OCR A-Level Chemistry Unified chemistry (03), June 2025: Question 3

11 marks · Hard difficulty · Extended Response

Explain the conditions for maximum equilibrium yield and calculate [H₂] using Kc, then determine orders of reaction, rate equation, rate constant, and deduce mechanism steps from experimental rate data.

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Question text

3 This question is about equilibrium and rates.

(a) Methanol, CH3OH, can be made industrially by the reaction of carbon monoxide, CO, with

hydrogen, H2, in the presence of a catalyst. This is a reversible reaction.

The equilibrium below is set up.

CO(g) + 2H (g) CH OH(g) ∆H = –91 kJ mol–1

The concentrations of CO(g) and CH3OH(g) in the equilibrium mixture are shown in the table

below.

CO(g) CH3OH(g)

equilibrium concentration / mol dm–3 0.27 0.11

The value of the equilibrium constant, K , for this reaction is 1.8 dm6 mol–2.

• Explain the conditions of pressure and temperature that would give the maximum equilibrium

yield of CH3OH(g).

• Calculate the concentration of H2(g) in the equilibrium mixture to an appropriate number of

significant figures.

… [5]

(b)* A reaction takes place when aqueous solutions of propanone, CH3COCH3(aq), and bromine,

Br (aq), are mixed in the presence of dilute acid, H+(aq).

The overall equation for this reaction is shown below.

CH3COCH3 + Br2 CH3COCH2Br + HBr Reaction 3.1

• A student carries out three experiments to investigate the rate of this reaction.

• In each experiment, the student investigates the effect of changing the concentration of one

of the three chemicals whilst keeping the concentrations of the other chemicals constant.

The student’s results are shown below.

Experiment 1

Only [CH3COCH3(aq)] is varied.

Initial concentrations

Initial rate

CH COCH (aq) Br (aq) H+(aq)

33 2 / mol dm–3 s–1

/ mol dm–3 / mol dm–3 / mol dm–3

1.50 × 10–3 2.00 × 10–1 2.50 × 10–1 1.70 × 10–7

4.50 × 10–3 2.00 × 10–1 2.50 × 10–1 5.10 × 10–7

Experiment 2

Only [Br2(aq)] is varied.

[Br2(aq)]

/mol dm[Br(aq)]–3

/mol dm–3

00 Time/s

0 Time/s

Experiment 3

Only [H+(aq)] varied.

Initial rate

/mol dmInitial rate–3s–1

/mol dm–3s–1

0 [H+(aq)]

0 /mol dm[H+(aq)]–3

/mol dm–3

Use the student’s results to determine the order with respect to each of the three chemicals, the

rate equation and rate constant for Reaction 3.1. Explain your reasoning.

Explain how the student’s results can be used to show that the mechanism of Reaction 3.1 has

more than one step.

… [6]

Extra answer space if required.

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

3 (a) Conditions of pressure and temperature 2 marks 5 FULL ANNOTATIONS MUST BE USED

Pressure: For moles, ALLOW molecules/particles

Right-hand side has fewer (gaseous) moles/ ORA for reverse reaction

3 (gaseous) moles form 1 (gaseous) moles DO NOT ALLOW gaseous atoms

AND

High pressure ✓ ALLOW increase pressure

Temperature: ALLOW Reverse reaction is endothermic/

(Forward reaction) is exothermic/ gives out heat/ ∆H –ve takes in heat/ ∆H +ve

AND

Low temperature ✓ ALLOW decrease temperature

=========================================== ====================================

Calculation 3 marks

FIRST CHECK ANSWER

IF [H ] = 0.48 (mol dm–3) award all 3 marks for calc

IF [H ] = 0.476 (mol dm–3) award 2 marks for calc

------------------------------------------------------------------------ Square brackets required in Kc expression

Kc expression

[CH OH] [CO] [H ]2

Kc = [CO] [H ]2 DO NOT ALLOW inverted Kc: [CH OH]

0.11 0.11 ALLOW ECF for calculation from inverted Kc expression

OR 1.8 = 2 OR 1.8 = 2 ✓

0.27 [H2] 0.27 x 0.11 1.8

i.e. [H2] = ( 0.27 ) OR 0.856….. ✓

Calculation of [H2]

