AQA A-Level Chemistry

WORKBOOK ANSWERS

AQA A-level Chemistry

Physical chemistry 2

This Answers document provides suggestions for some of the possible answers that might be given for the questions asked in the workbook. They are not exhaustive and other answers may be acceptable, but they are intended as a guide to give teachers and students feedback.

Physical chemistry

Topic 8 Thermodynamics

Born–Haber cycles

1Answer is C 

A represents the enthalpy of atomisation of sodium. B represents the first ionisation enthalpy of sodium. C is the first electron affinity of oxygen. D is the second electron affinity of oxygen. Atomisation is an endothermic process. Ionisation is an endothermic process. The first electron affinity of any element is an exothermic process but the second electron affinity is an endothermic process. The second electron affinity is endothermic as the negative electron is being added to a negative ion so there are repulsions between the particles.

2aAnswer is C 

2bAnswer is E 

2cAnswer is B 

2dAnswer is F 

It is important that you can identify the enthalpy changes in a given Born‒Haber cycle. Look at each of the levels carefully and you can see what changes are occurring and whether the change relates to potassium or chlorine or the compound. If you are asked to name them, for example name change C, the answer you should give is the enthalpy of atomisation of potassium. It is not enough to simply say enthalpy of atomisation as there is the potential for this to be confused with chlorine. The ones not asked are A and D. A is either the enthalpy of atomisation of chlorine or ½ the bond dissociation enthalpy of chlorine and D is the enthalpy of formation of potassium chloride.

2eE = –D + C + B + A + F

703 = −(−437) + 89 + 420 + 121 + F 

703 = 1067 + F 

F = –364 kJ mol−1

The most common error in this type of calculation is not changing the sign of D when going round the cycle in this way. It is worth putting −364 back in and checking that you get +703 for the lattice enthalpy.

3a

iMg(g) → Mg+(g) + e−

iiMg+(g) → Mg2+(g) + e−

iiiMg(s) + ½O2(g) → MgO(s) 

ivO–(g) + e−→ O2−(g) 

vMgO(s) → Mg2+(g) + O2−(g) 

These equations are very important as they are a major part of understanding the cycle. The states are vital to the answer so make sure you include them. If you learn the definitions of the enthalpy changes carefully you can work out the equations from them as the states and species are given.

3bFrom left to right:

iMg2+(g) + O(g) + 2e−

Mg2+(g) + O−(g) + e−

Mg2+(g) + O2−(g) 

Again think sensibly about the changes which are occurring in the cycle and you should be able to complete any cycle. The levels may also not be given to you but remember to go down for an exothermic change and up for an endothermic change.

iiLH= −fH+aH + IE1H + IE2H + ½BDEH+ EA1H + EA2H

LH = +602 + 150 + 736 + 1450 + ½(496) + (–142) + 844 

LH = +3888 kJ mol−1

Each error will be penalised by 1 mark. The most common errors are with the bond dissociation enthalpy/atomisation enthalpy of the non-metal and with the electron affinities. Make sure you can calculate any value in the cycle using +3888 as the lattice enthalpy.

4a

All correct = 4 marks

2 errors or 1 omission = 3 marks

3 errors or 2 omissions = 2 marks

4 errors or 3 omissions = 1 mark

>4 errors or >3 omissions = 0 marks

It is important to be able to recognise all the enthalpy changes in more complex Born–Haber cycles such as this one. They are not in any expected order and you may have to label them as in this question or complete missing pieces of information on the cycle. Try drawing out this cycle and completing it for iron(iii) oxide instead of aluminium oxide.

4bΔlattH = –ΔfH + 2ΔaH + 2ΔIE1H + 2ΔIE2H + 2ΔIE3H + 3ΔaH + 3ΔEA1H + 3ΔEA2H

+15179 = –(–1676) + 2(324) + 2(580) + 2(1800) + 2ΔIE3H + 3(248) + 3(–141) + 3(+790) 

2ΔIE3H = 5404

ΔIE3H = 2702 kJ mol–1

The calculation here is more complex as you have to remember to divide by 2 at the end to determine the value for the third ionisation enthalpy of aluminium. Be careful with the values for oxygen as 3 × atomisation and 3 × first electron affinity and 3 × second electron affinity are required.

