For a given reaction, presence of catalyst reduces the energy of activation by `2` kcal at 27^(@)C`. Thus rate of reaction will be increased by:

Options

A. 20 times
B. 14 times
C. 28 times
D. 2 times

Explanation

`k_1= Ae^(-E_a//RT) and k_2=Ae^(-[E_a-2]//RT)`
`:. k_2/k_1=(e^((-[E_a-2])/(RT)))/((-E_a)/(e^(RT)))= e^((-[E_a-2])/(RT)+E_a/(RT))=e^(2/(RT))`
`:. k_2/k_1=e^((2xx 10^(-3)cal)/(2xx 300))=e^(1000/300)=e^3.34=28.27`
(Where `R = 2 cal K^(-1)mol^(-1)and T= 300K`)