The time required for `10%` completion of a first order reaction at `298 K` is equal to that required for its `25%` completion at `308 K`. If the pre-exponential factor for the reaction is `3.56 xx 10^(9) s^(-1)`, calculate its rate constant at `318 K` and also the energy of activation.

Explanation

`t_(1)=(2.303)/(k_(1))log""(a)/(a-0.10a),t_(2)=(2.303)/(k_(2))log""(a)/(a-0.25a)`
As `t_(1)=t_(2),(2.303)/(k_(1))log""(a)/(0.90a)=(2.303)/(k_(2))log""(a)/(0.75a),(k_(2))/(k_(1))=(log(100//75))/(log(100//90))=2.73`
But `log(k_(2))/(k_(1))=(E_(a))/(2.303" R")((T_(2)-T_(1))/(T_(1)T_(2)))`
Putting `k_(2)//k_(1)=2.73,R=8.314" J K"^(-1)mol^(-1),T_(1)=298" K",T_(2)=308" K, we get "E_(a)=76.6" kJ mol"^(-1)`
Further, `k="Ae"^(-E_(a)//"RT")" or "logk=log A-(E_(a))/(2.303" RT")`
Putting `A=3.56xx10^(9)s^(-1),R=8.314xx10^(-3)"kJ K"^(-1)mol^(-1),E_(a)=76.6" kJ mol"^(-1),T=318" K",` we get
`k=9.3xx10^(-4)s^(-1)`