So far in this unit, we've been discussing the kinetics of only elementary reactions. These occur in a single step and typically involve only a single molecule or a group of atoms. These are the most basic types of chemical reactions and a starting point for understanding more complex reactions.
Now that we've got the basics down, let's get into the kinetics of complex reactions (which are more of what happens in real life).
What is a Mechanism?
Reaction mechanisms go more in-depth regarding chemical reactions than just a net equation. For example, a reaction between an acid and a base must happen in the presence of water, yet water is not included as a reactant. The actual reaction happens in two steps.
Elementary Steps
A mechanism basically takes a full reaction and splits it up into elementary steps. These steps essentially show what actually happens in a reaction. Let's take a look at the reaction mechanism for the decomposition of hydrogen peroxide (H₂O₂):
Rather than trying to understand the reaction by looking at the net equation, we can break it down to the level of its elementary steps. You can think of elementary steps as the smallest unit of a reaction that can be studied, as they are simple and individual steps that make up a reaction that result in a certain product.
As we can see, the decomposition of hydrogen peroxide occurs in two elementary steps:
- In the first step, hydrogen peroxide molecules react with iodide to form water molecules and iodite ions. We can see that one of the oxygen atoms from hydrogen peroxide reacts with the iodide to form iodite ions.
- In the next step, we have hydrogen peroxide reacting with the product of the last step: the iodite ions. In this process, another oxygen atom reacts, but with iodite this time, to form more water molecules, oxygen gas, and iodide. Try not to think of chemical reactions as mathematical equations. Imagine we are completing this reaction in a beaker. We have tons of hydrogen peroxide molecules reacting with tons of iodide ions, which is why hydrogen peroxide can be a reactant in both steps.
When we add up all of the individual elementary steps and cancel out the spectators, we end up with the overall balanced chemical equation (or net equation).
Why is this important, you may ask? Well, if we find the total reaction by adding the elementary steps, we see that IO⁻ and I⁻ cancel out to leave 2H₂O₂ → 2H₂O + O₂. So wait...why are the iodite and iodide ions even there? Let's get into that!
Catalysts and Intermediates
Since elementary steps show what is happening at a molecular level, their components may involve intermediates and catalysts as well as reactants and products.
Iodide as a Catalyst
In the reaction above, iodide acts as a catalyst. We'll discuss how catalysts work in chemical reactions in more depth later in this unit, but let's cover the basics. A catalyst is essentially a substance that increases the rate of a chemical reaction without being consumed in the process. It speeds up the reaction without being affected by it, as it goes in and comes out the exact same way.
Since the catalyst is not in the overall balanced equation, we can see that it does not impact the reaction products, but rather impacts the mechanism by changing the way the reaction occurs. The natural decomposition of H2O2 actually occurs, but it is incredibly slow. In a lab, we can't always wait, so sometimes we use catalysts to speed things up.
Iodite as an Intermediate
So, iodide is a catalyst, but what is iodite? The iodite ion produced in the first elementary step, and then consumed in the next, is an intermediate. Intermediates are species that are formed during a reaction and then go on to participate in further reactions. They are not reactants or products, but are present in the reaction mixture in significant concentrations only while the reaction itself is occurring.
Mechanisms and Rate Laws
Something we haven't discussed yet is that each elementary step has its own rate constant and activation energy. What can this tell us when looking at a reaction mechanism?
Rate-Determining Steps
Often when dealing with mechanisms, you will see one elementary step labeled "slow" and the others labeled "fast" since they have their own respective rate constants. The slow step is also called the rate-determining step as it defines the rate law of the overall reaction. This should make sense! A reaction can only proceed as fast as its slowest step.
💡Remember: The rate-determining step is the slowest step in a reaction mechanism and controls the overall rate of the reaction. The overall rate of a reaction is equal to the rate of the rate-determining step.
For a conceptual understanding, in a multi-step reaction, the rate-determining step is typically the step that has the highest activation energy. Since the rate of a reaction is directly proportional to the frequency of successful collisions between reactant molecules with enough energy to overcome the activation energy barrier, a step with a higher activation energy will have a lower rate constant and, therefore, will be slower than other steps. Thus, the rate-determining step sets the pace for the entire reaction and all other steps must keep up with it.
