Chemical Kinetics - Result Question 51
####52. 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:
[2012 M]
(a) $\ln \frac{k_2}{k_1}=-\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$
(b) $\ln \frac{k_2}{k_1}=-\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})$
(c) $\ln \frac{k_2}{k_1}=-\frac{E_a}{R}(\frac{1}{T_2}+\frac{1}{T_1})$
(d) $\ln \frac{k_2}{k_1}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$
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Solution:
- (b, d) According to Arrhenius equation
$ \begin{aligned} & \ln \frac{k_2}{k_1}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2}) \\ & \ln \frac{k_2}{k_1}=-\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1}) \end{aligned} $
