To learn how to write out rate laws, let's refresh ourselves on what exactly a rate law is. Given a reaction A → B, the rate law for this reaction 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 us that rate is directly proportional to the concentration of the reactants, or in this case just the single reactant.
Rate laws are simply what they look like: mathematical expressions to find the rate of a reaction. However, given any reaction, is there a way to find exactly what the rate law is? Let's find out!
👉 Want to review more about rate laws before moving on? Check out this study guide all about them.
Rate Laws Found Experimentally
Unfortunately, the only true way that rate laws can be found is through an experiment. It's a common mistake of chemistry students to look at the stoichiometric coefficients of the reactants and use those as reaction orders, but you cannot do that!
The reaction 2A → B is not necessarily second order, it could be, but we won't know until we run an experiment. The experiment run to find a rate law is quite simple. All it involves is running multiple trials of a reaction with different concentrations of each reactant, typically doubling one, and then seeing how the rate reacts to this change in concentration.
One important stipulation is that these reactions must and I mean must be run at the same temperature. If the temperature fluctuates, the rate can change dramatically, giving incorrect results. This is because k, the rate constant, is temperature dependent. Once you find the orders of a reaction, you can plug in values from an experiment to find k.
How Does Concentration Affect Rate?
You might be wondering, how does doubling the concentration change the rate? Well, let's take a look at a general example. Let's say that when the concentration of the reactant is 1 M, the rate of the reaction is 1 mol/Ls, but when the concentration of the reactant is 2 M, the rate of reaction, at the same temperature, is 4 mol/Ls. This means that by doubling the concentration, the rate quadrupled. This is a quadratic effect. Essentially, if R = k[A]ⁿ and doubling [A] leads to a quadrupling in rate, n must be two. We can see this through some simple algebra:
1 = k[A]ⁿ and 4 = k(2[A])ⁿ ==> 4 = k(4[A]²) ==> 1 = k[A]²
As a general rule of thumb: if doubling [] leads to rate doubling, it's first order, quadrupling rate means second order, octupling rate means 3rd order, etc., though it's rare to see more than 2nd order reactions.
Example Problem
Find the rate law for the reaction 2NO + 2H₂ → N₂ + 2H₂O given the reaction data below at 1280 °C.
Finding Order of NO
Between experiments 1 and 2, [NO] doubles and [H₂] remains constant, so we can use these to find the order of the reaction with respect to NO. When [NO] is doubled, the rate increases by a factor of 5.0/1.25, which is 4 (we can cancel out the 10⁻⁵ here). Therefore, the reaction is second order with respect to NO.
Finding Order of H₂
Similarly, between experiments 2 and 3, [NO] remains constant and [H₂] doubles. Therefore, we can do the same as before: [H₂] doubles, and the rate increases by a factor of 1 * 10⁻⁴/5 * 10⁻⁵ = 2. Thus, the reaction is first order with respect to H₂.
Finding k and putting it all together
We now know the rate law is in the form R = k[NO]²[H₂]. Now, by plugging in values you can find k and write out the full rate law. Give it a try! The first two steps really are the hard parts.
Elementary Reactions
Okay, but wait...this key topic is titled "Elementary Reactions," but I don't even know what those are yet. An elementary reaction is a chemical reaction that occurs in a single step and involves only a single molecule or a group of atoms. It is the most basic type of chemical reaction and is the starting point for understanding more complex reactions. Everything in this study guide applies to an elementary reaction, and we will discuss more complex reactions throughout the rest of this unit.
Key Properties of Elementary Reactions
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Rate laws from stoichiometry: Unlike most reactions, the rate law of an elementary reaction CAN be inferred from the stoichiometry of the particles participating in a collision. This is a special exception! For elementary reactions:
- A → products has rate = k[A]
- A + B → products has rate = k[A][B]
- 2A → products has rate = k[A]²
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Molecularity limitations: Elementary reactions involving the simultaneous collision of three or more particles are rare. This is because:
- The probability of three particles colliding at the exact same time and place is very low
- Most reactions proceed through a series of bimolecular (two-particle) steps instead
- Termolecular (three-particle) elementary reactions do exist but are uncommon
As we've seen, elementary reactions can be either first-order or second-order, depending on whether the rate of the reaction is dependent on the concentration of one species or two. Some specific examples of elementary reactions include the reaction of hydrogen and oxygen to form water, the decomposition of ozone, and the ionization of a gas.
AP Practice Question - 2017 #2
The following question was posted by College Board and is part of question two on the free-response section of the 2017 AP Chemistry exam.
