• # question_answer Given is the information: Reaction 1 :       $k=A{{e}^{-{{E}_{1}}/RT}}$ Reaction 2 :       $k'=A{{e}^{-{{E}_{2}}/RT}}$ If ${{E}_{1}}=2{{E}_{2}},$ then the ratio $\ell n\left[ \frac{{{k}_{T}}+10k}{{{k}_{T}}} \right]$ for the reaction will be A) Equal to that of reaction 2 B) greater than that of reaction 2 C) lesser than that of reaction 2 D) may be greater or lesser than reaction 2 depending upon the temperature T.

 $\ell n{{k}_{T}}=\ell nA-\frac{{{E}_{1}}}{RT}$ or         $\ell n{{k}_{T}}+10k=\ell nA-\frac{{{E}_{1}}}{R(T+10k)}$ or         $\ell n{{k}_{T}}+10k-\ell n\,{{k}_{T}}=\frac{{{E}_{1}}}{R}\left[ \frac{1}{T}-\frac{1}{T+10k} \right]$ or         $\ell n\left[ \frac{{{k}_{T}}+10k}{{{k}_{T}}} \right]=\frac{{{E}_{1}}(10k)}{RT(T+10k)}$ similarly, $\ell n\left( \frac{{{k}_{T}}+10k}{{{k}_{T}}} \right)=\frac{{{E}_{2}}(10k)}{RT(T+10k)}$ Dividing both, we get, $\frac{\ell n\left( \frac{{{k}_{T}}+10k}{{{k}_{T}}} \right)}{\ell n\left( \frac{{{k}_{T}}+10k}{{{k}_{T}}} \right)}=\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{2{{E}_{2}}}{{{E}_{2}}}=2$