Physical Chemistry
7 marks · EasyThis question is about the interpretation of reaction profile diagrams and the properties of equilibrium.
The diagram below shows an energy profile for a reaction without a catalyst.
Draw on the diagram a line to show how the profile changes when a catalyst is used. Label the activation energy for the catalysed reaction.
Explain why the overall enthalpy change (ΔH) for the reaction does not change when a catalyst is used.
In a reversible reaction at equilibrium, the following is observed: the concentrations of reactants and products stay the same, even though both the forward and reverse reactions are occurring. Explain why the concentrations remain unchanged at equilibrium, but the reaction is still said to be 'dynamic'.
A student suggests that adding a catalyst will increase the amount of product at equilibrium. Do you agree? Explain your answer.
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.i) A lower energy curve between reactants and products, starting and ending at same points as original profile, with reduced peak labelled 'Activation energy (with catalyst)' or similar. · must show catalysed pathway between A and C with lower maximum y-value, correct label of catalysed activation energy
(a.ii) Because a catalyst provides an alternative pathway with lower activation energy, but does not change the energies of reactants or products, so ΔH stays the same. · must show ΔH unchanged, reason (relative energies unaffected)
(b) At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, so concentrations of all substances remain constant. The system is dynamic because both reactions continue to occur, but at the same rate. · must show equal rates, constant concentrations, both reactions still happening (dynamic)
(c) No, adding a catalyst does not increase the amount of product at equilibrium. It only increases the speed at which equilibrium is reached by speeding up both the forward and reverse reactions equally. · must show no change to equilibrium position, catalyst speeds both reactions equally
Common mistakes: Drawing the catalysed pathway with different ΔH; Saying a catalyst increases yield at equilibrium
Show mark scheme and worked solution
Marking points
M1 (a(i)) · 1 mark · Draws a lower hump/pathway on the diagram with correct start and end points, showing the catalysed reaction route.
M2 (a(i)) · 1 mark · Labels activation energy for catalysed reaction clearly.
M3 (a(ii)) · 1 mark · Correctly states that ΔH does not change because energies of reactants and products do not change—only the activation energy changes.
M4 (b) · 1 mark · States that equal rates of forward and reverse reactions cause concentrations to remain constant.
M5 (b) · 1 mark · Explains the reaction is dynamic because both reactions continue to happen.
M6 (c) · 1 mark · States that adding a catalyst does not change the amount of product at equilibrium.
M7 (c) · 1 mark · Explains catalyst only increases speed to reach equilibrium, not position/yield.
Worked solution
a.i) The catalysed reaction profile should be drawn as a lower curve starting at the reactants and finishing at the products, but with a smaller maximum (peak) than the uncatalysed route. It should join the original endpoints and the label 'Activation energy (with catalyst)' or similar should indicate the new lower peak.
a.ii) The catalyst gives an alternative reaction route with a lower activation energy but does not affect the energy levels of reactants or products. So the enthalpy change (ΔH) remains the same.
b) At equilibrium, the rate of the forward and reverse reactions is equal so the concentrations of reactants and products remain unchanged. The system is dynamic because both forward and reverse reactions are still taking place, just at equal rates.
c) No, adding a catalyst does not increase the amount of product at equilibrium. It only makes the reaction reach equilibrium faster by speeding up both reactions equally; it does not affect the position of equilibrium.