Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 2 / A7 / b

The slope is $-5130$ K, so the activation enthalpy isFor $c^{\circ }=1\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$, using $k_2$ in $\mathrm{d}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$ gives the numerical prefactor $k_{\mathrm{B}}/h=2.08366\times 10^{10}{\mathrm{K}}^{-1}{\mathrm{s}}^{-1}$. The intercept therefore yieldsThe large logarithm must be subtracted; identifying $R\times 3.26$ with the activation entropy would omit the transition-state attempt-frequency factor. Changing rate units without changing the standard-concentration prefactor consistently would also give a spurious entropy.

The large negative activation entropy is consistent with a strongly organized bimolecular cycloaddition. Two separately translating and rotating molecules must approach with a restricted orientation to form a compact ring-forming transition state. Losing relative positional and orientational freedom lowers entropy.

Internal conformational restrictions and solvent rearrangement can contribute too. In a nonpolar solvent, a simple association picture is more plausible than assigning the whole effect to strong ionic solvation, but the numerical value is not a direct count of molecular orientations alone.

The activation enthalpy is obtained from the slope across the measured $1/T$ interval. The activation entropy comes from the intercept at $1/T=0$, far outside that interval, so a small slope uncertainty can produce a much larger uncertainty in the extrapolated intercept. Slope and intercept errors are consequently strongly correlated.

The entropy is also sensitive to the assumed prefactor and standard concentration, whereas a constant multiplicative calibration error shifts the intercept without changing the slope. This explains why its inferred precision is often poorer. Since enthalpy and entropy have different units, a claim about which absolute numerical uncertainty is larger requires a stated comparison convention.

Use the same transition state and consistent standard states in both directions. Subtracting their definitions givessoRing-forming association normally has ${\Delta }_{\mathrm{r}}{S}^{\circ }<0$. Hence the reverse activation entropy should be more positive, or less negative, than the forward value: a single adduct approaches a bond-breaking configuration rather than two molecules losing their relative motion. Its precise value and even whether it crosses zero cannot be deduced without the reaction entropy. The reverse elementary reaction is first order, so its Eyring equation has the corresponding first-order standard-state prefactor.