The rate constants $\mathrm{k}_1$ and $\mathrm{k}_2$ for two different reactions are $10^{16} \cdot…
The rate constants $\mathrm{k}_1$ and $\mathrm{k}_2$ for two different reactions are $10^{16} \cdot \mathrm{e}^{-2000 / T}$ and $10^{15} \cdot \mathrm{e}^{-1000 / T}$ respectively. The temperature at which $\mathrm{k}_1=\mathrm{k}_2$ is
$\frac{1000}{2.303} \mathrm{~K}$
$1000 \mathrm{~K}$
$\frac{2000}{2.303} \mathrm{~K}$
$2000 \mathrm{~K}$
Solution
$K_1=10^{16} \mathrm{e}^{-\frac{2000}{T}}$
or $\log \mathrm{K}_1=16-\frac{2000}{2.303 \mathrm{~T}}$
$K_2=10^{15} e^{\frac{1000}{T}}$
or $\log K_2=15-\frac{1000}{2.303 \mathrm{~T}}$
If $\mathrm{T}=\frac{1000}{2.303} \mathrm{~K}$, then $\mathrm{K}_1=\mathrm{K}_2$