Reversible Reactions and Equilibria
6 marks · NormalHydrogen iodide is formed in a reversible reaction:
Explain, in terms of exothermic and endothermic reactions, what happens to the position of equilibrium if the temperature is decreased.
Explain why adding a catalyst to the sealed reaction vessel does not change the position of equilibrium.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) Decreasing temperature favours the exothermic direction. Since the forward reaction is endothermic (ΔH positive), the reverse reaction is exothermic. So, decreasing temperature shifts equilibrium towards the left (reactants).
(b) A catalyst speeds up both the forward and reverse reactions equally. It only affects the rate at which equilibrium is reached, not the position of equilibrium itself.
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Marking points
M1 (a) · 1 mark · Recognises that lowering temperature favours exothermic direction
A1 (a) · 1 mark · Identifies which direction (reverse/left) is exothermic here
C1 (a) · 1 mark · Relates ΔH to endothermic forward and exothermic reverse
M2 (b) · 1 mark · States that a catalyst increases the rates of both forward and reverse reactions equally
A2 (b) · 1 mark · States that the position of equilibrium is unchanged
C2 (b) · 1 mark · Explains that catalyst only speeds up reaching equilibrium
Worked solution
(a) The forward reaction is endothermic (ΔH = +52 kJ mol⁻¹), so the reverse is exothermic.
Decreasing temperature favours the exothermic direction.
Therefore, equilibrium shifts towards the reverse reaction, or to the left (more reactants).
(b) A catalyst increases the rate of both forward and reverse reactions equally.
This means it has no effect on the position of equilibrium. It only brings equilibrium about faster.