Question

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 1

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)`