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WORKBOOK ANSWERSAQA A-level ChemistryPhysical 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 cycles1 Answer 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.

2a Answer is C

2b Answer is E

2c Answer is B

2d Answer 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.

© Alyn G. McFarland and Nora Henry 2016 Hodder Education 1

2e E = –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

i Mg(g) → Mg+(g) + e−

ii Mg+(g) → Mg2+(g) + e−

iii Mg(s) + ½O2(g) → MgO(s)

iv O–(g) + e− → O2−(g)

v MgO(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.

3b From left to right:

i Mg2+(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 + ½BDEH + 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.

4b ΔlattH = –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.

ii ΔsolH = –lattice enthalpy (formation) + ΣΔhydH

= +2492 + (–1920) + 2(–364)

= –156 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.

iii Lower 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 question6a

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

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

The unknown in this expression is ΔfH.

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

Name of enthalpy change Enthalpy change H/kJ mol−1

Atomisation of Mg Mg(s) → Mg(g) +150

First ionisation energy of Mg Mg(g) → Mg+(g) + e− +736

Second ionisation energy of Mg Mg+(g) → Mg2+(g) + e− +1450

Bond dissociation enthalpy of F2 F2(g) → 2F(g) +158

First electron affinity of F F(g) + e− → F−(g) −348

Lattice dissociation enthalpy of MgF2

MgF2(s) → Mg2+(g) + 2F−(g) +2883

Solution enthalpy of MgF2 MgF2(s) → Mg2+(aq) + 2F−(aq) −20

Enthalpy of hydration of Mg2+ Mg2+(g) → Mg2+(aq) −1891

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 + 2EA1H

+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+) + 2hydH(F−)

−20 = +2883 + (−1891) + 2hydH(F−)

2hydH = –1012

hydH = –506 kJ mol−1

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


6b

i Ions 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.

ii Magnesium 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)1 Answer 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 ΔS < 0’. This is the same question.

2 Answer 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.

3 Answer 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.

4a It 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 = ΣfH(products) − ΣfH(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 − TS

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 indicates that the reaction is not feasible.

4e When G = 0, H = TS

+602 = T(0.8666)

= 694.7 K

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).

5 When G = 0, H = TS

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 = ΣfH(products) – ΣfH(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, as ΔH 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.

iii ΔG = 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.

6b ΔG = 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.

6c ΔH = +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. ΔS is calculated 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.


1200

T/K

ΔG/k

J mol

–1

400 800 1200 16000

50

100

150

200

250

–50

x

x

x

–100

–150

–200

300

1200

x

7c 218

7d +218 kJ mol−1

ΔH is equal to the intercept on the ΔG axis 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.

7f 1180 K

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 equations1 Answer 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.

2 Answer 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

3a Order 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.

3b rate = 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.

3c 1.24 × 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 andcalculate 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.

4a k 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.

4b k 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.

ii ln A = 44.0

e44.0 = 1.29 × 1019

Ea = 24 500 × 8.31 = 203 595 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.

5b ln 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.


5c The gradient of the line is = = −19 000

= –19 000, so Ea = 19 000 × 8.31 = 157 890 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

ii Zero order would be a straight line.

6b

i At 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.

ii Concentration 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.

7a From 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

Rate is 2.20 × 10–3 × 6 = 0.0132

An alternative method would be to calculate a value for the rate constant using experiment 1 and use this to calculate the missing data in the expression. However, the calculation of the rate constant is later in the question and it is good practice to be able to work out the factor by which the concentration has changed and how this affects rate.

7b From experiment 1 to experiment 3:

rate increases by a factor of 4.5

[B] increases by a factor of b

[C] increases by a factor of 1.5


b × 1.52 = 4.5

b × 2.25 = 4.5 so b = 2

concentration of B is 0.0230 × 2 = 0.0460

The factors by which the concentrations are changing are very important as they dictate the effect on the rate. In this example b is used as the unknown factor by which [B] changes. This is calculated using the idea that as the reaction is first order with respect to B then a change of b will change the rate by b.

