The rate constants of the reaction at 100 K and 200 K are $0.01\, s^{-1}$ and $0.02\, s^{-1}$, respectively. The value of activation energy (E) will be

$(\log 2=0.3010)$

1152.65 J

The correct answer is Option (4) → 1152.65 J

Use the Arrhenius equation for two temperatures:

$\log\left(\frac{k_2}{k_1}\right) = \frac{E}{2.303\,R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)$

Given:

$k_1 = 0.01\,\text{s}^{-1},\quad k_2 = 0.02\,\text{s}^{-1}$

$T_1 = 100\,\text{K},\quad T_2 = 200\,\text{K}$

$\log 2 = 0.3010,\quad R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$

Substitution:

$0.301 = \frac{E}{2.303 \times 8.314}\left(\frac{100}{100 \times 200}\right)$

$E = 2.303 \times 8.314 \times 0.301 \times \frac{100 \times 200}{100}$

$E \approx 1152.65\ \text{J mol}^{-1}$