Pre-exponential factors of two different reactions of same order are identical. Let activation energ

Question

Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by $20 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If $\mathrm{k}_1$ and $\mathrm{k}_2$ are the rate constants of first and second reaction respectively at 300 K , then $\ln \frac{\mathrm{k}_2}{\mathrm{k}_1}$ will be $\_\_\_\_$ . (nearest integer) $\left[\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$

Professor PC Thomas Classes · Answer

See detailed solution for explanation.