A first order reaction is 50% completed in 20 minutes at ${27}^{0}C$ and in 5 minutes at ${47}^{0}C$ . The total energy of activation of the reaction is :

Determining the values of k₁ and k₂.

$$\begin{gathered} {\mathrm{k}}_1=\frac{2.303}{20 \min }\log \frac{100}{50}=0.035 {\min }^{-1} \\ {\mathrm{k}}_2=\frac{2.303}{5 \min }\log \frac{100}{50}=0.139 {\min }^{-1} \end{gathered}$$

Considering Arrhenius equation:

$$\begin{gathered} \log \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left[\frac{{\mathrm{T}}_2-{\mathrm{T}}_1}{{\mathrm{T}}_1{\mathrm{T}}_2}\right] \\ \log \frac{0.139 {\min }^{-1}}{0.035 {\min }^{-1}}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left[\frac{320-300}{320\times 300}\right] \\ {\mathrm{E}}_{\mathrm{a}}=\frac{0.6\times 2.303\times 8.314\times 300\times 320}{20}=55144 \mathrm{J}/\mathrm{mol}=55.144 \mathrm{kJ}/\mathrm{mol} \end{gathered}$$