Chemistry - Chemical Kinetics Question with Solution | TestHub

 Activation energy is the minimum amount of energy that must be provided for compounds to result in a chemical reaction.

$\mathrm{A}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{B}\left(\mathrm{g}\right)\mathrm{ };\mathrm{ }∆\mathrm{H}=-40$ kJ/mole

$∆\mathrm{H}={\mathrm{E}}_{\mathrm{a}\left(\mathrm{f}\right)}-{\mathrm{E}}_{\mathrm{a}\left(\mathrm{b}\right)}=-40\mathrm{ }\mathrm{k}\mathrm{J}$

$\frac{{\mathrm{E}}_{\mathrm{a}\left(\mathrm{f}\right)}}{{\mathrm{E}}_{\mathrm{a}\left(\mathrm{b}\right)}}=\frac{2}{3}$

$\Rightarrow \mathrm{ }{\mathrm{E}}_{\mathrm{a}\left(\mathrm{f}\right)}=\frac{2}{3}{\mathrm{E}}_{\mathrm{a}\left(\mathrm{b}\right)}$

$\Rightarrow \frac{2}{3}\mathrm{ }{\mathrm{E}}_{\mathrm{a}\left(\mathrm{b}\right)}-{\mathrm{E}}_{\mathrm{a}\left(\mathrm{b}\right)}=-\frac{{\mathrm{E}}_{\mathrm{a}\left(\mathrm{b}\right)}}{3}=-40\mathrm{ }\mathrm{k}\mathrm{J}$

${\mathrm{E}}_{\mathrm{a}\left(\mathrm{b}\right)}=+120\mathrm{ }\mathrm{k}\mathrm{J}$

${\mathrm{E}}_{\mathrm{a}\left(\mathrm{f}\right)}=\frac{2}{3}\times 120=80\mathrm{ }\mathrm{k}\mathrm{J}$