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?

The activation energy of a reaction is 75.24 kJ "mol"^(-1) in the abse

1.	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?
Answer» SOLUTION :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)` ` "orIn "k_(2)-"In "k_(1)=(1)/(RT) (E_(a_(1))-E_(a_(2)))` ` "orlog"(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)`.