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

`3.467 xx 10^(12)` collisions `s^(-1)`

Explanation

`k= 2.418 xx 10^(-5) s^(-1) , T = 546 K`
`R= 8.314 JK^(-1) mol^(-1) , E_a = 179.9 KJ mo^(-1)`
According to Arrhenius equation `k=Ae^(-E_a //RT) `
`log k= log A - (E_a)/( 2.303 R).(1)/(T)`
` or log A = log k + (E_a)/( 2.303 RT)`
`= log ( 2.418 xx 10^(-5) ) + ( 179.9 xx 10^3 )/( 2.303 xx 8.314 xx 546 )`
`=17.208 - 4.668 = 12.54`
taking antilog ,
` A = ` Antilog ` 12.54 = 3.467 xx 10^(12) ` collisions `s^(-1)`