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

Explanation

At `10%` completion x = 0.10a.
Substituting the values in the first order equation, we get
`k_(298K) =(2.303)/(t_(1))"log"(a)/(a-0.10a)= (2.303)/(t)"log"(10)/(9)=(2.303)/(t)(0.0458)=(0.1055)/(t)`
or `t_(1)=(0.1055)/(k_(298K))`
At 25% completion x = 0.25 a.
Substituting the values in the first order equation, we get
`k_(308K)=(2.303)/(t_(2))"log"(a)/(a-0.25a)=(2.303)/(t_(2))"log"(4)/(3)=(2.303)/(t_(2))(0.125)=(0.2879)/(t_(2))`
or `t_(2)=(0.2879)/(k_(308K))`
But `t_(1)=t_(2)`
Hence, `(0.1055)/(k_(298K))=(0.2879)/(k_(308K)) or =(k_(308K))/(k_(298K))=2.7289`
Substituting the values in Arrhenius equation, we get
`"log"(k_(308K))/(k_(298K))=(E_(a))/(2.303R) ((T_(2)-T_(1))/(T_(1)T_(2)))`
`therefore log(2.7289)=(E_(a))/(2.303xx8.314J K^(-1)"mol"^(-1)) xx ((308-298)K)/(298K xx 308K)`
or `0.4360=(E_(a))/(2.303xx8.314)xx(10)/(298xx308)`
or `E_(a)=76.623"kJ mol"^(-1)`.
Calculation of k at 318 K.
`logk=logA-(E_(a))/(2.303RT)=log(4xx10^(10))-(76623JK^(-1)"mol"^(-1))/(2.303xx8.314"J K"^(-1)"mol"^(-1)xx318K)`
`=10.6021-12.5843= -1.9822`
or `k="Antilog"(-1.9822)="Antilog"(bar2.0178)=1.042xx10^(-2)s^(-1)`.