Q. The activation energies of two reactions are $E_{1}$ and $E_{2}(E_{1} > E_{2})$. If the temperature of the system is increased from $T_{1}$ to $T_{2}$, the rate constant of the reactions changes from $k_{1}$ to $k_{1}$ in the first reaction and $k_{2}$ and $k_{2}$ in the second reaction. Predict which of the following expression is correct?
A) $\frac{k_{1}^{'}}{k_{1}^{'}} = \frac{k_{2}^{'}}{k_{2}^{'}}$
B) $\frac{k_{1}^{'}}{k_{1}^{'}} > \frac{k_{2}^{'}}{k_{2}^{'}}$
C) $\frac{k_{1}^{'}}{k_{1}^{'}} < \frac{k_{2}^{'}}{k_{2}^{'}}$
D) $\frac{k_{1}^{'}}{k_{1}^{'}} = \frac{k_{2}^{'}}{k_{2}^{'}} = 1$
Explanation
(b)
For first reaction,
$E_{1} = \frac{2.303RT_{1}T_{2}}{(T_{1} - T_{2})}\log\frac{k_{1}^{'}}{k_{1}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
For second reaction,
$E_{2} = \frac{2.303RT_{1}T_{2}}{(T_{1} - T_{2})}\log\frac{k_{2}^{'}}{k_{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Given, $E_{1} > E_{2}$
$\Rightarrow \frac{2.303RT_{1}T_{2}}{(T_{1} - T_{2})}\log\frac{k_{1}^{'}}{k_{1}} > \frac{2.303RT_{1}T_{2}}{(T_{1} - T_{2})}\log\frac{k_{2}^{'}}{k_{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$
$\therefore\frac{k_{1}^{'}}{k_{1}} > \frac{k_{2}^{'}}{k_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.