• # question_answer A reactant [a] forms two products $A\xrightarrow{{{k}_{1}}}B$ Activation energy ${{E}_{{{a}_{1}}}}$ $A\xrightarrow{{{k}_{2}}}C$ Activation energy $E{{\,}_{{{a}_{2}}}}$ If ${{E}_{{{a}_{2}}}}=2{{E}_{{{a}_{1}}}},$ then ${{k}_{1}}$ and ${{k}_{2}}$ will be related as A) ${{k}_{2}}={{k}_{1}}{{e}^{-{{E}_{{{a}_{1}}}}/RT}}$       B) ${{k}_{2}}={{k}_{1}}{{e}^{-{{E}_{{{a}_{2}}}}/RT}}$ C) ${{k}_{1}}={{k}_{2}}{{e}^{-{{E}_{{{a}_{1}}}}/RT}}$       D) ${{k}_{1}}=2{{k}_{2}}{{e}^{{{E}_{{{a}_{2}}}}/RT}}$

 ${{k}_{2}}=A{{e}^{-{{E}_{a2}}/RT}}$ ${{k}_{1}}=A{{e}^{-{{E}_{at}}/RT}}$ $\frac{{{k}_{2}}}{{{k}_{1}}}={{e}^{-{{E}_{at}}/RT}}{{k}_{2}}={{k}_{1}}{{e}^{-Eat/RT}}$ (Since, ${{E}_{a2}}=2{{E}_{at}}$) or         $\frac{{{k}_{2}}}{{{k}_{1}}}={{e}^{-{{E}_{at}}/RT}}{{k}_{2}}={{k}_{1}}{{e}^{-{{E}_{at}}/RT}}$