Activation energy ( $E_{a}$ )and rate constants ( $K_{1}$ and $K_{2}$ ) of a chemical reaction at two different temperatures ( $T_{1}$ and $T_{2}$ ) are related by

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

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

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

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

answer is B, D.

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Detailed Solution

Arrhenius equation

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

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

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