Elementary Reactions: Study Guide (Chemistry 5.4)

One collision, one step

A balanced equation tells you the starting materials and products, and in what ratio. It usually does not tell you how the change happens. Many reactions go through a series of simple steps. Each simple step is called an elementary reaction (or elementary step).

An elementary reaction is a single event at the particle level:

• one particle breaks apart or rearranges on its own, or
• two (occasionally three) particles collide, and in that one collision bonds break and form to give the products.

The equation for an elementary reaction shows exactly the particles that take part. If the step is 2 NO₂ → NO₃ + NO, then two NO₂ molecules hit each other, and in that collision an O atom moves from one to the other. Nothing else is involved.

An overall reaction is the sum of the steps. In Topic 5.2 you saw that the orders in an overall rate law have to be found by experiment. They cannot be read from the coefficients. The new idea in this topic is that for an elementary step, they can.

Why the coefficients become the orders

For two particles to react, they must first meet. So the rate of a bimolecular step is proportional to how often the right particles collide, and that depends on how many of each there are in a given volume: their concentrations.

Figure 1 shows the counting idea. Each A particle can meet each B particle, so the number of possible A–B pairs is (number of A) × (number of B). Double the number of A, and there are twice as many pairs. Double B as well, and there are four times as many. That gives

A + B → products: rate = k[A][B]

If the two colliding particles are the same substance, the same counting gives a square. Doubling [A] doubles the number of A particles, and each one now has twice as many A partners to hit:

A + A → products (written 2A → products): rate = k[A]²

A unimolecular step needs no collision partner of a particular kind. Each particle has the same chance of reacting in the next second, so the rate is proportional to how many there are, as in radioactive decay (Topic 5.3):

A → products: rate = k[A]

The rule is general: for an elementary step, the order in each reactant equals its coefficient in the step, and only reactants appear in the rate law.

Molecularity

The molecularity of an elementary step is the number of reactant particles that come together in it.

Molecularity	General form	Rate law	Overall order	Units of k (rate in M s⁻¹)
Unimolecular	A → products	k[A]	1	s⁻¹
Bimolecular	A + B → products	k[A][B]	2	M⁻¹ s⁻¹
Bimolecular	2A → products	k[A]²	2	M⁻¹ s⁻¹
Termolecular	A + B + C → products	k[A][B][C]	3	M⁻² s⁻¹
Termolecular	2A + B → products	k[A]²[B]	3	M⁻² s⁻¹

“Particles” can be molecules, atoms, ions or free radicals. A step such as 2A → products is bimolecular even though only one substance is involved: two particles collide.

Where the units of k come from

The rate of any step is measured in M s⁻¹. The concentrations in the rate law bring in one factor of M for each particle that collides. k has to cancel the extra ones:

• unimolecular: k × M = M s⁻¹, so k is in s⁻¹
• bimolecular: k × M² = M s⁻¹, so k is in M⁻¹ s⁻¹
• termolecular: k × M³ = M s⁻¹, so k is in M⁻² s⁻¹

So the units of k tell you the overall order, and for an elementary step that is also the molecularity. This is a useful check on any rate law you write: if the units do not come out as M s⁻¹, the setup is wrong. It also works the other way. If a question gives k = 3.0 × 10⁻² s⁻¹ for an elementary step, the step must be unimolecular.

Why three-particle collisions are rare

For a termolecular step, three particles must arrive at the same place at the same instant, and still have enough energy and the right orientation. Two-particle collisions happen all the time. But for a third particle to arrive during the extremely short moment that two others are in contact is far less likely. So termolecular steps are rare.

Steps with four or more particles colliding at once are so unlikely that they are not used in reaction mechanisms. Most elementary steps are unimolecular or bimolecular.

This gives you a quick test. If an overall equation has four or more reactant particles on the left, it cannot be a single elementary step. It must happen in several steps.

Worked example 1: writing rate laws for elementary steps

Question. Treat each equation below as an elementary step. For each, state the molecularity, write the rate law and give the units of k (rate in M s⁻¹).

