The rate of reaction doubles when its temperature changes from 300K to 310 K. Activation energy of such a reaction will be:

\((R = 8.314 \text{ J K}^{-1}\text{ mol}^{-1} \text{ and log 2 = 0.301})\)

\(53.6 \text{ kJ mol}^{-1}\)

Given,

\(T_1 = 27^oC = 300 K\)

and \(T_2 = 37^oC = 310 K\)

Let, \(k_1 = k\)

so, according to the question, \(k_2 = 2k\)

We know,

\(\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R }\left|\frac{1}{T_1} − \frac{1}{T_2}\right|\)

or,  \(\left(\frac{2k}{k}\right) = \frac{E_a}{2.303 × 8.314 }\left|\frac{1}{300} − \frac{1}{310}\right|\)

or,  \(k = \frac{E_a}{2.303 × 8.314 }\left|\frac{310 −300}{300 × 310}\right|\)

or,  \(k = \frac{E_a}{2.303 × 8.314 }\left|\frac{10}{300 × 310}\right|\)

or,  \(E_a = 53598.6 J\)

or,  \(E_a ≈ 53.6 \text{kJ mol}^{-1}\)