5a

1 mark for each correct line

Make sure you can follow the logic of the species that are already there in the diagram. The top three lines are very important, as these are the changes happening to the chlorine during the cycle.

5b

i–2492 = –2(–364) – 2ΔaH – 1450 – 736 – 150 + (–642) 

2ΔaH = 242

ΔaH = +121 kJ mol–1

The table contains information on hydration enthalpies of the magnesium and chloride ions. This may be confusing in this calculation as these values are not required until part (ii). 2 × atomisation of chlorine and 2 × first electron affinity are required. The final step is dividing by 2 to calculate the enthalpy of atomisation per mole of chlorine atoms.

iiΔsolH = –lattice enthalpy (formation) + ΣΔhydH

= +2492 + (–1920) + 2(–364) 

= –155 kJ mol–1

The enthalpy of lattice dissociation is used to calculate the enthalpy of solution, so the negative of the value in the table is used. Also 2 × the enthalpy of hydration of chloride ions is required in the calculation as 1 mol of MgCl2 contains 2 mol of Cl– ions.

iiiLower as Ca2+ is larger/Mg2+ is smaller/Ca2+has lower charge density.

Ca2+is less strongly attracted to water molecules.

The Ca2+ ion has a lower charge density, as it has the same charge as the Mg2+ ion but the Ca2+ ion is larger. This means that water molecules are not as attracted to it. In other questions watch out for a comparison between Na+ and Mg2+, as the Mg2+ is smaller and also has a higher charge so has a larger charge density than the Na+ ion.

Exam-style question

6a

iThe Born‒Haber cycle for magnesium fluoride would give the calculation as:

LH = −fH + H + IE1H + IE2H + ½BDEH + 2EA1H

The unknown in this expression is ΔfH.

It is important to be able to identify all the enthalpy changes in the table:

The last two are not used in this part of the question but may be used later in the question.

LH = −fH+ aH+ IE1H + IE2H + BDEH + 2EA1H

+2883 = −fH + 150 + 736 + 1450 + 158 + 2(–348) 

fH = 1798 − 2883 = –1085  kJ mol−1

The most common errors in this style of question involve the bond dissociation enthalpy of fluorine and the first electron affinity of fluorine. Make sure you understand what these changes mean and if they need to be multiplied in the calculation. As two fluoride ions are required,
F2→ 2F is needed, as is 2F + 2e−→ 2F−. Base your decision on the number of each ion required in the formula of the compound.

iisolH = LH + hydH(Mg2+) + 2hydH(F−) 

−20 = +2883 + (−1891) + 2hydH(F−)

2hydH = –1012 

hydH = –506 kJ mol−1

Remember that there are 2 mol of fluoride ions in 1 mol of magnesium fluoride.

6b

iIons are point charges/ions are perfect spheres/no additional covalent bonding.

Any one 

The perfect ionic model assumes that the ions are point charges and perfect spheres. It also assumes that there is not additional covalent bonding in the compound.

iiMagnesium fluoride is ionic/no additional covalent bonding.

Theoretical value = experimental value 

The experimental value for lattice enthalpy is determined from a Born‒Haber cycle. The theoretical value is determined by the perfect ionic model. If the experimental value is greater than the theoretical value, then the compound is not purely ionic but has some additional covalent bonding.

Gibbs free-energy (G) and entropy change (S)

1Answer is B 

This is a typical entropy question. Solids are more ordered than solutions/liquids, which are more ordered than gases. Look for reactions that produce a gas as these reactions often show an increase in entropy so can be ruled out. The question could have asked ‘for which reaction is ΔS0’. This is the same question.