The ammonium salt of isocyanic acid is a product of the decomposition of urea, CO(NH₂)₂, represented by the following chemical equation: CO(NH₂)₂ (aq) ⇌ NH₄⁺ (aq) + OCH⁻ (aq)
A student studying the decomposition reaction runs the reaction at 90°C. The student collects data on the concentration of urea as a function of time, as shown by the data table and the graph below.
| Time (hours) | [CO(NH₂)2] |
|---|---|
| 0 | 0.1000 |
| 5 | 0.0707 |
| 10 | 0.0500 |
| 15 | 0.0354 |
| 20 | 0.0250 |
| 25 | 0.0177 |
| 30 | 0.0125 |
(e) The student proposes that the rate law is rate = k[CO(NH₂)₂]. (i) Explain how the data supports the student's proposed rate law. (ii) Using the proposed rate law and the student's results, determine the value of the rate constant, k. Include units in your answer.
(f) The student learns that the decomposition reaction was run in a solution with a pH of 13. Briefly describe an experiment, including the initial conditions that you would change and the data you would gather, to determine whether the rate of the reaction depends on the concentration of OH⁻(aq).
Answer to FRQ 2017 #2e-f
Part e, i
For the first part of part e, we are trying to find evidence of this reaction being a first-order reaction. There are two ways we can do that:
- Look for a graph of ln[CO(NH₂)₂] vs. time and identify it as linear, with a slope of -k.
- Identify the half-life of the reaction as constant. Both of these come from content that we reviewed in the last study guide, so be sure to check that out!
Here, we are given a graph, but it is not linear. Therefore, we have to take a look at the half-life. To recall, the half-life is defined as the time it takes for the concentration of a substance to decrease by half. The easiest way to do this is to think:
- Okay so before the experiment started, the concentration of CO(NH₂)₂ was 0.1000 M. After one half-life, there would be 0.0500 M of CO(NH₂)₂ remaining.
- How long did it take for there to be 0.0500 M of CO(NH₂)₂? According to the provided graph, it took 10 hours.
- To check if something is constant, we have to see if the next half-life is 10 hours as well. Half of 0.0500 M is 0.0250 M, which is the concentration of CO(NH₂)₂ left after 20 hours. It took 10 hours to half its concentration again, meaning the half-life of the reaction is constant! This confirms the student's rate law.
The scoring guidelines posted by College Board gave one point for a correct explanation. Below is a sample response:
"From inspecting the data table or the graph, it is evident that the decomposition reaction has a constant half-life, which indicates that the reaction is a first-order reaction.
Part e, ii
Now that we've got our confirmed rate law, we have to solve for k. In order to do this, you can use the information you've gathered from the last part: this reaction is first-order and its half-life is 10.
Remember, t1/2 = 0.693/k, so all we have to do is plug 10 in for t1/2 and solve for k. There is another way to do this problem just in case you forget about this method. You can rather solve for k by using the integrated rate law for a first-order reaction. Both of these equations are on the given reference sheet, but they are not labeled for you.
Once you get your answer, be sure to add a unit! Since this is a first-order reaction, you know that the units should be s⁻¹. But this is where they trick many students. The time, in this question, was in the units of hours, so your units must be hours⁻¹, or h⁻¹. Below is a sample response provided by College Board.
Part f
This part may seem intimidating since it is asking you to describe an experiment, but you only have to do so briefly. As long as you give them what they ask for, you can gain full credit. The following is a sample response that would earn a full point of credit:
"Perform the experiment at a different concentration of OH⁻ (aq) and measure how the concentration of CO(NH₂)₂ changes over time. Other variables, such as temperature, should be held constant."
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| collision | The event in which reactant particles come together with sufficient energy and proper orientation to form products. |
| elementary reaction | A single-step reaction that represents one molecular event in a reaction mechanism, with a specific rate law determined by its molecularity. |
| rate law | A mathematical expression that relates the reaction rate to the concentrations of reactants, with each concentration raised to a power (order). |
| stoichiometry | The quantitative relationship between reactants and products in a balanced chemical equation that determines the rates of change of their concentrations. |
Frequently Asked Questions
What is an elementary reaction in chemistry?
