WJEC A-Level Chemistry Unit 3, June 2025: Question 9

19 marks · Hard difficulty · Structured Questions

Calculate reaction enthalpy using formation data, explain the common ion effect on ethanoic acid, calculate pH and plan buffer preparation, and determine Kc and reaction thermochemistry for a ruthenium-catalysed equilibrium.

Practise this question

Question

Question 9 presents the industrial production of ethanoic acid from iodomethane and carbon monoxide via CH3COI. Part (a) asks to calculate standard enthalpy change for reaction 1 using bond energies or enthalpies of formation tables and justify the choice. Part (b) explores the weak acid CH3COOH and strong acid HI: (i) asks the effect of HI on CH3COOH dissociation, (ii) asks to calculate the pH of 0.250 mol dm⁻³ HI, and (iii) is a 6-mark QER question to describe how to prepare a buffer solution of pH 4.45 from 100 cm³ of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³) and explain its action. Part (c) provides a table of Kc against temperature for a ruthenium-catalysed equilibrium [RuL5CH3] + CO ⇌ [RuL5COCH3], asking (i) what energy changes can be deduced and (ii) to calculate Kc and the corresponding reaction temperature given initial and equilibrium concentrations.
Question text

9. Ethanoic acid can be produced in a reaction between iodomethane and carbon monoxide,

followed by treatment with water.

CH3l + CO CH3COI Reaction 1

CH3COI + H2O CH3COOH + HI Reaction 2

(a) Use data from one of the tables below to calculate the more accurate value of the

standard enthalpy change for reaction 1. Give a reason for your choice of data. [3]

Bond energy Standard enthalpy change of

Bond –1 Substance –1

/kJmol formation/kJ mol

C H 413 CH3I –14

C C 346 CO –111

C I 240 CH3COI –162

C O 745

C O

in carbon 1072

monoxide

Enthalpy change for reaction 1 = kJ mol–1

(b) Reaction 2 produces a mixture of the weak acid CH3COOH and the strong acid HI.

(i) Suggest what effect the presence of HI will have on the dissociation of CH3COOH

12 in this mixture compared to an aqueous solution containing only CH3COOH.

Give your reasoning. [2]

(ii) Calculate the pH of an aqueous solution of the strong acid HI of concentration

0.250 mol dm–3. [2]

14 pH =

(iii) A student is given 100 cm3 of a solution of ethanoic acid of concentration

0.100 mol dm–3 and wishes to produce a buffer solution of pH 4.45. He has access

to the full range of chemicals in a laboratory.

Describe how the buffer solution could be formed, calculating the amounts of any

other substances required. Explain how the buffer solution works to keep the pH

of the solution constant.

[6 QER]

K for CH COOH = 1.74 × 10–5 mol dm–3

a 3

13 © WJEC CBAC Ltd. (1410U30-1)

(c) The complex [Ru(CO)2(P(CH3)3)2CH3I] catalyses the reaction of iodomethane with

carbon monoxide. One stage in the process is the reversible reaction below, with five of

the ligands shown as L.

14 © WJEC CBAC Ltd. (1410U30-1)

[RuL5CH3] + CO [RuL5COCH3]

The values of the equilibrium constant for this process at different temperatures in

methylbenzene solvent are listed in the table.

Temperature / °C K / mol–1 dm3

c

34 1220

42 694

49 406

56 290

64 196

74 17 120

(i) Suggest what information this table allows us to deduce about the energy

changes during this reversible reaction. [2]

(ii) A solution containing a mixture of complex [RuL5CH3] of concentration

1.25 × 10–3 mol dm–3 and carbon monoxide of concentration 7.55 × 10–4 mol dm–3

is placed in a sealed tube, heated to a set temperature and the mixture allowed

to reach equilibrium. The equilibrium mixture contains 1.37 × 10–4 mol dm–3 of

[RuL5COCH3].

Find the value of Kc and hence suggest the temperature used for the experiment.

[4]

16 © WJEC CBAC Ltd. (1410U30-1)

K = mol–1 dm3

c

Temperature = °C

Mark scheme

Show the mark scheme Mark scheme for Question 9. (a) ΔH = -162 - (-111 + -14) = -37 kJ mol⁻¹, with justification that formation enthalpies give more accurate values because bond energies are average values (3 marks). (b)(i) Less dissociation of CH3COOH as HI releases H⁺ shifting equilibrium to the left (2 marks). (b)(ii) pH = -log(0.250) = 0.60 (2 marks). (b)(iii) Indicative content for buffer: addition of sodium ethanoate or NaOH, equilibrium equations, explanation of resistance to pH change upon adding acid/base, and calculation showing [H⁺] = 3.548 × 10⁻⁵ mol dm⁻³, [CH3COONa] = 0.0490 mol dm⁻³, mass = 0.402 g (6 marks). (c)(i) Exothermic because Kc decreases as temperature increases (2 marks). (c)(ii) Equilibrium concentrations: [CO] = 6.18 × 10⁻⁴ mol dm⁻³, [RuL5CH3] = 1.113 × 10⁻³ mol dm⁻³, giving Kc = 199 mol⁻¹ dm³, corresponding to temperature range 60-64 °C (4 marks). Total: 19 marks.

