Chemistry - Chemical Kinetics Question with Solution | TestHub

Key Idea Rate constants at two different temperatures is given by Arrhenius equation as follows: $$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right],T_2>T_1$$ Given, ${\mathrm{E}}_{\mathrm{a}}=0,{\mathrm{T}}_1=200\mathrm{K},{\mathrm{k}}_1=1.6\times 10^6{\mathrm{s}}^{-1}$ $${\mathrm{T}}_2=400\mathrm{K},\mathrm{R}=8.314{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$$ According to Arrhenius equation, $$\log \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left[\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right]$$ On substituting the given values in above equation, we get $$\begin{array}{rl}& \log \frac{{\mathrm{k}}_2}{1.6\times 10^6}=\frac{0}{2.303\times 8.314}\left[\frac{\mathrm{l}}{200}-\frac{\mathrm{l}}{400}\right]\\ & \log \left(\frac{{\mathrm{k}}_2}{1.6\times 10^6}\right)=0\\ & \frac{{\mathrm{k}}_2}{1.6\times 10^6}=10^0=1\\ & {\mathrm{k}}_2=1.6\times 10^6{\mathrm{s}}^{-1}\text{ at }400\mathrm{K}\end{array}$$ Thus, option (2) is correct.