A first order reaction is `50%` completed in `30 min` at `27^(@)C` and in `10 min` at `47^(@)C`. Calculate the reaction rate constants at `27^(@)C` and the energy of activation of the reaction in `kJ mol^(-1)`.

Answer

`0.0231 "mimn"^(-1),43.84 kJ`

Explanation

`k(27^(@))=(0.6932)/((t_(1//2))_(27))`
The reaction at `47^(@)C` is 3 times faster than that at `27^(@)C` .Use Equation 12 , put `R = 8.314 xx10^(-3) kJ `.