Example of Rate-Determining Step
Let's go through this two-step reaction mechanism to better understand this concept and calculate the rate law of the overall reaction.
The first step to calculating the rate law of the overall reaction is to actually figure out what the overall reaction is. If you add up both elementary steps and then cancel out the intermediate (HI), you should get H₂ + 2ICl → I₂ + 2HCl.
The next step is to figure out which elementary step is the rate-determining step of the reaction. This is pretty easy; just find the elementary step that is slow! In this mechanism, it is the first elementary step.
Now, it is time to calculate the rate law of the first elementary step since it is the rate-determining step. When dealing with elementary steps (and only elementary steps), we can use the stoichiometric coefficients to tell us the order for each reactant. Thus, looking at the slow reaction, we know that the rate law of the overall reaction is: R = k[H₂][ICl]
You can see why this concept can lead to lots of errors. If a student were to forget to use the rate law of the rate-determining step, and rather calculated it by looking at the overall balanced equation, they would have had an exponent of two for ICl.
When the Slow Step Has Intermediates
An important thing to note is that sometimes the rate-determining step will have an intermediate in it. In this case, you will need to use some math to make a substitution since you cannot have an intermediate in your rate law. This math involves a topic in chemistry that you most likely learned called equilibrium. If you have, what you do is you essentially use the Keq of one of the elementary steps (typically one will be in "fast equilibrium") and use some substitutions. To see what this looks like, check out the next study guide!
Example Mechanism
The following mechanism was actually part of the 2019 AP Chemistry examination. Let's try to find the rate law of the overall reaction!
To find the rate law for this mechanism, we look to the slow, rate-limiting step. We find the rate-limiting step to be step one. By using the stoichiometric coefficients (which again, we can ONLY do with elementary steps), we find the rate law to be R = k[NO₂][NO₂], or once you simplify it, R = k[NO₂]².
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| balanced chemical equation | A chemical equation where the number of atoms of each element is equal on both the reactant and product sides. |
| catalyst | A substance that increases the rate of a chemical reaction without being consumed in the reaction. |
| elementary reaction | A single-step reaction that represents one molecular event in a reaction mechanism, with a specific rate law determined by its molecularity. |
| product | Substances formed as a result of a chemical reaction. |
| reactant | Substances that are consumed in a chemical reaction to form products. |
| reaction intermediate | A species that is produced in one elementary step of a reaction mechanism and consumed in a subsequent step, not appearing in the overall reaction. |
| reaction mechanism | The sequence of elementary steps that describes how a reaction proceeds at the molecular level. |
Frequently Asked Questions
What is a reaction mechanism and why do we need to know about it?
A reaction mechanism is the step-by-step sequence of elementary reactions that converts reactants to products—each step shows which bonds break/form, what intermediates and catalysts appear, and the molecularity (uni-, bi-, termolecular) of that step. We care because the mechanism explains the observed rate law and which step is rate-determining (RDS); elementary steps let you connect molecular collisions and activation energies to the macroscopic rate. Intermediates are produced and consumed during the mechanism (so they only exist while the reaction proceeds), and detecting them experimentally provides evidence for one mechanism over another (though detection data won’t be tested on the AP Exam, per the CED). Knowing mechanisms also helps understand catalytic cycles and how changing concentration, temperature, or catalysts will change rates. For a focused review, see the Topic 5.7 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK) or the Unit 5 overview (https://library.fiveable.me/ap-chemistry/unit-5), and practice problems at (https://library.fiveable.me/practice/ap-chemistry).
How do elementary steps work in chemistry reactions?
