Explain why the value of the equilibrium constant, K, for a given reversible reaction changes when the temperature is increased, but does not change when an inert gas is added to the equilibrium mixture at constant volume.

IB DP Chemistry Standard Level (2023 syllabus) — R2.3 How far? The extent of chemical change · Explain · 4 marks · View as Markdown

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

Model answer (4 marks)

K changes with temperature because a change in temperature alters the rate of the forward and reverse reactions by different amounts, changing the ratio of rate constants kf/kr. The new equilibrium position corresponds to a different ratio of product to reactant concentrations, so K (which is defined by that ratio) has a new value.

Adding an inert gas at constant volume does not change the partial pressures or molar concentrations of any reacting species. Because the concentrations/partial pressures of reacting species are unchanged, the equilibrium expression Kc (or Kp) retains the same numerical value, so K does not change.

Examiner tips

  • Use the term ‘ratio of rate constants’ to link temperature to K. Show that K is defined by the concentration ratio at equilibrium. State that an inert gas does not alter reacting species concentrations at constant volume. Mention Kc or Kp to cover both concentration and pressure forms.

Common mistakes

  • Confusing the effect of temperature on K with the effect on reaction rates alone. Saying the inert gas changes the equilibrium instead of noting it does not affect reacting species. Using the wrong equilibrium expression (e.g., Kp when Kc is required).

Mark scheme (4 marks)

  1. K changes with temperature because a change in temperature alters the rate of the forward and reverse reactions by different amounts / changes the ratio of rate constants kf/kr.
  2. The new equilibrium position corresponds to a different ratio of product to reactant concentrations, so K (which is defined by that ratio) has a new value.
  3. Adding an inert gas at constant volume does not change the partial pressures or molar concentrations of any reacting species.
  4. Because the concentrations/partial pressures of reacting species are unchanged, the equilibrium expression Kc (or Kp) retains the same numerical value, so K does not change.

Key terms in this question

reversible reaction

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