7c From experiment 1 to experiment 4:

rate increases by a factor of 5

[B] increases by a factor of 0.8 (decreases by 1.25 and 1/1.25 = 0.8)

[C] increases by a factor of c


0.8 × c2 = 5

c2 = 6.25, so c = 2.5

concentration of C is 0.0128 × 2.5 = 0.0320

The calculation is a little more complex because the order with respect to C is second order and also because the concentration of B is decreasing or changing by a factor of 0.8. So changing [B] by 0.8 changes the rate by 0.8. c is the factor by which [C] is changing and it is second order, so the rate changes by a factor of c2, which is used in the calculation. Again, if you are struggling with these calculations work out the value for k and substitute the values into the rate equation. You can also do this to check your answer is correct but only at the end if you have time.

7d rate = k[B][C]2


2.20 × 10−3 = k(0.0230)(0.0128)2

= 584 mol−2 dm6 s−1

Again substitute the values from experiment 1 into the rate equation and solve for k. The reaction is overall order 3 so the units are mol−2 dm6 s−1. Remember to give your answer to the number of significant figures asked for in the question.


Topic 10 Equilibrium constant (Kp) for homogeneous systems

1 Answer is B

The total number of moles is 0.0120 + 0.0740 + 0.114 = 0.2 mol. The mole fraction of nitrogen is 0.0120/0.2 = 0.06. The partial pressure is 0.06 × 100 = 6 kPa. Answer A is obtained by simply multiplying the moles of nitrogen by the total pressure. Answer C is the partial pressure of hydrogen and answer D is the partial pressure of ammonia. Check that you can calculate all these partial pressures, that the total of the mole fraction adds up to 1 and that the total of the partial pressures is equal to the total pressure (100 kPa).

2a

i

ii

iii

iv

With all these Kp expressions, make sure the products are on the top line and the reactants on the bottom line, raised to the power of the balancing numbers given in the equation. (iv) may seem unusual as there is a ½ power, but this is perfectly acceptable. Never use square brackets in a Kp expression as this may lose you marks. Square brackets mean concentration.

2b Answer is (ii)

The only Kp that would have no units is one that has equal numbers of moles of gas on the left-hand and on the right-hand side of the equation. This is only true in (ii).

2c An increase in pressure would move the position of equilibrium from left to right.

Smaller gas volume on right-hand side, so equilibrium moves to reduce the volume.

This minimises the effect of the increase in pressure.

A change in pressure has no effect on the value of Kp.

Remember that only changes in temperature will have an effect on an equilibrium constant. The explanation about the position of equilibrium comes from AS.


2d

The units of Kp are determined from the units of pressure. The Kp expression for (iv) has (pressure)/(pressure) , which gives pressure so as the pressure unit is Pa the unit must be

.

3a Moles of SO2 that reacted = 0.15 × 0.0240 = 3.6 × 10−3 mol

Moles of O2 that reacted = = 1.8 × 10−3 mol

Moles of SO2 remaining = 0.0240 – (3.6 × 10−3) = 0.0204 mol

Moles of O2 remaining = 0.0200 – (1.8 × 10−3) = 0.0182 mol

Moles of SO3 formed = 3.6 × 10−3 mol

This can be done using a table below the equation, as shown below:

Equation 2SO2 + O2 ⇌ 2SO3

Initial moles 0.0240 0.0200 0

Reacting moles −3.6 × 10−3 −1.8 × 10−4 +3.6 × 10−3

Equilibrium moles 0.0204 0.0182 3.6 × 10−3

The 15% tells us how much of the SO2 reacted. Using the balancing numbers in the equation we can work out how much of the oxygen reacts and how much of the SO3 is formed.

3b Total equilibrium moles of gas = 0.0204 + 0.0182 + 3.6 × 10−3 = 0.0422

Mole fraction of SO2 = = 0.483

Mole fraction of O2 = = 0.431

Mole fraction of SO3 = = 0.0853

Partial pressure of SO2 = 0.483 × 55 = 26.57 kPa

Partial pressure of O2 = 0.431 × 55 = 23.71 kPa

Partial pressure of SO3 = 0.0853 × 55 = 4.69 kPa

The mole fractions are calculated by dividing the moles by the total moles of gas present at equilibrium. The total of the mole fraction should be 1. In this case it is 0.9993, which is due to rounding. The partial pressures are calculated by multiplying the mole fraction of each gas by the total pressure (55 kPa in this case). The total of the partial pressures should equal the total pressure. In this case the partial pressures add up to 54.97 kPa, which is again due to rounding. Generally, work to 3 or 4 significant figures during a calculation.


