JEE Main 2023 — Chemical Kinetics Question with Solution

As per the question 

To determine the activation energy of the catalyzed backward reaction, we need to use the Arrhenius equation and the given information.

The Arrhenius equation is given by: 

$\mathrm{k}={\mathrm{Ae}}^{(-\mathrm{Ea}/\mathrm{RT})}$

${\mathrm{Ae}}^{-\frac{300\times {10}^3}{600\times \mathrm{R}}}={\mathrm{Ae}}^{\frac{-\mathrm{Ea}}{300\times \mathrm{R}}}$

$\Rightarrow \frac{{10}^3}{2}=\frac{\mathrm{Ea}}{300}$

$\Rightarrow {\mathrm{E}}_{\mathrm{a}}=150\times {10}^3\mathrm{J} {\mathrm{mol}}^{–1}$

$\Rightarrow {\mathrm{E}}_{\mathrm{a}}=150 \mathrm{kJ} {\mathrm{mol}}^{–1}$

$\therefore$ Activation energy of catalysed backward reaction

Energy of activation for backward reaction${\mathrm{E}}_{\mathrm{b}}={\mathrm{E}}_{\mathrm{a}}-∆\mathrm{H}$

$=150-20=130 \mathrm{kJ} {\mathrm{mol}}^{–1}$