The industrial manufacture of ammonia uses the reversible reaction shown. N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ/mol. Explain the effect on the equilibrium position and on the yield of ammonia when the pressure of the system is increased.

Cambridge International IGCSE Chemistry (0620) — 6.3 Reversible reactions and equilibrium · Explain · 4 marks · View as Markdown

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

Model answer (4 marks)

Increasing the pressure shifts the equilibrium to the right, towards the side with fewer moles of gas. The reaction has 4 moles of gas on the left and 2 moles on the right, so the system moves to the side with 2 moles to reduce the pressure.
The forward reaction rate increases more than the reverse rate, so the equilibrium position is displaced to the right.
As a result, the yield of ammonia increases.

Examiner tips

  • Use the Le Chatelier principle: pressure change → shift to fewer moles. Mention the mole numbers (4 → 2) to justify the shift. State the effect on yield explicitly.
  • common_mistakes
  • :
  • Saying the equilibrium shifts to the left or ignoring the mole difference. Failing to link the pressure change to the shift in equilibrium. Not mentioning the increase in ammonia yield.

Mark scheme (4 marks)

  1. Increasing pressure shifts the equilibrium position to the right / towards the side with fewer moles of gas
  2. Because the right-hand side has fewer moles of gas (2 moles) than the left-hand side (4 moles)
  3. The rate of the forward reaction increases more than the rate of the reverse reaction / system counteracts the increased pressure
  4. The yield of ammonia increases

Key terms in this question

reversible reaction · equilibrium position · yield · pressure

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