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

The arrhenius equation is ,
`k=Ae^(-E_(a)//R)`
In absence of catalyast, `k_(1)=Ae^(-100//RT)`
In presence of catalyst, `k_(2)=Ax^(-25//RT)`
So, `(k_(2))/(k_(1))=^(75//RT)or2.303 "log" (k_(2))/(k_(1))=(75)/(RT)`
`or 2.303 "log" (k_(2))/(k_(1))=(75)/(8.314xx10^(-3)xx293)`
`or "log" (k_(2))/(k_(1))=(75)/(8.314xx10^(-3)xx293xx2.303)`
or `(k_(2))/(k_(1))=2.314xx10^(13)`
As the things beings equal in presence or absence of a catalyst,
`(k_(2))/(k_(1))` must be =`("rate in presence of catalyst")/("rate in absence of catalyst")`
`i.e., (r_(2))/(r_(1))=(k_(2))/(k_(1))=2.34xx10^(13)`