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 `4xx10^(10)s^(-1)` Calculate k at 318 and `E_(a)`

Explanation

For a first order reaction, `k=(2.303)/(t)log.(a)/(a-x)`
Now, `k_(298)=(2.303)/(t_(1))log.(a)/(a-0.1a)=(2.303)/(t_(1))log.(10)/(9)`
or, `t_(1)=(0.1055)/(k_(298k))" or, "k_(308)=(2.303)/(t_(2))log .(a)/(a-0.25a)`
or, `t_(2)=(0.2878)/(k_(308))`
But `t_(1)=t_(2)" Hence", (0.1055)/(k_(298))=(0.2878)/(k_(308))" or, (k_(298))/(k_(298))=2.7280`
Applying Arrhenius equation , `log.(k_(308))/(k_(298))=(E_(a))/(2.303R)[((308-298))/(298xx308)]`
`:.log(2.7280)=(E_(a))/(2.303xx8.314)[((308-298))/(298xx308)]`
`thereforeE_(a)=76.623"kJ.mol"^(-1)`
`thereforek " at" 318K ` will be,
`logk=logA-(E_(a))/(2.303RT)`
`=log(4xx10^(10))-(76623)/(2.303xx8.314xx318)`
`=10.6021-12.5843= "antilog"(-1.9822)`
`=1.042xx10^(-2)s^(-1)`