If a catalyst drops the activation energy of a reaction at `25^(@)C` form `80 kJ "to" 40 kJ`, what is the magnitude of the affect on the forward reaction rate and reverse rate?

Explanation

`E_(a)= 80-40= 40KJ`
a. `k_(1)=Ae^(Ea//RT)`
`k_(2)=Ae^(Ea(c )//RT)`
`log.(k_(2))/(k_(1))=(E_(a)-RE_(a)-(c ))/(2.3xx8.314xx298K)`
`=(40xx10^(3)J)/(2.3xx8.314xx298K)= 7.0`
`:.(k_(2))/(k_(1))= 10^(7)`
The reaction rate increase by `10^(7)` times.
b. Since catalyst increases the rate of reacation but does not affect the equilibrium of the reaction. Therefore, `r_(f)=r_(b)`.