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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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