Explanation: Solve: আমরা জানি, \( \ln\left(\frac{K_2}{K_1}\right) = \frac{E_a}{R}\left[\frac{T_2-T_1}{T_1T_2}\right] \) এখানে, \(T_1 = 35^\circ C = 308 K\), \(T_2 = 45^\circ C = 318 K\), \(K_1 = 1\), \(K_2 = 2\), \(E_a = ?\). \( \text{So, } \ln\left(\frac{2}{1}\right) = \frac{E_a}{8.314} \times \frac{[318-308]}{[318 \times 308]} \implies \ln(2) = \frac{E_a}{8.314} \times 1.021 \times 10^{-4} \). \( \therefore E_a = \frac{\ln(2) \times 8.314}{1.021 \times 10^{-4}} = 56443.95 \, \text{J/mol} = 56.44 \, \text{kJ/mol} \). Ans. (B)