Chemistry - Chemical Kinetics Question with Solution | TestHub

Using Arrhenius equation, $\begin{array}{l}K=A\cdot e^{-\frac{E_a}{Rt}},\text{ we get }\\ \log k=\log A-\frac{E_a}{2.303RT}\therefore \end{array}$ $\log k_1=\log A-\frac{E_{a_{(1)}}}{2.303RT}$ and $\log k_2=\log A-\frac{E_{a_{(2)}}}{2.303RT}$ or $\log \frac{k_2}{k_1}=\frac{1}{2.303RT}\left[E_{a(1)}-E_{a(2)}\right]$ (from (i) and (ii)) $=\frac{1}{2.303\times 8.314\times 300}(76000-57000)$ or $\log \frac{k_2}{k_1}=\frac{19000}{2.303\times 8.314\times 300}$ $=\frac{190}{6.9\times 8.314}$ or $\frac{k_2}{k_1}=2000[$ taking antilog $]$