In general, it is observed that the rate of a reaction becomes double with every rise in…

Given: $T_1 = 295\text{ K}$, $T_2 = 305\text{ K}$, $\frac{k_2}{k_1} = 2$, $R = 8\cdot314\text{ J K}^{-1}\text{ mol}^{-1}$. Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2\cdot303 R} \left[\frac{T_2 - T_1}{T_1 T_2}\right]$ $\log 2 = \frac{E_a}{2\cdot303 \times 8\cdot314} \left[\frac{10}{295 \times 305}\right]$ $0\cdot3010 = \frac{E_a}{19\cdot147} \times \frac{10}{89975}$ $E_a = \frac{0\cdot3010 \times 19\cdot147 \times 89975}{10} = 51854\cdot8\text{ J mol}^{-1} = 51\cdot85\text{ kJ mol}^{-1}$ Final answer: 51.85 kJ mol-1