3c

= = 1.31 × 10−3 kPa−1

This is a common type of calculation and it may be structured into different parts as shown here or it may all be in one question where the final answer is the value of Kp with its units if there are any. Work your way through all of the steps and work to 3 or 4 significant figures. You can try the calculation again with 50% of the SO2 reacting. The partial pressures are 17.37 kPa, 20.26 kPa and 17.37 kPa and the value of Kp is 0.0494 kPa−1.

4

2A ⇌ B + C

Initial moles 1.00 0 0

Reacting moles −2x +x +x

Equilibrium moles 1 − 2x x x

Mole fractions 1 − 2x x x

Partial pressure (1 − 2x)P xP xP

= 0.0400

Filling in the partial pressures:

x = 0.200(1 − 2x)

x = 0.2 − 0.4x

1.4x = 0.2

x = 0.143

Moles of A = 1 − 2x = 1 − 0.286 = 0.714 mol

Moles of B = 0.143

Moles of C = 0.143


Note that the total pressures cancel out in the expression. If Kp has no units, you can use either the equilibrium moles, the mole fractions or the partial pressures in the Kp expression. The total moles is often not in terms in x; in this case it was 1 − 2x + x + x = 1.

−2x was used for the moles of A reacting as it makes it easier not to have x/2 in the expression for B and C, which is what would have happened if you had used −x for A. Always make sure you know what is being asked in terms of x. The moles of A were 1 − 2x whereas the moles of B and C were simply x, so always apply the value of x you calculate.

5

2P + Q ⇌ 3R

Initial moles 0.247 0.173 0

Reacting moles −0.062 −0.031 +0.0930

Equilibrium moles 0.185 0.142 0.0930

Mole fractions 0.440 0.338 0.221

Partial pressure 46.2 35.5 23.2

= 0.165 no units

This is a standard question for determining the value of Kp from equilibrium moles and partial pressures. As 0.0930 mol of R are present at equilibrium this means that (0.0930/3) × 2 = 0.062 mol of P reacted and 0.031 mol of Q reacted. This means that the moles of P present at equilibrium are 0.247 − 0.062 = 0.185. The moles of Q present at equilibrium are 0.173 − 0.031 = 0.142. The total equilibrium moles are 0.185 + 0.142 + 0.0930 = 0.420. The mole fractions are then determined by dividing the moles by 0.42 and the partial pressures are determined by multiplying the mole fractions by the total pressure, which is 105 kPa. There are no units of Kp as there is pressure3/pressure3 in the Kp expression.


Topic 11 Electrode potentials and electrochemical cells

1 Answer is C

Magnesium is the most easily oxidised as the potential for Mg → Mg2+ + 2e− is +2.37 V, so magnesium gives up its electrons most easily. These electrons can be gained by another species, causing that species to become reduced. The strongest oxidising agent in the table would be copper(II) ions as these are the most easily reduced.

2a Zn → Zn2+ + 2e−

It is always important to remember that oxidation reactions occur at the negative electrode. (Use NEGATOX to help you remember that the NEGATive electrode is where OXidation occurs). When the cell is in use, electrons flow from the negative electrode (where an oxidation occurs) to the positive electrode (where a reduction occurs).

2b Mn oxidation state in MnO2 is +4

Mn oxidation state in MnO(OH) is +3

Decrease in oxidation state is reduction

The change in oxidation state of manganese in this equation is a common question, but you can also be asked to write this equation. Remembering that the manganese(IV) oxide, MnO2, reacts with water and that the oxidation state changes from +4 to +3 will help. The number of electrons gained is the same as the same as the change in the oxidation state and the OH− on the right-hand side balances the charge.

