Elementary Reactions

AP Chemistry· AP Chemistry CED — Kinetics· 14 min read

1. Definition and Key Properties of Elementary Reactions★★☆☆☆⏱ 2 min

An elementary reaction (or elementary step, when part of a larger mechanism) is a single-step reaction that occurs exactly as written, with no intermediate sub-steps between reactants and products. Unlike overall balanced reactions, which only describe the net result of multiple reaction events, elementary reactions represent individual collision events between reactant particles.

The key distinguishing feature of elementary reactions is that the stoichiometric coefficient of each reactant directly equals its reaction order in the rate law. This rule does not hold for overall reactions, making elementary steps the foundation for all multi-step reaction mechanism problems on the AP exam.

📘 Definition

Elementary Reaction

A single-step chemical reaction that occurs via a single collision event between reactant particles, with no intermediate sub-steps. The reaction order of each reactant equals its stoichiometric coefficient.

Example:

A unimolecular decay reaction, written and occurring as $A \to products$ , is elementary.

2. Molecularity of Elementary Reactions★★☆☆☆⏱ 3 min

Molecularity is defined as the number of reactant particles that collide and react in a single elementary step. Because it counts discrete particles, molecularity can only be a small positive integer: 1 (unimolecular), 2 (bimolecular), or 3 (termolecular). Termolecular steps are very rare, because the probability of three particles colliding simultaneously with correct orientation and sufficient energy is extremely low.

A common point of confusion is distinguishing molecularity from reaction order. Reaction order describes how the rate depends on concentration, and can be zero, fractional, or negative for overall reactions. Molecularity is only defined for elementary steps, and is always 1, 2, or 3. To find molecularity, count the total number of reactant particles on the left-hand side of the balanced elementary step.

📐 Worked Example

Identify the molecularity of each of the following elementary reactions: 1. $C_{4} H_{8} \to 2 C_{2} H_{4}$ 2. $C l + C H_{4} \to H C l + C H_{3}$ 3. $B r + B r + A r \to B r_{2} + A r$

1
Count the total number of reactant particles on the left side of each elementary equation.
2
Reaction 1 has only 1 reactant particle ( $C_{4} H_{8}$ ), so molecularity is 1 (unimolecular).
3
Reaction 2 has two distinct reactant particles ( $C l$ and $C H_{4}$ ), so molecularity is 2 (bimolecular).
4
Reaction 3 has three total reactant particles (two $B r$ and one $A r$ ), so molecularity is 3 (termolecular).

Exam tip:

AP MCQ often includes distractors with non-integer molecularity. If an option lists molecularity as 0, 1.5, or any non-integer, eliminate it immediately.

3. Rate Laws for Elementary Reactions★★☆☆☆⏱ 4 min

For overall balanced reactions, reaction orders cannot be determined from the balanced equation — they must be measured experimentally. However, because an elementary reaction is a single collision event, the rate of the reaction is directly proportional to the concentration of each reacting particle raised to the power of its stoichiometric coefficient. This is because the probability of all required reactant particles colliding at the same time is proportional to the product of their individual concentrations.

\text{rate} = k [A]^a [B]^b \quad \text{for a general elementary reaction } aA + bB \rightarrow \text{products}

where $k$ is the rate constant for the elementary step, $a$ is the reaction order with respect to $A$ , and $b$ is the reaction order with respect to $B$ . The overall order of the elementary step is simply the sum of $a$ and $b$ .

📐 Worked Example

Write the rate law for the elementary reaction $2 H I \to H_{2} + I_{2}$ , and state the overall order of the reaction.

1
Confirm the reaction is explicitly stated to be elementary, so we can use stoichiometric coefficients as reaction orders.
2
The only reactant is HI, with a stoichiometric coefficient of 2.
3
Write the rate law directly from the coefficients:
4
$latex": "\text{rate} = k [HI]^2$
5
Sum the reaction orders to get the overall order of 2.

Exam tip:

In AP FRQ questions asking for a rate law for a forward elementary step, never include product concentrations. Only reactants appear in the rate law for forward elementary steps, which is what you will be asked for 99% of the time on the exam.

4. Elementary Steps in Multi-Step Reaction Mechanisms★★★☆☆⏱ 5 min

Nearly all overall reactions are not single elementary steps — they proceed via a sequence of multiple elementary steps called a reaction mechanism. When you add all elementary steps in a mechanism together, you get the balanced overall reaction.