Marks available

Question Marking details

AO1 AO2 AO3 Total Maths Prac

9 (a) –162 – (–111 – 14) (1) 1

= –37 kJ mol–1 (1) 1

enthalpies of formation give more accurate value as bond energies 3 1

are average values (1) 1

award (2) for value of –19 kJ mol–1 based on bond energies

(b) (i) award (1) each for any two of following

• less dissociation of CH3COOH

• HI releases H+ ions so the weak acid equilibrium shifts to the

reactants

• CH COO– ions combine with H+ ions to form CH COOH

(ii) pH = – log 0.250 (1)

22 2

pH = 0.60 (1)

Marks available

AO1 AO2 AO3 Total Maths Prac

(iii) Indicative content

1. Buffer formed from mixture of a weak acid and a salt of the

weak acid

2. Add sodium ethanoate (or any other Group 1 metal ethanoate or

small amount of sodium hydroxide)

3. CH COOH ⇌ CH COO– + H+ (or equivalent explanation)

4. CH COONa → CH COO– + Na+ (or equivalent explanation)

5. Addition of a small amount of acid shifts equilibrium to left hand9

+ 2 2 2 6 2 2

side removing H

6. Addition of a small amount of base removes H+ so equilibrium

shifts to the right to replace H+ OR OH– ions react with

undissociated acid molecules forming ethanoate ions and water

7. Buffer with pH 4.45 has [H+] = 3.548 × 10–5 mol dm–3

[CH3COOH] −5 –3

8. [CH3COONa] = 3.548 × 10−5 × 1.74 × 10 = 0.0490 mol dm

9. Mass of CH3COONa = 0.0490 × 0.1 × Mr = 0.402 g

Marks available

AO1 AO2 AO3 Total Maths Prac

5-6 marks

Clear explanation of how the buffer works and required mass of

sodium ethanoate calculated; both equations included

There is a sustained line of reasoning which is coherent, relevant,

substantiated and logically structured. The information included in

the response is relevant to the argument.

3-4 marks

Basic description of how the buffer is made and what happens10

when either acid or base is added; one equation included

There is a line of reasoning which is partially coherent, largely

relevant, supported by some evidence and with some structure.

Mainly relevant information is included in the response but there

may be some minor errors or the inclusion of some information not

relevant to the argument.

1-2 marks

Substance needed to form a buffer is identified; attempt to describe

what happens when either acid or base is added

There is a basic line of reasoning which is not coherent, supported

by limited evidence and with very little structure. There may be

significant errors or the inclusion of information not relevant to the

argument.

0 marks

No attempt made or no response worthy of credit.

(c) (i) reaction is exothermic (must attempt reason to gain this mark) (1)

award (1) for either of following

• as temperature increases, equilibrium shifts to the left

• as temperature decreases, equilibrium shifts to the right

Marks available

AO1 AO2 AO3 Total Maths Prac

(ii) award (1) for equilibrium concentrations

[CO] = 6.18 × 10–4 mol dm–3 1

[RuL CH ] = 1.113 × 10–3 mol dm–3

[(RuL5COCH3)] 1

Kc = [(RuL CH )] [CO] (1)

Kc = 199 mol–1 dm3 (1) 4 3

award (1) for any temperature in the range 60-64°C 1

ecf possible for temperature if it fits in the range

ecf possible for Kc value only when there is an error in calculating

equilibrium concentrations of CO and [RuL5CH3]

Question 9 total 2 7 10 19 8 2

How to answer it

WJEC Chemistry A-Level • Unit 4 Focus • 19 Marks Total

Synthesis & Properties of Ethanoic Acid: Energetics, Acids & Equilibria

📋 What this question tests

This multi-topic question assesses core physical chemistry principles drawn from advanced energetics, acid-base equilibria, and dynamic equilibrium kinetics:

