Calculate the frequency factor and activation energy of a reaction , if the rate constants of the reaction at `50^(@)Cand100^(@)C" are"1.5xx10^(7)s^(-1)&4.5xx10^(7)s^(-1)` respectively.

Explanation

We know, `log.(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2))),`
where `k_(1) and k_(2)` are the rate constants of the reaction at temperatures `T_(1)andT_(2)(T_(2)gtT_(1)) and E_(a)` is activation energy.
Given , `T_(1)=(273+50)=323K`
`T_(2)=(273+100)=373K`
`k_(1)=1.5xx10^(7)s^(-1) and k_(2)=4.5xx10^(7)s^(-1)`
`thereforelog((4.5xx10^(7))/(1.5xx10^(7)))=(E_(a))/(2.303xx8.314)((373-323)/(323xx373))`
or , `log3=(E_(a))/(2.303xx8.314)xx(50)/(323xx373)`
or, `E_(a)=22.01xx10^(3)"J.mol"^(-1)=22.01"kJ.mol"^(-1)`
According to Arrhenius equation , `k=Ae^(-E_(a)//RT)`
At , `T=323K.k=1.5xx10^(7)s^(-1)and`
at `T=373,k=4.5xx10^(7)s^(-1)`
Substituting any one of the values of k ,
`1.5xx10^(7)=Ae.^((-22.01xx10^(3))/(8.314xx323))thereforeA=5.44xx10^(10)s^(-1)`
`and4.5xx10^(7)=Ae.^((-22.01xx10^(3))/(8.314xx373))thereforeA=5.44xx10^(10)s^(-1)`
`therefore ` The value of the frequency factor `(A)=5.44xx10^(10)s^(-1)`.