Ethanol vapour can be produced by the reaction of ethene with steam in a closed container. The equation for the reaction is: C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g). The reaction reaches equilibrium at a constant temperature. Explain how the equilibrium constant Kp for this reaction would change, if at all, if (i) the total pressure is increased at constant temperature, and (ii) the temperature is increased. In each case justify your answer.

AQA A-Level Chemistry (7405) — 3.1.10 Equilibrium constant Kp (A-Level only) · Explain · 5 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (5 marks)

(i) Kp does not change when the total pressure is increased at constant temperature, because Kp is a function of temperature only.

(ii) Kp decreases when the temperature is increased. The forward reaction is exothermic, so raising the temperature favours the reverse reaction, shifting the equilibrium to the left (towards reactants). This reduces the proportion of products, lowering Kp.

Examiner tips

  • Use the definition of Kp as temperature‑dependent only; mention Le Chatelier for temperature changes.
  • State the reaction is exothermic and that increasing T shifts equilibrium left, reducing Kp.
  • Keep answer concise and use correct chemical symbols and units.

Mark scheme (5 marks)

  1. Kp does not change when pressure is increased (at constant temperature)
  2. Kp only changes when temperature changes / Kp is only affected by temperature
  3. The forward reaction is exothermic (so increasing temperature favours the reverse reaction)
  4. Increasing temperature shifts the equilibrium position to the left / towards reactants
  5. Kp decreases when temperature is increased (because the equilibrium shifts to the left, giving a lower proportion of products)

Key terms in this question

equilibrium constant Kp · closed container

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