2c Zn + 2MnO2 + 2H2O → Zn2+ + 2MnO(OH) + 2OH−

The skill of being able to combine half equations was first met at AS and is commonly tested in questions on electrode potentials and electrochemical cells. Make sure you have one oxidation and one reduction equation and multiply them to balance out the electrons. Add them together and cancel out electrons and anything else that appears on both sides of the equations and can be cancelled down. Most often it is OH−, H+ or H2O, but not in this case.

2d Zinc casing used up

As the reaction Zn → Zn2+ + 2e− occurs, the zinc casing is used up in this cell. This comprises the cell and the paste may leak out.

3a E(Fe2+/Fe) > E(Zn2+/Zn)

Fe2+ would be reduced to Fe as Zn oxidised to Zn2+

The emf for oxidation of iron and reduction of Zn2+ is negative/−0.32 V

It is important to understand that a greater standard electrode potential means that the reduction reaction is more likely. Fe2+/Fe has a greater value, so it is reduced rather than oxidised. Zinc will protect the iron from being oxidised and this is the basis of how a galvanised object still protects the iron underneath even when it is scratched and how sacrificial protection works to prevent rusting.


3b E(Sn2+/Sn) > E(Fe2+/Fe)

Sn2+ would be reduced to Sn as Fe oxidised to Fe2+

The emf for oxidation of iron and reduction of Sn2+ is positive/+0.30 V

This time the electrode potential for the reduction of tin is greater than the electrode potential for the reduction of iron, so tin is reduced and iron is oxidised. Damaged tin cans can cause the iron or steel in the can to rust, which can spoil the food inside. This is why you are advised to avoid buying dented or damaged food cans.

3c Standard hydrogen electrode

Connected to iron half cell using an external circuit with a (high resistance) voltmeter

Using a salt bridge

Fe2+(aq) and Fe3+(aq) both 1.00 mol dm−3

Platinum electrode

The details of how to use a standard hydrogen electrode to measure standard electrode potentials are important. You can give more detail about the standard hydrogen electrode but the main focus of this answer is to detail how the half cells are connected and the contents of the iron half cell in terms of the concentrations of both ions in the solution and also the platinum electrode. You might even be expected to draw this half cell and label the main points from above.

3d

i E(Fe3+/Fe2+) > E(Sn2+/Sn)

E(Fe2+/Fe) < E(Sn2+/Sn)

The emf of Fe3+ reacting with Sn is positive/0.91 V

The emf of Fe2+ reacting with Sn is negative/−0.30 V

The third mark here is similar to the first but it is best to give all possible answers and it also makes sure you understand the various ways the question can be answered. The second mark is similar to the fourth as well. However, as the question asks for the emf it is best to explain it in terms of the potentials and also calculate the emf. Remember that a greater E means that that half cell will undergo reduction.

ii Sn|Sn2+||Fe3+,Fe2+|Pt

State symbols are not required unless specifically asked for in the question. If they were required the answer would be Sn(s)|Sn2+(aq)||Fe3+(aq),Fe2+(aq)|Pt. The | represents the phase boundary between the solid Sn and the Sn2+(aq). It is also used to separate the platinum electrode from the two iron ions in solution. The two iron ions in solution are separated by a comma as they are in the same phase. || represents the salt bridge in the cell. Marks are deducted for each error in the cell representation.

iii Sn + 2Fe3+ → Sn2+ + 2Fe2+

Balancing the electrons gives 2Fe3+ + 2e− → 2Fe2+ with Sn → Sn2+ + 2e−. Combining these will give the ionic equation with the electrons cancelled out.