(a) O₃ → O₂ + O (b) NO + O₃ → NO₂ + O₂ (c) 2 NO₂ → NO₃ + NO (d) 2 NO + Br₂ → 2 NOBr

Method. Count the reactant particles, then use each coefficient as the order. Ignore the products.

Step	Particles colliding	Molecularity	Rate law	Units of k
(a)	one O₃	unimolecular	rate = k[O₃]	s⁻¹
(b)	one NO + one O₃	bimolecular	rate = k[NO][O₃]	M⁻¹ s⁻¹
(c)	two NO₂	bimolecular	rate = k[NO₂]²	M⁻¹ s⁻¹
(d)	two NO + one Br₂	termolecular	rate = k[NO]²[Br₂]	M⁻² s⁻¹

Check the units. Rate is in M s⁻¹. In (d), k × M² × M must give M s⁻¹, so k must be in M⁻² s⁻¹.

Interpretation. Of the four, (d) would be the least likely to happen as a single step, because it needs three particles to collide at the same moment.

Worked example 2: using an elementary rate law

Question. An elementary step A + B → C has k = 2.5 × 10⁵ M⁻¹ s⁻¹. (a) Calculate the rate when [A] = 4.0 × 10⁻⁴ M and [B] = 1.2 × 10⁻³ M. (b) Find the new rate if [A] is tripled and [B] is halved.

(a) The step is elementary and bimolecular, so rate = k[A][B]:

rate = (2.5 × 10⁵ M⁻¹ s⁻¹)(4.0 × 10⁻⁴ M)(1.2 × 10⁻³ M) = 0.12 M s⁻¹

(b) The rate is first order in each reactant, so the rate changes by 3 × ½ = 1.5:

new rate = 1.5 × 0.12 M s⁻¹ = 0.18 M s⁻¹

Check. Substitute directly: (2.5 × 10⁵)(1.2 × 10⁻³)(6.0 × 10⁻⁴) = 0.18 M s⁻¹. If the step had been second order in A instead (for example 2A + B → products), tripling [A] would multiply the rate by 3² = 9, not 3.

Worked example 3: can the overall reaction be one step?

Question. (a) For an invented gas reaction A₂ + B₂ → 2 AB, experiments give rate = k[A₂][B₂]. A student says: “This proves the reaction is a single elementary step.” Evaluate the claim. (b) Could the combustion of propane, C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O, be a single elementary step?

(a) If the reaction were one bimolecular step, its rate law would be k[A₂][B₂]. The experimental rate law matches, so a single step is consistent with the data. But it is not proved. A sequence of several steps can produce the same rate law, as you will see with reaction mechanisms. Agreement means “possible”, not “certain”. More evidence, such as detecting an intermediate, would be needed to decide.

(b) No. Six reactant particles (one C₃H₈ and five O₂) would have to collide at the same instant. Even three-particle collisions are rare; six is not realistic. Propane combustion must happen through many elementary steps.

The general rule. If the experimental rate law differs from the rate law written from the coefficients, the reaction is definitely not a single elementary step. If it agrees, a single step is possible but not proved.

Common misconceptions

• “You can always write the rate law from the balanced equation.” Only for an elementary step. For an overall reaction the orders come from experiment.
• “2A → products is unimolecular because there is only one reactant.” Molecularity counts particles, not substances. Two A particles collide, so it is bimolecular, and rate = k[A]².
• “Termolecular steps are impossible.” They are rare, not impossible. Steps with four or more particles colliding at once are the ones that are not expected.
• “Products go in the rate law.” The rate law of a forward elementary step contains only the reactants that collide.
• “Molecularity and order are the same thing.” For an elementary step the overall order equals the molecularity. For an overall reaction, order is measured and can be zero or a fraction, while molecularity is not defined.
• “A matching rate law proves the reaction is elementary.” It only shows that a single step is possible.

Where this leads

This topic builds on the rate laws of Topic 5.3, Concentration Changes Over Time. Next, Topic 5.5, Collision Model, explains why only some collisions lead to reaction: particles need enough energy and the right orientation. Later in the unit you will join elementary steps together into reaction mechanisms and use their rate laws to test a proposed mechanism against experiment. Try the practice questions now, then use the revision notes and the checklist to consolidate.