Molecularity and elementary reactions

A reaction goes in steps

A ticket says "home to school". The trip itself has steps: you walk to the bus stop, ride the bus and walk again. The ticket shows only the start and the end. A balanced equation is the same. It shows the start and the end of a reaction, and not the steps in between.

Look at this reaction between nitrogen dioxide and fluorine:

Three molecules cannot be expected to meet at one instant. Experiments support a two step path:

• Step 1, slow: $\text{NO}_2 + \text{F}_2 \rightarrow \text{NO}_2\text{F} + \text{F}$.
• Step 2, fast: $\text{NO}_2 + \text{F} \rightarrow \text{NO}_2\text{F}$.

Add the two steps. Step 1 makes one F atom and step 2 uses it, so the F atoms cancel. What is left is $2\text{NO}_2 + \text{F}_2 \rightarrow 2\text{NO}_2\text{F}$, the equation we started with.

The F atom is an intermediate. It forms in one step and is used up in the next. It is never a reactant or a final product, so it does not appear in the overall equation.

An elementary reaction is a single step in which the particles collide and change in one event, exactly as the step is written. The full list of elementary steps is the mechanism of the reaction. A reaction that has more than one step is a complex reaction.

Molecularity

The molecularity of an elementary reaction is the number of atoms, ions or molecules that collide in that step:

• Unimolecular: one particle changes, as in the rearrangement of cyclopropane to propene.
• Bimolecular: two particles collide, as in $\text{NO} + \text{O}_3 \rightarrow \text{NO}_2 + \text{O}_2$.
• Termolecular: three particles meet at the same moment, as in $\text{O} + \text{O}_2 + \text{M} \rightarrow \text{O}_3 + \text{M}$. Here M is a third body that carries away energy.

![Three pictures of elementary steps drawn with circles for molecules. Unimolecular: one molecule A changes into two molecules B and C, with rate equal to k times the concentration of A, for example cyclopropane changing to propene. Bimolecular: two molecules A and B collide and join to form AB, with rate equal to k times [A] times [B], for example nitrogen monoxide and ozone forming nitrogen dioxide and oxygen. Termolecular: three molecules A, B and C meet at the same moment to form ABC, with rate equal to k times [A][B][C]. This is very rare, for example an oxygen atom, an oxygen molecule and a third body M forming ozone.](https://tuygolavdykfydmasiqc.supabase.co/storage/v1/object/public/discussion-images/57b80fd5-0d05-40b0-bb7c-fdadd190c00d/figures/ch23-chemical-kinetics/kn-k1-molecularity.png)

Four rules go with molecularity:

• It is always a whole number: 1, 2 or 3. It is never zero and never a fraction.
• It belongs to an elementary step only. A complex reaction has no molecularity as a whole. Each of its steps has one.
• Termolecular steps are rare. For three particles to meet at once, with enough energy and the right orientation, is very unlikely.
• Steps with four or more particles are not known. The chance of such a meeting is almost zero.

Why do the chances fall so fast? In a gas, when every concentration is halved, a bimolecular step slows down to a quarter. A termolecular step slows down to one eighth.

Rate law of an elementary step

An elementary step happens exactly as it is written. So for an elementary step only, we can write the rate law straight from the equation. Each molecule that takes part gives one concentration factor:

• Unimolecular, $\text{A} \rightarrow \text{products}$: rate $= k[\text{A}]$.
• Bimolecular, $\text{A} + \text{B} \rightarrow \text{products}$: rate $= k[\text{A}][\text{B}]$.
• Bimolecular, $2\text{A} \rightarrow \text{products}$: rate $= k[\text{A}]^2$.
• Termolecular, $\text{A} + \text{B} + \text{C} \rightarrow \text{products}$: rate $= k[\text{A}][\text{B}][\text{C}]$.

For these steps the order equals the molecularity. For a complex reaction this is not so.

Order and molecularity

These two words are easy to mix up. Compare them:

• Order comes from experiment. It is found for the whole reaction, from the rate law. It may be 0, a whole number or a fraction.
• Molecularity comes from theory. It is found for one elementary step, from the equation of that step. It is 1, 2 or 3.
• A reaction of order 2 might be a single bimolecular step, or a complex reaction that happens to have a second order rate law.

