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.

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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