The activation energy of a reaction is 24.0 kcal `mol^(-1)` at `27^(@)C` and the presence of catalyst changes its activation energy to one-fourth at the same temperature. The approximate ratio of rate in the presence of catalyst to rate in the absence of catalyst will be (use R = 2 cal `mol^(-1) K^(-1)` ).

Options

A. `1.0 xx 10^(4)`
B. `1.0 xx 10^(3)`
C. `2.07`
D. `1.0 xx 10^(13)`

Answer

D

Explanation

From Arrhenius equation, `k = A e^(-E_(a)//RT)`
Let the rate in the absence of catalyst be `k_(1) = Ae^(-24//RT)` and in the presence of catalyst is `k_(2) = A e^(-24//4RT)`
Therefore, `(k_(2))/(k_(1)) = e^((24-6)//RT) rArr ln((k_(2))/(k_(1))) = (18 xx 1000)/(2 xx 300) = 30`
`log((k_(2))/(k_(1))) = (30)/(2.303) ~= 13, (k_(2))/(k_(1)) = 1 xx 10^(13)`
Since, rate `prop` rate constant, So `(r_(2))/(r_(1)) = (k_(2))/(k_(1)) = 1.0 xx 10^(13)`