2Answer is B 

All enthalpy changes and ΔG have units of kJ mol−1. Absolute entropy values and ΔS have units of J K−1 mol−1. Units are very important to all these questions, so make sure you know what the units are for each quantity.

3Answer is D 

For a reaction in which ΔH is negative and ΔS is positive, ΔG will always be negative irrespective of the value of T. Reactions where ΔH is positive and there is a decrease in entropy (as in A) are not feasible at any temperature. Reaction B is feasible above certain temperatures and reaction C is feasible up to a certain temperature.

4aIt is an element in its standard state.

This is a common question. It shows that you understand the definition of enthalpy of formation in terms of formation from the elements in their standard state.

4bS= ΣS (products) − ΣS (reactants)

= 2(68.7) + 4(240) + 205 − 2(217.9) 

= +866.6 J K−1 mol−1

The entropy change in the reaction is calculated from the absolute entropy values. Note that the entropy values for solids are usually low, such as the value for PbO, but the value for Pb(NO3)2 is high, which suggests that it is a disordered solid, which can be the case for substances that undergo thermal decomposition. The reaction produces two different gases, so we would expect the reaction to show a positive change in entropy.

4cH = ΣfH(products) − ΣfH(reactants)

H = 2(−217.3) + 4(+33.2) − 2(−451.9) 

H = +602 kJ mol−1

Make sure you can carry out Hess’s law calculations to determine enthalpy changes from enthalpies of combustion (ΔH = ΣΔcH(reactants) − ΣΔcH(products)) or from enthalpies of formation (ΔH = ΣΔfH(products) − ΣΔfH(reactants)). As this is a thermal decomposition, you would expect to get a positive value for ΔH otherwise you should check your calculation again.

4dG = H − TS

G = 602 − 500(0.8666) 

G = +168.7 kJ mol−1

G ≥ 0 or G is positive 

The most common mistake in this type of question is not to divide the ΔS value by 100 before using it in the ΔG expression. Calculating the ΔG expression to prove if the reaction is feasible or not. The reaction is only feasible when ΔG ≤ 0, so a positive ΔG indicatesthat the reaction is not feasible.

4eWhen G = 0, H = TS

+602 = T(0.8666)



= 694.7K 

This is a common question. The reaction becomes feasible when ΔG = 0, which means that
ΔH = TΔS. Remember again that ΔS must be in kJ K−1 mol−1, so the value calculated previously is divided by 1000. The temperature obtained will be in kelvin (K).

5When G = 0, H = TS

S = 42.7 − 32.7 = 10 J K−1 mol−1

9 = T(0.01) 



For any change in state ΔG = 0. ΔS is the difference between the absolute entropy values for Mg(s) and Mg(l). The ΔS value should again be divided by 1000 when used in the expression.

6a

iH = ΣfH(products) – ΣfH(reactants) 

H = –1676 – 3(–242)  = –950  kJ mol–1

It is very important to realise that the enthalpy of formation of elements in their standard states is zero, so Al(s) and H2(g) have a zero enthalpy of formation. This reaction is exothermic, asΔH is negative.

iiS = ΣS (products) – ΣS (reactants) 

S = (51 + 3(131)) – (2(28) + 3(189)) = –179  J K–1 mol–1

The reaction shows a decrease in entropy due to the fact that there are 2 mol of solid on the left and 1 mol of solid on the right, with 3 mol of gas on each side.

iiiΔG = ΔH – TΔS

ΔG = –950 – 500(–0.179) = –860.5 kJ mol–1

ΔG is negative, so reaction is feasible 

Calculating ΔG using ΔG = ΔH – TΔS gives a negative value which means that the reaction is feasible at that temperature. Remember to divide ΔS by 1000 before using it in the expression.

