The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the pre-exponential factor for the reaction is `3.56xx10^(9)s^(-1)` , calculate it's rate constant at 318k and also the energy of activation.

Explanation

For a first order reaction , time required
`t_(1)=(2.303)/(k_(1))log((100)/(100-10)) and t_(2)=(2.303)/(k_(2))log.((100)/(100-25))`
or, `t_(2)=(2.303)/(k_(2))log((100)/(75)) and t_(2)=(2.303)/(k_(2))log.((4)/(3))`
According to the question, `t_(1) =t_(2)`
`therefore(2.303)/(k_(1))log ((10)/(9)) = (2.303)/(k_(2))log ((4)/(3)) " or, " (k_(1))/(k_(2))= 0.366 " or, " (k_(2))/(k_(1))=2.73`
`log((k_(2))/(k_(1)))=(E_(a))/(2.303)((1)/(T_(1))-(1)/(T_(2)))` [From Arrhenius equation]
or, `log2.73=(E_(a))/(2.303xx8.314)((1)/(298)-(1)/(308))`
or, `E_(a)=76.6xx10^(3)"J.mol"^(-1)`
From the Arrhenius equation, `k=Ae^(-E_(a)//RT)`
`therefore logk=logA-(E_(a))/(2.303RT)`
or, `logk=log(3.56xx10^(9))-(76.6)/(2.303xx8.314xx10^(-3)xx318)`
`therefore k = 9.3xx10^(-4)s^(-1)`