If the rate law order is three, does that mean that three bodies collide at some elementary step?

Question

I am doing a chemistry review about rates, and I get asked "Which of these statements is false?"

The answer is

In reactions that are second order in one reactant and first order in another, the slow step generally involves a three-body collision of the reactants.

I imagine it looks like

$$ rate = k[A]^2[B]^1 $$

So at the slowest elementary step there should be $$ 2A + B = something $$

Which is a three way collision between two A molecules and a B.

What am I doing wrong?

EDIT:

Question number 2 (http://www.mesacc.edu/~paudy84101/CHM152F2005/Exam1%20Key.pdf)

This is not my review, in case you are wondering. I just have the same question.

Answer 1

Let's disprove that statement by counterexample. Here, the slow step is a collision of some intermediate and another thing, but the maths work out that the overall reaction order is third.

Let's have a two step reaction with the stoichiometry:

$$\ce{2A + B -> C}$$

The mechanism looks like this. The first step is fast and reversible and rapidly establishes equilibrium between $\ce{A + B}$ and an intermediate $\ce{I}$.

$$\ce{A + B <=>[k_1][k_{-1}] I}$$

The second step is slow, not so reversible and involves the intermediate $\ce{I}$ reacting with the other equivalent of $\ce{A}$.

$$\ce{I + A ->[k_2] C}$$

Now the rate of the reaction shall be written as the rate of appearance of $\ce{C}$, because it makes our lives a little easier.

$$\mathrm{rate}=\dfrac{d[\ce{C}]}{dt}=k_2 [\ce{A}][\ce{I}]$$

However, it's messy to have the intermediate in the rate law. Since the first step was fast and in equilibrium, we can get an expression for $\ce{[I]}$ in terms of $[\ce{A}]$ and $[\ce{B}]$.

$$K_{eq}=\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{I}]}{[\ce{A}][\ce{B}]}\\ [\ce{I}]=K_{eq}[\ce{A}][\ce{B}]$$

A little substituting and we get:

$$\mathrm{rate}=\dfrac{d[\ce{C}]}{dt}=\dfrac{k_2 k_1}{k_{-1}} [\ce{A}]^2 [\ce{B}]=k_{obs}[\ce{A}]^2 [\ce{B}]$$