  • Energetics: Comparing mean bond enthalpies versus standard enthalpies of formation (ΔfH°) and calculating ΔH of reaction using Hess's Law cycles.
  • Common Ion Effect: Explaining Le Chatelier shifts in weak acid dissociation equilibria upon adding a strong acid.
  • Acid-Base Calculations: Computing strong acid pH and carrying out a 6-mark QER response on acidic buffer action, component equations, and required salt mass.
  • Equilibrium & Le Chatelier: Deducing the sign of ΔH from the variation of Kc with temperature.
  • Quantitative Equilibrium: Setting up an equilibrium ICE table to calculate concentrations, computing Kc, and interpolating temperature from experimental data.
Part (a) • 3 Marks

Standard Enthalpy Change Calculation & Data Selection

Reaction 1: CH₃I + CO → CH₃COI

📐 Step-by-Step Calculation

  1. Select the most accurate dataset: Use Standard Enthalpies of Formation (ΔfH°).
  2. State the Hess's Law formula:
    ΔH = ΣΔfH°(products) - ΣΔfH°(reactants)
  3. Substitute the values:
    ΔH = [-162] - [(-14) + (-111)]
    ΔH = -162 - (-125)
  4. Calculate the final value:
    ΔH = -37 kJ mol⁻¹

✅ Mark Scheme Breakdown

  • Mark 1: Correct substitution using formation data: -162 - (-111 - 14) .
  • Mark 2: Correct final value: -37 kJ mol⁻¹ .
  • Mark 3 (Reason): Enthalpies of formation give a more accurate value because bond energies are average values (taken across many different molecules).
Note: Calculating -19 kJ mol⁻¹ using bond energies could only score a maximum of 2 marks and lost the explanation mark.

❌ Common Errors

  • Mixing up ΔH formulas: Using "Reactants - Products" for formation values (which is only for bond energies).
  • Double-negative slip: Calculating -162 - 125 = -287 kJ mol⁻¹ instead of -162 - (-125) = -37 kJ mol⁻¹.
  • Vague justification: Writing "formation data is experimental" without explicitly noting that bond energies are averages.

🧠 Exam Technique

Whenever an exam offers both bond enthalpies and formation enthalpies, always pick formation enthalpies for accuracy. Bond energies are mean values measured across various environments, whereas standard enthalpies of formation apply directly to the specific substances involved.

Part (b)(i) & (b)(ii) • 4 Marks Total

Weak Acid Dissociation & Strong Acid pH

Understanding the Common Ion Effect and Calculating Strong Acid pH

✅ (b)(i) Correct Answer (2 Marks)

Award 1 mark each for any two points:

  • There will be less dissociation of CH₃COOH.
  • HI is a strong acid and fully dissociates, releasing a high concentration of H⁺ ions; by Le Chatelier's principle, this shifts the weak acid dissociation equilibrium to the left / towards the reactants.
  • CH₃COO⁻ ions combine with added H⁺ ions to regenerate CH₃COOH molecules.

📐 (b)(ii) pH Calculation (2 Marks)

  1. Identify acid type: HI is a monoprotic strong acid, so it fully dissociates:
    [H⁺] = [HI] = 0.250 mol dm⁻³
  2. Apply pH expression:
    pH = -log₁₀[H⁺] = -log₁₀(0.250) (1 mark)
  3. Evaluate to 2 decimal places:
    pH = 0.60 (1 mark)

🧠 Exam Technique: pH Decimal Places

In A-Level Chemistry, pH values must always be quoted to at least 2 decimal places (e.g. 0.60 , never just 0.6 ). The number of decimal places in a logarithm reflects the significant figures of the original concentration.

💡 The Equilibrium Shift

Weak acid equilibrium: CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) .
Flooding the solution with H⁺ from HI forces the position of equilibrium to shift left to oppose the increase, effectively suppressing the ionisation of ethanoic acid.

Part (b)(iii) • 6 Marks (QER)

Buffer Solution Preparation & Buffer Action

Extended Response: Quantitative Preparation and Mechanism of Buffering

📐 Calculation: Mass of Sodium Ethanoate Needed

Step 1: Calculate target [H⁺]
[H⁺] = 10-pH = 10-4.45
[H⁺] = 3.548 × 10⁻⁵ mol dm⁻³
Step 2: Rearrange Ka expression
Ka = ([H⁺][CH₃COO⁻]) / [CH₃COOH]
[CH₃COO⁻] = (Ka × [CH₃COOH]) / [H⁺]
[CH₃COO⁻] = (1.74 × 10⁻⁵ × 0.100) / (3.548 × 10⁻⁵)
[CH₃COO⁻] = 0.0490 mol dm⁻³
Step 3: Moles in 100 cm³
V = 100 cm³ = 0.100 dm³
n = C × V = 0.0490 × 0.100
n = 4.90 × 10⁻³ mol
Step 4: Mass of CH₃COONa
Mr(CH₃COONa) = 24.0 + 3.0 + 32.0 + 23.0 = 82.0 g mol⁻¹
mass = n × Mr = 0.00490 × 82.0
Mass = 0.402 g (or 0.40 g)

