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For a first order reaction A rarr P, the...

For a first order reaction `A rarr P`, the temperature `(T)` dependent rate constant `(k)` was found to follow the equation `log k = -2000(1//T) + 6.0`. The pre-exponential factor `A` and the activation energy `E_(a)`, respective, are

A

`1.0xx10^(6)s^(-1) and 9.2 kJ mol^(-1)`

B

`6.0 s^(-1) and 16.6kJ mol^(-1)`

C

`1.0xx10^(6)s^(-1) and 16.6 kJ mol^(-1)`

D

`1.0xx10^(6) s^(-1) and 38.3 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D

According to Arrhenius equation
`K=Ae^(-E_(a)//RT)`
Taking logarithm of both the sides, we get
log `k=logA-(E_(a))/(2.303RT)`
We are given
log `k=6.0-(2000)(1)/(T)`
comparing the two, we obtain
log `A=6.0`
or `A=10^(6)sec^(-1)`
`(E_(a))/(2.303R)=2000`
or `E_(a)=(2000)(2.303)(8.14JK^(-1)mol^(-1))`
`=38.3KJmol^(-1)`
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