Chemistry - Chemical Kinetics Question with Solution | TestHub

$\mathrm{K}=A{e}^{-E_a/RT}$ After taking In both side $\ln K=\ln A-\frac{E_a}{RT}$ $\mathrm{In}{\mathrm{K}}_1=\ln \mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_1}$ at temp. ${\mathrm{T}}_1$...(i) $\mathrm{In}{\mathrm{K}}_2=\ln \mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_2}$ at temp. ${\mathrm{T}}_2$,,(ii) (ii) - (i) $\mathrm{In}{K}_2-\mathrm{InK}{K}_1=\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$ $\ln \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{500}-\frac{1}{700}\right]$ $\ln \frac{0.14}{0.04}=\frac{E_a}{R}\left[\frac{700-500}{500\times 700}\right]$ $\ln \frac{14}{4}=\frac{E_a}{R}\left[\frac{200}{500\times 700}\right]$ $\log 3.5=\frac{E_a}{2.303\times R}\left[\frac{1}{250\times 7}\right]$ $0.5441=\frac{E_a}{2.303\times 8.31}\left[\frac{1}{250\times 7}\right]$ $E_a=0.5441\times 8.31\times 250\times 7\times 2.303$ $=0.5441\times 83.1\times 25\times 7\times 2.303$ $=18222.65$ $\approx 18219\mathrm{J}$