A student is studying the production of methanol, which is manufactured industrially using the reversible reaction: CO(g) + 2H₂(g) ⇌ CH₃OH(g). The forward reaction is exothermic. The industrial process uses a copper-based catalyst and operates at a temperature of around 250 °C and a pressure of around 50–100 atmospheres. Explain how using a catalyst and a high pressure each affect the rate of reaction and the position of equilibrium in this process.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Methanol is an important industrial chemical used as a fuel and as a starting material for making other chemicals. It is produced from carbon monoxide and hydrogen gases in the presence of a copper-based catalyst.
Model answer (5 marks)
A catalyst provides an alternative reaction pathway with a lower activation energy, so the rate of both the forward and reverse reactions increases.
The catalyst does not change the position of equilibrium; it merely speeds up the attainment of equilibrium.
Increasing the pressure brings the gas molecules closer together, increasing the frequency of collisions and therefore the rate of reaction.
Because the reaction has 3 moles of gas on the left and 1 mole on the right, higher pressure shifts the equilibrium toward the product side (the side with fewer gas moles).
Thus, a high pressure increases both the rate of reaction and the yield of methanol.
The catalyst does not change the position of equilibrium; it merely speeds up the attainment of equilibrium.
Increasing the pressure brings the gas molecules closer together, increasing the frequency of collisions and therefore the rate of reaction.
Because the reaction has 3 moles of gas on the left and 1 mole on the right, higher pressure shifts the equilibrium toward the product side (the side with fewer gas moles).
Thus, a high pressure increases both the rate of reaction and the yield of methanol.
Examiner tips
- Mention the catalyst lowers activation energy and speeds up both directions equally.
- Explain that higher pressure favours the side with fewer gas moles.
- Show the reaction stoichiometry (3→1) to justify the pressure effect.
- Link the pressure effect to increased product yield.
Common mistakes
- Saying the catalyst shifts equilibrium to the product side.
- Confusing the effect of pressure on rate with its effect on equilibrium.
- Using the wrong side of the reaction (e.g. 1→3) when describing pressure shift.
Mark scheme (5 marks)
- The catalyst increases the rate of reaction by providing a different reaction pathway with a lower activation energy.
- The catalyst does not change / shift the position of equilibrium (it speeds up both the forward and backward reactions equally).
- Increasing pressure means gaseous particles are closer together, so they collide more frequently, leading to a higher rate of successful collisions and a faster rate of reaction.
- Increasing pressure shifts the equilibrium position towards the side with the smaller number of moles of gas, which is the product side (right-hand side) in this reaction.
- Therefore, high pressure increases both the rate of reaction and the yield of methanol (more product formed at equilibrium).
Key terms in this question
catalyst · pressure · position of equilibrium
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