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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:
A
$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$
B
$\ln\frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$
C
$\ln\frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$
D
$\ln\frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_2} + \frac{1}{T_1}\right)$
Detailed Solution
Arrhenius equation: $\ln k = \ln A - \frac{E_a}{RT}$
At $T_1$: $\ln k_1 = \ln A - \frac{E_a}{RT_1}$; at $T_2$: $\ln k_2 = \ln A - \frac{E_a}{RT_2}$
Subtracting: $\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$
This is the same as $-\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$, which is just another way of writing it.
At $T_1$: $\ln k_1 = \ln A - \frac{E_a}{RT_1}$; at $T_2$: $\ln k_2 = \ln A - \frac{E_a}{RT_2}$
Subtracting: $\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$
This is the same as $-\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$, which is just another way of writing it.