂ displaced 2I¯ + Cl₂ 2Cl¯ + I₂

The Reactions of Br₂ (aq) with Cl¯ (aq) and I¯ (aq)

Halide Ion Observations Conclusion Equation

Cl¯ (aq) No change Cl₂ not displaced No reaction

I¯ (aq)Brown Colour and/or

Black PrecipitateI₂ displaced

2I¯ + Br₂ 2Br¯ + I₂

The Reactions of I₂ (aq) with Br¯ (aq) and Cl¯ (aq)

Halide Ion Observations Conclusion Equation

Br¯ (aq) No change Cl₂ not displaced No reaction

I¯ (aq) No change I₂ not displaced No reaction

F₂ > Cl ₂ > Br ₂ > I ₂

The trend in the reducing power of the halide ions is shown in the reaction of solid halide salts with concentrated sulphuric acid. The oxidation state of sulphur in sulphuric acid is +6. This can be reduced to +4, 0 or -2 depending on the reducing power of the halide ion.

NaX ObservationsProduct

sType of Reaction

NaF Steamy fumes HF Acid-base (F⁻ acting as a base)

NaCl Steamy fumes HCl Acid-base (Cl¯ acting as a base)

NaBrSteamy fumesColourless gasBrown fumes

HBrSO₂Br₂

Acid-base (Br¯ acting as a base)Redox (reduction product is

H₂SO₄)Redox (oxidation product of Br¯)

NaI

Steamy fumesColourless gas

Yellow solidSmell of bad eggsBlack solid, purple

fumes

HISO₂S

H₂SI₂

Acid-base (I¯ acting as a base)Redox (reduction product is

H₂SO₄)Redox (reduction product is

H₂SO₄)Redox (reduction product is

H₂SO₄)Redox (oxidation product of I¯)

These results indicate that:•Iodide ions can reduce the sulphur in H₂SO₄ from oxidation state +6 to +4 in SO₂, then to 0 as the element sulphur, and finally to -2 as H₂S•Bromide ions can reduce the in H₂SO₄ from oxidation state +6 to +4 in SO ₂•Fluoride and chloride ions cannot reduce the sulphur in H₂SO₄ under these conditions

Results found by using silver nitrate and ammonia solution

Halide Ion

Precipitate

Colour

Solubility of precipitate in ammonia solution

F⁻ None - Precipitate doesn’t form

Cl¯ AgCl White Soluble in dilute NH₃ (aq)

Br¯ AgBr CreamSparingly soluble in dilute NH₃ (aq)Soluble in concentrated NH₃ (aq)

I¯ AgI Yellow Insoluble in concentrated NH₃ (aq)

These results show that the solubility of the silver halides in ammonia solution decreases in the following order;

AgF > AgCl > AgBr > AgI

Chlorine is a pale green toxic gas that dissolves in water giving a pale green solution. However, chloride is very reactive and undergoes reaction with the water molecules forming a mixture of hydrochloric acid and hypochlorous acid (chlorate (I) acid). This solution is usually called chlorine water:

Cl2 + H2O  HCl + HOCl This is an example of a disproportionation reaction in which the chlorine gets simultaneously

oxidised and reduced. Chlorine as an element is in the zero oxidation state. In hydrogen chloride it is in the -1

oxidation state and in chlorate (I) acid it is in the +1 oxidation state.

1½Cl2 + e-  Cl-

2½Cl2 + H2O  HOCl + H+ + e-

The first equation shows chlorine being reduced and the second shows chlorine being oxidised

If chlorine water is tested with indicator paper, it first registers the red colour of the hydrochloric acid, but then is rapidly decolourised by the action of the hypochlorous acid, which is a good bleaching agent.

The poisonous nature of chlorine is taken advantage of in water treatment. Humans are tolerant to small doses of chlorine, while small microorganisms are killed at even low chlorine concentrations.

This allows water treatment plants to add chlorine to the process, killing microorganisms and making the water fit for human consumption.

