A first order reaction is 50% complete in 40 minutes at 300K and in 20 minutes at 320K. Calculate the activation energy of the reaction.
(Given : log2= 0.3010 , log4 = 0.6021 , R= 8.314 J/K/mol)

$$\begin{gathered} \begin{array}{r}{\mathrm{T}}_1=300\mathrm{K},{\mathrm{t}}_{1/2}=40\text{ minutes }\\ {\mathrm{T}}_2=320\mathrm{K},{\mathrm{t}}_{1/2}=20\text{ minutes }\end{array} \\ \mathrm{For}, \mathrm{first} \mathrm{order} {\mathrm{t}}_{1/2}=0.693/\mathrm{k} \end{gathered}$$
$\begin{array}{l}\log \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\log \frac{{\left({\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}_1}{{\left({\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}_2}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right)\\ \log \left(\frac{40}{20}\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left(\frac{1}{300}-\frac{1}{320}\right)\\ 0.301=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left(\frac{1}{300}-\frac{1}{320}\right)\\ {\mathrm{E}}_{\mathrm{a}}=\frac{0.301\times 2.303\times 8.314\times 320\times 300}{20}\\ =27663.790\mathrm{J}/\mathrm{mole}\\ =27.67\mathrm{kJ}/\mathrm{mole}\end{array}$