Chemistry - Chemical Kinetics Question with Solution | TestHub

To be solved with the help of formula,

$\text{ log}\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{1}}}\text{=}\frac{{\text{E}}_{\text{a}}}{\text{2}\text{.303 R}}\left[\frac{{\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}}{{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}\right]$

${\text{T}}_{\text{1}}=273+27=300 \text{K}$

${\text{T}}_{\text{2}}=273+67=340 \text{K}$

$\text{log}\frac{6.9\times {10}^{-3}}{3.45\times {10}^{-5}}=\frac{{\text{E}}_{\text{a}}}{2.303\times 8.31}\left[\frac{340-300}{340\times 300}\right]$

$\text{log}200=\frac{{\text{E}}_{\text{a}}}{19.1379}\times \frac{40}{102000}$

$2.3010=\frac{{\text{E}}_{\text{a}}}{19.14}\times \frac{4}{10200}$

${\text{E}}_{\text{a}}=\frac{19.14\times 10200\times 2.3010}{4}$

$\text{=}\text{112304}\text{.907 J}{\text{mol}}^{-\text{1}}\text{=}\text{112}\text{.3 kJ}{\text{mol}}^{-\text{1}}$