OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2020: Question 20

9 marks · Medium difficulty · Structured Questions

Investigate the reaction kinetics and equilibrium constant for the esterification of ethanoic acid with methanol using graphical analysis and concentration data

Practise this question

Question

Exam question about the reaction between ethanoic acid and methanol, showing a concentration-time graph with data points plotted from 0 to 180 minutes, followed by sub-questions (a)(i) explaining excess reagent, (a)(ii) determining order from half-life, (a)(iii) determining initial rate, and (b) calculating Kc from equilibrium moles.
Question text

20 A student investigates the reaction between ethanoic acid, CH3COOH(l) and methanol, CH3OH(l),

in the presence of an acid catalyst. The equation is shown below.

CH3COOH(l) + CH3OH(l) CH3COOCH3(l) + H2O(l)

(a) The student carries out an experiment to determine the order of reaction with respect to

CH3COOH.

The student uses a large excess of CH3OH. The temperature is kept constant throughout the

experiment.

The student takes a sample from the mixture every 20 minutes, and then determines the

concentration of the ethanoic acid in each sample.

From the experimental results, the student plots the graph below.

5.0

4.0

3.0

2.0

1.0

0 20 40 60 80 100 120 140 160 180

time/min

(i) Explain why the student uses a large excess of methanol in this experiment.

… [1]

(ii) Use the half-life of this reaction to show that the reaction is first order with respect to

CH3COOH.

Show your working on the graph and below.

… [2]

(iii) Determine the initial rate of reaction.

initial rate = … mol dm−3 min−1 [2]

(b) The student carries out a second experiment to determine the value of Kc for this reaction.

The student mixes 9.6 g of CH3OH with 12.0 g of CH3COOH and adds the acid catalyst.

When the mixture reaches equilibrium, 0.030 mol of CH3COOH remains.

Calculate Kc for this equilibrium.

Kc = … [4]

Mark scheme

Show the mark scheme Mark scheme showing accepted answers for question 20. Part (a)(i) awards 1 mark for keeping methanol constant, (a)(ii) awards 2 marks for half-life determination and order deduction, (a)(iii) awards 2 marks for gradient calculation from graph tangent, and part (b) awards 4 marks for calculating initial moles, equilibrium moles, and the equilibrium constant Kc.

AO

Question Answer Marks Guidance

element

20 (a) (i) To keep [CH3OH] (effectively) constant 1 3.3 ALLOW Change in [CH3OH] is negligible

OR ALLOW rate is independent of [CH3OH]

Zero order with respect to CH3OH

OR IGNORE Methanol doesn’t run out/is not limiting

To ensure equilibrium is far to the right reagent.

(ii) One half-life t½ between 102 and 110 (mins) 2 3.1

Two half-lives calculated OR evidence on the graph of 3.2 ALLOW any two combinations of positions, e.g.

two half-lives 5 and 2.5 AND 4 and 2 AND 3 and 1.5

AND

constant half-life/values (means first order)

(iii) Using gradients 2 3.1×1

Evidence of tangent at t = 0 and intercept between

100 -140 (min)

3.2×1

Correctly calculated gradient in the range of

2.9 × 10–5 to 4.0 × 10–5 (mol dm–3 min–1)

OR

Using half-life ALLOW ECF from value of t½ in (a)(ii)

ln 2 –1

For t½ = 106 min, k = = 0.00654 (min )

t½

rate = 0.00654 × 5 × 10–3

= 3.27 × 10–5 (mol dm–3 min–1)

AO

18 element

(b) FIRST CHECK THE ANSWER ON ANSWER LINE 4 ALLOW minimum of 2SF throughout

If answer = 7.4 award 4 marks

-----------------------------------------------------------------

Initial moles of reactants 1 mark

9.6

n(CH3OH) initial = = 0.3 (mol)

AND

n(CH3COOH) initial = = 0.2 (mol) 1.2×1

Equilibrium moles 2 marks

n(CH3COOH) reacted = 0.2 – 0.03 = 0.17 (mol) ALLOW ECF from initial moles

AND

n(CH3OH) equil = 0.3 – 0.17 = 0.13 (mol) 2.8×3

n(CH3COOCH3) equil = 0.17 (mol) ALLOW ECF from equilibrium moles

AND Use of V not required but Kc expression must be

n(H2O) equil = 0.17 (mol) correct

Kc calculation 1 mark ALLOW up to calculator answer of 7.41025641

0.17/V × 0.17/V

Kc = = 7.4

0.13/V × 0.03/V

Total 9

How to answer it

Kinetics and Equilibrium Study Guide

OCR A-Level Chemistry • Exam Question Breakdown

What this question tests

This question assesses core Physical Chemistry concepts across modules 5.2 (Rates of Reaction) and 5.3 (Equilibria). Specifically, it tests your ability to isolate variables using a large excess of a reactant, interpret concentration-time graphs to determine reaction order via half-lives, calculate initial rates of reaction, and construct ICE tables to determine the equilibrium constant ( K_c ).

