OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2020: Question 17

13 marks · Hard difficulty · Calculations

Predict equilibrium conditions for methanol production, state green chemistry benefits of catalysts, calculate Gibbs free energy change and feasibility at a given temperature, and determine Kp and its units.

Practise this question

Question

An A-Level Chemistry question about the industrial production of methanol from carbon monoxide and hydrogen given by the equilibrium equation CO(g) + 2H2O(g) <=> CH3OH(g) with Delta H = -91 kJ mol-1. Part (a) asks to predict pressure and temperature conditions for maximum yield with an explanation (3 marks). Part (b) asks to state two ways catalysts help make chemical processes more sustainable (2 marks). Part (c) provides a table of standard entropy values and asks to calculate whether methanol production is feasible at 525 K (5 marks). Part (d) gives an equation linking Delta G, R, T, and Kp, asking to calculate Kp at 298 K to 3 significant figures and give units (3 marks).
Question text

17 Methanol, CH3OH, can be made industrially by the reaction of carbon monoxide with hydrogen, as

shown in equilibrium 1.

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

(a) Predict the conditions of pressure and temperature that would give the maximum equilibrium

yield of CH3OH in equilibrium 1.

Explain your answer.

… [3]

(b) A catalyst is used in the production of methanol in equilibrium 1.

State two ways that the use of catalysts helps chemical companies to make their processes

more sustainable and less harmful to the environment.

1 …

2 …

[2]

(c) Standard entropy values are given below.

Substance CO(g) H2(g) CH3OH(g)

Sө / J K−1 mol−1 198 131 238

A chemist proposed producing methanol at 525 K using equilibrium 1.

Explain, with a calculation, whether the production of methanol is feasible at 525 K.

… [5]

(d) At 298 K, the free energy change, ∆G, for the production of methanol in equilibrium 1 is

−2.48 × 104 J mol−1.

∆G is linked to Kp by the relationship: ∆G = −RT lnKp.

R = gas constant

T = temperature in K.

Calculate Kp for equilibrium 1 at 298 K.

Give your answer to 3 significant figures.

Kp = … units … [3]

Mark scheme

Show the mark scheme The mark scheme provides detailed answers for parts (a) through (d). Part (a) awards marks for stating high pressure and low temperature with reasoning about moles and exothermicity. Part (b) accepts lower energy demand and lower fossil fuel/CO2 emissions. Part (c) details the calculation steps for entropy change, Gibbs free energy change, and feasibility determination. Part (d) shows the logarithmic rearrangement for Kp, substitution of values, final answer with 3 significant figures, and correct units of atm^-2.

AO

Question Answer Marks Guidance

element

17 (a) High pressure AND low temperature 3 1.2×1 Marks are independent

Right-hand side has fewer (gaseous) moles/molecules ORA throughout

OR

left-hand side has more (gaseous) moles/molecules 1.1×2 ALLOW RHS

ALLOW suitable alternatives for RHS

(Forward) reaction is exothermic/gives out heat e.g. product side

OR reverse reaction is endothermic/takes in heat

(b) (Reaction can be carried out at) lower temperatures / 2 1.1×2 ALLOW lower pressures as alternative to lower

lower energy demand temperature

Less (fossil) fuels burnt / less CO2 emissions ALLOW reduced carbon footprint as alternative to

less fuels burnt

ALLOW different reactions can be used with

greater atom economy / less waste

ALLOW can reduce use of toxic substances

AO

10 element

(c) FIRST CHECK THE ANSWER ON ANSWER LINE 5 2.2×4

If answer = 25.55 kJ mol–1 OR 25550 J mol–1 award

first 4 marks

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

∆S = 238 – (198 + 2 × 131)

ALLOW ECF

= –222 (J K–1 mol–1) OR –0.222 (kJ K–1 mol–1)

IGNORE units at this stage

∆G = ∆H – T∆S

OR

∆G = –91 – (525 × –0.222) 3.2×1

OR

∆G = –91000 – (525 × –222)

= 25.55 kJ mol–1 OR 25550 J mol–1 Units for ∆G required

ALLOW 26 kJ mol–1 OR 26000 J mol–1 up to

calculator value.

(Reaction is) not feasible AND ∆G > 0

AO

Question Answer 11 Marks Guidance

element

(d) FIRST CHECK THE ANSWER ON ANSWER LINE 3 ALLOW ECF for transcription errors in first sum

If answer = 2.22 × 104 award first 2 marks

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

2.48 × 104

ln Kp = –∆G/RT = = 10.01 3.1×2 ALLOW 10 up to calculator value of 10.00979992

8.314 × 298

Kp = 2.22 × 104 (3SF required) ALLOW 22200

1.2×1 ALLOW 2.20 × 104 OR 22000 (use of 10)

ALLOW alternatives (k)Pa–2 OR

N-2 m4 OR mmHg–2 OR PSI–2 OR bar -2

Units = atm–2

Common errors for 1 mark:

22400 (use of 8.31)

4.50 x 10-5 (use of -10.01)

Total 14

How to answer it

Industrial Methanol Production & Equilibrium Study Guide

What this question tests

This multi-part physical chemistry question assesses your understanding of Le Chatelier's principle, industrial sustainability via catalysts, Gibbs free energy feasibility calculations (ΔG = ΔH - TΔS), and the quantitative link between free energy change and the equilibrium constant ( ΔG = -RT ln Kₚ ). You will need to balance moles of gas, convert between joules and kilojoules, and handle logarithmic rearrangements.

