The rate constants k_1 and k_2 for two different reactions are 10^...

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