4a

i Cd + 2NiO(OH) + 2H2O → Cd(OH)2 + 2Ni(OH)2

Knowing that the cadmium cell was the negative electrode in this question is essential, as this tells us that the cadmium half cell is the oxidation half cell (remember NEGATOX). The cadmium equation is reversed and the reduction (bottom) equation multiplied by 2 before adding them together and cancelling down the 2e– and also 2OH–. Each error in the equation would be penalised by 1 mark.

ii 1.40 = 0.52 − x

x = −0.88 V

The negative sign before x (which is the unknown electrode potential) is so that it is used as an oxidation in the calculation, but the value turns out to be negative as would be expected. If you were tempted to write +0.88 V remember that would mean that the cadmium would be reduced rather an oxidised as it would be greater than the E value of the other half cell. This is always a good way to check.

iii Cd(OH)2 + 2Ni(OH)2 → Cd + 2NiO(OH) + 2H2O

When a secondary cell is recharged the reaction is simply the reverse of the reaction when it is being discharged. This is a common question for secondary cells.

4b H2 + 2OH− → 2H2O + 2e−

2H2 + 4OH− → 4H2O + 4e− (accepted)

In this question you need to work out what must be added to the reaction at the positive electrode to obtain the overall equation.

Positive electrode (reduction): O2 + 2H2O + 4e− → 4OH−

Negative electrode (oxidation): to be determined

To get overall equation: 2H2 + O2 → 2H2O

There is no H2 on the left-hand side of the + electrode equation, so 2H2 is needed. O2 is present. Also the OH− have to cancel out so 4OH− is required on the left. On the right you need 4e– and also 4H2O so that the H2O cancels out.

This gives 2H2 + 4OH− → 4H2O + 4e−, which cancels down to H2 + 2OH− → 2H2O + 2e−. Either would be accepted.


Topic 12 Acids and bases1 Answer is B

The conjugate acid of a substance is the species that acts as an acid (proton donor) to form the substance in question. In this example HPO4

2− would be formed from H2PO4− when it donates a

single H+. The conjugate base of HPO42− is PO4

3− as PO43− would act as a base (proton acceptor) to

form HPO42−. Remember that a proton is the same as H+.

2 Answer is A

Sulfuric acid is a diprotic acid so [H+] = 2 × [acid]. So [H+] = 2 × 2.14 = 4.28 mol dm–3. Using pH = −log10[H+] we can determine the pH to be −log10(4.28) = −0.63. A common mistake would be not to multiply the concentration of the acid by 2, which would give B as an answer. C and D are there as distractors to those who may not think pH can have a negative value.

3 Answer is A

Kc may have units of mol dm–3 but its units do vary. Kp will either have no units or pressure units. Kw always has units of mol2 dm–6. Ka has concentration2 on the top and concentration on the bottom line of the expression so it will always have units of mol dm–3.

4a [H+] = 0.0125 mol dm−3

pH = −log10[H+] = −log10(0.0125) = 1.90

Again a strong acid and hydrochloric acid is monoprotic, so its [H+] is equal to the concentration of the acid.

4b [OH−] = 0.510 mol dm−3

1.96 × 10−14 mol dm−3

pH = –log10[H+] = −log10(1.96 × 10−14) = 13.71

As NaOH is a strong base/alkali its pH is determined using Kw. [H+] = Kw/[OH−]. The value for Kw at this temperature (298 K) was given at the start of the question.

4c

[H+] = 2.11 × 10–3 mol dm–3

pH = −log10[H+] = −log10(2.11 × 10–3) = 2.68

[H+] for a weak acid is determined using . pH is determined in the usual

way. Make sure you give all answers in this question to 2 decimal places. This is common with pH. Solution concentrations should be quoted to 3 significant figures.


5 Moles of H2SO4 = = 6.255 × 10–3 mol

Moles of H+ is 2 × 6.255 × 10–3 = 0.01251 mol

New concentration of H+ = 10–1.5 = 0.0316 mol dm–3

×1000 = 0.0316

v = 396 cm3

Volume of water added = 396 − 15 = 381 cm3

This is quite a complex calculation. The moles of H+ are determined in the initial solution (0.01251) and [H+] in the new solution is calculated from the pH (0.0316). The final volume is the unknown in the fourth step in this calculation and it is represented by v. Dividing the moles of H+

by v and multiplying by 1000 will calculate [H+]. Remember to subtract 15 at the end as the questions asks for the volume of water added and not the total final volume. You can check this is correct by doing the calculation the other way round using the final volume to determine the concentration of H+:

Moles of H+ = 0.01251

New [H+] = × 1000 = 0.0316 mol dm−3

pH = −log10[H+] = −log10(0.0316) = 1.5

6a

i A is 2-chloropropanoic acid.

