The time taken by a first order reaction to reach 10% completion at 298K is the same as the time taken to reach 30% completion at 308K . If value of `A=4xx10^(11)s^(-1)` , calculate the rate constant at 308K.

Explanation

`(2.303)/(k_(1))log.(a)/(a-0.1a)=(2.303)/(k_(2))log.(a)/(a-0.3a)`
or,
`(k_(2))/(k_(1))xx0.0457=0.1549 therefore (k_(2))/(k_(1))=3.39`
Again, `log.(k_(2))/(k_(1))= (E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
or, `log3.39=(E_(a))/(2.303xx8.314)xx((308-298))/(308xx298)`
`therefore E_(a) = 93.17 "kJ.mol"^(-1)`
`therefore k = Ae ^(-(E_(a))/(RT))=4xx10^(11)s^(-1).e^((-93170)/(8.314xx318)) `
`=1.98xx10^(-4)s^(-1)`