Activation energy $(E_{a})$ and rate constant $(k_{1} {and k}_{2})$ of a chemical reaction at two different temperatures $(T_{1} {and T}_{2})$ are related by :

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$In \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$In \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{1}})$

$In \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R} (\frac{1}{T_{2}} + \frac{1}{T_{1}})$

$In \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$

answer is B, D.

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Detailed Solution

According to Arrhenius equation

$In \frac{k_{2}}{k_{1}} \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$= - \frac{E_{a}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$

$In \frac{k_{1}}{k_{2}} = - \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

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