Which of the following would be a rate law for the reaction?

Question

The proposed mechanism for a reaction is

1. A(g) + B(g) ⇄ X(g) Fast

2. X(g) + C(g) ⇄ Y(g) Fast

3. Y(g) → D(g) Slow

Why is it, rate = k[A][B][C]?

Answer 1

Since the slow step is NOT the first step, you must look at steps ahead of the slow step, and because both X and Y are intermediates, they cannot appear in the rate equation. With these facts in mind, we proceed as follows in order to eliminate both X and Y. This takes some mathematical maneuvering, so hopefully you can follow.

1. A(g) + B(g) ⇄ X(g) Fast

rate forward = rate reverse

k₁[A][B] = k_-1[X]

[X] = k₁/k_-1[A][B] = k'[A][B] ... eq. A

2. X(g) + C(g) ⇄ Y(g) Fast

rate forward = rate reverse

k₂[X][C] = k_-2[Y] and [Y] = k₂/k_-2[X][C] = k''[X][C] ... eq. B

k₂[x]/k_-2[y] = 1/[C]

k₂k1/k_-1[A][B] = 1/[C]

Y(g) → D(g) Slow

rate = k[Y] and substituting eq. B we get...

rate = k'' [X][C] and [X] = k1/k-1[A][B] from eq. A so substituting this and combining rate constants to k'''...

rate = k'''[C][A][B] or simply rate = k[A][B][C]