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

Given, `t=2.303/K_(298)"log"_(10)100/90`
`=2.303/K_(308)"log"_(10)100/75`
`:. K_(308)/K_(298)=2.73`
Also `2.303"log"_(10)K_(308)/K_(298)=E_(a)/R([T_(2)-T_(1)])/(T_(1)T_(2))`
`2.303"log"_(10) 2.73=E_(a)/8.314xx([308-298])/(298xx308)`
`:. DeltaE=76.6227 kJ mol^(-1)=18.33 kcal mol^(-1)`
Now `K=Ae^(-E_(a)//RT)`
`K_(318)=3.56xx10^(9)xxe^(-76622.7//(8.314xx318))`
`=3.56xx10^(9)xx2.59xx10^(-13)`
`=9.22xx10^(-4) sec^(-1)`