 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2024/paper-2/a7/a/ii/solution

k_{2} = \frac{k _{B} T}{h c ^{\circ}} exp (\frac{Δ S ^{‡\circ}}{R}) exp (- \frac{Δ H ^{‡\circ}}{RT}) .

Taking logarithms of numerical values in fixed, consistent units gives

ln \frac{k _{2}}{T} = - \frac{Δ H ^{‡\circ}}{R} \frac{1}{T} + ln \frac{k _{B}}{h c ^{\circ}} + \frac{Δ S ^{‡\circ}}{R} .

Thus for slope

s

and intercept

b

Δ H^{‡\circ} = - R s, Δ S^{‡\circ} = R [b - ln \frac{k _{B}}{h c ^{\circ}}] .

The logarithms imply reference units, as made explicit in the supplied numerical plot. Approximate linearity requires the activation parameters and transmission coefficient to vary little over the temperature interval; a significant activation heat-capacity difference produces curvature.

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