Activation energy $E_a$ and rate constant ($k_1$ and $k_2$) of a chemical reaction at two different temperatur

Question

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:

Accepted Answer

According to the Arrhenius equation, the rate constant $k$ is given by $k = A e^{-E_a/RT}$. Taking the natural logarithm on both sides yields $\ln k = -\frac{E_a}{RT} + \ln A$. 
For two different temperatures $T_1$ and $T_2$, the equations are:
$\ln k_1 = -\frac{E_a}{RT_1} + \ln A$ 
$\ln k_2 = -\frac{E_a}{RT_2} + \ln A$ 
Subtracting the first equation from the second gives:
$\ln k_2 - \ln k_1 = -\frac{E_a}{RT_2} - \left(-\frac{E_a}{RT_1}\right)$
$\ln \frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ 
This is also commonly written as $\ln \frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$.

Step-by-Step Solution

Practice All CHEMISTRY Questions