0.11 = 0.86 to 2 SF ✓

[H2] = ( ) OR 0.476 ✓

0.27 1.8

Calculator: 0.4757493548 IF no square root,

0.11

–3 [H2] = = 0.226337…. x

= 0.48 (mol dm ) AND 2 SF ✓ 0.27 1.8

2 SF is ‘appropriate’ with 2 SF values in question = 0.23 to 2 SF ✓ ECF

NOTE: 0.48 subsumes mark for 0.476 NO other ECFs

Question Answer 14 Marks Guidance

(b) Please refer to the marking instructions on page 5 of this mark 6 Indicative scientific points may include:

scheme for guidance on how to mark this question. PAGE 8 NEEDS SEEN

Order from Experiment 1

Level 3 (5–6 marks) • CH3COCH3 is 1st order

All three student’s results are used to determine ALL • [CH3COCH3] 3, rate 3

correct orders

AND correct rate equation AND correct rate constant, k Order from Experiment 2

AND explanation for why the mechanism has more than • Br is zero order

one step.

• constant gradient OR constant rate

There is a well-developed line of reasoning which is clear OR rate independent of concentration

and logically structured. OR linear negative gradient

Level 2 (3–4 marks) Order from Experiment 3

Three orders correct AND explanation for why the • H+ is 1st order

mechanism has more than one step. • straight line through 0,0

OR OR rate proportional to [H+]

Two orders correct AND rate equation consistent with any OR linear increase

incorrect orders +

OR doubling [H ] leads to doubling of rate

One order correct AND attempts to calculate rate constant

Rate equation and rate constant

consistent with any incorrect orders AND explanation for +

• rate = k [CH3COCH3] [H ]

why the mechanism has more than one step. 0 1 + 1

ALLOW rate = k [Br2] [CH3COCH3] [H ]

There is a line of reasoning presented with some structure. rate 1.70 10–7

• k = [ CH COCH ][H+] OR –3 –1

33 1.50 10 2.50 10

Level 1 (1–2 marks) • k = 4.53 10–4

One order correct AND rate equation consistent with any 3 –1 –1

• IGNORE units: (dm mol s )

incorrect orders –1 3 –1

OR two orders correct (Any order, e.g. mol dm s )

The information is basic and communicated in an Communication

unstructured way. Aspects of the communication statement might typically

have been met when:

0 marks there is a link between evidence and orders is

No response or no response worthy of credit. clear for three students results.

Possible explanation of multi-step mechanism

e.g.

• Br2 in overall equation; not in rate equation

• H+ in rate equation; not in overall equation

• OR CH COCH + H+ → slow /rate-determining step

• OR stoichiometry of overall equation different from

rate equation

IGNORE attempts at full mechanism

How to answer it

Equilibrium Yield, K_c Calculations & Reaction Kinetics

📌 What this question tests

This question assesses fundamental concepts across Physical Chemistry (Equilibria and Rates):

Le Chatelier’s principle: Predicting and justifying optimal conditions of temperature and pressure for exothermic gaseous reactions.
Equilibrium constant (K_c): Setting up correct mathematical expressions and rearranging them (taking roots) to solve for an unknown concentration to appropriate significant figures.
Graphical and tabular kinetics: Deducing reactant orders from initial rates data, concentration–time curves, and rate–concentration graphs.
Rate equation & constant: Combining orders into a rate law, calculating k, and deducing multi-step mechanism details from kinetic data.

Part (a) • 5 Marks

Methanol Synthesis Equilibrium & K_c

CO(g) + 2H₂(g) ⇌ CH₃OH(g) ΔH = −91 kJ mol⁻¹

💡 Key Knowledge: Le Chatelier Explanations

Always state the condition first, followed by the exact comparative justification:

Pressure: High pressure favors the side with fewer gaseous moles (3 moles of gas on the left produce 1 mole of gas on the right).
Temperature: Low temperature favors the exothermic direction (forward reaction has negative ΔH) to release heat.

❌ Common Errors in Part (a)

Writing "fewer gaseous atoms" instead of gaseous moles or molecules (immediate mark loss).
Inverting the K_c expression (putting reactants on top).
Forgetting to square the [H₂] term in K_c, or forgetting to take the square root at the very end.
Giving the final answer to 3 or more significant figures instead of 2 SF.

📐 Step-by-Step Calculation: Finding [H₂]

Given: [CO] = 0.27 mol dm⁻³, [CH₃OH] = 0.11 mol dm⁻³, K_c = 1.8 dm⁶ mol⁻²

1 Write the K_c expression:

K_c = [CH₃OH] / ([CO][H₂]²)

2 Substitute known values into the expression:

1.8 = 0.11 / (0.27 × [H₂]²)

3 Rearrange to make [H₂]² the subject:

[H₂]² = 0.11 / (0.27 × 1.8) = 0.11 / 0.486 = 0.226337... mol² dm⁻⁶

4 Take the square root:

[H₂] = √(0.226337...) = 0.475749... mol dm⁻³

5 Apply appropriate significant figures:

Values in the question (0.27, 0.11, 1.8) are all given to 2 significant figures.
Therefore, [H₂] = 0.48 mol dm⁻³ (2 SF).

