For a first order reaction the rate constant at 500 k is `8xx10^(-4)s^(-1)`. Calculate the frequency, if the energy of activation for the reaction is `190 "kJ mol"^(-1)`.

Explanation

`k=8xx10^(-4)s^(-1)`
`T = 500 k`
Ea `= 190 "kJ mol"^(-1)=190xx10^(3)"J mol"^(-1)`
According to Arrhenius equation,
`k = Ae^((-E_(a))/(RT))`
`rArr log k = log A - (E_(a))/(RT)`
`rArr log k = log A - (E_(a))/(2.303 RT)`
`rArr log A = log k + (E_(a))/(2.303 RT)`
`= log (8xx10^(-4)s^(-1))+(190xx10^(3+)"J mol"^(-1))/(2.303xx8.314JK^(-1)mol^(-1)xx500K)`
`=(0.9030-4)+19.8463 = 16.7493`
`rArr A =` Antilog (16.7493)
`= 5.6xx10^(16)s^(-1)` (approximately)