The rate-determining step

When a reaction has several steps, which one controls the speed? Think of a road with a narrow bridge. Cars race along the open road, and then queue at the bridge. The number of cars that cross each hour depends on the bridge. A wider road does nothing to help.

In a mechanism, the slowest step is the rate-determining step. The overall reaction cannot go faster than that step. The rate law of the overall reaction is the rate law of the slow step.

Go back to $2\text{NO}_2 + \text{F}_2 \rightarrow 2\text{NO}_2\text{F}$. The slow step is $\text{NO}_2 + \text{F}_2 \rightarrow \text{NO}_2\text{F} + \text{F}$, which is bimolecular. So:

The reaction is second order. The balanced equation has three molecules of reactant, and a reader who trusts the equation would write $k[\text{NO}_2]^2[\text{F}_2]$ and expect third order. Experiment shows second order. This is the order that the mechanism explains.

In the simulation, the average time for one molecule to pass both steps is the time in the first step plus the time in the second step. Raise the fast step as far as you like: the total time barely changes. Raise the slow step: the total time falls at once.

A reaction in which a reactant is not in the rate law

Tertiary butyl bromide reacts with hydroxide ions:

The balanced equation has two reactants. Yet experiments show $\text{rate} = k[(\text{CH}_3)_3\text{CBr}]$, and doubling the hydroxide concentration changes nothing. The mechanism explains this:

• Step 1, slow: $(\text{CH}_3)_3\text{CBr} \rightarrow (\text{CH}_3)_3\text{C}^+ + \text{Br}^-$.
• Step 2, fast: $(\text{CH}_3)_3\text{C}^+ + \text{OH}^- \rightarrow (\text{CH}_3)_3\text{COH}$.

Hydroxide reacts only in the fast step, after the slow step has finished. So the rate does not depend on it. The order in hydroxide is zero.

Testing a mechanism

A mechanism is a proposal. To test one, ask two questions:

1. Do the steps add up to the overall equation? The intermediates must cancel.
2. Does the rate law of the slow step match the experiment?

If the answers are yes, the mechanism is possible. It is not proved, because another mechanism could give the same rate law. If either answer is no, the mechanism is wrong.

Practise with new numbers

Try it in Python

The program has three candidate rate laws for the reaction of $\text{NO}_2$ and $\text{F}_2$ and three measured runs. A law fits when the value of $k$ is the same in every run.

Common mistakes

• Writing the rate law from the overall equation. Do this only for an elementary step.
• Giving a molecularity to the whole of a complex reaction. Only the single steps have one.
• A fractional or zero molecularity. Molecularity is 1, 2 or 3. A fractional order is possible. A fractional molecularity is not.
• Putting an intermediate in the final rate law. The slow step may use an intermediate, but the overall law is written with the reactants.
• Thinking that the fastest step controls the rate. It is the slowest step.
• Saying that a fitting mechanism is proved. A match is evidence. It is not proof.

Remember these

• An elementary reaction is one step. A complex reaction is a sequence of elementary steps, its mechanism.
• Molecularity: the number of particles that collide in an elementary step. It is 1, 2 or 3 and never a fraction.
• For an elementary step, the rate law follows from the equation, and order = molecularity.
• The slowest step is the rate-determining step. It gives the rate law of the reaction.
• A mechanism must add up to the overall equation and must give the measured rate law.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Can the molecularity of an elementary step be $2.5$? Can the order of a reaction be $2.5$?

2. For the elementary step $2\text{A} + \text{B} \rightarrow \text{C}$, write the rate law and say what the molecularity is.

3. A reaction has two steps. Step 1 is slow and step 2 is fast. A scientist makes step 2 ten times faster. What happens to the overall rate?

4. A reaction has the balanced equation $\text{A} + 2\text{B} \rightarrow \text{C}$. A student says the rate law is $k[\text{A}][\text{B}]^2$ and the reaction is third order. What is wrong with this?

5. An intermediate is not in the overall equation. How can you tell, from the steps, which species is an intermediate?