The time required for 10% completion of first order reaction at 298K is equal to that required for its 75% completion at 308K. If the pre-expontential factor for the reaction is 3.56xx10^(9)s^(-1), calculate its rate constant at 318K and also the energy of activation.

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

1.	The time required for 10% completion of first order reaction at 298K is equal to that required for its 75% completion at 308K. If the pre-expontential factor for the reaction is 3.56xx10^(9)s^(-1), calculate its rate constant at 318K and also the energy of activation.
Answer» Solution :For first ORDER reactions, `t=(2.303)/(K)log.(N_(0))/(N_(1))` At `298K`, `t=(2.303)/(k_(298))log.(100)/(90)` At `309k, t=(2.303)/(k_(308))log.(100)/(25)` Since time is same `(2.303)/(k_(298))log.(100)/(90)=(2.303)/(k_(308))log.(100)/(25)` or `(0.0458)/(k_(298))=(0.1249)/(k_(308))1` or `(k_(308))/(k_(298))=(0.1249)/(0.0458)=2.73` According to ARRHENIUS equation `2.303log.(k_(308))/(k_(298))=(E_(a))/(8.314)[(1)/(298)-(1)/(208)]` or `2.303log2.73=(E_(a))/(8.314)[(10)/(298xx308)]` `E_(a)=76.65kJ` and `K_(318)=9.306xx10^(-4)s^(-1)`