 Solution

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.