In chemistry, when discussing kinetics, we know that when the concentration of a reactant rises, the rate of the reaction similarly increases. This makes sense since it logically follows that if we have more reactants in the same volume compared to a lesser amount of reactant, there will be a quicker reaction. However, how do we quantitatively determine how much faster the rate will be? Well, this is where a rate law comes into play.
👉 Be sure to review reaction rates and factors that influence it, such as concentration as we spoke about here.
What is a Rate Law?
In chemistry, a rate law is an equation that describes the relationship between the rate of a chemical reaction and the concentrations of the reactants. A rate law is defined by saying: R = k[A]^n[B]^m... where:
- R is the rate of the reaction (sometimes also notated as Δ[]/Δt, which we will delve more in-depth into in the next section),
- k is the rate constant,
- [A] and [B] represent the concentrations of reactants, and
- n and m are reaction orders for each reactant (A, B, etc).
This rate law is generalized, which is why there is a ... following. A reaction could hypothetically have 3, 4, or 5 reactants, though for the AP exam you often won't see more than 2. It's actually quite rare for a reaction with 3+ reactants since it would require three atoms/molecules to bump into each other just right for a reaction to take place. It occurs, but not often and not quickly.
What Does Reaction Order Mean?
n and m define what is called each reactant's reaction order. Reaction order brings us back to the initial question we asked: how do we determine quantitatively how concentration changes the rate of reaction? The reaction order is the answer! It describes how the rate of the reaction changes as the concentration of each reactant changes.
For example, let's say the imaginary rate law for the reaction A + B → C is R = k[A]²[B]¹. This can tell us that as we increase the concentration of A (assuming a constant [B]), the rate will increase quadratically. For example, if we double the concentration of A, the rate will quadruple. Similarly, if we double the concentration of B, the rate will double, since the order of B is 1. The same applies for orders of 3, 4, etc. (if we double [], R goes up by 8 times and 16 times respectively).
The overall reaction order for the full reaction is the sum of the orders for each reactant. In our imaginary example, the overall reaction order would be 3, since the reaction order of reactant A is 2 and the reaction order of reactant B is 1.
Using Experiments to Determine a Rate Law
There's one important thing to note about rate laws: they can only be determined experimentally. What a chemist will do is run a ton of tests at different concentrations and then find the corresponding rates for each test. With this data, they can determine the rate of the reaction. Let's take a look at an example to find out how we mathematically figure this out:
Here we have a reaction: 2NO + 2H₂ → N₂ + 2H₂O. We can see three experiments done with different concentrations for each reactant. Let's take a look at the first two experiments where [NO] changes.
- We see the concentration of [NO] double from the first experiment to the second, and then we see the rate increase from 1.25 * 10⁻⁵ to 5.00 * 10⁻⁵. This is a change by a factor of 4 (5 * 10⁻⁵/1.25 * 10⁻⁵ = 4).
- Therefore, since doubling concentration makes the rate quadruple, the reaction is 2nd order with regard to NO. Note that in order to find the reaction order with respect to NO, we had to choose two experiments where the [NO] changed, but [H₂] was constant.
- Let's take a look at H₂ now. For experiments 2 and 3, the concentration of H₂ doubles like it did for NO, but the rate increases from 5 * 10⁻⁵ to 1 * 10⁻⁴, by a factor of 2 (1 * 10⁻⁴/⁵ * 10⁻⁵ = 2).
- Therefore, the reaction is first order in H₂. Now, we can put together the rate law by putting all of this together: R = k[NO]²[H₂]. As an exercise, pick one of the experiments and plug in the correct numbers to figure out the value of k, and then read the next section and figure out the right units for k. (You should get k = 250 M⁻²s⁻¹).
If you were to assume the reaction orders just by looking at the chemical equation, you'd get a rate law of R = k[NO]²[H₂]². This is why it is important to use experimental data to write the correct rate law: R = k[NO]²[H₂].
