1.

The rate constant, the activation energy and frequency factor of a chemical reaction at 25^(#)C are 3.0xx10^(-4)s^(-1), 104.4"kJ mol"^(-1) and 6.0xx10^(14)s^(-1) respectively. What is the value of the rate constant when T rarr oo ?

Answer»

Solution :Formula :
`k_(1)=Ae^(-E_(a//RT))`
Given :
Rate CONSTANT, `k_(1)-3.0xx10^(-4)s^(-1)`
FREQUENCY factor, `A = 6.0xx10^(14)s^(-1)`
Activation energy, `E_(a)=104.4 kJ`
`= 104400 J`
Temperatures, `T_(1)=25^(@)C , T_(2) = oo`
`k_(2)=A`
`k_(2)=A.e^(-(E_(a))/(RT))`
`k_(2)=6.0xx10^(14)xx e^(-1(104400)/(8.314xx oo)`
`k_(2)=6.0xx10^(14)xx1 (therefore e (-E_(a))/(R oo)=1)`
`therefore` The rate constant at `T rarr oo` is `6.0xx10^(14)s^(-1)`.


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