# 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.

> OCR A-Level Chemistry A (H432) — 5.1 Rates, equilibrium and pH · Explain · 5 marks

> 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.

## Mark scheme (5 marks)

1. Increasing pressure shifts the equilibrium position towards the side with the smaller number of moles of gas
2. There are 3 moles of gas on the left and 1 mole on the right, so high pressure favours the forward reaction / product formation
3. Increasing temperature shifts the equilibrium position in the direction of the endothermic reaction
4. The forward reaction is exothermic, so increasing temperature shifts equilibrium to the left, reducing the yield of methanol
5. 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

- [closed system](https://www.gradenine.co.uk/glossary/closed-system)

## Related

- [Revision notes for OCR A-Level Chemistry A (H432)](https://www.gradenine.co.uk/learn)
- [How to answer "Explain" questions](https://www.gradenine.co.uk/tools/command-word-cheatsheet)
- [Practice this with AI marking (free)](https://www.gradenine.co.uk/start)

---
Source: [GradeNine](https://www.gradenine.co.uk/q/the-industrial-production-of-methanol-involves-21a5ada3) · Published by Druglandscape Ltd.