1.

The rate constant for the decomposition of hydrocarbons is 2.418xx10^(-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»

SOLUTION :The FOLLOWING data is provided:
`K=2.418xx10^(-5)s^(-1), E_(a)=179.9kJ" mol"^(-1), T=546K`
Applying Arrhenius equation,
`k=Ae^(-E_(a)//RT)" or In k"="In A"-(E_(a))/(RT)`
or `log k= LOGA -(E_(a))/(2.303RT) "or log A"=log k+(E_(a))/(2.303RT)`
Substituting the values, we get
`logA=log (2.418xx10^(-5)s^(-1))+(179.9"kJ mol"^(-1))/(2.303xx8.314xx10^(-3)"kJ K"^(-1)"mol"^(-1)xx546K)`
`=(-5+0.3834) s^(-1)+17.2081=12.5924s^(-1)`
or `A="Antilog"(12.5924)s^(-1)=3.912xx10^(12)s^(-1)`.


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