Ozone gas is also used for water treatment, but it has the disadvantage of being difficult to store and more expensive. It does however have the advantage of not adding taste to the water.

As a solution of chlorine water is a mixture of acids it is logical that it should react with bases. Sodium hydroxide neutralises chlorine water forming salts from the two acids present:

2NaOH + HOCl + HCl  NaCl + NaOCl + 2H2O If chlorine is bubbled through cold sodium hydroxide solution a different reaction occurs:

3Cl2 + 6NaOH --> NaClO3 + 5NaCl + 3H2O This is an example of a disproportionation reaction in which the chlorine is simultaneously

oxidised and reduced. Chlorine as an element is in the zero oxidation state. In sodium chlorate(V) it is in the +5

oxidation state and in sodium chloride it is in the -1 oxidation state. The two half equations representing each process are:

½ Cl2 + e-  Cl-

½ Cl2 + 6OH-  ClO3- + 3H2O + 5e-

To add the two half-equations together the electrons must first be equalised by multiplying the first equation by five:

2½Cl2 + 5e- 5Cl-

½ Cl2 + 6OH-  ClO3- + 3H2O + 5e-

3Cl2 + 6OH-  ClO3- + 3H2O + 5Cl-

Sodium chlorate(V) is a useful weedkiller and oxidising agent.

Trends in Physical PropertiesTrends in Chemical Properties

ElementSymbo

lAtomic No.

Outer electro

ns

Metallic

Radius (nm)

Melting

Point (K)

Boiling

Point (K)

1st + 2nd IEs(kJmol⁻¹)

Density (gcm⁻³)

Magnesium Mg 12 2 0.160 922 1380 738+1451=2189 1.74

Calcium Ca 20 2 0.197 1112 1757 590+1145=1735 1.54

Strontium Sr 38 2 0.215 1042 1657 550+1064=1614 2.60

Barium Ba 56 2 0.224 998 1913 503+965=1468 3.51

Trend - -

Electron arrangement: the elements all have two electrons in an outer s-orbital. This s-orbital becomes further away from the nucleus as we go down the group

The sizes of the atoms: the atoms get bigger as we go down the group. The atomic (metallic) radii increase because each element has an extra filled main level of electrons compared with the one above it

Melting points: Group 2 elements are metals with high melting points, typical of a giant metallic structure. As we go down the group, the electrons in the ‘sea’ of delocalised electrons are further away from the positive nuclei. As a result, the strength of the metallic bonds decreases as we go down the group. For this reason, the melting points of Group 2 elements decrease slightly as we go down the group, starting with Ca. Magnesium, with the lowest melting point, does not fit this trend. This is because the lattice arrangement of atoms is different from that of the elements below, which makes them slightly easier to separate

Ionisation energies: in all their reactions, atoms of elements in Group 2 lose their outer electrons to form ions with two positive charges.M M¹⁺ + 2e⁻So, an amount of energy equal to the sum of the first and the second ionisation energies is needed for complete ionisationM M¹⁺ + e⁻ + M¹⁺ M²⁺ + e⁻Both the first ionisation energy and the second ionisation energy decrease as we go down the group; it takes less energy to remove the electrons as they become further and further away from the positive nucleus. The nucleus is more effectively shielded by more inner shells of electrons. In all their reactions, the metals get more reactive as we go down the group.

Oxidation is the loss of electrons so in all their reactions, the Group 2 metals are oxidised. The metals go from oxidation state 0 to oxidation state +2. These are redox reactions

Reaction with water With water, we see the same trend in reactivity; the metals get more reactive

as we go down a group. The basic reaction is as follows, where M is any Group 2 metal:

M(s) + 2H₂O(l) M(OH)₂(aq) + H₂(g) Magnesium hydroxide is ‘milk of magnesia’ and is used in indigestion remedies

to neutralise excess stomach acid. Magnesium reacts very slowly with cold water but rapidly with steam to form an

alkaline oxide and hydrogen.Mg(s) + H₂O(g) MgO(s) + H₂(g)

Calcium reacts in the same way but more vigorously, even with cold water. Strontium and barium react more vigorously still. Calcium hydroxide is sometimes called ‘slaked lime’ and is used to treat acidic soil. Most plants have an optimum level of acidity or alkalinity in which they thrive. For example, grass prefers a pH of around 6 so if the soil has a pH much below this, then it will not grow as well as it could. Crops such as wheat, corn, oats and barley prefer soil that is nearly neutral.

