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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

`1.0 xx 10^(6) s^(-1)` and `9.2 kJ mol^(-1)`

B

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

C

`1.0 xx 10^(6) s^(-1)` and `16.6 kJ mol^(-1)`

D

`1.0 xx 10^(6) s^(-1)` and `38.3 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
`k = Ae^(E_(a)//RT)`
or `log k = logA - (E_(a))/(2.303 RT)`
`log k = -2000((1)/(T)) + 6.0`
On comparison, we get
`log A = 6`
or `A = 10^(6)s^(-1)`
`:. (E_(a))/(2.303R) = 2000`
or `E_(a) = 2.303 xx 8.314 xx 2000`
`~~ 38.3 kJ mol^(-1)`
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