A first order reaction is 50% complete in 30 minutes at `27^(@)C` and in 10 minutes at `47^(@)C`. The reaction rate constant at `27^(@)C` and the energy of activation of the reaction are respectively

Explanation

`k=0.0231" min"^(-1), 43.848" kJ mol"^(-1)`
`k=(0.693)/(t_(1//2)), (k_(2))/(k_(1)) =(E_(a))/(2.303) [(T_(2)-T_(1))/(T_(1) cdot T_(2))]`
Substituting the value at the two given conditions `k_(1)=k_(300) =(0.693)/(30) =0.0231" m in"^(-1) rArr k_(2)=k_(320) =(0.693)/(10) =0.0693" m in"^(-1)`
We also know that log `(k_(2))/(k_(1)) =(Ea)/(2.303 R) xx(T_(2)-T_(1))/(T_(1)xxT_(2))`
or `E_(a) =(2.303 R xx T_(1) xx T_(2))/(T_(2)-T_(1))" log "(k_(2))/(k_(1)) =(2.303xx8.314xx10^(-3) xx300xx320)/(320-300)xx log (0.0693)/(0.0231) =43.848" kJ mol"^(-1)`