A first order reaction is $\displaystyle 50$%complete in $\displaystyle 30$ minutes at $\displaystyle 300$ K and in $\displaystyle 10$ minutes at $\displaystyle 320$ K. Calculate activation energy $\displaystyle \left(\mathrm{E}_{\mathrm{a}}\right)$ for the reaction. $\displaystyle \left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$ $\displaystyle 3$ [Given: $\displaystyle \log 2=0.3010, \log 3=0.4771, \log 4=0.6021$ ]

\[\begin{aligned} & \mathrm{k}=\frac{0.693}{\mathrm{t}_{1 / 2}} \\ & \mathrm{k}_{1}=\frac{0.693}{30} \mathrm{~min}^{-1} \\ & \mathrm{k}_{2}=\frac{0.693}{10} \mathrm{~min}^{-1} \\ & \log \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{Ea}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right] \\ & \log 3=\frac{\mathrm{Ea}}{2.303 \times 8.314}\left[\frac{1}{300}-\frac{1}{320}\right] \\ & \mathrm{Ea}=\frac{0.4771 \times 19.147 \times 300 \times 320}{20} \\ & \mathrm{Ea}=43848 \mathrm{~J} \mathrm{~mol}^{-1} \text { or } 43.848 \mathrm{KJ} \mathrm{~mol}^{-1} \text { or } 43.85 \mathrm{k} \mathrm{~J} \mathrm{~mol}^{-1} \\ & \text { (Deduct }^{1 / 2} \text { mark for incorrect or no unit) } \end{aligned} \]