Chemistry - Chemical Kinetics Question with Solution | TestHub

At 400 K the rate constant for a reaction is $2\times 10^{-4}{\sec }^{-1}$ and the graph between $\ln \mathrm{k}$ and $\frac{1}{\mathrm{T}}$ gives constant negative slope with magnitude 2303 K .

The rate constant at 600 K found to be $\mathrm{M}\times 10^{-3}{\sec }^{-1}$.

The value of ' M ' will be (In nearest integer) [given $\log 6.4=0.8$ ]

[Use : $\frac{1}{400}-\frac{1}{600}=0.0008$ ]