Question

If the uncatalyzed reaction proceeds at a rate of 2.6 x 10-5 s-1 and the enzymatic reaction rate is 50 s-1 , what is the difference in the height of the activation barrier for the enzymatic reaction? If the average H-bond in the chorismate mutase active site releases 18.5 kJ/mol of energy, how many H-bonds would have to form to account for the rate enhancement?

Answer 1

Height of activation barrier means activation energy (E_a)

rate (R) = K [concentration of the reactant]

from arhenious equation

K = A. e^-E_a/RT

so, R₁/R₂ = K₁/K₂ = e^-E_a1/RT /  e^-E_a2/RT

Or, 2.610^-5 50 = e^E_a2-E_a1/RT

or, e^E_a2-E_a1/RT = 0.052 10^-5

or, E_a2-E_a1 /RT = ln (0.052 10^-5)

or, E_a2-E_a1 /RT = -12.16

OR,  E_a1-E_a2 = 12.16RT

Therefore, difference in the height of activation barrier between uncatalysed and enzymatic reaction is 12.16RT Joule/mole

optiumum temperature of enzyme activity is 37.5 ⁰c = 273+37.5 = 310.5 K

hence difference in the height of activation barrier = 12.168.314310.5 = 31391 Joule/mole = 31.931 KJ/mole

total no of H bond would have to form for the rate enhancement = 31.931/18.5 = 1.72 2