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 per-exponential factor for the reaction is 3.56xx10^(9)s^(-1), calculate its rate constant at 318 K and also the energy of activation.

The time required for 10% completion of a first order reaction at 298

1.	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 per-exponential factor for the reaction is 3.56xx10^(9)s^(-1), calculate its rate constant at 318 K and also the energy of activation.
Answer» SOLUTION :`t_(1)=(2.303)/(k_(1))log""(a)/(a-0.10a),t_(2)=(2.303)/(k_(2))log""(a)/(a-0.25a)` As `t_(1)=t_(2),(2.303)/(k_(1))log""(a)/(0.90a)=(2.303)/(k_(2))log""(a)/(0.75a),(k_(2))/(k_(1))=(log(100//75))/(log(100//90))=2.73` But `log(k_(2))/(k_(1))=(E_(a))/(2.303" R")((T_(2)-T_(1))/(T_(1)T_(2)))` Putting `k_(2)//k_(1)=2.73,R=8.314" J K"^(-1)mol^(-1),T_(1)=298" K",T_(2)=308" K, we get "E_(a)=76.6" KJ mol"^(-1)` Further, `k="Ae"^(-E_(a)//"RT")" or "logk=log A-(E_(a))/(2.303" RT")` Putting `A=3.56xx10^(9)s^(-1),R=8.314xx10^(-3)"kJ K"^(-1)mol^(-1),E_(a)=76.6" kJ mol"^(-1),T=318" K",` we get `k=9.3xx10^(-4)s^(-1)`