Chemistry - Chemical Kinetics Question with Solution | TestHub

Key Idea : The Arrhenius equation is represented as $k=A{e}^{-E_a/RT}$ In the given equations, first take log and then compare them. $\begin{array}{rl}& k_1=10^{16}{e}^{-2000/T}\\ & k_2=10^{15}{e}^{-1000/T}\end{array}$ On taking log, we get $\begin{array}{rl}& \log k_1=\log 10^{16}-\frac{2000}{2.303T}\\ & \log k_2=\log 10^{15}-\frac{1000}{2.303T}\\ & \because k_1=k_2\end{array}$ Hence, from Eqs (i) and (ii) $T=\frac{1000}{2.303}\mathrm{K}$