Introduction to Rate Law
Infer empirical orders from controlled initial-rate comparisons and check rate-constant units.
AP Chemistry Topic 5.2 · CED printed p. 95 / PDF p. 102. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Balanced equations and stoichiometric ratios
- Molarity and concentration versus amount
- Bond rearrangement and energy conservation
- Slopes, logarithms and reciprocal graphs
First decide what the instrument measures. If a colored reactant is the only significant absorber at the chosen wavelength and Beer’s law is valid, absorbance is proportional to its concentration at fixed path length. Its declining signal can therefore track disappearance. Gas pressure can track a species amount when temperature, volume, and gas composition are accounted for. Use early measurements for initial rates and run controls for background signal; the instrument does not magically measure rate just because its display changes.
An empirical rate law, , summarizes how the measured rate responds to concentrations under the tested conditions. The exponents come from experimental dependence for an overall reaction. Do not copy them from the balanced overall equation. A species can participate in the net reaction and still have zero order over the observed range, often because some other part of the route controls the rate.
The order in A is m; the overall order is m+n. If m =2, doubling A with B fixed multiplies rate by four. If n =0, changing B in that regime leaves rate unchanged. Order is a description of concentration sensitivity, not a count of how many elementary steps occur. In later mechanism work, elementary-step stoichiometry can determine that step’s order, but that exception must be stated explicitly.
The rate constant k belongs to a specified reaction model and temperature. If rate units are M s−1, total order z gives units M1−z s−1 for k. First order gives s−1; second order gives M−1 s−1; third order gives M−2 s−1. Increasing concentration changes r in the law, not k, while changing temperature usually changes k. A numerical k without its units and temperature context is incomplete.
Compare two initial-rate trials that differ in only one concentration. Divide rates so k and fixed factors cancel, then solve the remaining power. If both concentrations change, use already-known order information or another pair; do not attribute the whole rate change to one reactant. Replicates and a wide enough concentration range help distinguish a real power dependence from noise. A fitted rate law is supported within its tested range; it is not a proof of a unique mechanism.
Let's work through it
At one temperature, trials give ([A], [B], r): (0.100 M,0.100 M,2.00×10−4 M s−1), (0.200,0.100,8.00×10−4), (0.100,0.300,6.00×10−4).
- Comparing trials 1 and 2 holds B fixed: fourfold rate from twofold A means 2m =4, so m =2.
- Comparing trials 1 and 3 holds A fixed: threefold rate from threefold B means n =1. Overall order is three.
- k =r/([A]2[B]) =2.00×10−4/(0.1002×0.100) =0.200 M−2 s−1. Recheck with trial 2: it gives the same k.
A calibrated colored-reactant signal has absorbance 0.500 at 0.0200 M and 0.450 after 10.0 s. Assume only this reactant absorbs, fixed path length, linear Beer behavior, constant volume.
- Because concentration is proportional to absorbance, final concentration is 0.0200×0.450/0.500 =0.0180 M.
- Average disappearance rate is (0.0200−0.0180)/10.0 =2.00×10⁻⁴ M s⁻¹. An initial rate would require the early tangent slope or a sufficiently short interval.
- Use the same calibration/background correction across controlled trials; signal decrease alone is not yet a rate.
Make the picture move
Interactive concept model
Use data to identify a rate law.
Predict first. Compare the initial-rate rows. What changes when initial concentration doubles? Then compare the half-life dependence in the two scenarios.
rate = k[A]; [A]ₜ = [A]₀ exp(−kt)
- First-order concentration
- Labeled selected or reference state
On a small screen, scroll inside the graph or diagram to read its labels. The values and explanation also appear below.
For the specified first-order law, doubling initial [A] multiplies initial rate by 2. At 20 s, [A] = 0.147 mol L⁻¹.
Read the data and comparisons
| Initial [A] (mol L⁻¹) | Initial rate (mol L⁻¹ s⁻¹) | Rate relative to selected initial rate |
|---|---|---|
| 0.2 | 0.01 | 0.5 |
| 0.4 | 0.02 | 1 |
| 0.8 | 0.04 | 2 |
Macroscopic / data. Initial rate is a concentration change per unit time; it differs from the concentration itself.
Particle view. Changing particle concentration can change encounter frequency, but a rate-law exponent alone does not prove a molecular mechanism.
Symbolic view. The supplied exponent is 1; k units are s⁻¹.
Why it works. Order is established from rate evidence, not read from an overall balanced equation. The table compares otherwise identical hypothetical initial-rate trials. The curve follows the integrated law for disappearance of A; the plotted curve is not the table of initial rates. A straight ln([A]/1 M) versus time line has slope −k for first order; a straight 1/[A] versus time line has slope +k for second order. The concentration reference makes the logarithm dimensionless without changing its slope.
