5.4 Elementary Reactions

An elementary reaction happens in a single step, so its rate law can be read directly from the stoichiometry of the colliding particles. For an elementary step, coefficients become reaction orders, which is not true for an overall reaction unless the mechanism supports it. For AP Chemistry, use this shortcut only when the step is identified as elementary.

Elementary Reactions Summary

Elementary reactions are single-step reactions, so their rate laws can be written directly from the stoichiometry of the particles in that collision. If an elementary step is A + B → products, the rate law is rate = k[A][B]. If an elementary step is 2A → products, the rate law is rate = k[A]².

AP Chemistry Topic 5.4 is about that exception. For an elementary reaction, coefficients in the elementary step become exponents in the rate law. For an overall reaction, you still need experimental data because the overall equation may hide multiple elementary steps.

Why This Matters for the AP Chemistry Exam

This topic trains you to represent an elementary reaction as a rate law using stoichiometry, and to tell the difference between an elementary step and an overall reaction. That distinction shows up when you analyze kinetics data, write rate laws, and later evaluate reaction mechanisms. On free-response questions you may need to explain why data supports a proposed rate law, calculate a rate constant with correct units, or describe an experiment that tests whether a species affects the rate. Getting comfortable with elementary reactions now sets you up for collision theory, energy profiles, and mechanisms in the rest of Unit 5.

Key Takeaways

• For an elementary reaction, the rate law comes directly from the stoichiometry of the particles in the collision (the coefficients become the exponents).
• For an overall reaction, you cannot use coefficients as orders. Orders must be found from experiment.
• Molecularity describes how many particles collide in a step: unimolecular (one), bimolecular (two), termolecular (three).
• Elementary steps with three or more particles colliding at once are rare because that simultaneous collision is unlikely.
• The rate constant k depends on temperature, so all trials used to find a rate law must be run at the same temperature.
• Always include units on k, and match the time unit given in the problem.

Quick Review: What a Rate Law Is

Given a reaction A → B, the rate law is R = k[A]ⁿ, where R is the reaction rate, k is the rate constant, and n is the order of the reaction in A. The rate law shows that rate is proportional to reactant concentration raised to a power.

What an Elementary Reaction Is

An elementary reaction is a chemical reaction that occurs in a single step through a single collision event. It is the simplest type of step and the building block for understanding multistep reactions later in the unit.

The big idea for this topic: for an elementary reaction, the rate law can be inferred directly from the stoichiometry of the particles in the collision. This is a special exception, not the general rule.

• A → products has rate = k[A]
• A + B → products has rate = k[A][B]
• 2A → products has rate = k[A]²

Molecularity

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

• Unimolecular: one particle, rate = k[A]
• Bimolecular: two particles, rate = k[A][B] or k[A]²
• Termolecular: three particles

Elementary reactions involving the simultaneous collision of three or more particles are rare. Three particles meeting at the same place, same time, with the right orientation is unlikely, so most reactions proceed through a series of one- and two-particle steps instead.

The Key Warning: Overall Reactions Are Different

For an overall reaction (one that may actually proceed through several elementary steps), you cannot read the rate law from the coefficients. The only reliable way to find an overall rate law is through an experiment.

The reaction 2A → B is not automatically second order. It might be, but you only know after running trials. The standard method: run multiple trials at different reactant concentrations (usually doubling one at a time) and watch how the rate responds.

One firm rule: every trial must be run at the same temperature. If temperature changes, the rate can shift dramatically and give you wrong orders. This is because k is temperature dependent. Once you know the orders, plug in data from one trial to solve for k.

How Concentration Affects Rate

Suppose at 1 M the rate is 1 mol/(L·s), and at 2 M (same temperature) the rate is 4 mol/(L·s). Doubling the concentration quadrupled the rate. Using R = k[A]ⁿ, doubling [A] and quadrupling the rate means n = 2:

1 = k[A]ⁿ and 4 = k(2[A])ⁿ ==> 4 = k(4[A]²) ==> 1 = k[A]²

Worked Example: Finding an Overall Rate Law

Find the rate law for 2NO + 2H₂ → N₂ + 2H₂O using initial-rate data taken at 1280 °C.

