In the presence of catalyst, the activation energy of the reaction is lowered by 2 kcal at `27^(@)C`. The rate of reaction will increased by

Options

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

Answer

C

Explanation

`k_1=Ae^(-(E_(a)-2)//RT)`
`k_2=Ae^(-(E_(a)-2)//RT)=Ae^(-E_(a)//RT_(e^(2//RT)))`
`:." "k_(2)/k_(1)=e^2//RT" or in"k_2/k_1=2/(RT)`
Putting `R=2xx10^(-3)`kcal `K^(-1) mol^(-1)`
and `T=300 K,` we get
`log""k_2/k_1=2/(2*303xx2xx10^-3xx300
)=1*4473`
`:." " k_2/k_1=28" or "k_(2)=28xxk_1`