How much faster would a reaction proceed at $25^{\circ} \mathrm{C}$ than at $0^{\circ} \mathrm{C}$ if the activation energy is $65 \mathrm{~kJ}$?

The Arrhenius equation relates the rate constants $k_1$ and $k_2$ at temperatures $T_1$ and $T_2$ as follows:

$\ln\left(\dfrac{k_2}{k_1}\right) = \dfrac{E_a}{R} \left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$

Given $E_a = 65000 \text{ J/mol}$, $T_1 = 273 \text{ K}$, $T_2 = 298 \text{ K}$, and $R = 8.314 \text{ J/(mol K)}$.

Substituting the values:

$\ln\left(\dfrac{k_2}{k_1}\right) = \dfrac{65000}{8.314} \left(\dfrac{1}{273} - \dfrac{1}{298}\right)$

$\ln\left(\dfrac{k_2}{k_1}\right) = 7818.14 \times \left(\dfrac{298 - 273}{273 \times 298}\right)$

$\ln\left(\dfrac{k_2}{k_1}\right) = 7818.14 \times \left(\dfrac{25}{81354}\right)$

$\ln\left(\dfrac{k_2}{k_1}\right) = 7818.14 \times 0.00030729 \approx 2.4025$

Taking the exponential of both sides:

$\dfrac{k_2}{k_1} = e^{2.4025} \approx 11.05$

The reaction proceeds approximately 11 times faster.

Answer: 11 times