Question

The time required for `10%` completion of a first order reaction at 298 K is equal to that required for its `25%` completiion at `308K.` If the value of A is `4xx10^(10)s^(-1).` calculate k at 318K and `E_(a),`

Answer 1

Calculation of activation energy `(E_(a))`
For `1^(st)` order reaction :
`k=(2.303)/(t)log ""([A]_(0))/([A])`
`At 298K`
`K_(1)=(2.303)/(t) log """"(100)/(90)-(i)`
At 308 K
`K_(2)=(2.303)/(t)log ""(100)/(75) -(ii)`
Dividing eq (ii) by (i)
`(k_(2))/(k_(1))=(log ""(100)/(75))/(log ""(100)/(90))=(0.1249)/(0.0458)=2.73`
According to Arrhenius theory
`log ""(k_(2))/(k_(1))=(E_(a))/(2.303 R)xx(T_(2)-T_(1))/(T_(1)T_(2))`
`log 2.73 =(E_(a))/(2.303R)((308-298)/(298xx308))`
`E_(a)=(0.4361xx2.303xx(8.314J mol ^(-1))xx298xx308)/(10)`
`E_(a) k =log A (-E_(a))/(2.303 xx(8.314J mol^(-1) K^(-1))xx(318K))`
`log k =10.6021 -12.5870 =-1.9849`
k = Antilog `(-19849)=` Antilog `(bar2.0151)=1.035xx10^(-2)s^(-1)`
`E_(a)=76.640kJ mol ^(-1)`
`k=1.035 xx10^(-2)s^(-1)`