A first order reaction is $50\%$ completed in 30 min at ${27}^{°}\text{C}$ and in 10 min at  

 ${47}^{°}\text{C}$ .Calculate the energy of activation of the reaction in ${\text{KJ mol}}^{-1}$

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

                  $k=\frac{0.693}{t\frac{1}{2}}$

Here, $t_{1/2}$ is a half life of a reaction and k is rate constant.

${\text{At 27°C, rate constant k}}_{27°\text{C}}=\frac{0.693}{30\min }=0.0231/{\min }^{-1}$

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

According Arrhenius equation ,  

$\log \frac{k_2}{k_1}=\frac{-{\text{E}}_{\text{a}}}{2.303\times \text{R}}\left[\frac{{\text{T}}_2-{\text{T}}_1}{{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}\right]$

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

$\log \frac{0.0693}{0.0231}=\frac{-{\text{E}}_{\text{a}}}{2.303\times 8.314}\left[\frac{320-300}{320\times 300}\right]$   $\left(\text{Since}{\text{T}}_{\text{K}}=273+°\text{C}\right)$

By solving the equation we will get the activation energy, ${\text{E}}_{\text{a}}=43.8{\text{KJmol}}^{-1}$ .