The rate constant for the decomposition of hydrocarbons is `2.418 × 10^(–5)s^(– 1)` at 546 K. If the energy of activation is `179.9 kJ//mol`, what will be the value of pre-exponential factor.

Answer

`3.9xx10^(12)s^(-10`

Explanation

`k=2.418xx10^(-5)s^(-1),E_(a)=179.9kJ mol^(-1),T=546K,`
Using Arrhenius equation,
`k=Ae^(-Ea//RT)`
`ln k=lnA-(E_(a))/(RT)`
`or log k=log A-(E_(a))/(2.303RT)`
`or log A= log k +(E_(a))/(2.303RT)`
`=log(2.418xx10^(-5)s^(-1))+(179kJ mol^(-1))/(2.303xx8.314xx10^(-3)kJ mol^(-1)xx546K)`
`or A= Antilog(12.5924)s^)-1)=3.912xx10^(12)s^(-1)`