Explain why a mixture of SO₂(g), O₂(g) and SO₃(g) in which Qc > Kc is not at equilibrium, and describe the direction in which the reaction will proceed to reach equilibrium.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
The industrial production of sulfuric acid involves the reversible reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). At a given temperature, the equilibrium constant Kc has a fixed value.
Model answer (4 marks)
Qc is the reaction quotient, calculated with the same expression as Kc but using the current concentrations. If Qc > Kc the ratio of product to reactant concentrations is too high for the equilibrium position. Therefore the mixture is not at equilibrium. The reaction will shift in the reverse (backward) direction – to the left – to consume SO₃ and produce SO₂ and O₂. As the reaction proceeds in this direction the concentration of SO₃ decreases and the concentrations of SO₂ and O₂ increase until Qc equals Kc, at which point equilibrium is re‑established.
Examiner tips
- Define Qc and compare it with Kc; state the condition Qc>Kc clearly.
- Explain the direction of shift (reverse) and the effect on concentrations.
- Show that equilibrium is restored when Qc=Kc.
- Use correct chemical equations and stoichiometry.
Mark scheme (4 marks)
- Qc is the reaction quotient, calculated using the same expression as Kc but with the current (non-equilibrium) concentrations.
- When Qc > Kc, the ratio of product concentrations to reactant concentrations is too large compared to the equilibrium position.
- The reaction proceeds in the reverse/backward direction (to the left) to reach equilibrium.
- As the reaction proceeds in reverse, product concentration(s) decrease and reactant concentration(s) increase until Qc equals Kc, at which point equilibrium is re-established.
Related
- All IB DP Chemistry Standard Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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