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

`1.0 xx 10^(5) s^(-1)` and `9.2 kJ"mol"^(-1)`

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

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

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

Text Solution

Verified by Experts

The correct Answer is:

Given, log K `=6- 2000/T`
Since, log K = log A `-(Ea)/(2.303 RT)` So, `A = 10^(6) "sec"^(-1)` and `Ea = 38.3` kJ/mole.

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