An elementary step is one single molecular event in a reaction mechanism—think of it as one collision or one unimolecular change. Each step is written like a balanced chemical equation and has a molecularity: unimolecular (involves 1 species), bimolecular (2), or termolecular (3, very rare). Elementary steps combine in sequence to give the overall balanced equation (CED 5.7.A.2). Intermediates are formed in some steps and consumed in later ones (CED 5.7.A.3); catalysts appear and are regenerated. Importantly, the rate law for an elementary step follows directly from its molecularity (e.g., a bimolecular step → rate ∝ [A][B]). The slowest elementary step is the rate-determining step and controls the observed rate law. Experimental detection of an intermediate supports a mechanism, though detecting them isn’t tested on the AP exam (CED note). For a clear walkthrough and examples, see the Topic 5.7 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK) and more practice at Fiveable (https://library.fiveable.me/practice/ap-chemistry).
What's the difference between a reaction intermediate and a regular product?
A reaction intermediate is a species that’s formed in one elementary step and then consumed in a later step of the mechanism—it exists only while the reaction is occurring. A “regular” product is formed in the mechanism and remains as a final product of the overall balanced equation. In CED terms: intermediates appear in the sequence of elementary reactions but cancel out when you add the steps to get the overall equation (EK 5.7.A.3); products survive to the end and show up in the overall stoichiometry. Intermediates can affect the mechanism and rate law (they’re often involved in the rate-determining step), and experimentally detecting an intermediate gives evidence for a proposed mechanism (though the AP exam won’t test you on collecting that detection data). For a quick refresher on mechanisms and intermediates, see the Topic 5.7 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
I'm confused about how elementary steps add up to make the overall equation - can someone explain?
Think of a mechanism as a short story told in scenes (elementary steps). To get the overall reaction you simply add the steps algebraically: write each elementary equation, cancel any species that appear on both sides (those are intermediates), and add remaining reactants/products. The summed equation must match the overall balanced equation (CED 5.7.A.2). Example: Step 1: A + B → I ; Step 2: I + C → D. Add them, cancel I, overall: A + B + C → D. Coefficients matter—multiply whole steps if needed to make things cancel stoichiometrically. Remember: intermediates are produced then consumed (CED 5.7.A.3), catalysts appear then are regenerated, and the slowest elementary step (rate-determining step) often controls the observed rate law. Mechanisms must be consistent with both stoichiometry and experimental kinetics (detection of an intermediate helps support a mechanism, CED 5.7.A.4). For extra examples and practice, see the Topic 5.7 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK) and more practice questions (https://library.fiveable.me/practice/ap-chemistry).
Why do reaction intermediates disappear during the reaction?
Intermediates disappear because they’re created in one elementary step and then used up in a later step. By definition (CED 5.7.A.3), a reaction intermediate is produced transiently and consumed before the overall reaction finishes, so it only exists while the mechanism is happening. That’s different from a transition state (an instantaneous high-energy configuration)—intermediates are real, lower-energy species you could sometimes detect. Experimental detection of an intermediate gives evidence for a particular mechanism (CED 5.7.A.4), though detection details aren’t tested on the AP exam. If you need to practice identifying intermediates or writing mechanisms, check the Topic 5.7 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK) and pump problems from the Unit 5 page (https://library.fiveable.me/ap-chemistry/unit-5) or the 1000+ practice questions (https://library.fiveable.me/practice/ap-chemistry).
What are all the components that can be in a reaction mechanism?
A reaction mechanism can include: the sequence of elementary steps (each an elementary reaction), the reactants and final products, reaction intermediates (formed in one step and consumed in another), catalysts (including a catalytic cycle), and the rate-determining step. Other important components/ideas you’ll see in mechanisms are molecularity of elementary steps (unimolecular, bimolecular, termolecular), transition states/activated complexes, activation energies, and reaction coordinate diagrams. Mechanistic rate laws, the steady-state and pre-equilibrium approximations, and the idea of stoichiometric consistency also connect mechanisms to observed kinetics. Note: experimental detection of an intermediate helps support a mechanism, but collection of that data is excluded from AP exam assessment per the CED. For more on this topic see the Topic 5.7 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK), unit overview (https://library.fiveable.me/ap-chemistry/unit-5), and tons of practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do scientists actually detect reaction intermediates in real experiments?
