1.

The rate constant for the decomposition of a hydrocarbon 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 :Here, `k=2.418xx10^(-5)s^(-1),E_(a)=179.9" kJ mol"^(-1),T=546" K".`
ACCORDING to Arrhenius equation,
`k="A E"^(-E_(a)//RT)" or ln "k="ln "A-(E_(a))/(RT)" or "logk=logA-(E_(a))/(2.303" RT")`
or `logA=logk+(E_(a))/(2.303" RT")=log(2.418xx10^(-5)s^(-1))+(179.9" kJ mol"^(-1))/(2.303xx8.314xx10^(-3)" kJ K"^(-1)" mol"^(-1)xx546" K")`
`=(-5+0.3834)s^(-1)+17.2081=12.5924" s"^(-1)`
or `A=" ANTILOG "(12.5924)s^(-1)=3.912xx10^(12)s^(-1).`


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