6bΔG = 0 = ΔH – TΔS 

ΔH = enthalpy of change of state of MgO(s) → MgO(l)

ΔH = x – (–601) = x + 601 

ΔS = 48 – 27 = 21 J K–1 mol–1





3125 × 0.021 = x + 601

65.625 = x + 601

x = 65.625 – 601 = –535.4 kJ mol–1

This is a more complex calculation. The enthalpy of fusion (melting) of magnesium oxide can be calculated from the enthalpy of formation of MgO(s) and MgO(l). So for the change MgO(s) → MgO(l), ΔH = ΔfH MgO(l) – ΔfH MgO(s) = x – (–601) where x is the value to be calculated. When a substance changes state, ΔG = 0. ΔS is calculated as usual and used with the melting temperature in K to calculate the unknown x.

6cΔH = +242 kJ mol–1

ΔS= 131 + ½(205) – 189  = +44.5 J K–1 mol–1

T = ΔH/ΔS = 242/0.0445 = 5438.2 K 

The reaction is the reverse of the formation of water vapour, so ΔH = –(242) = +242 kJ mol–1. ΔSiscalculated as before using ½ mol of O2 in the same way as whole numbers are used. T is calculated from ΔG = 0 when the reaction becomes feasible. This is common for endothermic reactions (often decompositions), which show an increase in entropy.This is true of many thermal decomposition reactions, where a gas would be released and the reaction is endothermic.

7a

2 marks for all points plotted correctly

1 mark for a best-fit line

7b −  0.185 

Any sensible answer in this range will be allowed and, even if you do not draw the graph correctly, you will be marked on the calculation of the gradient of the graph you did draw.

7c218 

7d+218  kJ mol−1

ΔH is equal to the intercept on the ΔGaxis in this type of graph. The + is important here as this shows that it is an endothermic reaction, which is what you would expect for a thermal decomposition.

7e+ 185  J K−1 mol−1

The gradient is equal to −ΔS. ΔS is quoted in J K−1 mol−1 and the value obtained from the graph has units of kJ K–1 mol–1. The value from the graph must have its sign changed and be multiplied by 1000 to correctly give a value for ΔS. The reaction shows a positive entropy change, which would be expected when there is a gas released on heating a solid.

7f1180K 

The reaction becomes feasible when ΔG≤0. This happens when the line cuts the temperature axis and the value of ΔG = 0 and at all temperatures above this.

Topic 9 Rate equations

1Answer is B 

Rate has units of mol dm−3 s−1. The units of the rate constant are determined using the overall order of the reaction, which is 2. The equation for the reaction is used here as a distractor because if you confuse the balancing numbers with the order of reaction you would get answer C for a reaction with overall order 3. A would be the answer for a reaction which has overall order 1 and D is the answer for a reaction which has overall order 4.

2Answer is D 

As X is not in the rate equation, it is a zero-order reactant and so does not take part in the rate-determining step (slow step). The only option which does not involve X in the slow step is D. All mechanisms would cancel down to give the overall equation for the reaction

3aOrder with respect to P is 1 

Order with respect to Q is 0 

Order with respect to R is 2 

From experiment 1 to experiment 2 the concentration of P is the only one changing and as it doubles from 0.154 to 0.308, the rate of reaction also doubles, so the order with respect to P is 1.

From experiment 1 to 3 the concentrations of P and Q are changing. Both [P] and [Q] increase by a factor of 1.5 and the rate increases by a factor of 1.5, so as the order with respect to P is first order, Q must be a zero-order reactant.

From experiment 3 to experiment 4, [Q] increases by a factor of 1.33 but it is a zero-order reactant, so this will have no effect on the rate. [R] increases by a factor of 2 and overall the rate increases by a factor of 4, so the order with respect to R is 2.

3brate = k[P][R]2

Q can be left out of the rate equation as it is zero order. Even if you cannot work this out make an attempt and then use your rate equation to do any subsequent calculations.

3c1.24 × 10−4 = k(0.154)3

k = 0.0340  mol−2 dm6 s−1

Simply slot in the values from experiment 1 into the rate equation you have written and
calculate a value for k. The reaction has overall order 3, so this means the units are calculated as the units of rate divided by the total units of all the concentrations in the rate equation, so this is mol dm−3 s−1/(mol dm−3)3 which is mol dm−3 s−1/mol3 dm−9 = mol−2 dm6 s−1. Check you can work these out or simply learn them for different overall orders.