Intermediates are often short-lived and low in concentration, so chemists use fast, sensitive methods to catch them. Common approaches: time-resolved spectroscopy (ultrafast lasers can follow events in femto–picoseconds; stopped-flow/flash photolysis works in ms–μs), cryogenic trapping (freeze the mixture to “trap” species), rapid mixing and quench experiments, mass spectrometry (often with soft ionization), ESR for radicals, and NMR or IR if the intermediate lives long enough. Kinetic evidence also helps: measured rate laws, pre-equilibrium or steady-state analyses, and isotope labeling can show an intermediate must exist to match observed kinetics (CED 5.7.A.3–5.7.A.4). Remember: on the AP exam you won’t be asked to collect intermediate-detection data, but knowing these techniques helps you evaluate mechanisms (see the Topic 5.7 study guide for more: https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK). For extra practice on kinetics and mechanism questions, try the practice set (https://library.fiveable.me/practice/ap-chemistry).
Can you give me a simple example of a reaction mechanism with intermediates?
A simple textbook example is the two-step mechanism for the reaction: A + B → C (overall). Step 1 (fast equilibrium, bimolecular): A + B ⇌ I (I is an intermediate) Step 2 (slow, unimolecular): I → C Here I is produced in step 1 and consumed in step 2, so it’s a reaction intermediate (CED 5.7.A.3). The elementary steps add to the overall balanced equation (CED 5.7.A.2). If step 2 is slower, it’s the rate-determining step and controls the observed rate law—often the mechanistic rate law uses the concentration of I, so you’d use the pre-equilibrium or steady-state idea to eliminate I and get a rate law in terms of A and B (CED keywords: elementary reaction, unimolecular/bimolecular, intermediate, rate-determining step). Experimental detection of I supports a mechanism (CED 5.7.A.4), though detection methods aren’t tested on the AP exam. For more examples and practice, see the Topic 5.7 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK) and the unit review (https://library.fiveable.me/ap-chemistry/unit-5).
What's the difference between a catalyst and a reaction intermediate?
A reaction intermediate is a species that’s formed in one elementary step and then consumed in a later step—it exists only while the reaction is happening and does not appear in the overall balanced equation (CED 5.7.A.3). A catalyst, by contrast, speeds up a reaction without being used up: it may appear in a step, change form in the mechanism, then be regenerated by a later step so it’s present at the start and end (think catalytic cycle). Key differences: intermediates are produced then consumed; catalysts are regenerated. Both show up in proposed mechanisms (elementary steps must add to the overall equation, CED 5.7.A.1–2). Experimentally detecting an intermediate helps support a mechanism (useful for AP reasoning), but note that detailed experiments detecting intermediates won’t be tested on the AP exam (CED exclusion). For a quick review, see the Topic 5.7 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK) and practice problems at (https://library.fiveable.me/practice/ap-chemistry).
I don't understand how to identify which compounds are intermediates vs products - help?
Think of a mechanism as a little story made of elementary steps. Products are species in the overall balanced equation that remain after the reaction finishes. Intermediates are formed in one elementary step and consumed in a later step—they only exist while the reaction is happening. Two quick tests: - Stoichiometric check: combine all elementary steps; anything that cancels out (appears on both sides) is an intermediate. - Temporal behavior: if a species is produced then later used up within the mechanism, it’s an intermediate; if it’s produced and stays in the final overall equation, it’s a product. Also watch for catalysts: they appear as reactants in an early step and are regenerated later (so they’re not products). The CED definition (5.7.A.3) says intermediates are present only while the reaction is occurring—experimental detection can support a mechanism but won’t be tested on the AP exam. For extra practice and examples, check the Topic 5.7 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK) and more Unit 5 resources (https://library.fiveable.me/ap-chemistry/unit-5).
How do you know if a proposed reaction mechanism is correct or not?
