Chemistry - Chemical Kinetics Question with Solution | TestHub

According to the Arrhenius equation $\log K=\log A-\frac{E_a}{2.303RT}$....(i) $\frac{\log K_{30B}}{\log K_{29B}}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_2}\right]\dots (ii)$ $\begin{array}{rl}& \frac{\log K_{308}}{\log K_{298}}=\frac{9.0}{2.303\times 2\times 10^{-3}}\left[\frac{308-298}{308\times 298}\right]\\ & or\frac{\log K_{308}}{\log K_{298}}=0.2129\end{array}$ $\begin{array}{rl}{K}_{308}& =1.632K_{298}\dots (iii)\\ \%\text{ increase }& =\frac{K_{308}-K_{298}}{K_{298}}\times 100\\ & \cong 63\%\end{array}$