Physical Chemistry
7 marks · EasyThis question is about catalysts and the effect of temperature on reversible reactions.
Draw a reaction profile diagram for an exothermic reaction. Clearly label the following on your diagram:
- Reactants - Products - Activation energy (with and without catalyst) - ΔH (enthalpy change)
A reversible reaction takes place:
Explain what happens to the position of equilibrium if the temperature is increased. State your answer in terms of exothermic and endothermic reaction direction.
State whether adding a catalyst will change the position of equilibrium in this reaction. 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.
Show answer
(a) — · must show vertical axis: energy, horizontal axis: progress of reaction, curve showing higher peak (uncatalysed) and lower peak (catalysed route), reactants at higher energy than products (exothermic), arrows: activation energy (with/without catalyst), ΔH, labels for all above
(b) Position of equilibrium shifts to the left, favouring the endothermic (reverse) reaction when temperature increases. Increasing temperature favours the endothermic direction, so more SO₂ and O₂ will form. · must show explains shift to endothermic direction, links to equilibrium position (left/right), relates to reaction equation
(c) Adding a catalyst does not change the position of equilibrium. It speeds up the forward and reverse reactions equally, only allowing equilibrium to be reached faster. · must show states position unchanged, explains both directions sped equally
Must show: diagram shows exothermic profile, labels for reactants/products/activation energies/ΔH
Common mistakes: Not labeling both activation energies; Mixing up endothermic/exothermic directions; Stating catalyst changes position of equilibrium
Show mark scheme and worked solution
Marking points
D1 (a) · 1 mark · Diagram shows an exothermic profile with correct shape: reactants higher than products.
D2 (a) · 1 mark · Both activation energies are shown/labelled (with and without catalyst).
D3 (a) · 1 mark · ΔH (enthalpy change) correctly labelled from reactants to products.
E1 (b) · 1 mark · States that increasing temperature shifts equilibrium to left/towards the endothermic direction.
E2 (b) · 1 mark · Binds explanation to exothermic and endothermic reaction directions, referencing equation if possible.
S1 (c) · 1 mark · States that a catalyst does not change the position of equilibrium.
S2 (c) · 1 mark · Explains that both forward and reverse reactions are sped equally, so only rate to equilibrium is changed.
Worked solution
a) Diagram:
b) The reaction is exothermic in the forward direction (ΔH negative). Increasing temperature favours the endothermic (reverse) reaction, so equilibrium shifts to the left, increasing the amounts of SO₂ and O₂ and decreasing SO₃.
c) Adding a catalyst will not change the position of equilibrium. The catalyst increases the rates of both the forward and reverse reactions equally, so equilibrium is reached faster, but the final proportions remain unchanged.