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

We know that,
`k=(2.303)/(t)log_(10).((a)/(a-x))`
`At 298 K, x=10,a=100` ltbr `k_(298)=(2.030)/(t_(1))log_(10).(100)/(90)`
At 305K, a=100, x=25, (a-x)=75
`k_(308)=(2.303)/(t_(2))log_(10).((100)/(75))`
`t_(1)=t_(2)`, dividing eq. (ii) by eq.
`:. (k_(308))/(k_(298))=2.73`
`log.(k_(308))/(k_(298))=(E)/(2.303R)((1)/(T_(1))-(1)/(T_(2)))`
`log 2.73=(E)/(2.303xx8.314)((1)/(308)-(1)/(298))`
E=76.622 kJ/ mol
Similarly, we can solve for `k_(318)` which is equal to `9.22xx10^(-4) s^(-1)`