Edexcel A-Level Chemistry AS Paper 2, June 2025: Question 7

12 marks · Medium difficulty · Open Response

Explain the choice of temperature and pressure for methanol synthesis using equilibria and kinetics, sketch a Maxwell–Boltzmann distribution showing catalysed and uncatalysed activation energies, and explain the catalyst's effect.

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Question

Question 7 regarding the industrial preparation of methanol: 3H2(g) + CO2(g) ⇌ CH3OH(g) + H2O(g) with ΔH = -49 kJ mol^-1. Part (a)(i) asks to explain why 700 K is chosen rather than 400 K or 1000 K for 3 marks. Part (a)(ii) asks why 90 atm is used rather than 40 atm or 140 atm for 3 marks. Part (b)(i) provides blank axes and asks to sketch a Maxwell-Boltzmann distribution of molecular energies showing labelled axes and activation energies for catalysed and uncatalysed reactions for 4 marks. Part (b)(ii) asks how the graph shows the effect of a catalyst for 2 marks.

Mark scheme

Show the mark scheme Mark scheme for Question 7 totaling 12 marks. 7(a)(i) awards 3 marks for higher T giving faster rate, lower T giving higher yield due to exothermic forward reaction, and 700 K being a compromise between rate and yield. 7(a)(ii) awards 3 marks for higher pressure giving faster rate, higher pressure giving higher yield as there are fewer moles of gas on RHS, and 140 atm being too expensive, making 90 atm a compromise. 7(b)(i) awards 4 marks for vertical axis label, horizontal axis label, correct asymmetric curve starting at origin and asymptotic to the x-axis, and correctly positioned catalysed and uncatalysed activation energies. 7(b)(ii) awards 2 marks for stating a greater proportion/area under the curve beyond the catalysed activation energy and that more molecules have sufficient energy to react.

How to answer it

Industrial Equilibrium & Kinetics: Methanol Synthesis

Edexcel AS Chemistry | Rates, Equilibria & Maxwell–Boltzmann Distribution

What this question tests

  • Le Chatelier's Principle & Chemical Kinetics: Explaining why industrial compromise conditions (temperature and pressure) are chosen to balance equilibrium yield, rate of reaction, and plant operating costs.
  • Maxwell–Boltzmann Energy Distributions: Correctly drawing and fully labelling molecular energy distribution curves from the origin with asymptotic decay.
  • Catalytic Action: Showing and explaining how a catalyst provides an alternative pathway with a lower activation energy, increasing the proportion of successful collisions.
3H₂(g) + CO₂(g) ⇌ CH₃OH(g) + H₂O(g)   ΔH = -49 kJ mol⁻¹   [Conditions: 700 K, 90 atm, catalyst]
Part (a)(i) — 3 Marks

Explaining the Compromise Temperature (700 K)

Why 700 K is chosen rather than 400 K (too low) or 1000 K (too high).

✅ Correct Answer (Mark Scheme)

  • Mark 1 (Rate) A higher temperature gives a faster reaction because a greater proportion of particles have energy greater than the activation energy, Eₐ (or more frequent/successful collisions).
  • Mark 2 (Yield) A lower temperature gives a greater equilibrium yield because the forward reaction is exothermic (equilibrium shifts right).
  • Mark 3 (Compromise) 700 K is used as a compromise temperature between reaction rate and equilibrium yield.

💡 Key Knowledge

  • Kinetic factor: Increasing temperature increases kinetic energy, collision frequency, and exponentially increases particles with E ≥ Eₐ.
  • Thermodynamic factor: Since ΔH is negative (-49 kJ mol⁻¹), raising temperature favours the endothermic backward reaction, reducing methanol yield.
  • Trade-off: 400 K gives high yield but reaction is too slow; 1000 K is fast but gives negligible yield.

🧠 Exam Technique

  • Always state which temperature you are referencing (e.g. "higher temp gives..." or "at 400 K...").
  • Explicitly use the word compromise and state clearly what two things are being compromised: rate vs. yield.

❌ Common Errors to Avoid

  • Stating "equilibrium shifts right" without explicitly stating that the yield increases loses Mark 2.
  • Claiming the compromise for temperature is about "cost" or "safety". Mark 3 is only awarded for the compromise between rate and yield.
Part (a)(ii) — 3 Marks

Explaining the Compromise Pressure (90 atm)

Why 90 atm is chosen rather than 40 atm (too low) or 140 atm (too high).