There is a clear trend in the solubilities of the hydroxides, as we go down the group: they become more soluble. The hydroxides are all white solids.

Magnesium hydroxide, Mg(OH)₂ (milk of magnesia), is almost insoluble. It is solid as a suspension in water, rather than a solution.

Calcium hydroxide, Ca(OH)₂, is sparingly soluble and a solution is used as lime water.

Strontium hydroxide, Sr(OH)₂, is more soluble Barium hydroxide, Ba(OH)₂, dissolves to

produce a strongly alkaline solution:Ba(OH)₂(s) + aq Ba²⁺(aq) + 2(OH)⁻(aq)

The solubility trend in the sulphates is exactly the opposite, they become less soluble as we go down the group. So, barium sulphate is virtually insoluble. This means that it can be taken by mouth as a ‘barium meal’ to outline the gut in medical X-rays. (The heavy barium atoms is very good at absorbing X-rays). This test is safe, despite the fact that barium compounds are very toxic, because barium is so insoluble.

The insolubility of barium sulphate is also used in a simple test for sulphate ions in solution. The solution is first acidified with nitric or hydrochloric acid. Then, barium chloride solution is added to the solution under test and if a sulphate is present a white precipitate of barium sulphate formed.

Ba²⁺(aq) + SO₄²⁻(aq) BaSO₄(s) The addition of acid removes carbonate ions as

carbon dioxide. (Barium carbonate is also a white insoluble solid, with would be indistinguishable from barium sulphate)

Compound Uses

Magnesium hydroxide Used in medicine.Milk of magnesia which is used to neutralise excess stomach acid and treat indigestion

Calcium hydroxide Used in agriculture to neutralise acidity of soil

Barium sulphate “Barium meal”Used in medicine.Very insoluble & appears opaque on X-Rays so is used to diagnose problems with the digestive system

Principles of Metal ExtractionEnvironmental Aspects of Metal Extraction

The production of metals is an important part of the chemical industry. The way in which metals are produced from natural resources involves an understanding of both social and economic aspects of the processes, as well as an appreciation of the underlying chemistry.

Of the possible engineering materials, only Al and Fe are very abundant, with Ti being the next most common. These three metals are also widely distributed, and so are metals we might expect to be used whenever possible.

Some metals in everyday use, e.g. Cu and Ni, are relatively rare in the Earth’s crust. Fortunately, they sometimes occur in high-grade ores in a few specific locations. To be economically viable, ores should have a high concentration of the desired metal and be free of impurities.

It should be noted that ores of expensive commercial metals, such as copper, may have a low concentration of the metal whereas ores of lower value metals, such as iron, must contain very high concentrations of the metal to be economically viable.

Metals usually occur in combination with oxygen or sulphur. The process of extraction is therefore one of reduction – often of oxides.

As these oxides are very stable compounds, energy usually has to be put in to reduce them to the metal i.e. extraction reactions are endothermic (unless a very powerful reducing agent such as Al metal is used)

Sulphide ores are not usually reduced directly to the metals. They are first roasted in air to produces oxides; this liberates sulphur dioxide and so gives rise to the potential pollution hazard of acid rain.

