The energy of activation for a reaction is `100 KJ mol^(-1)`. The peresence of a catalyst lowers the energy of activation by `75%`. What will be the effect on the rate of reaction at `20^(@)C`, other things being equal?

Explanation

`k=Ae^(-E_(a)//RT)`
Case-I: `k_(1)=Ae^(-100//RT),k_(2)= Ae^(-25//RT)`
`(k_(1))/(k_(2))=(e^(-100//RT))/(e^(-25//RT))=e^(-75//RT)`
`log.(k_(2))/(k_(1))= 75//RT=(75xx10^(3))/(8.314xx293)= 30.788`
`(k_(2))/(k_(1))= 2.35xx10^(30)`
Case-II: `r=k[A]^(n)`
`square:.n and [A]` are same for cases I and II.
`:. (r_(2))/(r_(1))=(k_(2))/(k_(1))= 2.35xx10^(30)`