# What is the rate law for the following mechanism in terms of the overall rate constant k? Step 1: A + B leftrightarrow C (fast) Step 2:  B + C -> D (slow)

Jul 16, 2016

The rate law is "rate" = k["A"]["B"]^2.

#### Explanation:

The reaction is

$\text{A + B"color(white)(l) ⇌ "C}$ (fast)
$\text{B + C" color(white)(l)stackrelcolor(blue)(k_2color(white)(m))(→) "D}$ (slow)

The rate of the reaction is governed by the slow step.

This gives the rate equation

"rate" = k["B"]["C"]

The problem is that $\text{C}$ isn't one of the reactants.

We must express $\text{[C]}$ in terms of $\text{[A]}$ and $\text{[B]}$.

We can do that because the first step is an equilibrium.

$K = \text{[C]"/"[A][B]}$

$\left[\text{C"] = K["A"]["B}\right]$

Substituting into the original rate expression, we get

"rate" = k_2["B"] × K["A"]["B"] = k_2K["A"]["B"]^2

Let $k = {k}_{2} K$, and the rate law becomes

"rate" = k["A"]["B"]^2