Question #44452

The Cannizzaro reaction of benzaldehyde ($\ce{PhCHO}$) proceeds via a hydride shift mechanism. At moderate base concentrations, the reaction involves a mono-anion intermediate, following the rate law $r = k_{obs}[\ce{PhCHO}]^x[\ce{OH^-}]^y$. At extremely high base concentrations, the mechanism shifts to involve a dianion intermediate, following the rate law $r' = k'_{obs}[\ce{PhCHO}]^x[\ce{OH^-}]^z$.

The rate-determining step (RDS) in both pathways is the hydride transfer. In a high-dielectric solvent ($\epsilon = 80$), the charge-delocalized transition state of the mono-anion pathway is significantly stabilized compared to a low-dielectric solvent ($\epsilon = 40$). This stabilization lowers the activation energy by $\Delta E$, causing the observed rate constant $k_{obs}$ to increase by a factor of $e^2$ at $300\text{ K}$.

Assuming ideal kinetic behavior and using $R = \frac{25}{3}\text{ J K}^{-1}\text{ mol}^{-1}$, calculate the value of the expression $(x + y + z) \times \Delta E$ in $\text{kJ/mol}$.