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

Step I. Calculation of activation energy
For a first order reaction, `k= 2.303/t log 100/90`
At 308 K, `k_(308 K) =2.303/t log 100/75` , `k_(308K)/K_(298 K)`
`=(log)(100/75)/(log)(100/90) = 0.1249/0.04558 = 2.73`
According to Arrheneius theory, `(log) (k_(308K))/(k_(298 K)) = E_(a)/(2.303 R) [(T_(2)-T_(1))/(T_(1)T_(2))]`
`log 2.73 = E_(a)/(2.303 R) [(208-298)/(298 xx 308)]`
`E_(a) = (0.4361 xx 2.303 xx (8.314 J mol^(-1)) xx 298 xx 308)/(10)`
`=76640 J mol^(-1) = 76.640 kJ mol^(-1)`
Step II. Calculation of rate constant
According to Arrheneius equation, log k `= log A-(E_(a))/(2.303 RT)`
`log k = log(4 xx 10^(10))-(76640 J mol^(-1))/(2.303 xx (8.314 J mol^(-1) K^(-1)) xx (318 K))`
log k `= 10.6021 - 12.5870 = -1.9849`
k=Antilog `(-1.9849)`=Antilog(`bar2.0151) = 1.035 xx 10^(-2)s^(-1)`