The activation energy of a reaction is 22.5 k Cal `mol^(-1)` and the value of rate constant at `40^(@)C` is `1.8xx10^(-5)s^(-1)`. Calculate the frequency factor, A.

Explanation

Here, we are given that
`E_(a) = 22.5 "kcal mol"^(-1)=22500 "cal mol"^(-1)`
`T=40^(@)C=40+273=313 K`
`k = 1.8xx10^(-5)sec^(-1)`
Substituting the values in the equation.
`log A = log k +((E_(a))/(2.303 RT))`
`log A = log(1.8xx10^(-5))+((22500)/(2.303xx1.987xx313))`
`log A = log(1.8)(-5)+(15.7089)`
`log A = 10.9642`
A = antilog (10.9642)
`A = 9.208xx10^(10)"collisions s"^(-1)`