Understanding k, the Rate Constant
The rate constant, k, is a tricky thing to understand. Essentially, it serves as a proportionality constant for the reaction to take place. It makes a bit more sense if you understand the calculus behind kinetics (which we will describe in the next section, though it is by no means required for the AP exam), but essentially all you need to know is that k is a constant that quantifies the rate of each reaction and that it is temperature specific. This means that for the same reaction at different temperatures, the rate constant is different!
Another important aspect of the rate constant is that its units change depending on the overall reaction order. Let's see if we can figure some of them out. Rate is always in M/s, and concentration is always in M (M = mol/L). Thus it follows that for certain reactions:
Zeroth Order
If the overall reaction order is 0:
- The rate law is R = k[A]⁰, which simplfies to R = k
- k is in units M/s.
First Order
If the overall reaction order is 1:
- R = k[A]¹, which you can think of as M/s = k * M
- k is in s⁻¹ (per seconds)
Second Order
If the overall reaction order is 2:
- R = k[A]², which you can think of as M/s = k * M²
- k is in M⁻¹*s⁻¹ (1/Ms)
🎥 Watch AP Chemistry teacher Mónica Gracida review reaction rates and rate laws in unit five of AP Chemistry: Kinetics.
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| initial rates method | An experimental technique for determining reaction order by comparing the initial rates of a reaction under different initial concentrations of reactants. |
| overall order | The sum of all the individual reaction orders (powers) for each reactant in the rate law expression. |
| rate constant | The proportionality constant in a rate law expression that relates reaction rate to reactant concentrations; its value depends on temperature. |
| rate law | A mathematical expression that relates the reaction rate to the concentrations of reactants, with each concentration raised to a power (order). |
| reactant concentration | The amount of a reactant present in a given volume of solution, typically expressed in molarity (mol/L). |
| reaction order | The power to which the concentration of a reactant is raised in the rate law expression; indicates how the reaction rate depends on that reactant's concentration. |
| reaction rate | The speed at which reactants are converted to products per unit of time in a chemical reaction. |
Frequently Asked Questions
What is a rate law and how do I write one?
A rate law is an equation that links reaction rate to reactant concentrations: rate = k[A]^m[B]^n… Here k is the rate constant (temperature dependent), m and n are the reaction orders for A and B (they’re exponents you find from data). To write one from experiments, use the method of initial rates: run experiments with different starting concentrations, compare how the initial rate changes, and solve for the exponents. Example: if doubling [A] doubles rate (with [B] constant), m = 1; if doubling [B] quadruples rate, n = 2. The overall order is m + n. Units of k depend on overall order (e.g., M·s⁻1 for first order, M⁻1·s⁻1 for second order). The AP exam expects you to represent experimental data with a consistent rate-law expression and use initial-rate comparisons (CED 5.2.A). For a focused review and examples, see the Topic 5.2 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ), the Unit 5 overview (https://library.fiveable.me/ap-chemistry/unit-5), and extra practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why do we need to determine reaction order experimentally instead of just using the balanced equation?
Because the balanced equation just shows overall stoichiometry, not how molecules actually collide and react. The rate law depends on the reaction mechanism—which elementary steps and collisions make the slow (rate-determining) step—so the exponents (orders) come from molecular-level steps, not coefficients in the balanced equation except for elementary reactions. That’s why you must determine orders experimentally (e.g., method of initial rates, spectrophotometric or conductometric monitoring) and find the rate law r = k[A]^m[B]^n. The rate constant k is temperature dependent, and the units of k tell you the overall order (CED 5.2.A.2–5.2.A.4). On the AP exam you’ll be asked to use initial-rate data or propose mechanisms that match observed orders, so practice those skills (see the Topic 5.2 study guide: https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ). For more practice, try the unit problems (https://library.fiveable.me/ap-chemistry/unit-5) or the large practice set (https://library.fiveable.me/practice/ap-chemistry).
How do you find the rate constant k from experimental data?