Sulphur dioxide dissolves in water in the clouds to form sulphurous acidSO₂ (g) + H₂O (l) H₂SO₄ (aq)

Some of the sulphur dioxide is oxidised to sulphur trioxide2SO₂ (g) + O₂ (g) 2SO₃ (g)

This gas dissolves in water to form sulphuric acidSO₃ (g) + H₂O (l) H₂SO₄ (aq)

These acids can then fall as acid rain, damaging plants and polluting lakes Oxide ores are often reduced directly to the metal. However, the use of fossil

fuels e.g. coke from coal, may give a rise to environmental problems through emission of carbon dioxide, a greenhouse gas

The method chosen for industrial scale for each metal depends on:◦ The cost of the reducing agent: a reducing agent that is naturally

available such as carbon (from coal) or hydrogen (from natural gas) is cheaper than one, such as aluminium, which has to be prepared by a separate and often more costly processCarbon is usually cheaper than hydrogen and easier to store. Carbon is readily available as coke while hydrogen has to be prepared from methane or water.

◦ The energy costs for the process: a process, which requires less energy by operating at a lower working temperature usually, has an economical advantage over a process that requires greater energy input

◦ The required purity of the metal. The costs of the reductant and the energy required have to be

weighed against each other. If a metal is required with very high purity (and in small quantity) for a particular use, then a relatively expensive process may be used. The extra cost being necessary to satisfy demand for high purity and justified by the product’s market value. If high purity is not essential, then the cheapest method of producing saleable metal is used.

Carbon is a cheap and plentiful reducing agent – it occurs in coal (which when heated in the absence of air gives coke, a solid with a very high carbon content) and in charcoal (which is obtained from wood, a renewable resource)

All metal oxides in theory can be reduced by carbon if the temperature is high enough. In practice, temperatures above 2000˚C are uneconomic and impractical

The most important example of carbon reduction is the manufacture of iron from high quality haematite, Fe₂O₃

Iron oxides are reduced in the blast furnace using coke.

This is a continuous process (minimises costs) in which iron (III) oxide, Fe₂O₃ ,coke and limestone are fed in at the top of the furnace and hot air is blown in near the bottom

Molten iron collect at the bottom of the furnace and is run off. The product is very impure iron which typically contains ~4% carbon, as well as manganese, silicon, phosphorus and sulphur in amounts which depend on the operating conditions and ore used.

1. Coke reacts with the hot air blast in a strongly exothermic reactionC (s) + O₂ (g) CO₂ (g)This reaction produces the heat energy needed for the reduction of the iron (III) oxide

2. The carbon dioxide formed reacts at lower temperatures with unreacted coke to form carbon monoxideCO₂ (g) + C (s) 2CO (g)

3. The carbon monoxide reduces most of the iron (III) oxide at around 1200˚CFe₂O₃ (s) + 3CO (g) 2Fe (l) + 3CO₂ (g)

4. In the hotter part of the furnace, coke also reacts directly with the iron oxideFe₂O₃ (s) + 3C (s) 2Fe (l) + 3CO (g)

The other product in the process is slag; this contains impurities from the ore combined with lime, and is largely calcium silicate, CaSiO₃; it decomposes in the heat of the furnace to form calcium oxide (lime) and carbon dioxideCaCO₃ (s) CaO (s) + CO₂ (g)

Calcium oxide is a basic oxide and combines with the acidic oxides in the furnace to form the slag. Sand and soil contain the acidic silica, which reacts to give calcium silicateCaO (s) + SiO₂ (s) CaSiO₃ (l)

Slag is used to make ‘breeze blocks’ for use in the construction industry

Many metals are extracted by reduction of their oxides with carbon. However, some metals form metal carbides rather than the metal itself, so it is not a practical method for extracting these metals

This route can be used to make carbides, which are important engineering materials and potential catalysts. When titanium (IV) oxide is heated with carbon, the following reaction occurs:

TiO₂ + 3C TiC + 2CO Oxides of tungsten give carbides when they are

heated with carbon. Tungsten is prepared by reduction of its oxides with hydrogen

The manufacture of aluminium is carried out by the electrolysis of purified bauxite, Al₂O₃, which is dissolved in molten cryolite, Na₃AlF₆. The melting point of aluminium oxide is >2000˚C, but by dissolving it in molten cryolite, the temperature of the melt is reduced to about 970˚C. The electrodes are made of carbon and the reactions at the electrodes are:◦ At the cathode: Al³⁺ + 3e⁻ Al◦ At the anode: 2O²⁻ O₂ + 4e⁻

