Chemistry - Chemical Kinetics Question with Solution | TestHub

The activation energy of reverse reaction will depend upon whether the forward reaction is exothermic or endothermic. As $\Delta \mathrm{H}={\mathrm{E}}_{\mathrm{a}\text{ (forward reaction) }}-{\mathrm{E}}_{\mathrm{a}(\text{ backward reaction) }}$ For exothermic reaction $\Delta \mathrm{H}=-\mathrm{ve}$ $\begin{array}{rl}& \therefore -\Delta \mathrm{H}={\mathrm{E}}_{\mathrm{a}(\mathrm{f})}-{\mathrm{E}}_{\mathrm{a}(\mathrm{b})}\\ & \text{ or }{\mathrm{E}}_{\mathrm{a}(\mathrm{f})}={\mathrm{E}}_{\mathrm{a}(\mathrm{b})}-\Delta \mathrm{H}\\ & \therefore {\mathrm{E}}_{\mathrm{a}(\mathrm{f})}<{\mathrm{E}}_{\mathrm{a}(\mathrm{b})}\end{array}$ for endothermic reaction $\begin{array}{rl}& \Delta \mathrm{H}=+\mathrm{ve}\\ & \therefore \Delta \mathrm{H}={\mathrm{E}}_{\mathrm{a}(\mathrm{f})}+{\mathrm{E}}_{\mathrm{a}(\mathrm{b})}\text{ or }{\mathrm{E}}_{\mathrm{a}(\mathrm{f})}=\Delta \mathrm{H}+{\mathrm{E}}_{\mathrm{a}(\mathrm{b})}\\ & \therefore {\mathrm{E}}_{\mathrm{a}(\mathrm{f})}>{\mathrm{E}}_{\mathrm{a}(\mathrm{b})}\text{. }\end{array}$