First find the reaction’s rate law (the orders) from experiments, then solve for k. 1) Use the method of initial rates (CED 5.2.A.5): compare experiments where one reactant’s concentration changes while others are constant to determine each reactant’s order (m, n). 2) Once you have the form rate = k[A]^m[B]^n, pick one experimental trial and plug in the measured initial rate and concentrations. Solve k = rate / ([A]^m[B]^n). Repeat with other trials and average k—k should be the same for all trials at the same temperature. 3) Check units: units of k depend on overall order (e.g., M s⁻¹ for zero-order, s⁻¹ for first-order, M⁻1 s⁻¹ for second-order). Remember k depends on temperature (CED 5.2.A.4). 4) Alternate approach: if you know the reaction is zero-, first-, or second-order in one reactant, use integrated rate laws (plot [A] vs t, ln[A] vs t, or 1/[A] vs t)—the slope gives k directly. For more practice and the AP-aligned summary, see the Topic 5.2 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ) or the unit overview (https://library.fiveable.me/ap-chemistry/unit-5). For lots of problems, try Fiveable practice (https://library.fiveable.me/practice/ap-chemistry).
What's the difference between reaction order and overall order?
Reaction order (or “order with respect to” a reactant) is the exponent on that reactant’s concentration in the rate law—it tells you how the rate changes when that reactant’s concentration changes. Overall order is the sum of those exponents for all reactants in the rate law. Example: rate = k[A]^2[B]^1 → order with respect to A = 2, with respect to B = 1, overall order = 3. The orders come from experiment (e.g., method of initial rates); they’re not just the stoichiometric coefficients unless the step is elementary (CED 5.2.A.2–5.2.A.5). The units of k depend on the overall order (for a third-order rate constant, units are M^-2 s^-1). On the AP exam you’ll be asked to write consistent rate laws from data and use initial rates to find orders (Topic 5.2). For a quick refresher see the Topic 5.2 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
I'm confused about how to use initial rates to find the order of reaction - can someone explain this step by step?
Start with the rate law form: rate = k [A]^m [B]^n. Using initial-rate experiments you keep things simple: compare two experiments where only one reactant concentration changes. Then the ratio of initial rates equals the concentration ratio raised to the unknown exponent. Step-by-step: 1. Pick two trials where [B] is the same but [A] changes. Write rate2/rate1 = ([A]2/[A]1)^m. 2. Plug numbers and solve for m. If rate doubles when [A] doubles, m = 1; if rate quadruples, m = 2; if rate doesn’t change, m = 0. 3. Repeat for B using trials where [A] is constant to find n. 4. Once m and n are known, find k using k = rate / ([A]^m [B]^n) with any initial-rate data. Calculate units of k from overall order (CED: 5.2.A.2–5.2.A.4). 5. If concentration changes for multiple reactants at once, use rate ratios and take logs: log(rate2/rate1) = m·log([A]2/[A]1)+ n·log([B]2/[B]1). This is the AP “method of initial rates” (CED 5.2.A.5). For guided examples and practice, check the Topic 5.2 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ) and try problems at Fiveable’s practice page (https://library.fiveable.me/practice/ap-chemistry).
How do you know what units the rate constant should have?
You get the units of k from the rate law and the fact that rate is in concentration/time (usually M s⁻¹). Write rate = k [A]^m [B]^n, where overall order = m + n. Solve for k: k = (rate) / ([A]^m[B]^n). So the units are M·s⁻¹ divided by M^(m+n) = M^(1−(m+n))·s⁻¹. Quick examples: - Zero order (overall 0): units = M·s⁻¹ (e.g., M/s) - First order (overall 1): units = s⁻¹ - Second order (overall 2): units = M⁻¹·s⁻¹ Remember k is temperature dependent and its units always reflect the overall reaction order (CED 5.2.A.4). If you want practice identifying units from given rate laws, check the Topic 5.2 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ) or more unit review at (https://library.fiveable.me/ap-chemistry/unit-5). For extra problems try Fiveable’s practice set (https://library.fiveable.me/practice/ap-chemistry).
What does it mean when a reaction is zero order vs first order vs second order?
