The rate of reaction becomes double when its temperature is raised from $300 \mathrm{~K}$ to $330 \mathrm{~K}$. The activation energy of the reaction is (Given: $\log 2=0.30$ )

Question

Aicharya · Accepted Answer

We have
$$
\begin{array}{c}
E_{\mathrm{a}}=\frac{\left\{\log \left(k_{2} / k_{1}ight)ight\}(2.303 R)\left(T_{1} T_{2}ight)}{T_{2}-T_{1}}=\frac{(\log 2)(2.303)\left(8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}ight)(300 \mathrm{~K})(330 \mathrm{~K})}{(30 \mathrm{~K})} \\
=\frac{(0.30)(2.303)(8.314)(300)(330)}{(30)} \mathrm{J} \mathrm{mol}^{-1}=18956 \mathrm{~J} \mathrm{~mol}^{-1}=18.96 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{array}
$$

The rate of reaction becomes double when its temperature is raised from $300 \mathrm{~K}$ to $330…

Solution