📘 Definition

Key Mechanism Species

Two types of species do not appear in the overall balanced reaction: Reaction intermediates are produced in an early elementary step and consumed in a later step. Catalysts are consumed in an early step and regenerated in a later step.

Example:

In the ozone decomposition mechanism, $C l O$ is an intermediate and $C l$ is a catalyst.

For a mechanism to be valid, two conditions must hold: 1) the sum of elementary steps matches the experimental overall reaction, and 2) the rate law derived from the mechanism matches the experimentally determined rate law for the overall reaction.

📐 Worked Example

A reaction mechanism for the conversion of ozone to oxygen is given below: Step 1 (fast): $C l + O_{3} \to C l O + O_{2}$ Step 2 (slow): $C l O + O \to C l + O_{2}$ Identify the reaction intermediate and the catalyst, and confirm the sum of elementary steps gives the overall reaction $O_{3} + O \to 2 O_{2}$ .

1
Track where each non-overall species is produced and consumed: Cl is consumed in step 1 and produced in step 2; ClO is produced in step 1 and consumed in step 2.
2
By definition: a catalyst is consumed first then produced, so Cl is the catalyst. An intermediate is produced first then consumed, so ClO is the reaction intermediate.
3
Add the two steps to get total reactants and products: Left side: $C l + O_{3} + C l O + O$ , Right side: $C l O + O_{2} + C l + O_{2}$ .
4
Cancel species that appear on both sides: Cl and ClO cancel, leaving $O_{3} + O \to 2 O_{2}$ , which matches the given overall reaction.

Exam tip:

Always double-check the order of production/consumption to avoid confusing intermediates and catalysts on FRQ questions — this is one of the most commonly missed points on mechanism problems.

5. AP-Style Practice Problems★★★★☆⏱ 5 min

📐 Worked Example

Which of the following correctly gives the rate law and molecularity for the elementary reaction $2 N O + O_{2} \to 2 N O_{2}$ ?
A) Rate = $k [N O] [O_{2}]$ , unimolecular
B) Rate = $k [N O]^{2} [O_{2}]$ , bimolecular
C) Rate = $k [N O]^{2} [O_{2}]$ , termolecular
D) Rate = $k [N O_{2}]^{2}$ , termolecular

1
For elementary reactions, reaction orders equal the stoichiometric coefficients of reactants, so the rate law here must be $k [N O]^{2} [O_{2}]$ . This eliminates options A and D.
2
Molecularity counts the total number of reactant particles: 2 $N O$ + 1 $O_{2}$ = 3 total particles, so molecularity is termolecular. This eliminates option B.
3
The correct answer is C.

📐 Worked Example

The overall reaction $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ follows the three-step mechanism below:
Step 1 (fast): $N_{2} O_{5} ⇌ N O_{2} + N O_{3}$
Step 2 (slow): $N O_{2} + N O_{3} \to N O + N O_{2} + O_{2}$
Step 3 (fast): $N O + N O_{3} \to 2 N O_{2}$

(a) Identify all reaction intermediates in this mechanism.
(b) Write the rate law for the overall reaction consistent with this mechanism.
(c) What is the overall order of the reaction based on this mechanism?

1
(a) Track production and consumption: $N O_{3}$ is produced in step 1 and consumed in steps 2 and 3; $N O$ is produced in step 2 and consumed in step 3. Both are produced early and consumed late, so $N O_{3}$ and $N O$ are the reaction intermediates.
2
(b) The slow (rate-determining) step is elementary, so its rate law is:
3
$latex": "\text{rate} = k_2 [NO_2][NO_3]$
4
From the equilibrium in step 1, forward rate equals reverse rate: $k_{1} [N_{2} O_{5}] = k_{- 1} [N O_{2}] [N O_{3}]$ , so rearranged, $[N O_{2}] [N O_{3}] = \frac{k _{1}}{k _{- 1}} [N_{2} O_{5}]$ .
5
Substitute into the rate law for the slow step:
6
$latex": "\text{rate} = \frac{k_1 k_2}{k_{-1}} [N_2O_5] = k [N_2O_5]$
7
(c) The only reaction order is 1 for $N_{2} O_{5}$ , so the overall order of the reaction is 1.