Order with respect to NO

Between experiments 1 and 2, [NO] doubles while [H₂] stays constant. The rate increases by a factor of 4 (the 10⁻⁵ factors cancel). A doubling that quadruples the rate means the reaction is second order with respect to NO.

Order with respect to H₂

Between experiments 2 and 3, [H₂] doubles while [NO] stays constant. The rate increases by a factor of 2. A doubling that doubles the rate means the reaction is first order with respect to H₂.

Solving for k

The rate law is R = k[NO]²[H₂]. Plug values from one experiment into this expression to solve for k, then write out the full rate law with units. The hard part is finding the orders; the rest is substitution.

How to Use This on the AP Chemistry Exam

Free Response

When a question gives concentration-versus-time data and asks you to support a proposed rate law, look for the signature of each order:

• First order: a plot of ln[A] versus time is linear, and the half-life is constant.
• Second order: a plot of 1/[A] versus time is linear.
• Zeroth order: a plot of [A] versus time is linear.

For finding k, use the matching integrated rate law or the first-order half-life relationship t₁/₂ = 0.693/k. These relationships are on the reference sheet, but they are not labeled, so know what each one does.

Common Trap

Watch your units on k. The order sets the units, and the time unit must match the problem. If the data is in hours, your k is in h⁻¹, not s⁻¹.

Worked FRQ Example (2017 #2)

This comes from question 2 on the free-response section of the 2017 AP Chemistry exam.

Decomposition of urea: CO(NH₂)₂ (aq) ⇌ NH₄⁺ (aq) + OCN⁻ (aq), run at 90 °C. A student collects concentration data over time:

(e) The student proposes rate = k[CO(NH₂)₂]. (i) Explain how the data supports this. (ii) Determine k, with units.

(f) The reaction was run at pH 13. Describe an experiment, including the initial conditions you would change and the data you would gather, to test whether the rate depends on [OH⁻].

Part e, i

To support a first-order rate law, you can either show that a plot of ln[CO(NH₂)₂] versus time is linear (slope −k), or show that the half-life is constant.

Here the given graph is not linear, so check the half-life. The half-life is the time for a concentration to drop by half.

• Start: 0.1000 M. After one half-life, 0.0500 M remains.
• From the data, 0.0500 M occurs at 10 hours, so the first half-life is 10 hours.
• Half of 0.0500 M is 0.0250 M, which occurs at 20 hours. That is another 10 hours, so the half-life is constant.

A constant half-life indicates a first-order reaction, which supports the proposed rate law. The College Board scoring guidelines awarded one point for this explanation. A sample response:

Part e, ii

Since the reaction is first order with half-life 10 hours, use t₁/₂ = 0.693/k and solve for k. You could also use the first-order integrated rate law. Both equations are on the reference sheet.

Add units. First order means inverse time, and since the data is in hours, k is in h⁻¹, not s⁻¹. This unit mismatch trips up many students.

Part f

You only need a brief description. A response that supports a stronger score:

Common Misconceptions

• Coefficients always give the rate law. They do for an elementary step, but not for an overall reaction. Overall orders must be found experimentally.
• Molecularity and reaction order are the same thing. Molecularity describes how many particles collide in a single step. Order is the experimentally determined exponent in a rate law. They match for an elementary step, but not in general.
• k is just a fixed number. k depends on temperature, which is why trials must be run at the same temperature to find correct orders.
• Three-particle elementary steps are common. They are rare because a simultaneous, correctly oriented three-particle collision is very unlikely.
• Units of k are always s⁻¹. The units depend on the overall order and the time unit in the data. Match the problem's time unit and the reaction order.
• A linear concentration-versus-time graph means first order. A linear [A] versus time plot indicates zeroth order. First order is linear for ln[A] versus time; second order is linear for 1/[A] versus time.

Related AP Chemistry Guides

• Unit 5 Overview: Kinetics
• 5.1 Reaction Rates
• 5.2 Introduction to Rate Law
• 5.3 Concentration Changes Over Time
• 5.5 Collision Model
• 5.6 Reaction Energy Profile