A chemist is investigating the production of ethanol by the reversible reaction between ethene and steam: C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g). The forward reaction is exothermic. The reaction is carried out in a closed system using a catalyst. Explain how changing the temperature and pressure would each affect the position of equilibrium in this reaction, and state one advantage of using a catalyst.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
The industrial production of ethanol uses the reversible reaction: C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g). The forward reaction is exothermic. The reaction reaches dynamic equilibrium in a closed system.
Model answer (5 marks)
Decreasing the temperature favours the exothermic forward reaction, so more ethanol is formed.
Increasing the temperature favours the endothermic reverse reaction, so less ethanol is produced.
Because the left side contains 2 mol of gas and the right side 1 mol, increasing pressure shifts the equilibrium toward the side with fewer moles of gas – the forward direction – giving more ethanol.
Decreasing pressure shifts the equilibrium toward the side with more gas – the reverse direction – giving less ethanol.
A catalyst provides an alternative pathway with a lower activation energy, increasing the rate of both forward and reverse reactions but leaving the equilibrium composition unchanged.
Increasing the temperature favours the endothermic reverse reaction, so less ethanol is produced.
Because the left side contains 2 mol of gas and the right side 1 mol, increasing pressure shifts the equilibrium toward the side with fewer moles of gas – the forward direction – giving more ethanol.
Decreasing pressure shifts the equilibrium toward the side with more gas – the reverse direction – giving less ethanol.
A catalyst provides an alternative pathway with a lower activation energy, increasing the rate of both forward and reverse reactions but leaving the equilibrium composition unchanged.
Examiner tips
- Use the word ‘favours’ or ‘shifts’ to show direction; link temperature to the exothermic nature. Mention the gas‑mole count when explaining pressure effects. State that the catalyst only changes the rate, not the equilibrium position.
- Use the exact wording from the mark scheme – e.g. ‘exothermic forward reaction’ – to secure full marks.
Common mistakes
- Confusing the effect of temperature (mixing up exothermic and endothermic directions). Forgetting that the catalyst does not alter the equilibrium composition. Using vague terms like ‘more’ or ‘less’ without specifying the direction of shift.
Mark scheme (5 marks)
- Decreasing temperature shifts the equilibrium position in the direction of the exothermic reaction (forward reaction), producing more ethanol.
- Increasing temperature shifts the equilibrium position in the direction of the endothermic reaction (backward reaction), producing less ethanol.
- There are 2 moles of gas on the left and 1 mole of gas on the right, so increasing pressure shifts the equilibrium towards the side with fewer moles of gas (right/forward direction), producing more ethanol.
- Decreasing pressure shifts the equilibrium towards the side with the greater number of moles of gas (left/backward direction), producing less ethanol.
- A catalyst increases the rate of reaction by providing a different reaction pathway with a lower activation energy, but does not change the position of equilibrium.
Key terms in this question
Related
- All OCR A-Level Chemistry A (H432) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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