The rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} e^{-2000/T}$ and $10^{15} e^{-1000

Question

The rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} e^{-2000/T}$ and $10^{15} e^{-1000/T}$, respectively. The temperature at which $k_1 = k_2$ is:

Accepted Answer

Given,
$k_1 = 10^{16} e^{-2000/T}$
$k_2 = 10^{15} e^{-1000/T}$
When $k_1 = k_2$:
$10^{16} e^{-2000/T} = 10^{15} e^{-1000/T}$
$\frac{10^{16}}{10^{15}} = \frac{e^{-1000/T}}{e^{-2000/T}}$
$10 = e^{1000/T}$
Taking natural logarithm ($\ln$) on both sides:
$\ln(10) = \frac{1000}{T}$
We know that $\ln(10) = 2.303 \log_{10}(10) = 2.303 	imes 1 = 2.303$.
Therefore, $2.303 = \frac{1000}{T} \implies T = \frac{1000}{2.303} 	ext{ K}$.

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