The activation energy of a reaction is 75.2 kJ `mol^(-1)` in the absence of a catalyst and 50.14 kJ `mol^(-1)` in the presence of a catalyst. How many times will the reaction grow in the presence of catalyst if the reaction proceeds at `25^(@)`C ?

Explanation

Let the rate constant in the absence of catalyst = `k_(1)`
Let the rate constant in the presence of catalyst = `k_(2)`
Activation energy in the absence of catalyst `(E_(1)) = 75.2 kJ mol^(-1)`
Activation energy in the presence of catalyst `(E_(2)) = 50.14 kJmol^(-1)`
According to Arrhenius equation,
`k_(1) = Ae^(-E1//RT), k_(2)= Ae^(-E2//RT)`
`k_(2)/k_(1) = e^((E_(1)-E_(2)//RT))` or ln `k_(2)/k_(1)= (E_(1)-E_(2))/(RT)`
`log K_(2)/K_(1) = (E_(1)-E_(2))/(RT)`
`=((75.3-50.14) xx (10^(3)Jmol^(-1)))/(2.303 xx (8.314 JK^(-1)mol^(-1) xx 298K)) = 4.391`
`k_(2)/k_(1)= "Antilog" 4.391 = 24604` or `k_(2) = 24604 k_(1)`
Thus, the reaction rate has increased nearly by 24604 times.