According to the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$ Given that the rate doubles, $k_2 = 2k_1$. $T_1 = 20^\circ\text{C} = 20 + 273 = 293 \text{ K}$ $T_2 = 35^\circ\text{C} = 35 + 273 = 308 \text{ K}$ $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$ Substituting the values: $\log(2) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{308 - 293}{293 \times 308} \right]$ $0.3010 = \frac{E_a}{19.147} \left[ \frac{15}{90244} \right]$ $E_a = \frac{0.3010 \times 19.147 \times 90244}{15}$ $E_a \approx 34673 \text{ J mol}^{-1}$ $E_a \approx 34.7 \text{ kJ mol}^{-1}$