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2024 H2 Chem P1 Q11 — Kinetics Solution

Kinetics 2024 A-Level H2 Chemistry · Paper 1 · Q11

Question

A Boltzmann distribution for a sample of a reacting gas at a constant temperature, \(T^\circ C\), is shown.
The activation energy, EA, for the reaction is marked.
The temperature of the sample of gas is decreased to \((T - 50)^\circ C\).
Which point could show the intercept of the new Boltzmann distribution curve with the value of the activation energy of the reaction at \(T - 50^\circ C\)?Question diagram
  1. AA
  2. BB
  3. CC
  4. DD
Correct answer: D

Video Solution

Explanation

The answer is D.
• The activation energy of a reaction does not change with temperature, so the new intercept must still lie on the dashed \(E_\text{A}\) line. This rules out A and C, which are not on the \(E_\text{A}\) line.
• At the lower temperature, \((T - 50)^\circ\text{C}\), the peak of the Boltzmann distribution moves to lower energy and the curve at high energies falls. Fewer particles have energy equal to or greater than \(E_\text{A}\), so the new curve crosses the \(E_\text{A}\) line lower than the present curve does.
• D lies on the \(E_\text{A}\) line below the present intercept, so D is correct.
• B is wrong: it lies on the \(E_\text{A}\) line above the present intercept, which would mean more particles at \(E_\text{A}\), as happens when the temperature is raised.

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