Zero-, first-, and second-order describe how the rate depends on concentration in the rate law: rate = k[A]^n (CED 5.2.A.2–3). - Zero order (n = 0): rate = k. The rate is constant and independent of [A]. Units of k are M·s⁻¹. A plot of [A] vs. time is linear. - First order (n = 1): rate = k[A]. Rate is proportional to [A]; doubling [A] doubles the rate. Units of k are s⁻¹. The integrated form is ln[A] = −kt + ln[A]0, so ln[A] vs. time is linear; the half-life t½ = 0.693/k is constant. - Second order (n = 2): common forms are rate = k[A]^2 or k[A][B]. Doubling [A] quadruples the rate (for A^2). Units of k are M⁻¹·s⁻¹. For A^2, 1/[A] vs. time is linear and t½ depends on [A]0. To find orders experimentally use the method of initial rates (CED 5.2.A.5) or test linearity of the integrated plots. For more practice and the Topic 5.2 study guide, see Fiveable (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ) and the Unit 5 overview (https://library.fiveable.me/ap-chemistry/unit-5). For lots of problems, try Fiveable practice (https://library.fiveable.me/practice/ap-chemistry).
Why does temperature affect the rate constant but not the reaction order?
Reaction order (the exponents in a rate law) is an experimental description of how rate depends on concentrations—you determine it by comparing initial rates (CED 5.2.A, method of initial rates). It reflects the reaction mechanism (which elementary steps involve which species) and so is set by how collisions/steps control rate, not by how fast those steps happen. Temperature changes how often and how energetically molecules collide and how many collisions exceed the activation energy. That shows up in the rate constant k (CED 5.2.A.4) via the Arrhenius relation k = A e^(−Ea/RT): raising T increases k exponentially. k’s value (and units—which depend on overall order) is temperature dependent; the numeric reaction orders (exponents) stay the same unless the mechanism itself changes at a different T. For AP review: practice applying initial-rates data to find orders and remember k varies with T (see the Topic 5.2 study guide) (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ). For more unit review and tons of practice problems go to (https://library.fiveable.me/ap-chemistry/unit-5) and (https://library.fiveable.me/practice/ap-chemistry).
How do I compare initial rates to determine the order with respect to each reactant?
Compare initial rates by holding one reactant’s concentration effectively the same between two experiments and seeing how the rate changes—that gives the order for the reactant that changed. Steps (method of initial rates): 1. Pick two experiments where [B], [C], etc. are constant and only [A] changes. 2. Write the rate law form: rate = k[A]^m[B]^n... 3. Divide the two rate expressions to cancel k and constant concentrations: rate2/rate1 = ([A]2/[A]1)^m. 4. Solve for m (the order in A). Repeat for each reactant. Quick rules-of-thumb: - If doubling [A] doubles rate → first order (m = 1). - If doubling [A] leaves rate unchanged → zero order (m = 0). - If doubling [A] quadruples rate → second order (m = 2). After finding each exponent, sum them for overall order and use one experiment to solve for k. This is exactly what the AP wants you to do on initial-rate problems (Topic 5.2). For walkthroughs and practice, see the Topic 5.2 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ) and more practice questions (https://library.fiveable.me/practice/ap-chemistry).
What's the relationship between the exponents in the rate law and the coefficients in the balanced equation?
Short answer: there’s no guaranteed connection—the exponents in a rate law (the orders) come from experiments, not from the balanced stoichiometric coefficients, unless the reaction step shown is an elementary (single-step) reaction. Why: the CED (Topic 5.2) says the rate law is rate = k[A]^m[B]^n and m, n are determined by data (method of initial rates), not by the overall balanced equation. Only for elementary reactions do the molecularities equal the exponents (e.g., a bimolecular elementary step gives 2nd order). The sum of the exponents gives the overall order and affects k’s units and how you analyze initial-rate problems on the AP exam. For practice: review the Topic 5.2 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ) and try method-of-initial-rates problems from the unit page (https://library.fiveable.me/ap-chemistry/unit-5) or the practice bank (https://library.fiveable.me/practice/ap-chemistry).
Can you explain how to set up a rate law experiment in the lab?