An elementary reaction is a single-step process where reactant particles collide and convert to products in one event—its stoichiometry tells you the rate law directly. For a unimolecular step A → products, rate = k[A]; for a bimolecular step A + B → products, rate = k[A][B]; termolecular (three-body) collisions are very rare (CED 5.4.A). Molecularity = the number of particles that must collide simultaneously (unimolecular, bimolecular, termolecular). Collision theory and transition-state ideas explain why rate constants include activation energy and frequency factors: only collisions with proper orientation and enough energy form the activated complex. On the AP exam you may be asked to write rate laws for elementary steps or relate molecularity to order (use stoichiometry of that elementary step). For more examples and practice, see the Topic 5.4 study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg), the Unit 5 overview (https://library.fiveable.me/ap-chemistry/unit-5), and practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do you write a rate law for an elementary reaction?
For an elementary reaction the rate law comes directly from the stoichiometry (molecularity) of that single step—the exponent for each reactant equals the number of molecules of that species in the collision. So: - Unimolecular (A → products): rate = k[A] (first order) - Bimolecular (A + B → products): rate = k[A][B] (second order overall) - Bimolecular (2A → products): rate = k[A]^2 Termolecular (three-body) elementary steps are very rare, so you won’t usually see rate = k[A][B][C] on the exam. This matches CED Essential Knowledge 5.4.A.1–.2: infer rate laws from collision stoichiometry and note termolecular rarity. For AP practice, review the Topic 5.4 study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg) and try problems in the Unit 5 overview (https://library.fiveable.me/ap-chemistry/unit-5) or the 1000+ practice questions (https://library.fiveable.me/practice/ap-chemistry).
What's the difference between elementary reactions and overall reactions?
Elementary reactions are single-step collisions or transformations—an event where reactant particles collide and directly form products or an activated complex. For an elementary step you can write its rate law from the stoichiometry (molecularity): unimolecular -> rate = k[A], bimolecular -> rate = k[A][B] or k[A]^2, and termolecular steps are rare (CED 5.4.A). Each elementary step has its own activation energy and frequency factor (collision/transition-state ideas). An overall reaction is the sum of all elementary steps in a mechanism. The stoichiometric coefficients in the overall equation do NOT necessarily give the rate law; the observed rate law depends on the mechanism and usually the rate-determining step (slow step) controls the kinetics. So: elementary = single step, rate law from stoichiometry; overall = sum of steps, rate law must be determined experimentally or from the mechanism (using the slow step). For more review and examples see the Topic 5.4 study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg) and Unit 5 overview (https://library.fiveable.me/ap-chemistry/unit-5). Practice problems are at (https://library.fiveable.me/practice/ap-chemistry).
Why can you use stoichiometry to find the rate law for elementary reactions but not for overall reactions?
You can use stoichiometry for an elementary reaction because an elementary step describes a single collision event—its molecularity tells you how many molecules collide simultaneously, and those same species appear in the rate law. For example, a bimolecular elementary step A + B → products gives rate = k[A][B]; the exponents match the collision stoichiometry (CED 5.4.A.1, keywords: elementary reaction, molecularity). Termolecular steps (three-body collisions) are rare, so you usually see unimolecular or bimolecular steps (CED 5.4.A.2). An overall balanced equation, however, can hide the mechanism: it’s the net result of one or more elementary steps. The overall stoichiometric coefficients don’t tell you which step is rate-determining or whether intermediates are involved, so you can’t infer exponents from them. To get the overall rate law you need the mechanism or experimental data (or identify the slow step and use steady-state/pre-equilibrium approximations). For a quick refresher, see the Topic 5.4 study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg) and grab practice problems at (https://library.fiveable.me/practice/ap-chemistry).
I'm confused about how to go from the balanced equation of an elementary reaction to its rate law - can someone explain?
If the step is truly elementary, its rate law comes straight from its molecularity (the stoichiometry of colliding particles). For an elementary reaction like A + B → products, rate = k[A][B] (first order in A and B). For 2A → products, rate = k[A]^2 (second order overall, because two A collide). Unimolecular steps (A → products) give rate = k[A]. Termolecular (three-body) steps are rare, so don’t expect them often. This rule is only valid for single elementary steps—for multi-step mechanisms you can’t assume the overall stoichiometric coefficients give the rate law; experiments or steady-state/equilibrium assumptions are needed. The AP CED expects you to infer rate laws from elementary-step stoichiometry (5.4.A). For more examples and quick review, see the Topic 5.4 study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg) and lots of practice problems at (https://library.fiveable.me/practice/ap-chemistry).
Why are elementary reactions with three or more particles colliding at the same time so rare?
