Explanation: Hints: \( \log K = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \) অথবা, \( \ln k = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \) Solve: \( \ln 8 = \frac{E_1}{R} \left( \frac{1}{275} - \frac{1}{375} \right) \) ........... (i) \( \ln 4 = \frac{E_2}{R} \left( \frac{1}{275} - \frac{1}{375} \right) \) ........... (ii) (ii) ÷ (i) \[ \frac{E_2}{E_1} = \frac{\ln 4}{\ln 8} = \frac{2}{3} \implies E_2 = \left( \frac{2}{3} \times 3 \right) = 2 \, \text{kJ mol}^{-1} \] Ans. (C) ব্যাখ্যা: আরহেনিয়াস সমীকরণ: \( \ln K = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right) \) \[ \begin{aligned} \ln k_1 &= \ln A - \frac{E_a}{R} \times \frac{1}{T_1} .......... (i) \\ \ln k_2 &= \ln A - \frac{E_a}{R} \times \frac{1}{T_2} .......... (ii) \text{(ii) - (i)} \implies \ln \left( \frac{k_2}{k_1} \right) &= \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] \[ \log K_1 = \log A - \frac{E_a}{2.303 R} \times \frac{1}{T_1} .......... (i) \log K_2 = \log A - \frac{E_a}{2.303 R} \times \frac{1}{T_2} .......... (ii) \] \[ \text{(ii) - (i)} \implies \log \left( \frac{K_2}{K_1} \right) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]