Some of the oxygen evolved reacts with the anodes at high temperature, forming carbon monoxide and carbon dioxide:

2C + O₂ 2CO and C + O₂ CO₂ The process uses large amounts of electricity (because electricity is needed to

melt the cryolite and decompose the Al₂O₃) and is only economic when electricity is relatively cheap. The process is continuous but regular additions of aluminium oxide are needed, and the carbon electrodes need replacing as they are consumed. There is potential environmental problems through waste cryolite causing fluoride pollution

Aluminium is strong, light and can be made into thin sheets. It has a low density, and is an excellent conductor of heat and electricity. Although reactive, it does not corrode easily. If exposed to air, a clean surface reacts quickly with oxygen forming a protective coating of aluminium oxide.

When the purity of the metal is of prime importance, and contamination by carbon or oxygen cannot be tolerated, then reduction of a metal halide with a reactive metal becomes a desirable route despite the costs associated

Titanium is very abundant in the Earth’s crust. It is also a very desirable engineering metal, having a low density, high strength and high resistance to corrosion. However, unlike iron, traces of carbon, oxygen or nitrogen have unwelcome effects (e.g. the metal is brittle). Titanium is thus extracted from its chloride by reduction with an active metal

The ore rutile, impure titanium (IV) oxide, is converted into titanium (IV) chloride using chlorine and coke at about 900˚C

TiO₂ + 2C + 2Cl₂ TiCl₄ + 2CO Titanium (IV) chloride is a colourless liquid which fumes in moist air because of

hydrolysis. It is purified from other chlorides (e.g. those of iron, silicon and chromium) by fractional distillation under argon or nitrogen. In the UK, the chloride is then reduced by sodium in the exothermic reaction

TiCl₄ + 4Na Ti + 4NaCl The sodium is initially held at about 550˚C but the temperature rises to ~1000˚C

during the reaction. An inert atmosphere of argon is used to prevent contamination of the metal with oxygen or nitrogen. The sodium chloride by-product is washed out, leaving titanium as a granular powder.

Elsewhere in the world, magnesium is used as the reducing agent in a similar reduction process (called the Kroll process). The magnesium chloride by-product is removed from the titanium by vacuum distillation at high temperature.

TiCl₄ + 2Mg Ti + 2MgCl₂ This is a batch process – one batch at a time, which is more costly than a continuous

process. Costs are high because:◦ Chlorine and sodium have to be produced first (reducing agents not readily available)◦ High temperatures are involved in both stages of production◦ Precautions have to be taken in handling TiCl₄, which reacts violently even with cold water◦ An argon atmosphere has to be maintained to prevent oxidation

Despite extensive searches, other methods for producing titanium have not yet been able to compete because of cost and purity considerations. Thus use is limited, despite its desirable properties and a high natural abundance, due to the high cost of production.

Tungsten cannot be extracted using carbon because a carbide is formed It is useful because of its high boiling point It is reduced using hydrogen to extract it

WO₃ + 3H₂ W + 3H₂O

Recycling of metals is carried out extensively, and would be environmentally friendly if all scrap was returned to the metal works and recycled. Unfortunately, because scrap metal is often widely spread and many miles from metal-producing plants, the scrap must be collected and transported. This procedure creates an energy cost which must be carefully calculated and offset against the savings in extraction

Iron (as steel) is the most extensively recycled metal with as much as 40% of the world’s iron and steel production coming from scrap metal. The world reserved of iron ore are vast and commercial ores have high iron content – so why recycle?

Percentage iron: the scrap iron contains a higher percentage of iron than a commercial ore. Many alloy steels are produced using only scrap metal in the Electric Arc Process. Also, in the BOS steel making process, about 30% of scrap metal is added before adding the impure liquid iron

Environmental issues: without recycling, a serious environmental would result from all the scrap metal that is discarded. About 1.5 million cars a year are discarded in the UK; with about 75% of a car’s mass is recycled and only about 0.3% going to landfill. Steel cans are also recycled as they are magnetic and easily separated from other waste.

