Question

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

• A. 43.8 KJ $mo{l}^{-1}$
• B. 53.8 KJ $mo{l}^{-1}$
• C. 60 KJ $mo{l}^{-1}$
• D. 70 KJ $mo{l}^{-1}$

Answer 1

The correct option is A 43.8 KJ $mo{l}^{-1}$

$\begin{array}{c}{k}_{{27}^{\circ }}=\frac{\ln 2}{30}{min}^{-1}\\ k_{{47}^{\circ }}=\frac{\ln 2}{10}{min}^{-1}\\ \frac{k_{47}}{k_{27}}=3\\ \log \frac{k_{47}}{k_{27}}=0.4771\end{array}$

$\ln \frac{k_2}{k_1}=\frac{Ea}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

$\begin{array}{c}0.4771=\frac{E_a}{2.303\times 8.314}\left(\frac{20}{300\times 320}\right)\\ E_a=\frac{0.4771\times 2.303\times 8.314\times 300\times 320}{20}\\ Ea=43.84kJmol\end{array}$