Because an elementary step happens only when the reacting particles actually collide at the same time with the right orientation and enough energy, the chance that three (or more) particles meet those conditions simultaneously is extremely small. Collision theory + molecularity: bimolecular (two-body) collisions are common enough to control rates, but termolecular collisions require three-body encounters—much lower collision frequency, plus an extra constraint on orientation and activation energy for forming the activated complex. Practically, most “termolecular” steps are really two fast bimolecular steps with a short-lived intermediate. On the AP exam you should name molecularity (uni-, bi-, termolecular), relate it to the rate law for an elementary step, and explain collision frequency/orientation/activated-complex ideas (CED 5.4.A.1–2). For a quick topic review, see the Topic 5.4 study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do you know if a reaction mechanism step is elementary or not?
An elementary step is one single collision/event—so you can tell it’s elementary if the step’s chemical equation shows the actual particles that collide and the stoichiometric coefficients equal the molecularity. For an elementary step: - Write the rate law from stoichiometry (unimolecular: rate = k[A]; bimolecular: rate = k[A][B] or k[A]^2). This is exactly what the CED expects (5.4.A). - If a step would require three or more species colliding simultaneously, it’s probably not elementary (termolecular steps are very rare). - Check mechanism consistency: an elementary step shouldn’t create or consume a stable intermediate without showing it, and the proposed elementary-step rate law must match the experimentally determined rate law. If the predicted rate law disagrees with experiments, that step isn’t elementary or the mechanism is wrong. For more examples and practice (recommended for the AP exam), see the Topic 5.4 study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg), the Unit 5 overview (https://library.fiveable.me/ap-chemistry/unit-5), and hundreds of practice problems (https://library.fiveable.me/practice/ap-chemistry).
What does it mean when they say elementary reactions involve particle collisions?
Saying elementary reactions “involve particle collisions” means the step actually happens when the reacting molecules (or atoms/ions) physically collide: for that collision to produce products it must have the right orientation and enough kinetic energy to get over the activation energy (collision theory/transition state). For an elementary step the stoichiometry tells you the molecularity—e.g., unimolecular (A → products) comes from a single-particle event, bimolecular (A + B → products or A + A → products) from a two-particle collision, and termolecular (three-body) events are very rare. Because an elementary step is a single collision event, its rate law follows directly from molecularity (rate = k[A] for unimolecular, rate = k[A][B] or k[A]^2 for bimolecular), which is exactly what LO 5.4.A in the CED expects you to use on the exam. For a quick topic review, see the Topic 5.4 study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
Can you have an elementary reaction where four molecules collide simultaneously?
Short answer: practically no—a simultaneous four-molecule collision as a true elementary step is effectively impossible in most conditions. Why: elementary-step molecularity equals the number of particles that collide in that elementary event. Unimolecular and bimolecular steps are common; termolecular (three-body) collisions are already very rare because collision theory says the probability of three particles meeting with correct orientation and enough energy at the same instant is tiny. A four-body collision would be orders of magnitude less likely. For reactions that look overall like four reactants, the mechanism is almost always a sequence of unimolecular/bimolecular elementary steps (so the rate law must be determined experimentally unless the step is truly elementary). For AP exam framing: CED 5.4.A.2 notes three-or-more particle collisions are rare—expect unimolecular, bimolecular, and occasional termolecular examples on the exam. For a quick review, see the Topic 5.4 study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg) and more practice at (https://library.fiveable.me/practice/ap-chemistry).
Why do most elementary reactions only involve one or two particles?
Most elementary steps are unimolecular or bimolecular because of collision statistics and the collision model. An “elementary reaction” is a single collision event, so its molecularity equals the number of particles that must collide simultaneously. Two-body collisions are common because the chance that two molecules hit each other with correct orientation and enough energy (≥ Ea) is reasonably high; three-body (termolecular) collisions require three particles to meet at once with correct geometry and energy, which is extremely unlikely, so those steps are rare. That’s why for elementary steps the rate law follows the stoichiometry (unimolecular → rate ∝ [A]; bimolecular → rate ∝ [A][B] or [A]^2), but you almost never assume third-order elementary steps on the AP exam. For more on molecularity, collision frequency, and examples, see the Topic 5.4 study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg). For extra practice, check Unit 5 resources and practice questions (https://library.fiveable.me/ap-chemistry/unit-5 and https://library.fiveable.me/practice/ap-chemistry).
How do I write the rate law expression if I know the stoichiometry of an elementary reaction?
