Topic 5.4 Notes – Elementary Reactions

1. What an Elementary Reaction Is

An elementary reaction happens in one single step at the molecular level. The balanced equation shows the actual collision that occurs.

That means:

• No hidden steps
• No intermediates
• No mechanism behind the scenes

If you see

$$\ce{A + B -> products}$$

and it’s labeled elementary, that equation literally represents one collision between A and B.

This is very different from an overall reaction, which is usually made of multiple elementary steps (a mechanism). For overall reactions, the balanced equation does not tell you the rate law.

Here’s the big idea for this topic:

For an elementary reaction, the coefficients of the reactants become the exponents in the rate law.

Only reactants matter. Never include products in a rate law.

2. Molecularity and Types of Elementary Reactions

Molecularity is the number of particles that collide in a single elementary step.

It only applies to elementary reactions. You would never say “the molecularity of the overall reaction.”

Unimolecular

One particle rearranges or decomposes.

$$\ce{A -> products}$$

Rate law:

$$\text{rate} = k[A]$$

• First order in A
• First order overall

Only one molecule is involved in the rate-determining event.

Bimolecular

Two particles collide.

Two common forms:

$$\ce{A + B -> products}$$

$$\text{rate} = k[A][B]$$

or

$$\ce{2A -> products}$$

$$\text{rate} = k[A]^{2}$$

• First order in each reactant (if different)
• Second order overall

Most elementary reactions are unimolecular or bimolecular.

Termolecular

Three particles collide simultaneously.

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

$$\text{rate} = k[A][B][C]$$

• Third order overall

These are rare. The chance that three particles hit each other at the same time with correct orientation and enough energy is very small.

On the AP exam, if they give you a termolecular step, they will clearly label it as elementary.

Summary Table

3. Writing the Rate Law from an Elementary Reaction

This is the skill they expect you to do instantly.

If you are told:

$$\ce{2NO2(g) + F2(g) -> 2NO2F(g)} \quad \text{(elementary)}$$

The rate law is:

$$\text{rate} = k[\ce{NO2}]^{2}[\ce{F2}]$$

Why?

• Coefficient 2 in front of $\ce{NO2}$ → exponent 2
• Coefficient 1 in front of $\ce{F2}$ → exponent 1
• Products do not appear

That direct connection only works because it’s elementary.

If they don’t say it’s elementary, you cannot assume this.

That’s a very common trap in multiple choice.

4. How This Connects to Experimental Rate Laws

Most rate laws are found experimentally, not from stoichiometry.

Why? Because most reactions are multi-step mechanisms.

For overall reactions:

• The balanced equation does not reveal reaction orders
• Orders must be determined by comparing trials
• Temperature must stay constant because $k$ depends on temperature

Elementary reactions are the exception. Their stoichiometry matches the rate law because the equation reflects the actual collision.

On free response, they sometimes give a mechanism and ask which step determines the rate. If that slow step is elementary, its reactant coefficients become the rate law exponents.

5. Big Patterns to Lock In

• Molecularity only applies to elementary steps.
• Most elementary steps are unimolecular or bimolecular.
• Termolecular steps are rare because three-body collisions are unlikely.
• Reactions higher than third order are essentially never elementary.

When more particles must collide at once, the probability drops, so the step tends to be slower.