Question

The activation energy of a reaction is `94.14 kJ mol^(-1)` and the value of rate constant at 313 K is `1.8 xx 10^(-5)s^(-1)`. Calculate the frequency factor A.

Answer 1

According to the avialable data:
`E_(a) = 94.14 kJ mol^(-1)=94140 J mol^(-1), T=313K, k=1.8 xx 10^(-5)s^(-1)`
According to Arrhenius equation:
`logk = (-E_(a))/(2.303 RT) + log A` or `log A= log k + E_(a)/(2.303 RT)`
`log A = log(1.8 xx 10^(-5)) + (94140 J mol^(-1))/(2.303 xx 8.314 J K^(-1)mol^(-1) xx 313K)`
`=[0.2553 -5]+[15.7082]=10.9635`
A = Antilog`(10.9635)=9.194 xx 10^(10)` collisions `s^(-1)`.