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

Answer

C

Explanation

`k=Ae^(-E_(a)//RT)`
Let with catalyst, rate constant
`k' = Ae^(-E_(a)//RT)`
`therefore (k')/(k)=(e^(-E_(a)//RT))/(e^(-E_(a)//RT))=e^((E_(a)-E'_(a))//RT)`
Here, `E_(a)-E'_(a)=2000 cal`
`R=2 cal K^(-1)mol^(-1)`
`T=27+273=300 K`
`therefore (K')/(K)=e^((2000)/(2xx300))=e^((10)/(3))`
`=281 " " [because e=2.7 ("approx")]`
and `(2.7)^(3.33) ~~ 28`
`therefore (R')/(R )=(k')/(k)=28`
`therefore = 28 R`