The activation energy of a reaction is `80"kj.mol"^(-1)` . The activation energy reduces by 75% in presence of a catalyst. If the other parameters are kept constant , compare the rates of the reaction in presence and absence of the catalyst at `25^(@)C`.

Explanation

According to Arrhenius equation, `k=Ae^(-(E_(a))/(RT))`
In absence of catalyst , `k_(1)=Ae^(((-E_(a))/(RT)))=Ae^(((-80)/(RT)))`
`E_(a)=80-[80xx(75)/(100)]=20"kJ.mol"^(-1)`
`k_(2)=Ae^(((-20)/(RT)))`
`therefore(k_(2))/(k_(1))=e^(((60)/(RT)))=3.29xx10^(10)`