If a step is truly elementary, its stoichiometry equals its molecularity, so the rate law follows directly from the number of reacting particles in that step (CED 5.4.A). Practically: - Unimolecular: A → products → rate = k[A] (first order) - Bimolecular: A + B → products → rate = k[A][B] (second order overall) - Bimolecular (same species): 2A → products → rate = k[A]^2 - Termolecular: A + B + C → rate = k[A][B][C] (rare in reality) Molecularity = number of molecules colliding simultaneously and determines the algebraic form of the rate law for that elementary step. If a mechanism has multiple steps, only the rate-determining elementary step controls the observed rate law. This is exactly what AP expects you to do for Topic 5.4 (see CED 5.4.A). For a quick refresher, check the Topic 5.4 study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg) and grab extra practice problems at (https://library.fiveable.me/practice/ap-chemistry).
What's the connection between reaction mechanisms and elementary reactions?
A reaction mechanism is just the step-by-step story made of elementary reactions. An elementary reaction is a single molecular event (unimolecular, bimolecular, rarely termolecular) whose rate law follows directly from its stoichiometry (CED 5.4.A: rate ∝ [reactants]^(molecularity)). A mechanism strings those elementary steps together; the observed overall rate law usually reflects the slowest step (the rate-determining step), not the overall balanced equation. So, to propose or test a mechanism you (1) write plausible elementary steps, (2) use stoichiometry of those steps to write their rate laws, and (3) compare the predicted overall rate law to experimental data—if they match, the mechanism is consistent. Remember collision/transition-state ideas explain why termolecular steps are rare and why activation energy and frequency factors matter. For more practice and AP-specific review of Topic 5.4, see the Fiveable study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg) and tons of practice problems (https://library.fiveable.me/practice/ap-chemistry).
I don't understand why elementary reactions are different from the reactions we usually study - what makes them special?
Elementary reactions are special because they show a single collision or event—one step in a mechanism—so their rate law comes directly from the stoichiometry (molecularity). For an elementary unimolecular step A → products, rate = k[A]; for a bimolecular step A + B → products, rate = k[A][B]. That’s exactly what the CED expects you to use (5.4.A.1). By contrast, the overall balanced equation you usually see can hide multiple elementary steps, so you can’t infer its rate law from overall stoichiometry alone. Key reasons they’re different: they describe an actual collision/transition-state event (collision theory, activated complex, activation energy, frequency factor), and true termolecular (three-body) elementary steps are very rare (5.4.A.2) because simultaneous three-particle collisions are unlikely. On the AP, be ready to write rate laws for elementary steps from stoichiometry and to recognize when a reaction is a multi-step mechanism. Review the Topic 5.4 study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg) and the Unit 5 overview (https://library.fiveable.me/ap-chemistry/unit-5); practice problems are at (https://library.fiveable.me/practice/ap-chemistry).
How do you determine the order of an elementary reaction from its equation?
If the step is elementary, its rate law comes straight from the stoichiometry of the reacting particles (molecularity). Count how many molecules collide in that single step—that tells you the reaction order for that step. - Unimolecular (A → products): rate = k[A] (first order) - Bimolecular (A + B → products): rate = k[A][B] (second order overall) - Bimolecular (2A → products): rate = k[A]^2 (second order) - Termolecular (A + B + C → …): rate = k[A][B][C] (third order)—but termolecular steps are rare (CED 5.4.A.2). Important: this only applies to elementary steps (CED 5.4.A.1). For an overall balanced equation that isn’t an elementary step, you can’t infer the rate law from stoichiometry—you need experimental data or a proposed mechanism with an elementary rate-determining step. For more on Topic 5.4, see the AP study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg) and grab practice problems at (https://library.fiveable.me/practice/ap-chemistry).
Why do we even need to know about elementary reactions for the AP exam?
You need elementary reactions because the AP expects you to connect microscopic steps to macroscopic rate laws and mechanisms. For an elementary step, the rate law follows directly from its stoichiometry (molecularity): unimolecular → rate ∝ [A], bimolecular → rate ∝ [A][B] or [A]^2. That lets you (a) write/interpret rate laws from mechanisms, (b) spot which steps are plausible (termolecular are rare), and (c) link collision/transition-state ideas to activation energy and frequency factor—all CED keywords. Questions on Unit 5 (Kinetics) test these skills on both MC and FR items (use science practices like Models & Representations and Mathematical Routines). Review the Topic 5.4 study guide (https://library.fiveable.me/ap-chemistry/unit-5/elementary-reactions/study-guide/SPsFzzECb4aCre0wFrGg) and then drill problems from the practice bank (https://library.fiveable.me/practice/ap-chemistry) to get fluent at inferring rate laws and evaluating mechanisms.