A first order reaction is 50% completed in 30 min at 27$°$ C and in 10 min at 47$°$ C .Calculate the energy of activation of the reaction in  $\mathrm{KJ} {\mathrm{mol}}^{-1}$

For first order kinetics reaction, the relation between half-life and rate constant is given below.

$\mathrm{k}\frac{0.693}{{\mathrm{t}}_{1/.3}}$

Here , ${\mathrm{t}}_{1/2 }$is a half life of a reaction and k is rate constant.

At 27°C, rate constant ${\mathrm{K}}_{27°\mathrm{C}}=\frac{0.693}{30 \mathrm{mins}}=0.02321/ {\min }^{-1}$

At $47°\mathrm{C} {\mathrm{K}}_{{47}^{°}\mathrm{C}}=\frac{0.693}{10 \min }=0.0693 {\min }^{-1}$

According Arrhenius equation ,

$\log \frac{K_2}{K_1}=\frac{-E_a}{2.303\times R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

Put all known data in this equation and then the activation energy is

$\log \frac{0.0693}{0.0231}=\frac{-E_a}{2.303\times 8.314}\left[\frac{320-300}{320\times 300}\right] (Since{T}_K=273+°\mathrm{C})$

By solving the equation we will get the activation energy ,${\mathrm{E}}_{\mathrm{a}}=43.8\mathrm{KJ} {\mathrm{mol}}^{-1}$