The activation energy of a reaction is 75.2 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 the catalyst if the reaction proceeds at `25^@ C? [R = 8.314 JK^(-1) ` mole`""^(1)` ]

Explanation

According to Arrhenius equation,
` k= Ae^(-E_a//RT) or I n k= I n A-E_(a)//RT`
If `k_1` and `k_2` are the rate constants and `E_1` and `E_2` are the activation energies in the absence and presence of a catalyst, then
In ` K_1 = I n A-(E_1)/(RT)`
` I n k_2 = I n A - (E_2)/(RT)`
` therefore I n (k_2)/(k_1) = (E_1 )/(RT) - (E_2)/(RT) or I n (k_2)/(k_1) = (1)/(RT) [E_1 -E_2]`
` or log "" (k_2)/(k_1) = (1)/( 2.303 RT) [E_1 -E_2]`
Substituting the values in eq. (iii)
` log ""(K_2)/(K_1) = (1 )/(2.303 xx 8.314 xx 298 ) [75.2 - 50.14 ] xx 10^3 = 4.392`
` (k_2)/(k_1) =` Antilog ` 4.392 = 24660`