✅ Indicative Content for 5–6 Marks

  • How to form: Mix the weak acid with its conjugate salt by dissolving 0.402 g of sodium ethanoate (or adding a measured small amount of NaOH) into the 100 cm³ ethanoic acid solution.
  • Key equilibria / equations:
    CH₃COOH ⇌ CH₃COO⁻ + H⁺ (partially dissociated, large reservoir of CH₃COOH)
    CH₃COONa → CH₃COO⁻ + Na⁺ (fully dissociated, large reservoir of CH₃COO⁻)
  • On adding small amount of H⁺: Added H⁺ reacts with the salt conjugate base: CH₃COO⁻ + H⁺ → CH₃COOH . Equilibrium shifts left, removing excess H⁺ and maintaining pH.
  • On adding small amount of OH⁻: OH⁻ reacts with H⁺ ( OH⁻ + H⁺ → H₂O ). The ethanoic acid equilibrium shifts right to replenish H⁺ (or: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O ). pH remains constant.

🧠 QER Level Descriptors & Criteria

5–6 Marks: Clear explanation of buffer action (both acid and base addition) AND correct calculation of mass (0.402 g) AND both relevant chemical equations. Coherent, logical structure.
3–4 Marks: Basic description of buffer preparation, how it resists acid OR base, with at least one equation and partially correct calculation.
1–2 Marks: Identification of correct substance to add (e.g. sodium ethanoate) and brief attempt at buffer explanation.
Part (c)(i) & (c)(ii) • 6 Marks Total

Ruthenium Complex Catalysis & Equilibrium Constants

[RuL₅CH₃] + CO ⇌ [RuL₅COCH₃]

✅ (c)(i) Energy Change Deduction (2 Marks)

  • Enthalpy change: The forward reaction is exothermic (ΔH is negative). (1 mark)
  • Reasoning: As temperature increases (from 34 °C to 74 °C), the value of Kc decreases (from 1220 to 120), indicating that the position of equilibrium shifts to the left / towards reactants to absorb heat. (1 mark)
Note: The mark scheme stipulates you must attempt the reason to be awarded the mark for stating exothermic.

📐 (c)(ii) Calculation of Kc and Temperature (4 Marks)

Species Initial / mol dm⁻³ Change / mol dm⁻³ Equilibrium / mol dm⁻³
[RuL₅CH₃] 1.25 × 10⁻³ -1.37 × 10⁻⁴ 1.113 × 10⁻³
CO 7.55 × 10⁻⁴ -1.37 × 10⁻⁴ 6.18 × 10⁻⁴
[RuL₅COCH₃] 0 +1.37 × 10⁻⁴ 1.37 × 10⁻⁴
  1. Find equilibrium concentrations: (1 mark)
    [CO]eq = 7.55 × 10⁻⁴ - 1.37 × 10⁻⁴ = 6.18 × 10⁻⁴ mol dm⁻³
    [[RuL₅CH₃]]eq = 1.25 × 10⁻³ - 1.37 × 10⁻⁴ = 1.113 × 10⁻³ mol dm⁻³
  2. State the expression for Kc: (1 mark)
    Kc = [[RuL₅COCH₃]] / ([[RuL₅CH₃]] × [CO])
  3. Calculate value of Kc: (1 mark)
    Kc = (1.37 × 10⁻⁴) / (1.113 × 10⁻³ × 6.18 × 10⁻⁴)
    Kc = 199 mol⁻¹ dm³ (allow 199.2)
  4. Deduce the temperature: (1 mark)
    Looking at the data table, Kc = 196 at 64 °C.
    Therefore, estimated Temperature = 64 °C (allow any temperature in the range 60–64 °C).

❌ Common Mistakes in Part (c)

  • Using initial concentrations in Kc: Forgetting to subtract the reacted amount (1.37 × 10⁻⁴) from initial reactant concentrations.
  • Inverting Kc: Writing reactants over products instead of [products] / [reactants] .
  • Temperature misidentification: Choosing 74 °C because it's the bottom row, rather than finding which temperature corresponds to the calculated Kc value of ~199.

🧠 Error-Carried-Forward (ECF) Benefit

The mark scheme allows ECF for the temperature deduction provided your temperature fits logically within the table's range for your calculated Kc. Show every working line clearly to safeguard arithmetic marks!

Topics

Physical Chemistry · 1.7 Simple equilibria and acid-base reactions · 2.1 Thermochemistry · 3.8 Equilibrium constants · 3.9 Acid-base equilibria

Question and mark scheme from the WJEC A-Level Chemistry examination, Unit 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.