Part (a) — Equilibrium Yield (3 Marks)

Predicting Conditions for Maximum Yield

✅ Correct Answer

  • Pressure: High pressure
  • Temperature: Low temperature
  • Reasoning (Pressure): Fewer moles of gaseous molecules on the right-hand side (1 mole) compared to the left-hand side (3 moles).
  • Reasoning (Temperature): The forward reaction is exothermic ( ΔH = -91 kJ mol⁻¹ ).

🧠 Exam Technique

  • State both conditions clearly first (High pressure AND low temperature) to secure the direct condition marks.
  • Explicitly quote the mole ratio (3 moles vs 1 mole) rather than just saying "fewer moles".
  • Reference the sign of ΔH to justify the temperature effect.
🎯 Mark Breakdown: 1 mark for high pressure, 1 mark for low temperature, 1 mark for explaining the mole/exothermic reasoning correctly.
Part (b) — Green Chemistry (2 Marks)

Sustainability Benefits of Catalysts

✅ Correct Answers (Any 2)

  • Allows the reaction to be carried out at lower temperatures / lowers energy demand.
  • Less fossil fuels burnt, leading to reduced CO₂ emissions / smaller carbon footprint.
  • Enables reactions with higher atom economy or less waste.

❌ Common Errors

  • Vague statements like "it's better for the environment" without linking to energy or emissions.
  • Stating that catalysts "increase yield" (catalysts affect the rate of attainment of equilibrium, not the position of equilibrium or yield).
🎯 Mark Breakdown: 1 mark per valid sustainability point (max 2).
Part (c) — Entropy and Feasibility (5 Marks)

Calculating Reaction Feasibility at 525 K

📐 Step-by-Step Calculation

  1. Calculate ΔS°:
    ΔS° = ΣS°(products) - ΣS°(reactants)
    ΔS° = 238 - (198 + (2 × 131)) = -222 J K⁻¹ mol⁻¹
  2. Calculate ΔG using ΔG = ΔH - TΔS:
    Convert ΔH from kJ to J: -91 kJ mol⁻¹ = -91000 J mol⁻¹
    ΔG = -91000 - (525 × -222)
    ΔG = -91000 + 116550 = +25550 J mol⁻¹ (+25.55 kJ mol⁻¹)
  3. Feasibility Conclusion:
    Not feasible because ΔG > 0 .

💡 Key Knowledge & Traps

  • Unit Mismatch Trap: ΔH is given in kJ mol⁻¹ while ΔS is in J K⁻¹ mol⁻¹ . You must multiply ΔH by 1000 before combining them.
  • Feasibility Rule: A process is only feasible (spontaneous) when ΔG ≤ 0 . A positive ΔG means the reaction is not feasible.
🎯 Mark Breakdown: 1 mark for correct ΔS calculation, 1 mark for unit conversion / temperature substitution, 1 mark for correct ΔG value, 1 mark for correct units ( kJ mol⁻¹ or J mol⁻¹ ), 1 mark for stating reaction is not feasible because ΔG > 0.
Part (d) — Equilibrium Constant Kₚ (3 Marks)

Relating Free Energy to Kₚ

📐 Step-by-Step Calculation

  1. Rearrange the equation:
    ln Kₚ = -ΔG / (RT)
  2. Substitute values:
    ΔG = -2.48 × 10⁴ J mol⁻¹
    R = 8.314 J K⁻¹ mol⁻¹ , T = 298 K
    ln Kₚ = -(-24800) / (8.314 × 298) = 10.0098
  3. Evaluate Kₚ:
    Kₚ = e¹⁰·⁰⁰⁹⁸ = 22222... = 2.22 × 10⁴
  4. Units:
    atm⁻² (or kPa⁻² , Pa⁻² , bar⁻² based on pressure units used).

❌ Common Errors & Significant Figures

  • Sign Error: Forgetting that ΔG in the formula is negative, leading to a negative natural log and an incorrect exponential.
  • Significant Figures: The question explicitly requests the final answer to 3 significant figures ( 2.22 × 10⁴ ).
  • Units for Kₚ: For equilibrium CO + 2H₂ ⇌ CH₃OH , expression is Kₚ = p(CH₃OH) / (p(CO) × p(H₂)² ) , giving units of pressure⁻² .
🎯 Mark Breakdown: 1 mark for correct calculation of ln Kₚ or exponentiation, 1 mark for final value to 3 sf ( 2.22 × 10⁴ ), 1 mark for correct units ( atm⁻² ).

Topics

Module 5: Physical chemistry and transition elements · 5.1 Rates, equilibrium and pH · 5.2 Energy

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.