A chemical plant produces sulfuric acid using the Contact process. One stage involves the reversible reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The forward reaction is exothermic. The plant operates at a moderate temperature of around 450 °C rather than a very low temperature, and uses vanadium pentoxide as a catalyst. Explain how using a catalyst increases the rate of this reaction, and explain why the plant uses a moderate temperature rather than a very low temperature, even though a lower temperature would favour a greater yield of SO₃.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
The Contact process is used industrially to manufacture sulfuric acid. The key reversible reaction is: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The forward reaction is exothermic. Vanadium pentoxide is used as a catalyst at approximately 450 °C.
Model answer (5 marks)
A catalyst provides a different reaction pathway with a lower activation energy. Consequently, a greater proportion of the gas molecules now have at least the activation energy, so more collisions are successful and the rate of formation of SO₃ increases.
A lower temperature would shift the equilibrium position towards the exothermic forward reaction, giving a higher yield of SO₃. However, at a very low temperature the rate of reaction would be too slow. Therefore the plant operates at a moderate temperature of about 450 °C – a compromise that gives a sufficiently high rate of reaction while still maintaining an acceptable yield of SO₃.
A lower temperature would shift the equilibrium position towards the exothermic forward reaction, giving a higher yield of SO₃. However, at a very low temperature the rate of reaction would be too slow. Therefore the plant operates at a moderate temperature of about 450 °C – a compromise that gives a sufficiently high rate of reaction while still maintaining an acceptable yield of SO₃.
Examiner tips
- Mention the lower activation energy first, then the increased proportion of successful collisions. Explain the temperature trade‑off: equilibrium vs. rate. Use the exact terms ‘exothermic’, ‘activation energy’, ‘rate’ and ‘yield’.
Common mistakes
- Failing to state that the catalyst lowers the activation energy. Confusing the effect of temperature on equilibrium with its effect on reaction rate. Using vague wording such as ‘faster’ without linking to activation energy.
Mark scheme (5 marks)
- A catalyst provides a different reaction pathway with a lower activation energy
- A greater proportion of particles now have at least the activation energy, so more collisions are successful and the rate increases
- A lower temperature would shift the equilibrium position towards the exothermic (forward) reaction, giving a higher yield of SO₃
- However, at a very low temperature the rate of reaction would be too slow
- A moderate temperature is a compromise between a sufficient rate of reaction and an acceptable yield of SO₃
Key terms in this question
Related
- All OCR A-Level Chemistry A (H432) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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