4ak is the rate constant.

A is the Arrhenius constant.

e is a mathematical constant/base of natural log.

Ea is the activation energy.

R is the gas constant.

T is the temperature.

4bk increases.

The individual components of the Arrhenius equation should be understood. Mathematically, e is the base of ln (natural log) and it has a value of 2.718 to 3 decimal places. It is accessed on your calculator using the inverse or shift function of ln. R is the gas constant and this is usually given as 8.31 J K–1 mol–1. The temperature in these calculations should be in kelvin (K). As T is on the bottom line of the Ea/RT expression you would perhaps expect it to decrease the value of k, but the power is –Ea/RT, so the negative power means that an increase in the value of T increases the rate constant.

4c

i

This expression gives a straight line as it is of the form y = mx + c, where m is the gradient
(–Ea/R) and ln A is the intercept on the ln k axis.

iiln A = 44.0 

e44.0 = 1.29 × 1019



Ea =24500 ×8.31 = 203595 J mol–1

= 204 kJ mol–1

It is important to be able to manipulate the information from the graphical analysis of the Arrhenius equation. Make sure you know that the intercept on the vertical axis is ln A and the gradient is –Ea/R. The gradient would have units of K — the rise (no units for a log value) is divided by 1/T, which gives K. Multiplying by R, which has units of J K–1 mol–1, cancels out the K and this leaves units of J mol–1 for Ea. This can be converted to kJ mol-1 by dividing by 1000.

5a

This expression is derived by taking natural logs of the Arrhenius equation. The intercept on the ln k axis is ln A. The gradient is −Ea/R.

5bln A = 38 

A = e38 = 3.19 × 1016

The intercept of the line with the ln k axis equals ln A. Converting this back to A means using e. This is obtained by using the shift or inv function on your calculator with the ln button. Practise working both ways with this log function. ln (3.19 × 1016) should give you around 38 and e38 should give you 3.19 × 1016.

5cThe gradient of the line is = = −19 000 

= –19000, so Ea = 19000 × 8.31 = 157890 J mol−1

Ea = 158 kJ mol–1

The gradient is negative as the line slopes downwards. The gradient is −Ea/R and R is given as the gas constant. Remember that the value obtained will be in J mol−1 due to the units of the gas constant, so remember to divide this by 1000 to convert to kJ mol−1.

6a

i[A] = 0.009 mol dm–3

Time = 390 s 

iiZero order would be a straight line.

6b

iAt 0.012:gradient = 0.016/800 = 2 × 10–5 mol dm–3 s–1

At 0.018:gradient = 0.018/400 = 4.5 × 10–5 mol dm–3 s–1

The gradient of the line from a concentration–time graph gives a measure of the rate of reaction. The units are mol dm–3 s–1 as concentration is being divided by time in seconds. You may be asked to draw a tangent and work out its gradient. There would always be a little bit of give and take with the value for the gradient of a tangent you have drawn.

iiConcentration changes by = 1.5 

Rate changes by = 2.25 

Order with respect to A = 2  (as 1.52 = 2.25)

rate = k[A]2

= 0.139  mol–1 dm3 s–1

The first thing to notice is that, because this is a decomposition, A is the only reactant. The change in concentration is a factor of 1.5 between 0.012 and 0.018 and the rate changes by a factor of 2.25. If the order of reaction had been 1, then the rate would have changed by a factor of 1.5 as well. We knew it was not zero order from the shape of the curve, so second order would have been a good guess. The value of the rate constant and its units are determined in the usual way using the suggested concentration and rate. Make sure you can write the units of the rate constant even if you get the rate equation wrong as the units you write would be marked correct for the given rate equation.

7aFrom experiment 1 to experiment 2:

[B] increases by a factor of 1.5 

[C] increases by a factor of 2 

[A] no effect as zero order

Rate increases by a factor of 1.5 × (22) = 6 