Chemistry - Chemical Kinetics Question with Solution | TestHub

Using the Arrhenius equation,

$$k=A{e}^{-E_a/RT}$$

Taking natural logarithm on both sides, the Arrhenius equation becomes, $\ln \mathrm{k}=\ln \mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}$

$\ln \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right)$

On substituting all the given values in Eq. (ii) we get

$\ln \frac{{\mathrm{k}}_2}{10^{-5}}=4606\left(\frac{1}{400}-\frac{1}{500}\right)$

$\ln \frac{k_2}{10^{-5}}=2.303$

$\frac{{\mathrm{k}}_2}{10^{-5}}=10$

$\Rightarrow {\mathrm{k}}_2=10^{-4}{\mathrm{s}}^{-1}$

Therefore, rate constant for the reaction at 500 K is $10^{-4}{\mathrm{s}}^{-1}$.