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For a reaction, activation energy $E_a=0$ and the rate constant at 200 K is $1.6\times10^6$ s$^{-1}$. The rate constant at 400 K will be [Given that gas constant R = 8.314 J K$^{-1}$ mol$^{-1}$]
Detailed Solution
$\log\frac{K_2}{K_1}=\frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)=0$ since $E_a=0$. So $K_2=K_1=1.6\times10^6$ s$^{-1}$.