Explain how the Maxwell–Boltzmann distribution of molecular energies can be used to account for the observation that a small increase in temperature produces a disproportionately large increase in the rate of a chemical reaction.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (4 marks)
The Maxwell–Boltzmann distribution shows the spread of kinetic energies among molecules at a given temperature, with only a fraction of molecules possessing energy greater than or equal to the activation energy. When temperature increases, the distribution curve shifts to the right (towards higher energies) and flattens, so the peak moves to a higher energy value. The area under the curve to the right of the activation energy increases significantly, meaning the fraction of molecules with energy ≥ Ea increases markedly even for a small temperature rise. Because the rate depends exponentially on the fraction of molecules above Ea (as in the Arrhenius equation, k = Ae^(−Ea/RT)), even a modest temperature increase produces a disproportionately large increase in the rate constant and hence the reaction rate.
Examiner tips
- Use the phrase ‘fraction of molecules with energy ≥ Ea’ to link the distribution to the Arrhenius equation.
- Show the shift of the curve to the right and the increase in the area beyond Ea.
- Mention the exponential dependence in k = Ae^(−Ea/RT) to justify the large rate change.
- Keep the answer concise – 4 marks only.
Common mistakes
- Confusing the Maxwell–Boltzmann distribution with the kinetic energy of a single molecule.
- Failing to mention the exponential relationship in the Arrhenius equation.
- Giving a qualitative description without linking the distribution to the rate constant.
Mark scheme (4 marks)
- The Maxwell–Boltzmann distribution shows the spread of kinetic energies among molecules at a given temperature, with only a fraction of molecules possessing energy greater than or equal to the activation energy.
- When temperature increases, the distribution curve shifts to the right (towards higher energies) and flattens, so the peak moves to a higher energy value.
- The area under the curve to the right of the activation energy increases significantly, meaning the fraction of molecules with energy ≥ Ea increases markedly even for a small temperature rise.
- Because the rate depends exponentially on the fraction of molecules above Ea (as in the Arrhenius equation, k = Ae^(−Ea/RT)), even a modest temperature increase produces a disproportionately large increase in the rate constant and hence the reaction rate.
Key terms in this question
Maxwell–Boltzmann distribution
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