= Solution
Use the same transition state and consistent standard states in both directions. Subtracting their definitions gives
$$
\Delta S_f^{\ddagger\circ}=S_{\mathrm{TS}}^\circ-S_{\mathrm R}^\circ,
\qquad
\Delta S_r^{\ddagger\circ}=S_{\mathrm{TS}}^\circ-S_{\mathrm P}^\circ,
$$
so
$$
\boxed{\Delta S_r^{\ddagger\circ}=\Delta S_f^{\ddagger\circ}-\Delta_{\mathrm r}S^\circ.}
$$
Ring-forming association normally has $\Delta_{\mathrm r}S^\circ<0$. Hence \b[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.
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