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The activation energy of a reaction is 9...

The activation energy of a reaction is 94.14 KJ/mol and the value of rate constant at `40^@ C` is `1.8 xx 10^(-1) sec^(-1)`. Calculate the frequency factor A.

Text Solution

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Here, we are given that `E_(a)=94.14" kJ mol"^(-1)=94140" J mol"^(-1)`
`T=313" K",k=1.8xx10^(-5)sec^(-1)`
Substituting the values in the equation : `logk=(E_(a))/(2.303" RT")+log" A"`
or `" "log" A"=logk+(E_(a))/(2.303" RT")=log(1.8xx10^(-5)s^(-1))+(94140" J mol"^(-1))/(2.303xx8.314" JK"^(-1)mol^(-1)xx313"K")`
`=(log" "1.8)-5+15.7082=0.2553-5+15.7082=10.9635`
`"A"=" antilog "(10.9635)=9.194xx10^(10)s^(-1).`
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