The rate constant for the decompoistion of a certain reaction is described by the equation:
`log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T)`
What is the effect on the rate of reaction at `127^(@)C`, if in the presence of catalyst, energy of activation is lowered by `10 kJ mol^(-1)`?

Options

A. `5` times
B. `10` times
C. `7` times
D. `20` times

Answer

D

Explanation

`E_(a)` is lowered and thus rates are increased in exponent form
`k = Ae^(-E_(a)//RT)` …(i)
`k' = Ae^((-E_(a-x))//RT)` …(ii)
Divide Eq. (ii) by Eq. (i)
`(k')/(k) = e^(x//RT) = e^(10//8.314 xx 10^(-3) xx 400) = 20`