Sulfur dioxide reacts with oxygen in a closed vessel to form sulfur trioxide. The equation for the reaction is: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The reaction reaches equilibrium at a constant temperature. Explain how the expression for Kp is written for this equilibrium, and describe what happens to the value of Kp if the pressure of the system is increased at constant temperature.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
The industrial production of sulfuric acid relies on the equilibrium: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The position of this equilibrium and the equilibrium constant are both important to chemists.
Model answer (5 marks)
Kp is expressed as the ratio of the partial pressures of the products to those of the reactants, each raised to the power of its stoichiometric coefficient. For the reaction 2SO₂(g)+O₂(g)⇌2SO₃(g) the expression is
Kp = (pSO₃)² / [(pSO₂)²·pO₂]
The value of Kp has units that depend on the overall change in moles of gas; for this reaction it is Pa⁻¹ (or atm⁻¹). Increasing the total pressure of the system at constant temperature does not change the numerical value of Kp – only a change in temperature will alter Kp.
Kp = (pSO₃)² / [(pSO₂)²·pO₂]
The value of Kp has units that depend on the overall change in moles of gas; for this reaction it is Pa⁻¹ (or atm⁻¹). Increasing the total pressure of the system at constant temperature does not change the numerical value of Kp – only a change in temperature will alter Kp.
Examiner tips
- Write the full expression with stoichiometric exponents; include units or note that they depend on Δn.
- State that Kp is independent of pressure changes; emphasise temperature dependence.
Common mistakes
- Writing Kp with reactants in the numerator or forgetting to raise pressures to stoichiometric powers.
- Claiming that Kp changes when pressure is altered; confusing Le Chatelier with Kp behaviour.
Mark scheme (5 marks)
- Kp is written in terms of partial pressures of products over reactants (products in numerator, reactants in denominator)
- The partial pressures are raised to the power of their stoichiometric coefficients, i.e. (p SO₃)² in the numerator and (p SO₂)² × (p O₂) in the denominator
- Kp has units (accept: Pa⁻¹ or atm⁻¹ depending on unit used, or statement that units depend on the expression / overall change in moles of gas)
- Increasing pressure does NOT change the value of Kp
- Kp only changes if temperature changes / Kp is only affected by temperature
Key terms in this question
Related
- All AQA A-Level Chemistry (7405) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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