Chemistry - Chemical Kinetics Question with Solution | TestHub

The Arrhenius equation is,
${\text{log}}_{{10}^{ }}\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{1}}}=\frac{{\text{E}}_{\text{a}}}{\text{2}\text{.303×R}}\left[\frac{{\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}}{{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}\right]$
Given${\text{k}}_{\text{1}}=9.5\times {10}^{-5}{\text{s}}^{-\text{1}}\text{;}{\text{k}}_{\text{2}}=1.9\times {10}^{-3}{\text{s}}^{-1}$
$\text{R}=8.314{\text{ Jmol}}^{-1}{\text{K}}^{-1}$;
${\mathrm{T}}_1=407\mathrm{ }\mathrm{K}$and${\mathrm{T}}_2=420\mathrm{ }\mathrm{K}$
Substituting the values in Arrhenius equation.
$\text{lo}{\text{g}}_{10}\frac{1.9\times 10^{-4}}{9.5\times 10^{-5}}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left[\frac{420-407}{420\times 407}\right]$
${\mathrm{E}}_{\mathrm{a}}=75782.3\mathrm{ }\mathrm{J}\text{mo}{\text{l}}^{-1}$
Applying now$\log {\mathrm{k}}_1=\log \mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{ }\mathrm{R}{\mathrm{T}}_1}$
$\log 9.5\times {10}^{-5}=\log A-\frac{75782.3}{2.303\times 8.314\times 407}$
$\text{log}\frac{\text{A}}{9.5\times {10}^{-5}}=\frac{75782.3}{2.303\times 8.314\times 407}=9.7246$
So,$\mathrm{A}=5.0\times 10^5{\mathrm{s}}^{-1}$