The rate of a reaction doubles when temperature changes from $\displaystyle 27$°C to $\displaystyle 37$°C. Calculate energy of activation for the reaction. $\displaystyle \left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$ (Given : $\displaystyle \log 2=0 \cdot 3010, \log 3=0 \cdot 4771, \log 4=0 \cdot 6021$ )

\[\begin{aligned} \log \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}} & =\frac{\mathrm{Ea}}{2 \cdot 303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right] \\ \log \frac{2 \mathrm{k}_{1}}{\mathrm{k}_{1}} & =\frac{\mathrm{E} \mathrm{a}}{2 \cdot 303 \times 8 \cdot 314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}\left[\frac{1}{300}-\frac{1}{310}\right] \\ \mathrm{E}_{\mathrm{a}} & =\frac{0 \cdot 3010 \times 19 \cdot 147 \mathrm{~J} \mathrm{~mol}^{-1} \times 300 \times 310}{10} \\ \mathrm{E}_{\mathrm{a}} & =53598.2 \mathrm{~J} \mathrm{~mol}^{-1} \text { or } 53 \cdot 598 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { or } 53.6 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} \]