The Arrhenius equation is given by: $$ K = A e^{-E_a / RT} $$ Taking the natural logarithm on both sides, we get: $$ \ln K = \ln A - \frac{E_a}{RT} $$ For two different temperatures $T_1$ and $T_2$ with corresponding rate constants $K_1$ and $K_2$, the equation becomes: $$ \ln \frac{K_2}{K_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) $$ Given data: $T_1 = 0^\circ \text{C} = 273 \text{ K}$ $T_2 = 25^\circ \text{C} = 298 \text{ K}$ $K_1 = 0.237 \times 10^{-4} \text{ L mol}^{-1} \text{ s}^{-1}$ $K_2 = 2.64 \times 10^{-4} \text{ L mol}^{-1} \text{ s}^{-1}$ $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$ Substituting the values into the equation: $$ \begin{aligned} \ln \left( \frac{2.64 \times 10^{-4}}{0.237 \times 10^{-4}} \right) &= \frac{E_a}{8.314} \left( \frac{1}{273} - \frac{1}{298} \right) \\ \ln(11.139) &= \frac{E_a}{8.314} \times (0.003663 - 0.003356) \\ 2.4105 &= \frac{E_a}{8.314} \times 0.000307 \\ E_a &= \frac{2.4105 \times 8.314}{0.000307} \\ E_a &= 65318.5 \text{ J mol}^{-1} \\ E_a &\approx 65.32 \text{ kJ mol}^{-1} \end{aligned} $$ Thus, the activation energy is $65.32 \text{ kJ mol}^{-1}$.