$$A \rightarrow B$$
The rate constants of the above reaction at 200 K and 300 K are 0.03 min$$^{-1}$$ and 0.05 min$$^{-1}$$ respectively. The activation energy for the reaction is J (Nearest integer)
(Given: ln 10 = 2.3, R = 8.3 J K$$^{-1}$$ mol$$^{-1}$$, log 5 = 0.70, log 3 = 0.48, log 2 = 0.30)

Using the Arrhenius equation:

$$ \ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) $$

Given: $$k_1 = 0.03$$ min$$^{-1}$$ at $$T_1 = 200$$ K, $$k_2 = 0.05$$ min$$^{-1}$$ at $$T_2 = 300$$ K

$$ \ln\frac{k_2}{k_1} = \ln\frac{0.05}{0.03} = \ln\frac{5}{3} = 2.3(\log 5 - \log 3) = 2.3(0.70 - 0.48) = 2.3 \times 0.22 = 0.506 $$

$$ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{200} - \frac{1}{300} = \frac{300 - 200}{200 \times 300} = \frac{100}{60000} = \frac{1}{600} $$

$$ 0.506 = \frac{E_a}{8.3} \times \frac{1}{600} = \frac{E_a}{4980} $$

$$ E_a = 0.506 \times 4980 = 2520 \text{ J mol}^{-1} $$