Methanol can be manufactured from carbon monoxide and hydrogen according to the following reversible reaction: CO(g) + 2H₂(g) ⇌ CH₃OH(g) ΔH = −90 kJ mol⁻¹ At a certain temperature, the equilibrium constant, Kc, for this reaction has a value of 2.5 × 10⁴ mol⁻² dm⁶. When the temperature is increased to 400 °C, the value of Kc decreases significantly. Explain what the value of Kc tells us about the position of equilibrium, and explain why increasing the temperature causes Kc to decrease for this reaction.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (5 marks)
A large value of Kc (2.5 × 10⁴) shows that at equilibrium the concentration of methanol is much greater than that of CO and H₂, so the equilibrium lies to the right, towards the products.
The reaction is exothermic (ΔH = –90 kJ mol⁻¹). Adding heat therefore favours the endothermic reverse reaction. When the temperature is raised to 400 °C the equilibrium shifts to the left, producing less methanol and more CO and H₂. Consequently the new Kc value is smaller than the original value.
The reaction is exothermic (ΔH = –90 kJ mol⁻¹). Adding heat therefore favours the endothermic reverse reaction. When the temperature is raised to 400 °C the equilibrium shifts to the left, producing less methanol and more CO and H₂. Consequently the new Kc value is smaller than the original value.
Examiner tips
- State that a large Kc means the products dominate at equilibrium. Explain that the reaction is exothermic and that increasing temperature favours the reverse, endothermic direction. Show the link between the shift and the decrease in Kc.
- common_mistakes
- :
- Failing to mention that the large Kc indicates the equilibrium lies to the right. Confusing the effect of temperature on exothermic reactions (writing that heat favours the forward reaction). Not linking the shift to a decrease in the numerical value of Kc.
Mark scheme (5 marks)
- A large value of Kc (2.5 × 10⁴) means the equilibrium lies to the right / towards the products
- This means the concentration of products (methanol) is much greater than the concentration of reactants at equilibrium
- The forward reaction is exothermic (ΔH is negative / releases heat)
- Increasing temperature favours the endothermic (reverse) reaction / equilibrium shifts to the left
- This means less product and more reactants are present at the new equilibrium, so Kc decreases
Key terms in this question
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