📐 Worked Example

The radioactive decay of carbon-14, used for radiocarbon dating, is a unimolecular elementary first-order process with a rate constant $k = 1.21 \times 1 0^{- 4} year^{- 1}$ . A 10 g sample of ancient wood has an initial carbon-14 concentration of $1.65 \times 1 0^{- 10} mol/g$ . What is the initial rate of decay of carbon-14 in this sample, in moles per year?

1
Since decay is an elementary unimolecular reaction, the rate law is directly derived from stoichiometry: $rate = k [^{14} C]$ .
2
First find the total initial moles of C-14 in the 10 g sample: $10 g \times 1.65 \times 1 0^{- 10} mol/g = 1.65 \times 1 0^{- 9} mol$ .
3
Substitute into the rate law to get the initial rate:
4
$latex": "\text{rate} = (1.21 \times 10^{-4}\ \text{year}^{-1})(1.65 \times 10^{-9}\ \text{mol}) = 2.00 \times 10^{-13}\ \text{mol/year}$
5
This result matches the expected slow decay of ancient carbon-14 samples.

6. Common Pitfalls

Wrong move:

Using stoichiometric coefficients from an overall reaction to write the rate law, the same way you do for an elementary reaction.

Why:

Students generalize the rule for elementary reactions to all reactions, forgetting that only elementary steps have orders matching coefficients.

Correct move:

Always confirm the reaction is explicitly labeled as elementary before using coefficients to get reaction orders; for overall reactions, only use experimentally derived orders.

Wrong move:

Assigning a non-integer or zero molecularity to an elementary reaction.

Why:

Students mix up the definitions of molecularity (count of particles) and reaction order (can be any value).

Correct move:

Remember molecularity is only 1, 2, or 3 for elementary steps; eliminate any MCQ option with non-integer molecularity immediately.

Wrong move:

Leaving reaction intermediates in the final overall rate law derived from a mechanism.

Why:

Students forget intermediates are not stable species and must be substituted out using equilibrium expressions for fast pre-steps.

Correct move:

Always substitute out intermediate concentrations using expressions from fast equilibrium steps before writing the final rate law.

Wrong move:

Counting product particles to determine the molecularity of an elementary reaction.

Why:

Students count all particles in the equation instead of only reactants.

Correct move:

Only count the number of reactant particles on the left-hand side of the elementary step to find molecularity.

Wrong move:

Labeling a catalyst as a reaction intermediate in a multi-step mechanism.

Why:

Students mix up the order of production and consumption for the two species types.

Correct move:

Follow the rule: catalysts = consumed first, produced later; intermediates = produced first, consumed later.

7. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Elementary Step Rate Law	$rate = k \prod [X_{i}]^{n_{i}}, n_{i} = stoichiometric coefficient$	Only valid for elementary reactions; does not apply to overall reactions
Unimolecular Elementary Step	$A \to products, rate = k [A]$	Molecularity = 1, overall order = 1
Bimolecular (A + B)	$A + B \to products, rate = k [A] [B]$	Molecularity = 2, overall order = 2
Bimolecular (2A)	$2 A \to products, rate = k [A]^{2}$	Molecularity = 2, overall order = 2
Termolecular Elementary Step	$a A + b B + c C \to products, rate = k [A]^{a} [B]^{b} [C]^{c}$	Molecularity = 3, rare in mechanisms, overall order = 3
Molecularity	Count of reactant particles in an elementary step	Always 1, 2, or 3; never zero, negative, or fractional
Reaction Intermediate	Produced in early step, consumed in later step	Never included in final overall rate law
Catalyst	Consumed in early step, produced in later step	Not consumed overall; can appear in the rate law

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Identify molecularity of elementary step
2022 · FRQ
Derive overall rate law from mechanism
2021 · MCQ
Distinguish intermediates and catalysts

What's Next

Mastering elementary reactions is the foundational prerequisite for working with full reaction mechanisms, the next core topic in AP Chemistry Unit 5 Kinetics. Without understanding how to write rate laws for elementary steps and identify intermediates, you cannot derive the overall rate law for a multi-step mechanism, which is a common high-weight FRQ question on the AP exam. Beyond kinetics, understanding elementary steps helps you interpret collision theory and activation energy, because each elementary step has its own activation energy and Arrhenius behavior. This topic also builds the foundation for understanding biological and industrial catalysis, where catalysts work by providing a new sequence of elementary steps with lower activation energy.