The rate constant of a reaction `1.5xx10^(7)s^(-1) " at " 50^(@)C and 4.5xx10^(7)s^(-1) at 100^(@)C` Evaluate the Arrhenius parameters A and `E_(a)`

Explanation

According to the Arrhenius equation , `k=Ae^(_E_(a)//RT)`
`thereforelogk=logA-(E_(a))/(2.303RT) " or, " log ((k_(2))/(k_(1)))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
Given: `log((4.5xx10^(7))/(1.5xx10^(7)))=(E_(a))/(2.303xx8.314)=[(1)/(323)-(1)/(373)]`
or, `E_(a)=2.2xx10^(4)"J.mol"^(-1)`
From the Arrhenius equation, `k=Ae^(-E_(a)//RT)`
or, `logk=logA-(E_(a))/(2.303RT)`
or, `log(4.5xx10^(7))=logA-(2.2xx10^(4))/(2.303xx8.314xx373)`
or, `A=5.42xx10^(10)s^(-1)`