The activation energy of a reaction is `94.14 kJ mol^(-1)` and the value of rate constant at `313 K` is `18 xx 10^(-5) s^(-1)`. Calculate the frequency factor or pre-exponential factor, `A`.

Explanation

Given, `E_(a) = 94.14 kJ mol^(-1) = 94140 J mol^(-1)`,
`T = 313 K, k = 1.8 xx 10^(-5)s^(-1)`
or `log k = (-E_(a))/(2.303RT) + logA`
`logA = log(1.8 xx 10^(-5))+(94140)/(2.303 xx 8.314 xx 313)`
`= 25.33 - 5 + 15.7082 = 10.9635`
`A = "antilog"(10.9635) = 9.194 xx 10^(10) "colliison s"^(-1)`