B is 3-chloropropanoic acid.

ii A has a chiral centre/asymmetric centre/four different groups bonded to the same carbon atom.

CH3C*HClCOOH/correct identification of the chiral/asymmetric centre

iii 2CH2ClCH2COOH + Na2CO3 → 2CH2ClCH2COONa + CO2 + H2O

iv Cl atom withdraws electrons.

This weakens the O–H, so more H+ released.

The parts of (a) are synoptic questions taking in aspects of organic chemistry, as many weak acids are organic.


6b

i CH3CHClCOOH ⇌ CH3CHClCOO– + H+

ii Ka = = 10(–2.83) = 1.48 × 10–3 mol dm–3

[H+] = = = 5.09 × 10–3 mol dm–3

pH = –log10 [H+] = 2.29

Writing dissociation equations and calculating the pH of a weak acid are fundamental to acid equilibria. Make sure you include a reversible arrow if the acid is weak. Remember to use the log button on your calculator and not the ln button.

6c

i [OH–] = 0.0152 mol dm–3

6.58 × 10–13 mol dm–3

pH = –log10[H+] = 12.18

ii Acid A = = 4.375 × 10–4 mol

Moles of NaOH = 4.375 × 10–4 mol

Volume of NaOH = × 1000 = 28.8 cm3

iii Answer is C

iv If C was chosen in (iii) the answer is cresol purple.

The colour change is yellow to purple.

The indicator changes colour in the pH range of the vertical region of the curve.

The parts of (c) are common in titration questions. In (c)(i) make sure you use the Kw value given to calculate [H+] from the [OH–]. The initial pH is useful in deciding on the pH curve later in the question. The volume calculation is also very common and all the curves have their vertical region around 29 cm3. As the acid is a weak acid the pH changes rapidly between pH values of 6 and 10/11. Also, as alkali is added to acid, the curve should start in the acid region and the pH increase as alkali is added. The indicator chosen must change colour in the pH range of the vertical region of the curve you have chosen. Also make sure you use the colour going from lower pH to higher as this is the way this titration has been carried out.


7 Ka =

[HA] = 0.125 mol dm−3

[A–] = 0.210 mol dm−3

[H+] =

[H+] = = 1.43 × 10−4 mol dm−3

pH = −log10[H+] = −log10(1.43 × 10−4) = 3.85

Using the answer carried through on your calculator you would get 3.84, but both 3.84 and 3.85 would be accepted as long as you show your working out. The pH of a buffer is calculated from the [HA] and [A−]. These are determined in the question. As the total volume was 100 cm3, both moles are multiplied by 10 to calculate the concentration. The A− comes from NaA. The Ka expression is then used to calculate [H+] and pH is determined in the usual manner. The same answer is obtained using the actual moles but it is better to work out the concentration if you are given a volume. Check that you get the same answer with 0.0125 and 0.0210.

Alternatively you can use the Henderson‒Hasselbalch equation, which is:

pH = pKa + log10([A−]/[HA]) where pKa is −log10Ka

pKa = 3.62; so pH = 3.62 + log10(0.210/0.125) = 3.85

8 Moles of HCl added = = 5.2 × 10−4 mol

New HA = 0.0125 + 5.2 × 10−4 = 0.01302 mol

New A− =0.0210 − 5.2 × 10−4 = 0.02048 mol

New total volume = 105 cm3

New [HA] = × 1000 = 0.124 mol dm−3

New [A–] = × 1000 = 0.195 mol dm–3

[H+] =

[H+] = = 1.53 × 10−4 mol dm−3

pH =−log10[H+] = −log10(1.53 × 10−4) = 3.82

When acid is added to a buffer, HA increases by the same as the moles of H+ added. A− decreases by the same as the moles of H+ added. The new concentrations are worked out using the new total volume of 105 cm3 and the [H+] determined using Ka. pH is then determined. The


same answer for [H+] would be determined using the new moles but if a volume is given then use it as it is more correct to put concentrations into the Ka expression.

