A catalyst decreases `E_(a)` form `100 KJ mol^(-1)` to `80 KJ mol^(-1)` At what temperature the rate of reaction in the absence of catalyst at `500 K` will be equal to rate reaction in the presence of catalyst?

Options

A. `400 K`
B. `200 K`
C. `625 K`
D. None of these

Answer

A

Explanation

`(E_(a))/(RT_(1)) = (-E'_(a))/(RT_(2)) rArr (100)/(500) = (80)/(T)`
`:. T' = 400 K`