Question

A first order reaction is $50\%$ completed in 30 min at ${27}^{\circ }C$ and in 10 min at ${47}^{\circ }C$. The energy of activation of the reaction in $kJ$ $mo{l}^{-1}$ is: (Write answer in the form of 10x).

Answer 1

$k_{1}at{27}^{\circ }C$ (300 K) = $\frac{0.693}{30min}=0.0231mi{n}^{-1}$
$k_{2}at{47}^{\circ }C$ (320 K) = $\frac{0.693}{10min}=0.0693mi{n}^{-1}$
Using Arrhenius equation :
$log\left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
$log\left(\frac{0.0693}{0.0231}\right)=\frac{E_a}{2.303\times 8.314\times {10}^{-3}kJmo{l}^{-1}{K}^{-1}}\left(\frac{20}{300\times 320}\right)$
$E_a=43.85kJmo{l}^{-1}$