The extraction methods for both iron and aluminium produce carbon dioxide a ‘greenhouse gas’; re-melting does not produce this. Unlike extraction of iron, the extraction of aluminium is very expensive because electrolysis consumes vast amount of energy. The re-melting of aluminium cans, which are high-quality scrap, saves 95% of this extraction energy.

It should be noted that there are considerable energy costs in collecting and separating the scrap aluminium and these costs must be carefully balanced against the savings made in extraction.

Synthesis of ChloroalkanesNucleophilic SubstitutionElimination

Methane does not react with chlorine at room temperature, or in the dark. In the presence of ultraviolet light, however, a mixture of chlorine and methane will explode, forming hydrogen chloride and a mixture of chlorinated methanes. This is a free radical substitution reaction which occurs in steps:

1. Initiation◦ Cl₂ 2Cl∙◦ The ultraviolet light provides the energy needed to start the reaction i.e. to split some chlorine

molecules into atoms (radicals). This occurs first because the Cl-Cl bond in chlorine is weaker than the C-H bond in methane

2. Propagation◦ Cl∙ + CH₄ CH₃∙ + HCl◦ CH₃∙ + Cl₂ CH₃Cl + Cl∙◦ Overall reaction: CH₄ + Cl₂ CH₃Cl + HCl◦ In each step, a radical is used and a radical is formed so the process continues and leads to a chain

reaction. Each step is also exothermic, so the chain reaction can produce an explosion

3. Termination◦ If two radicals combine (i.e. unpaired electrons pair up to form a covalent bond) they form a stable

molecule, and the sequence of reaction stops. Possible termination steps include: CH₃∙ + Cl∙ CH₃Cl CH₃∙ + CH₃∙ CH₃CH₃

◦ Such termination steps lead to trace amounts of impurities, such as ethane, in the chloromethane

The reaction of a chlorine radical with methane extracts a hydrogen radical to form HCl

Chloromethane still contains three hydrogen atoms, so further pairs of propagation steps are possible, leading to CH₂Cl₂, dichloromethane, CHCl₃, trichloromethane and CCl₄, tetrachloromethane

CH₃Cl + Cl∙ ∙CH₂Cl + HCl ∙CH₂Cl + Cl₂ CH₂Cl₂ + Cl∙ CH₂Cl₂ + Cl∙ ∙CHCl₂ + HCl ∙CHCl₂ + Cl∙ CHCl₃ + Cl∙ CHCl₃ + Cl∙ ∙CCl₃ + HCl CCl₃ + Cl₂ CCl₄ + Cl∙

When haloalkanes are warmed with aqueous solution or potassium hydroxide, alcohols are formed

When haloalkanes are warmed with aqueous/alcoholic solutions of potassium cyanide, nitriles (RCN) are formed.Note: an extra carbon has been added to the chainNitriles are hydrolysed to carboxylic acids by heating under reflux either with aqueous alkali, or mineral acid; an amide intermediate forms

When haloalkanes are warmed with excess ammonia in a sealed container, primary amines are formedSince the acid HBr will immediately react with the base ammoniaThe excess ammonia minimises the chance of further reaction of the primary amines to form secondary, tertiary or quaternary ammonium saltsReduction of nitriles, using hydrogen, in the presence of a nickel catalyst, also produces primary amines

Structure, Bonding and ReactivityAddition ReactionsPolymerisation

Alkenes contain 2 hydrogen atoms fewer than their parent alkanes, and are said to be unsaturated. This is because they C=C double bond

The double bond results from overlap of the spare unbonded, singly-filled p-orbital present on each carbon atom in the bond. This overlap produces a cloud of electrons above and below the molecule (a Π bond). The two carbon atoms of the double bond, plus the fur atoms attached to the double bond must lie in the same plane, thus ethene is a planar molecule.

