What is the rate law of the following mechanism:?

E  +  S   <- ->   ES       (fast)
   ES         ->  E + P    (slow)

E is enzyme, S is substrate, P is product.

joeyeehu...

Best Answer - Chosen by Voters

let the rate constant of the forward reaction be k1;
let the rate constant of the backward reaction be k2;
let the rate constant of the forward reaction of the second eq be k3

For the first reversible equation:(fast)
forward reaction: rate = k1 [E][S]
backward reaction: rate = k2 [ES]

At equilibrium, rate of forward reaction = rate of backward reaction
k1[E][S] =k2[ES]

[ES] = { k1[E][S] } / k2 ......................(1)

For the slow reaction:

rate = k3 [ES]...................................(2)

Substitute (1) into (2)

rate = (k3) { k1[E][S] } / k2

rate = k [E] [S] ; where k = (k3)(k1) / k2