Question #44491

The reversible electrophilic aromatic substitution (e.g., sulfonation) of a monosubstituted benzene yields ortho ($O$) and para ($P$) products, exhibiting a classic kinetic vs. thermodynamic control dichotomy.

At $T_1 = 300\text{ K}$ (under strict kinetic control), the initial rate of formation of $O$ is $2e^2$ times that of $P$. At $T_2 = 600\text{ K}$ (under strict thermodynamic control), the equilibrium concentration of $P$ is $\frac{e^4}{2}$ times that of $O$.

The pre-exponential factors for the forward reactions satisfy $A_o = 2A_p$. The difference in standard entropy of reaction is $\Delta S^\circ_p - \Delta S^\circ_o = -R \ln 2$ (accounting for the statistical factor of two ortho positions vs. one para position).

Assuming activation energies and standard enthalpies are temperature-independent, what is the difference in the activation energies of the reverse reactions, $(E_{a,-o} - E_{a,-p})$?