The activation energy of a reaction is 75.24 kJ `"mol"^(-1)` in the absence of a catalyst and `50.14 "kJ mol"^(-1)` with a catalyst. How many times will the rate of reaction grow in the presence of a catalyst if the reaction proceeds at `25^(@)C`?

Explanation

Using Arrhenius equation in the following form and substituting the values, we get
In `k_(1)="In A"-(E_(a_(1)))/(RT) and "In "k_(2)="In A"-(E_(a_(2)))/(RT)`
` "or In "k_(2)-"In "k_(1)=(1)/(RT) (E_(a_(1))-E_(a_(2)))`
` "or log"(k_(2))/(k_(1))=(1)/(2.303xxRT) (75.24"kJ mol"^(-1)-50.14"kJ mol"^(-1))`
`=(1)/(2.303xxRT) (25.10" kJ mol"^(-1))`
`=(1)/(2.303xx8.314JK^(-1)"mol"^(-1)xx298K) xx25100 "J mol"^(-1)=4.40`
or `(k_(2))/(k_(1))="Antilog "4.40=2.5xx10^(4)=25000`
The rate or reaction will grow by 25000 times in the presence of catalyst.
The decrease in activation energy takes place by `25.1 "kJ mol"^(-1)`.