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

The rate constant for the decomposition of hydrocarbons is `2.418 xx10

1.	The rate constant for the decomposition of hydrocarbons is `2.418 xx10^(-5)s^(-1)` at 546 K. If the energy of activatin is `179.9 kJ//mol.` What will be the value of per-exponential factor ?
Answer» According to Arrhenius equation, `log k = log A -(E_(a))/(2.303RT)` `k =2.418 xx10^(-5)s^(-1)` `E _(a) =179.9 KJmol ^(-1) or 179900 J mol ^(-1)` `R= 8.314 JK ^(-1)mol ^(-1)` `T=546K` `log A =log k + (E_(a))/(2.303RT)` `=log (2.418 xx10^(-5)s^(-1))+ (179900 J mol ^(-1))/(2.303 xx(8.314JK^(-1))xx546k)` `=-4.6184+17.21 =12.5916` `A="Antilog" 12.5916=3.9xx10^(12)s^(-1)` `A=3.9xx10^(12)s^(-1)`

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