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The half time of a first order reaction ...

The half time of a first order reaction is `6.93 xx 10^(-3) min` at `27^(@)C`. At this temeprature, `10^(-8)%` of the reactant molecules are able to cross-over the energy barrier. The pre-exponential factor `A` in the Arrhenius equation is equal to

A

`10^(4) min^(-1)`

B

`10^(8) min^(-1)`

C

`10^(10) min^(-1)`

D

`10^(12) min^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
`k = (0.693)/(t_(1//2)) = (0.693)/(6.93 xx 10^(-3)) = 10^(2) min^(-1)`
`e^(-E_(a)//RT) = (10^(-8))/(100) = 10^(-10)`
Hence, `A = (k)/(e^(-E_(a)//RT)) = (10^(2))/(10^(-10)) = 10^(12) min^(-1)`
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