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
`4 × 10^(10)s^(–1)`. Calculate k at 318K and `E_(a)`.

Answer

`a.E_(a)=76.623kJ mol^(-1)`
`b. k=1.042xx10^(-2)s^(-1)`

Explanation

`a.` `k_(298K)=(2.303)/(t_(1))log((100)/(100-10))=(2.303)/(t_(1))log``(10)/(9)`
`=(2.303)/(t_(1))xx(0.0458)`
`=(0.1055)/(t_(1))`
`:.t_(1)=(0.1055)/(k_(28K)) ...(i)`
`b.` `k_(308K)=(2.303)/(t_(2))log((100)/(100-25))=(2.303)/(t_(2))log((4)/(3))`
`=(2.303)/(t_(2))xx(0.125)`
`=(0.2879)/(t_(2))`
`or t_(2)=(0.2879)/(k_(308K)) ....(ii)`
Since `t_(1)=t_(2)` hence `Eqs. (i)` and `(ii)` are equal.
`:. (0.1055)/9k_(298K)0=(0.2879)/(k_(308K))`
`or (k_(308K))/(k_(298K))=2.7289 ...(iii)`
Using Arrhenius equation,
`log``(k_(308K))/(k_(298K))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
From `Eq. (iii)`,
`:.log 2.7289=(E_(a))/(2.303xx8.314)xx(10)/(298xx308)`
`or E_(a)=76.623kJ mol^(-1)`
`c.` Calculation of `k` at `318 K:`
`logk=logA-(E_(a))/(2.303RT)`
`=log(4xx10^(10))s^(-1)`
`-(76.623xx1000J mol^(-1))/(2.303xx8.314J K^(-1) mol^(-1)xx318K)`
`=10.6021-12.5843`
`=-1.9822`
or `k= "Antilog"(-1.9822)="Antilogbar"(1).0178`
`=1.042xx10^(-1)s^(-1)`