The industrial production of methanol involves the following reversible reaction carried out in a closed system: CO(g) + 2H₂(g) ⇌ CH₃OH(g) The forward reaction is exothermic. Industrial chemists must choose carefully between using a high temperature or a low temperature, and between using a high pressure or a low pressure. Explain how temperature and pressure each affect the position of equilibrium in this reaction, and suggest why a moderate temperature is chosen in industry rather than a very low temperature.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Methanol is an important industrial solvent and fuel additive. It is manufactured on a large scale using carbon monoxide and hydrogen gases under carefully controlled conditions of temperature and pressure.
Model answer (5 marks)
Increasing pressure reduces the volume of the system. Because the forward reaction has fewer gas molecules (1 mol CH₃OH) than the reverse (3 mol CO + 2 H₂), Le Chatelier’s principle predicts that the equilibrium shifts to the product side.
Increasing temperature supplies heat. The forward reaction is exothermic, so heat is a product. Adding heat therefore favours the reverse, endothermic reaction, shifting equilibrium to the left and lowering the methanol yield.
In industry a very low temperature would give a high equilibrium yield, but the reaction rate would be extremely slow. A moderate temperature is chosen as a compromise, giving an acceptable rate while still maintaining a reasonable yield.
Increasing temperature supplies heat. The forward reaction is exothermic, so heat is a product. Adding heat therefore favours the reverse, endothermic reaction, shifting equilibrium to the left and lowering the methanol yield.
In industry a very low temperature would give a high equilibrium yield, but the reaction rate would be extremely slow. A moderate temperature is chosen as a compromise, giving an acceptable rate while still maintaining a reasonable yield.
Examiner tips
- Show the change in moles of gas for pressure effect; state the direction of shift. Explain temperature effect using exothermic/endothermic. Mention rate–yield trade‑off for the temperature choice. Use correct terminology (Le Chatelier, equilibrium, exothermic).
Common mistakes
- Confusing the direction of the temperature shift (claiming it moves to the forward side). Forgetting that the forward reaction has fewer gas molecules. Not linking the temperature choice to reaction rate. Using vague terms like "good" instead of "moderate".
Mark scheme (5 marks)
- Increasing pressure shifts the equilibrium position towards the side with the smaller number of moles of gas
- There are 3 moles of gas on the left and 1 mole on the right, so high pressure favours the forward reaction / product formation
- Increasing temperature shifts the equilibrium position in the direction of the endothermic reaction
- The forward reaction is exothermic, so increasing temperature shifts equilibrium to the left, reducing the yield of methanol
- A very low temperature gives a higher equilibrium yield but the rate of reaction would be too slow, so a moderate temperature is a compromise between yield and rate
Key terms in this question
Related
- All OCR A-Level Chemistry A (H432) revision notes →
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- Decode the mark scheme abbreviations →
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