Sulfur dioxide reacts with oxygen to form sulfur trioxide in a reversible reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The forward reaction is exothermic. Explain what happens to the position of equilibrium and the yield of sulfur trioxide when the temperature is increased.

AQA A-Level Chemistry (7405) — 3.1.6 Chemical equilibria, Le Chatelier and Kc · Explain · 5 marks · View as Markdown

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

Model answer (5 marks)

Increasing the temperature shifts the equilibrium to the left, favouring the reverse reaction. Because the forward reaction is exothermic, adding heat is equivalent to adding a reactant of the endothermic direction, so the system absorbs the extra heat by converting SO₃ back into SO₂ and O₂. Consequently the yield of SO₃ decreases, although the equilibrium remains dynamic with both forward and reverse reactions occurring.

Examiner tips

  • State the shift to the left and explain it in terms of heat being an ‘reactant’ for the endothermic reverse reaction.
  • Mention that the yield of SO₃ falls because the reverse reaction is favoured.
  • Use the correct chemical equations and keep the answer concise.

Mark scheme (5 marks)

  1. The position of equilibrium shifts to the left (reverse reaction is favoured)
  2. This is because increasing temperature favours the endothermic reaction / the system opposes the change by absorbing the extra heat energy
  3. The yield / amount of sulfur trioxide decreases
  4. The reverse reaction converts sulfur trioxide back into sulfur dioxide and oxygen
  5. The reaction is reversible / a dynamic equilibrium is established, so the equilibrium position changes but both forward and reverse reactions still occur

Key terms in this question

reversible reaction · position of equilibrium · exothermic

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