The rate of a reaction doubles, when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be $\left(\mathrm{R}=8.314{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}\right.$and $\log 2=0.301)$

From Arrhenius equation, $\log \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{-{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right)$
Given, $\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=2,{\mathrm{T}}_2=310 \mathrm{K}\text{ and }{\mathrm{T}}_1=300 \mathrm{K}$
On putting values,
           $\Rightarrow \log 2=\frac{-{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left(\frac{1}{310}-\frac{1}{300}\right)$
          $\Rightarrow {\mathrm{E}}_{\mathrm{a}}=53598.6\mathrm{J}/\mathrm{mol}\approx 53.6\mathrm{kJ}/\mathrm{mol}$