You judge a proposed mechanism by checking three AP-aligned things: 1. Stoichiometry: add the elementary steps and make sure they algebraically give the overall balanced equation (intermediates cancel; catalysts appear and disappear). 2. Kinetics: the mechanism must predict the experimentally determined rate law. Usually the slowest step (rate-determining step, RDS) controls the rate; use steady-state or pre-equilibrium approximations to derive the mechanistic rate law and see if it matches the observed form (e.g., rate = k[A][B]^2). 3. Physical plausibility: each elementary step’s molecularity must be reasonable (uni- or bimolecular common; termolecular rare), and proposed intermediates and catalytic cycles make chemical sense. Experimental detection of an intermediate strengthens a mechanism (though detection methods themselves won’t be on the AP Exam per the CED). For review and worked examples on these checks, see the Topic 5.7 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why do some reactions happen in multiple steps instead of just one step?
Most real reactions happen in multiple elementary steps because a single step that breaks and forms many bonds at once would need an extremely high activation energy and a very unlikely collision geometry. Mechanisms are built from elementary reactions (uni-, bi-, rarely termolecular) that each involve simple bond changes and have their own transition state and activation energy. Intermediates are produced and consumed along the pathway; one slow step (the rate-determining step) controls the overall rate and thus the observed rate law. Catalysts can provide an alternate multi-step pathway with lower Ea (a catalytic cycle). Experimental detection of intermediates or matching the sum of elementary steps to the overall balanced equation helps support a proposed mechanism (though detection details aren’t assessed on the AP Exam). For more on elementary steps, molecularity, and rate-determining steps, see the Topic 5.7 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK). For extra practice, try problems at (https://library.fiveable.me/practice/ap-chemistry).
What happens to intermediates after the reaction is over?
Intermediates are species formed in one elementary step and consumed in a later step of the mechanism—they exist only while the reaction is happening. Once the reaction goes to completion (or to a steady state/equilibrium), those intermediates have been converted to final products (or returned to reactants) and aren’t present as persistent substances in the overall balanced equation. Detecting an intermediate experimentally can support one proposed mechanism over another, but remember the CED: actually collecting data on intermediates won’t be tested on the AP Exam. If you want a clear walkthrough of how intermediates fit into multistep mechanisms and rate laws (including the steady-state and pre-equilibrium ideas), check the Topic 5.7 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK). For extra practice on related kinetics ideas, try the Unit 5 review (https://library.fiveable.me/ap-chemistry/unit-5) or the 1000+ practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do elementary reactions combine to give you the balanced equation?
Think of a mechanism as a list of elementary steps you add up algebraically. To get the overall balanced equation you: - Write each elementary step (they’re already molecularly balanced). - Add them together and cancel any species that appear on both sides (those are intermediates—formed in one step, used in another). - Make sure catalysts also cancel (present at start and regenerated later). - The net result (after canceling) must match the overall balanced equation—that’s the AP requirement (CED 5.7.A.2). Example in symbols: Step 1: A → I (fast) Step 2: I + B → P (slow, rate-determining) Add: A + B → P (I cancels)—overall equation. Note: the rate law is set by the rate-determining step and may involve intermediates via steady-state/pre-equilibrium reasoning. For review and examples, check the Topic 5.7 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).
If intermediates get consumed, how can we prove they were actually there?
You can’t see an intermediate at the end because it’s produced and then consumed in later elementary steps, but you can still build strong evidence it existed. Common ways chemists prove an intermediate was present (and these line up with CED 5.7.A.3–5.7.A.4): - Direct detection with fast spectroscopy (UV/IR, NMR) or rapid-quench techniques that catch the species before it’s consumed. - Trapping/isolation: add a reagent that reacts selectively with the intermediate to form a stable, isolable product. - Kinetic evidence: rate laws that match a proposed mechanism (using steady-state or pre-equilibrium approximations) and a rate-determining step that requires the intermediate. - Isotopic labeling or product-distribution studies that show the intermediate’s involvement. On the AP exam you won’t be asked to collect detection data, but you should be able to identify intermediates in mechanisms, explain how their existence supports a mechanism, and connect that to rate laws (see the Topic 5.7 study guide on Fiveable: https://library.fiveable.me/ap-chemistry/unit-5/intro-reaction-mechanisms/study-guide/ATIcHMtAfJw4fOwFzDGK). For more practice, check AP Chem problems at (https://library.fiveable.me/practice/ap-chemistry).