Part (a)(i): Excess Reagents in Kinetics

Explain why the student uses a large excess of methanol in this experiment. [1 mark]

✅ Correct Answer

To keep the concentration of CH₃OH effectively constant (so it does not affect the rate equation determination).

💡 Key Knowledge

  • Isolation Method: By putting one reactant in a massive excess, its concentration remains virtually unchanged throughout the reaction.
  • This pseudo-isolation allows determination of the order with respect to the other reactant ( CH₃COOH ).
Mark scheme note: Also accepts "zero order with respect to CH₃OH" or "to ensure equilibrium is far to the right". Ignore colloquial phrases like "methanol won't run out".

Part (a)(ii): Determining Reaction Order from Half-Life

Use the half-life of this reaction to show that the reaction is first order with respect to CH₃COOH. [2 marks]

✅ Correct Answer

  • First half-life ( t₁⸝₂ ) from 5.0 to 2.5 mol dm⁻³ is between 102 and 110 minutes .
  • A second half-life (e.g. from 4.0 to 2.0 , or 3.0 to 1.5 ) is calculated/shown to be approximately constant, proving first order.

🧠 Exam Technique

  • Show your working clearly: Draw dotted lines directly on the graph from concentration points down to the time axis to show your half-life measurements.
  • State at least two half-lives explicitly to earn the second mark. Constant half-life = first order.

❌ Common Errors

  • Stating "the half-life is constant" without quoting any numerical values from the graph.
  • Misreading the graph scale intervals (each small square on the time axis represents 2 minutes ).
Mark breakdown: 1 mark for calculating/reading a half-life between 102–110 min; 1 mark for showing a second half-life or stating the half-life is constant.

Part (a)(iii): Calculating Initial Rate

Determine the initial rate of reaction. [2 marks]

📐 Step-by-Step Calculation

  1. Method 1 (Tangent at t = 0): Construct a tangent to the curve at t = 0 where concentration = 5.0 mol dm⁻³ . Calculate the gradient ( Δy / Δx ). Expect a value in the range 2.9 × 10⁻⁵ to 4.0 × 10⁻⁵ mol dm⁻³ min⁻¹ .
  2. Method 2 (Using half-life): Since k = ln 2 / t₁⸝₂ , using t₁⸝₂ = 106 min gives k = 0.00654 min⁻¹ . Initial rate = k × [CH₃COOH] = 0.00654 × 5.0 = 3.27 × 10⁻⁵ mol dm⁻³ min⁻¹ .

❌ Common Errors

  • Forgetting to include or miscalculating standard rate units: mol dm⁻³ min⁻¹ (watch out for minutes vs seconds!).
  • Drawing a poorly proportioned tangent that does not accurately reflect the curve's steepness at time zero.
Mark breakdown: 1 mark for evidence of a tangent at t = 0 (or correct rate constant method); 1 mark for a correctly calculated numerical value within the allowed range with correct units.

Part (b): Calculating the Equilibrium Constant (K_c)

Calculate K_c for this equilibrium. [4 marks]

📐 Step-by-Step Calculation (ICE Table Method)

  1. Find Initial Moles:
    • n(CH₃OH) = 9.6 g / 32.0 g mol⁻¹ = 0.30 mol
    • n(CH₃COOH) = 12.0 g / 60.0 g mol⁻¹ = 0.20 mol
  2. Find Equilibrium Moles:
    • n(CH₃COOH)equiv = 0.030 mol (given)
    • n(CH₃COOH)reacted = 0.20 - 0.03 = 0.17 mol
    • n(CH₃OH)equiv = 0.30 - 0.17 = 0.13 mol
    • n(CH₃COOCH₃)equiv = 0.17 mol (1:1 stoichiometry)
    • n(H₂O)equiv = 0.17 mol (1:1 stoichiometry)
  3. Construct K_c Expression and Calculate:
    • Expression: K_c = ([CH₃COOCH₃][H₂O]) / ([CH₃COOH][CH₃OH])
    • Volume V cancels out because there are equal moles on both sides of the equation ( 2 moles ⇌ 2 moles ).
    • K_c = (0.17 × 0.17) / (0.13 × 0.030) = 7.41

🧠 Exam Technique & Top Answers

  • Final Answer: 7.4 (to 2 significant figures, matching the precision of the data provided). Up to calculator value accepted.
  • Volume cancellation saves time—recognise when mole ratios mean volume terms cancel out in the K_c expression!
Mark breakdown: 1 mark for initial moles of both reactants; 2 marks for correct equilibrium moles of all four species; 1 mark for correct final K_c evaluation.

Topics

Module 5: Physical chemistry and transition elements · Module 6: Organic chemistry and analysis · 5.1 Rates, equilibrium and pH · 6.1 Aromatic compounds, carbonyls and acids

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.