If the alkene is unsymmetrical and the molecule being added is also unsymmetrical then two possible products can form. The major product is the one formed via the more stable carbocation.

Tertiary > Secondary > Primary This is due to the inductive (electron

releasing) effect of the attached alkyl groups. The more alkyl groups around the carbocation, the more stable it is and more likely it is to be formed.

A curly arrow shows the

movement of a pair of electrons

NomenclatureEthanol ProductionClassification and ReactionsElimination

When naming an alkene derivative, look for the longest carbon chain in the skeleton

Look at slide 50 for a recap.

Alcohol is produced by the fermentation process that uses living yeast cells to convert sugars (e.g. glucose) into carbon dioxide and ethanol

C₆H₁₂O₆ 2C₂H₅OH + 2CO₂ The reaction is slow at low temperature as the enzymes

in yeast are inactivated. At higher temperatures, the yeast is killed as the enzymes are denatured. The process is therefore usually carried out at ~35˚C

Fermentation produces an aqueous solution of ethanol of concentration between 3-15%. Beers are usually about 3-7% and wines about 9-13%. Fermentation rarely produces higher ethanol concentrations as these kill the yeast. More concentrated alcohol solutions i.e. spirits such as whisky, brandy or gin, are about 40% alcohol, and are produced by fractional distillation of the fermentation products.

Ethanol is also produced industrially by the direct hydration of ethene using steam and a phosphoric acid catalyst at 300˚C and 6.5x10³ kPa (65 atmospheres) pressure

The direct hydration method is preferred at present for the production of ethanol for industrial use in the UK. However, as the method uses ethene as a raw material, it may become less popular compared with fermentation in the future as oil supplies begin to run out

MethodRate of

Reaction

Quality of

ProductRaw Material Type of Process

Hydration Fast PureEthene from Oil(finite resource)

Continuous(manpower cheap)

(equipment expensive)

Fermentation Slow ImpureSugars

(renewable resource)

Batch(manpower expensive)

(equipment cheap)

Alcohols are oxidised by strong oxidising agents. Potassium dichromate(VI) in the presence of dilute acid is the reagent of choice. The reaction is useful because the different types of alcohol behave differently.◦ 1º alcohols get oxidised to aldehydes◦ 2º alcohols are oxidised to ketones◦ 3º alcohols are not oxidised

Alcohol: ethanol 1st Step: H⁺/K₂Cr₂O₇; acidified potassium dichromate (VI); orange Product: ethanal (aldehyde); goes green 2nd Step: Tollen’s reagent/Fehlings solution Product: ethanoic acid (carboxylic acid); Tollen’s reagent silver

mirror; Fehlings solution brick red precipitate Methods:

Distillation(aldehyde)

Refluxing(carboxylic

acid)

CH₃CH₂OH + [O] CH₃COH + H₂OCH₃CH₂O + [O] CH₃COOH CH₃CH₂OH + 2[O] CH₃COOH + H₂O

Alcohol: propan-2-ol 1st Step: H⁺/K₂Cr₂O₇; acidified potassium dichromate (VI); orange Product: propanone (ketone); goes green 2nd Step: Tollen’s reagent/Fehlings solution Product: no change; no silver mirror; no brick-red precipitate; no

further oxidation Methods: use refluxing

CH₃CHCH ₃OH + [O] CH₃COH₃ + H₂O

Alcohol: 2-methylpropan-2-ol 1st Step: H⁺/K₂Cr₂O₇; acidified potassium

dichromate (VI); orange Product: no change; no oxidation state; stays

orange

The products formed depend on the conditions used: Alkenes are formed in the presence of H2SO4 (or H3PO4 better, as it doesn't produce as many by-products) and the correct temperature (hot for primary, warm for secondary and cool for tertiary) alcohols lose a water molecule.

One of the advantages of this method of preparing alkenes is that the starting materials, the alcohols, can be produced by biological means such as fermentation. This means that plastic manufacture does not have to depend on the petrochemicals industry.

