A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction.

(Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK^–1 mol^–1)

$$\begin{gathered} T_1 = 300 K\hspace{0.17em} \\ {K}_1 = \frac{0.693}{40} \\ {T}_2 = 320 K \\ {K}_2 = \frac{0.693}{20} \\ \log \frac{K_2}{K_1} = \frac{E_a}{2.303 R}\left(\frac{T_2-T_1}{T_1{T}_2}\right) \\ \log \frac{{\displaystyle \frac{0.693}{20}}}{{\displaystyle \frac{0.693}{40}}} = \frac{E_a}{2.303 x 8.314}\left(\frac{320-300}{320 x 300}\right) \\ \log \left(\frac{40}{20}\right) = \frac{E_a}{2.303 x 8.314}\left(\frac{20}{320 x 300}\right) \\ = \frac{0.3010 x 2.303 x 8.314 x 320 x 300}{20} = E_a \\ = 27764 J \\ = 27.76 KJ \end{gathered}$$