Conditions and limits. Hypothetical single-reactant disappearance with unit stoichiometric coefficient, constant volume and temperature (298 K), constant k = 0.05 s⁻¹, no reverse reaction over this interval. Supplied ideal data are model outputs, not measurements; rates and concentrations are rounded to three significant digits.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
For r =k[A]2[B], what if A doubles and B halves?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
For r =k[A]2[B], what if A doubles and B halves?
Worked answer and reason
Rate becomes 22×½ =2 times as large at the same temperature.
Multiply the independent concentration factors.
Retrieval 2
Can 2A + B → C alone establish r =k[A]2[B]?
Worked answer and reason
No, unless it is explicitly an elementary step. An overall rate law needs experimental evidence or a justified mechanism.
Net stoichiometry does not reveal the controlling collision events.
Catch the tempting mistake
- k increases when more reactant is added at fixed temperature.
- r changes through concentration factors; k stays fixed within the same model and conditions.
- Overall reaction coefficients are always rate exponents.
- Only explicitly elementary steps have that direct molecularity inference.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 5 problem 3 · ap application · data
- Unit 5 problem 4 · ap synthesis · verbal
- Unit 5 problem 5 · ap application · data
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Scientific calculator
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For the teacher
- Have students identify which comparison holds the other reactant fixed.
- Require a dimensional check of k and a rate prediction independent of the fitted trial.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 5.2.A
- Represent experimental data with a consistent rate law expression.
- 5.2.A.1
- Experimental methods can be used to monitor the amounts of reactants and/or products of a reaction over time and to determine the rate of the reaction.
- 5.2.A.2
- The rate law expresses the rate of a reaction as proportional to the concentration of each reactant raised to a power.
- 5.2.A.3
- The power of each reactant in the rate law is the order of the reaction with respect to that reactant. The sum of the powers of the reactant concentrations in the rate law is the overall order of the reaction.
- 5.2.A.4
- The proportionality constant in the rate law is called the rate constant. The value of this constant is temperature dependent and the units reflect the overall reaction order.
- 5.2.A.5
- Comparing initial rates of a reaction is a method to determine the order with respect to each reactant.
Science references for this teaching
- College Board AP Chemistry Course and Exam Description, Fall 2024 — Exact unit learning requirements and exclusions; supplied PDF is pinned by the integration catalog.
- OpenStax Chemistry 2e: Chemical Reaction Rates — Scientific check: reaction-rate conventions, concentration slopes and stoichiometric rates.
- OpenStax Chemistry 2e: Rate Laws — Scientific check: empirical orders, initial-rate comparisons and units of k.
Glossary and deeper reading · 7 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Stoichiometry
Stoichiometry uses a balanced chemical equation to relate amounts of reactants and products. Equation coefficients provide mole ratios rather than direct mass ratios. For example, forming water from hydrogen and oxygen consumes two moles of hydrogen molecules per mole of oxygen molecules.
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Molarity
Molarity is the amount of a solute in moles divided by the final solution volume in liters. It describes concentration rather than the total amount of solute. For example, 0.10 mole dissolved to make 0.50 liter of solution gives a molarity of 0.20 mole per liter.
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Beer–Lambert law
The Beer–Lambert law relates absorbance to concentration, light-path length, and molar absorptivity under suitable conditions. Comparisons require a controlled wavelength, consistent path length, and an appropriate calibration. For example, doubling concentration doubles absorbance in the law's linear range if the other factors remain fixed.
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Absorbance
Absorbance is a logarithmic measure of how much a sample reduces transmitted light at a specified wavelength. It depends on the ratio of incident to transmitted intensity rather than simply being the fraction absorbed. For example, absorbance 1 corresponds to transmission of one-tenth of the incident intensity.
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Reaction order
Reaction order describes the concentration dependence in a rate law. An order in one reactant is its exponent, and overall order is the sum of the concentration exponents. For example, a second-order dependence on one reactant makes quadrupling the rate follow from doubling that reactant's concentration under fixed conditions.
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Elementary reaction
An elementary reaction is one step in a proposed mechanism rather than the sum of multiple steps. For a specified elementary step, the rate-law concentration factors follow its reacting-particle event. For example, a bimolecular elementary collision can have a rate proportional to the concentrations of its two reactants.
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Rate constant
A rate constant is the proportionality factor in a specified rate law. Its value depends on conditions such as temperature, and its units depend on the law's overall order. For example, a first-order rate constant has units of inverse time when concentration is measured per unit volume.