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 value of `A` is `4xx10^(10)s^(-1)`, calculate `k` at `318K` and `E_(a)`.

Explanation

For a first order reaction ,
`t = (2.303)/(k) "log" (a)/(a-x)`
At 298 K , `t = (2.303)/(k) "log" (100)/(90)`
= `(0.1054)/(k)`
At 308 K , `t' = (2.303)/(k') "log" (100)/(75)`
= `(2.2877)/(k')`
According to the question .
t= t'
`implies (0.1054)/(k) = (0.2877)/(k')`
`implies (k')/(k) = 2.7296`
From Arrhenius equation , we obtain
log `(k')/(k) = (E_(a))/(2.303R) ((T'-T)/(T T'))`
log (2.7296) = `(E_(a))/(2.303 xx 8.314) ((308 - 298)/(298 xx 308))`
`E_(a) = (2.303 xx 8.314 xx 298 xx 308 xx "log" (2.7296))/(308 - 298)`
= `76640.096 J mol^(-1)`
`= 76.64 J mol^(-1)`
To calculate k at 318 K ,
It is given that , `A = 4 xx 10^(10) s^(-1) , T = 318 K`
Again from Arrhenius equation , we obtain
log k = log A - `(E_(a))/(2.303 RT)`
= log `(4 xx 10^(10)) - (76.64 xx 10^(3))/(2.303 xx 8.314 xx 318)`
= `(0.6021 + 10) - 12.5876`
= `-1.9855`
Therefore , k = Antilog `(-1.9855)`
= `1.034 xx 10^(-2) s^(-1)`