SATHEE: Chemical Kinetics - Result Question 51

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

(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

$\begin{aligned} \ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \\ \ln \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}}) \end{aligned}$

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Chemical Kinetics - Result Question 51

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