Question

The rate constant for the decomposition of hydrocarbons is `2.418xx10^(-5)s^(-1)` at `546K`. If the energy of activation is `179.9kJ mol^(-1)`, what will be the value of pre`-`exponential factor?

Answer 1

According to Arrheneius equation,
log K = log `A - E_(a)/(2.303 RT)`
`k=2.418 xx 10^(-5)s^(-1)`, `E_(a) = 179900 J mol^(-1)`, T=546 K
log A `= log K + (E_(a))/(2.303 RT)`
`=log(2.418 xx 10^(-5)s^(-1)) + (179900 J mol^(-1))/(2.303 J K^(-1)mol^(-1)) xx 546 K`
`-4.6184 + 17.21 = 12.5916`
A = Antilog 12.5916 = `3.9 xx 10^(12)s^(-1)`