✅ Correct Answer (Mark Scheme)

  • Mark 1 (Rate) A higher pressure gives a faster rate of reaction (particles are closer together, so more frequent collisions).
  • Mark 2 (Yield) A higher pressure gives a greater yield because the forward reaction produces fewer moles of gas (4 mol of gas on left → 2 mol of gas on right).
  • Mark 3 (Cost/Compromise) 140 atm is not used because of the high energy costs to maintain pressure OR high equipment costs to withstand pressure (90 atm is a compromise between rate/yield and cost).

💡 Key Knowledge

  • Gas stoichiometry: Left-hand side = 3 H₂ + 1 CO₂ = 4 moles. Right-hand side = 1 CH₃OH + 1 H₂O = 2 moles.
  • Le Chatelier's Principle dictates that increasing pressure shifts equilibrium toward the side with fewer gas moles (the products).
  • Unlike temperature, high pressure favours both rate and yield! The limiting factor is purely financial/structural.

🧠 Exam Technique

  • When discussing cost, give a specific reason: either energy required to compress/pump gases or thicker pipes/specialist vessels required.
  • Always justify equilibrium shifts with mole counts from the stoichiometric equation.

❌ Common Errors to Avoid

  • Vague statements like "high pressure is dangerous" or "pipes might explode". Examiners ignore safety/danger; the mark scheme strictly requires economic arguments (cost).
  • Simply stating "higher pressure gives higher yield" without referencing the number of gas moles loses Mark 2.
Part (b)(i) — 4 Marks

Sketching the Maxwell–Boltzmann Distribution Curve

Accurately drawing and annotating a single-temperature distribution with catalysed and uncatalysed activation energies.

📋 Visual Guide to Drawing the Curve:
  • Y-axis: "Number of molecules (with energy E)" or "Fraction of molecules". (Do not label as 'rate'!)
  • X-axis: "Energy" or "Kinetic Energy, E".
  • Origin: Curve must start at (0,0) because zero molecules have zero kinetic energy.
  • Curve Shape: Rises steeply to a peak, then tails off gradually to the right (asymmetrical/skewed right).
  • Right-hand Tail: Must approach the horizontal axis asymptotically. Crucial: It must NEVER touch the x-axis or curl back upwards.
  • Activation Energy Lines: Draw two vertical dashed lines to the right of the peak:
    • Left line: Eₐ(cat) (lower activation energy).
    • Right line: Eₐ(uncat) (higher activation energy).

✅ Mark Scheme Breakdown

  • Mark 1 Vertical axis correctly labelled: Number of molecules (with energy, E) / fraction of molecules.
  • Mark 2 Horizontal axis correctly labelled: (Kinetic) energy / E.
  • Mark 3 Correct curve shape: starts at origin, asymmetric, approaches energy axis asymptotically without touching it.
  • Mark 4 Both Eₐ(cat) and Eₐ(uncat) identified on the x-axis, with Eₐ(uncat) to the right of Eₐ(cat), and neither near the peak maximum.

❌ Common Examiner Traps

  • Drawing two distribution curves: The question asks for the effect of a catalyst at a single temperature. Adding a second curve (e.g. for higher temperature) caps your score at max 2 marks.
  • Touching the axis: The tail touching the horizontal axis loses Mark 3 immediately.
  • Placing Eₐ at the peak: Activation energy is situated to the right of the peak.
Part (b)(ii) — 2 Marks

Explaining Catalytic Effect Using the Graph

Linking the visual features of your sketch to kinetic theory.

✅ Correct Answer (Mark Scheme)

  • Mark 1 There is a greater proportion / greater area under the graph beyond the catalysed activation energy Eₐ(cat) compared to uncatalysed Eₐ.
  • Mark 2 (Therefore) more molecules have sufficient energy to react (energy E ≥ Eₐ).

🧠 Exam Technique

  • The question specifically states: "Explain how your Maxwell–Boltzmann graph shows...".
  • You must explicitly mention the area under the graph or proportion shown on the graph to earn the first mark.
  • Reciting "a catalyst provides an alternative pathway with lower activation energy" is correct chemistry, but scores 0 marks here because it does not refer to the diagram!

Topics

Physical Chemistry · Topic 9: Kinetics I · Topic 10: Equilibrium I

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.