Alternatively, you can use the Henderson‒Hasselbalch equation which is:

pH = pKa + log10([A−]/[HA])

where pKa is −log10Ka.

pKa = 3.62; so pH = 3.62 + log10(0.195/0.124) = 3.82

9a [OH–] = 0.242 mol dm−3

4.13 × 10−14 mol dm−3

pH = −log10[H+] = −log10(4.13 × 10−14) = 13.38

Potassium hydroxide is a strong base/alkali, so its [OH−] is converted into [H+] using Kw. Then pH is determined in the usual manner.

9b

i 2KOH + H2SO4 → K2SO4 + 2H2O

The most common mistake here is to write the formula of potassium sulfate incorrectly as KSO4 in haste, which means the very important 2:1 ratio of KOH:H2SO4 can be missed. This ratio is vital. Some questions may give you the equation so that you don’t mess it up but it is still a common question.

ii Moles of KOH = = 6.05 × 10−3 mol

Moles of H2SO4 = = 3.025 × 10−3 mol

Volume of H2SO4 required = = 21.6 cm3

This is a standard titration question at AS and also at A-level. Make sure you use the ratio in the equation to calculate the amount, in moles, of H2SO4. This will then allow you to calculate the ratio.


iii Moles of KOH = = 6.05 × 10−3 mol

Moles of H2SO4 = = 2.10 × 10−3 mol

Reactant in excess is KOH using 2:1 ratio

Moles of KOH in excess = 6.05 × 10−3 − (2 × 2.1 × 10−3) = 1.85 × 10−3 mol

New total volume = 15 + 25 = 40 cm3

New concentration of KOH = × 1000 = 0.04625 mol dm−3

[OH-] = 0.04625 mol dm–3

2.16 × 10−13 mol dm−3

pH = −log10[H+] = −log10(2.16 × 10–13) = 12.67

It is important to be able to use the ratio in the equation to work out which reactant is in excess but you have a clue from (b)(ii) as 21.6 cm3 of sulfuric acid were required for neutralisation (pH = 7) and you are asked to determine the pH when only 15.0 cm3 have been added, so the KOH should be in excess. As 1 mol of H2SO4 reacts with 2 mol of KOH, you must multiply the moles of H2SO4 by 2 to determine how many moles of KOH will have reacted. This is then subtracted from the initial moles of KOH. The final volume of the solution is 40 cm3. This is used to calculate the new concentration of KOH and the rest of the calculation is like the normal method of calculating the pH of a strong base/alkali.

9c Moles of CH3COOH = = 4.08 × 10−3 mol

Moles of KOH = = 2.42 × 10−3 mol

In buffer, moles of CH3COOH = 4.08 × 10−3 – 2.42 × 10–3 = 1.66 × 10−3 mol

New concentration of CH3COOH = × 1000 = 0.0332 mol dm−3

Moles of CH3COO− = moles of KOH = 2.42 × 10−3 mol

New concentration of CH3COO– = ×1000 = 0.0484 mol dm−3

Ka = so [H+] =

[H+] = = 1.21 × 10−5 mol dm−3

pH = −log10[H+] = −log10(1.21 × 10−5) = 4.92


Buffers are among some of the more complex calculations. It is important to be able to determine the moles of the weak acid and the moles of the salt in the solution. Once this is done these can be used in the Ka expression to calculate [H+] or they can be more correctly converted to concentrations using the total volume and the new concentrations used in the Ka expression. Check that you get the same answer using the moles of the acid and the salt (1.66 × 10−3 and 2.42 × 10−3, respectively). Alternatively you can use the Henderson‒Hasselbalch equation, which is:

pH = pKa + log10([A−]/[HA])

where pKa is −log10Ka.

pKa = 4.75; so pH = 4.75 + log10(0.0484/0.0332) = 4.91

The answer is rounded due to the rounding of the value of pKa.


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Hodder Education - Educational Resources for Students ...€¦ · Web viewThe strongest oxidising agent in the table would be copper(ii) ions as these are the most easily reduced