Mark Breakdown (5 marks):
• [1 mark] High pressure + 3 gaseous moles form 1 gaseous mole (or fewer gaseous moles on RHS).
• [1 mark] Low temperature + forward reaction is exothermic (ΔH is negative).
• [1 mark] Correct K_c expression with values substituted: 1.8 = 0.11 / (0.27 × [H₂]²).
• [1 mark] Correct rearrangement for [H₂] giving 0.476 mol dm⁻³.
• [1 mark] Final answer correctly rounded to 2 significant figures: 0.48 mol dm⁻³ (subsumes 0.476 mark).

Part (b)* • 6 Marks (Level of Response)

Kinetics & Mechanism of Propanone Bromination

CH₃COCH₃(aq) + Br₂(aq) → CH₃COCH₂Br(aq) + HBr(aq) (catalysed by H⁺)

🧠 Exam Technique: Deducing Each Order

Experiment 1 (Table):
Comparing runs 1 and 2: [CH₃COCH₃] increases from 1.50×10⁻³ to 4.50×10⁻³ (×3).
Rate increases from 1.70×10⁻⁷ to 5.10×10⁻⁷ (×3).
👉 Since rate is directly proportional to concentration, order with respect to CH₃COCH₃ = 1 (First order).
Experiment 2 (Graph: [Br₂] vs Time):
The graph is a straight downward line with a constant gradient.
A constant gradient means rate is independent of [Br₂].
👉 Order with respect to Br₂ = 0 (Zero order).
Experiment 3 (Graph: Rate vs [H⁺]):
The graph is a straight line through the origin (0,0).
Rate is directly proportional to [H⁺].
👉 Order with respect to H⁺ = 1 (First order).

💡 Why the Mechanism has More than One Step

Top-level responses explain the conflict between the overall equation and the rate equation clearly:

Role of Br₂: Br₂ appears in the overall stoichiometric equation, but its order is zero (it does not appear in the rate equation). Therefore, Br₂ cannot be involved in the slow, rate-determining step.
Role of H⁺: H⁺ appears in the rate equation but is not consumed in the overall equation (it acts as a catalyst).
Stoichiometry vs Orders: The species involved in the rate equation (1 CH₃COCH₃ and 1 H⁺) do not match the overall stoichiometric equation (1 CH₃COCH₃ + 1 Br₂).

📐 Calculation: Rate Equation & Value of k

1. Rate Equation:

rate = k[CH₃COCH₃][H⁺]

Note: Writing rate = k[CH₃COCH₃]¹[Br₂]⁰[H⁺]¹ is also fully acceptable.

2. Rearrange and substitute data from Experiment 1 (Run 1):

k = rate / ([CH₃COCH₃][H⁺])

k = 1.70 × 10⁻⁷ / (1.50 × 10⁻³ × 2.50 × 10⁻¹)

k = 1.70 × 10⁻⁷ / (3.75 × 10⁻⁴) = 4.53 × 10⁻⁴ (or 4.5 × 10⁻⁴)

Units: dm³ mol⁻¹ s⁻¹ (the mark scheme notes to IGNORE units, so no penalty for missing or wrong units!).

✅ How to Secure Level 3 (5–6 Marks)

All 3 orders correct with clear evidence linked:
• CH₃COCH₃: concentration × 3 causes rate × 3 (1st order).
• Br₂: graph shows constant gradient / rate is independent of [Br₂] (0 order).
• H⁺: straight line through origin / rate directly proportional to [H⁺] (1st order).
Correct rate equation: rate = k[CH₃COCH₃][H⁺]
Correct numerical rate constant: k = 4.53 × 10⁻⁴
Mechanism explanation: Br₂ is in the overall equation but not in the rate equation (or Br₂ does not participate in the rate-determining step).

Examiner Insight for 6-mark Level of Response Questions:
Structure your answer using subheadings for each experiment. Students often lose marks because they state the order without quoting the specific evidence from the graph (e.g., failing to explicitly state "constant gradient" for Experiment 2 or "straight line through (0,0)" for Experiment 3).

Topics

Module 5: Physical chemistry and transition elements · Module 3: Periodic table and energy · 5.1 Rates, equilibrium and pH · 3.2 Physical chemistry

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.