1.

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» `k=2.418xx10^(-5)s^(-1)`
` T = 546 k `
`E_(a)=179.9KJ mol^(-1) =179.9xx10^(3) J mol ^(-1)`
According to the Arrhenius equation,
`k=Ae^(-E)`
`rArr "In "k=In A-E_(0)/(RT)`
`rArr"log"k=log A-E_(0)/(2.303RT)`
`rArr"log"A=log K+E_(0)/(2.303RT)`
`="log"(2.418xx10^(-5)S^(-1))+(179.xx10^(3) J "mol"^(-1))/(2.303xx8.314 "JK"^(-1)"mol"^(-1)xx546K)`
=`(0.3835 - 5) + 17.2082`
12.5917
Therefore, A = antilog `(12.5917)`
`= 3.9 × 10^(12) s^(−1)` (approximately)


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