• question_answer For reaction $A\to B$the rate constant ${{k}_{1}}={{A}_{1}}{{e}^{-{{E}_{{{a}_{1}}}}/RT}}$ and for the reaction $X\to Y$ the rate constant ${{k}_{1}}={{A}_{2}}\,{{e}^{-{{E}_{{{a}_{2}}}}/RT}}$, if ${{A}_{1}}={{10}^{10}}$,${{A}_{2}}={{10}^{12}}$ and ${{E}_{{{a}_{1}}}}=800\,\text{cal/mol}$,${{E}_{{{a}_{2}}}}=1600\,\text{cal/mol}$, then temperature at which ${{k}_{1}}={{k}_{2}}$ is (given R = 2 cal/K-mol) A)  $\left( \frac{200}{0.693} \right)K$                         B)  $200\times 0.693K$ C)  $\left( \frac{400}{0.693} \right)K$                         D)  $400\times 0.693K$

Idea This problem includes concept of Arrhenius equation to calculate the value of temperature. To solve the problem write the Arrhenius equation at two different temperatures then equate both the equation you will get the final result. ${{k}_{1}}={{k}_{2}}$ ${{A}_{1}}{{e}^{-E{{a}_{1}}/RT}}={{A}_{2}}{{e}^{-E{{a}_{2}}/RT}}$ ${{10}^{10}}{{e}^{-\frac{800}{RT}}}={{10}^{-12}}{{e}^{-\frac{1600}{RT}}}$ $\frac{{{A}_{2}}}{{{A}_{1}}}={{e}^{\left( \frac{{{E}_{{{a}_{2}}}}-{{E}_{{{a}_{1}}}}}{RT} \right)}}$ ${{10}^{2}}={{e}^{\frac{800}{RT}}}$       $2\ln 10=\frac{800}{RT}$ $2\times 2.303\times 0.3010=\frac{800}{2T}$ $T=\frac{400}{0.613\times 2}=\frac{200}{0.693}=\left( \frac{200}{0.693} \right)K$ TEST Edge In JEE Main, questions related to collision theory and Arrhenius theory are also asked. Students are advised to understand the theory of Arrhenius equation and collision theory.
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