Pick a clean, easy reaction (e.g., A + B → products) and a measurable signal (absorbance, conductivity, gas volume). Run at least three trials that change only one reactant’s initial concentration while keeping temperature constant and other concentrations large/constant. Monitor concentration vs. time and determine the initial rate (slope at t ≈ 0 or during first ~10% conversion). Use the method of initial rates: compare how rates change when [A] or [B] changes to get orders m and n (rate ∝ [A]^m[B]^n). Once orders are known, solve for k from rate = k[A]^m[B]^n and report units matching overall order. Repeat trials for precision, calibrate your instrument, control temperature, and follow safety rules. AP tip: the exam expects using initial-rate comparisons to find orders and k (CED 5.2.A, method of initial rates). For a refresher and sample setups, see the Topic 5.2 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).
I don't understand how to calculate reaction order when the concentration doubles but the rate increases by 4 - what am I missing?
You're close—use the rate law idea: rate ∝ [A]^n. If [A] doubles and rate increases by 4, plug into the ratio: rate2 / rate1 = (2)^n = 4 → 2^n = 4 → n = 2. So the reaction is second order with respect to that reactant. This is exactly the method of initial rates (compare initial rates while changing one concentration)—orders are determined experimentally, not from the balanced equation. Remember: if doubling gave a 2× change, n = 1 (first order); if no change, n = 0 (zero order). Also note the overall order (sum of exponents) sets the units of k and is tested on the AP (Topic 5.2: method of initial rates). For more worked examples and AP-aligned practice, see the Topic 5.2 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ) and the AP practice problems (https://library.fiveable.me/practice/ap-chemistry).
What happens to the rate law if one of the reactants is in large excess?
If one reactant is present in large excess, its concentration stays essentially constant during the time you measure the rate. That lets you treat the rate law as if that reactant were part of the rate constant—a pseudo-order simplification. For example, if rate = k[A][B] and [B] ≫ [A], then rate ≈ k′[A] where k′ = k[B] (constant). The true reaction orders (the exponents in the real rate law) don’t change—you’re just measuring an apparent order that’s easier to analyze. This is why experiments often use an excess reactant to get pseudo-first-order kinetics and determine the order with respect to the limiting species using the method of initial rates (an AP CED technique). For more on this and practice problems, see the Topic 5.2 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ) and Unit 5 resources (https://library.fiveable.me/ap-chemistry/unit-5).
How do you determine if a reaction is first order or second order just by looking at the data?
Look at how rate changes when you change concentration (method of initial rates). If you double [A] and: - rate doubles → rate ∝ [A]^1 → first order in A. - rate quadruples → rate ∝ [A]^2 → second order in A. - rate doesn’t change → zero order. For more rigorous checks use linear plots of time-course data: - First order: ln[A] vs t is straight line (slope = −k). - Second order (one reactant): 1/[A] vs t is straight line (slope = +k). On the AP exam they expect you to compare initial rates or use these integrated-rate plots to identify orders (CED 5.2.A, method of initial rates). If you want step-by-step examples and quick practice, see the Topic 5.2 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ) and try problems at Fiveable’s practice page (https://library.fiveable.me/practice/ap-chemistry).
Why can't you determine the rate law just from looking at the balanced chemical equation?
You can’t get the rate law just from the balanced equation because the balanced equation only shows overall stoichiometry, not the microscopic steps that actually make the reaction happen. Rate laws depend on the mechanism: the elementary steps (and the slow, rate-determining step) determine how rate depends on concentrations. For many reactions the overall equation comes from multiple elementary steps, so the exponents (orders) in rate = k[A]^m[B]^n must be measured experimentally (e.g., by the method of initial rates). The rate constant k is also temperature-dependent and not shown by the equation. On the AP exam you’re expected to represent experimental data with a consistent rate law and use initial-rate comparisons to find orders (CED 5.2.A, keywords: rate law, reaction order, rate constant, method of initial rates). For a clear review, see the Topic 5.2 study guide (https://library.fiveable.me/ap-chemistry/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).