Fermentation  alcohols  dehydration  alkenes  polymerisation  polymers

Mass SpectrometryInfrared Spectroscopy

Mass spectrometry is the main method for finding the relative molecular mass of organic compounds.

See slide 7 for a recap The strength of the magnetic field required to

deflect the ion into the detector depends on the mass and charge of the ion. The output is then presented as a graph of relative abundance against mass/charge ratio. However, since the charge on the ions is usually +1, the mass/charge ratio is effectively relative mass. This graph is called a mass spectrum.

When ethanol is ionised it forms the ion CH₃CH₂OH⁺. This is called the molecular ion. Many of these ions will then break up; some of their bonds break as they are ionised, so we have other ions of smaller molecular mass. This process is called fragmentation. Each of these produces a line in the mass spectrum. These can provide information that will help to deduce the structure of the compound. They also act as a ‘fingerprint’ to help identify.

However there are normally a few ionised molecules remaining intact to give a peak corresponding to the relative molecular mass of the compound.

The main peak furthest to the right of the mass spectrum, corresponds to the molecular ion (it has the highest mass). The molecular ion peak for ethanol is at mass 46; this tells us the relative molecular mass of ethanol.

Mass spectrometry is the most important technique for measuring the relative molecular mass of organic compounds.

Many mass spectrometers can measure masses to three or sometimes four decimal places. This method allows us to work out the molecular formula of the parent ion. It makes use of the fact that isotopes of atoms do not have exactly whole number atomic masses (except for carbon-12).

For example, a parent ion of mass 200, to the nearest whole number could have the following molecular formulae: C₁₀H₁₆O₄;C₁₁H₄O₄;C₁₁H₂₀O₃

Adding up the accurate atomic masses gives the following Mr’s:◦ C₁₀H₁₆O₄ = 200.1049◦ C₁₁H₄O₄ =200.0110◦ C₁₁H₂₀O₃ =200.1413

These can easily be distinguished by high resolution mass spectrometry

A pair of atoms joined by a chemical bond is always vibrating. Stronger bonds vibrate faster (at higher frequency) and heavier atoms make the bond vibrate more slowly (at lower frequency). Every bond has its own unique natural frequency that is in the infra-red region of the electromagnetic spectrum

When we shine a beam of infra-red radiation (heat energy) through a sample, the bonds in the sample can absorb energy from the radiation and vibrate more. However, any particular bond can only absorb radiation that has the same frequency as the natural frequency of the missing frequencies that correspond to the bonds in the sample.

A beam of infra-red radiation containing a frequencies is passed through a sample

The radiation that emerges is missing the frequencies that correspond to the types of bonds found in the sample

The instrument plots a graph of the intensity of the radiation emerging from the sample, called the transmittance, against the frequency of radiation

The frequency is expressed as a wavenumber, measured in cm⁻¹

Source of infra-red radiation

Sample Detector

The dips in the infra-red spectrum (peaks) represent particular bonds. These can help us identify the functional groups present in a compound

Bond Wavenumber/cm⁻¹

C-H 2850-3300

C-C 750-1100

C=C 1620-1680

C=O 1680-1750

C-O 1000-1300

O-H (alcohols) 3230-3550

O-H (acids) 2500-3000

N-H 3300-3500

The area of an infra-red spectrum below about 1500cm usually has many peaks caused by a complex vibrations of the whole molecule. This shape is unique for any particular substance.

It can be used to identify the chemical, just as people can be identified by their fingerprints. It is therefore called the fingerprint region.

We can use a computer to match the fingerprint region of a sample with those on a database of compounds. An exact match confirms the identification of the sample.

Infra-red spectra can also be used to show up the presence of impurities. These may not be revealed by peaks that should not be there in the pure compound.

In the spectra below, we can compare pure and impure caffeine. The broad peak at around 3000cm in the impure sample is an O-H stretch caused by water in the sample that has not completely dried. Notice that there are no O-H bonds in caffeine.