Introduction to Rate Law - AP Chemistry

SECTION 1

Historical Context & Motivation

Chemistry in the nineteenth century was largely a descriptive science: chemists catalogued reactions and their products, but the speed at which those reactions occurred remained poorly understood. Industrial processes—from the manufacture of sulfuric acid to the fermentation of ethanol—demanded not just knowledge of what products form but how quickly they appear. The field of chemical kinetics arose to fill that gap, transforming chemistry from a qualitative art into a quantitative, predictive discipline. The concept of a rate law sits at the heart of this transformation, giving chemists a compact algebraic expression that links reaction speed to the concentrations of reactants.

1850 • Wilhelmy's Sucrose Hydrolysis

Ludwig Wilhelmy performed the first quantitative kinetics study by tracking the inversion of sucrose with a polarimeter, demonstrating that the rate of reaction was proportional to the concentration of sucrose remaining.

1864 • Guldberg & Waage's Law of Mass Action

Cato Guldberg and Peter Waage proposed that the 'driving force' of a reaction depends on the active masses (concentrations) of the reactants, laying the conceptual groundwork for the rate law.

1884 • van 't Hoff's Études de Dynamique Chimique

Jacobus van 't Hoff published the first systematic treatment of reaction order, distinguishing first-order and second-order kinetics and introducing integrated rate equations.

1889 • Arrhenius Equation

Svante Arrhenius connected the rate constant k to temperature through an exponential relationship involving activation energy, completing the modern kinetic picture.

These milestones converge on a central question that the rate law answers: for a given reaction, how does the rate change when you change the concentration of each reactant? Understanding the rate law is the first step toward controlling reaction speed—whether you are an engineer optimizing a catalytic converter or a biochemist modeling enzyme kinetics.

SECTION 2

Core Principles & Definitions

Before writing a rate law, you need a precise vocabulary. The rate of reaction is defined as the change in concentration of a reactant or product per unit time. For the generic reaction aA + bB → cC + dD, the rate can be written as −(1/a)(Δ[A]/Δt) or +(1/c)(Δ[C]/Δt), where the stoichiometric coefficients normalize the expression so every species gives the same numerical rate. The rate law is the experimentally determined equation Rate = k[A]^m[B]ⁿ that quantifies how concentration affects speed.

Rate Constant (k)

A proportionality constant specific to a given reaction at a given temperature. Its units depend on the overall reaction order and it increases with temperature according to the Arrhenius equation.

Reaction Order (m, n)

The exponents in the rate law that describe the sensitivity of the rate to each reactant's concentration. They are determined experimentally—not from stoichiometric coefficients—unless the reaction is elementary.

Overall Order

The sum of all individual orders (m + n + …). The overall order dictates the units of k and controls how rapidly the rate changes as concentrations vary.

Initial Rate Method

The most common experimental approach: measure the instantaneous rate at time zero for several trials in which only one reactant concentration is varied. Ratios of rates reveal each order.

✦

KEY TAKEAWAY

Think of a rate law like a recipe's sensitivity to its ingredients. Doubling the sugar in a cake batter might make it noticeably sweeter (first-order dependence), while doubling it could make the sweetness overwhelm everything else quadratically (second-order dependence). The exponents in the rate law tell you exactly how 'sensitive' the reaction rate is to each reactant—and, crucially, those exponents come from experiment, not from the balanced equation.

SECTION 3

Visual Explanation — Anatomy of a Rate Law

Each component of the rate law is color-coded. The rate constant k depends on temperature, while the reactant concentrations are raised to their respective experimentally determined orders.

The diagram above deconstructs the generic rate law expression into its constituent parts. Notice that the rate constant k is separated from the concentration terms; this is deliberate, because k carries all of the temperature dependence of the reaction (through the Arrhenius equation), whereas the concentration terms capture the dependence on how much reactant is present. The exponents m and n are integers or simple fractions determined by experiment—they are not derived from the balanced equation unless the reaction proceeds via a single elementary step.

SECTION 4

Mathematical Framework

The mathematical backbone of a rate law consists of the differential rate expression, the method for determining orders from experimental data, and the dimensional analysis of the rate constant. Mastering these three pieces allows you to write, interpret, and apply any rate law that appears on the AP Chemistry exam.

GENERAL DIFFERENTIAL RATE LAW

Rate = k [A]ᵐ [B]ⁿ

Rate = reaction rate (M·s⁻¹); k = rate constant; [A], [B] = molar concentrations; m, n = reaction orders with respect to A and B.

DETERMINING ORDER BY RATIO OF RATES

Rate₁ / Rate₂ = ([A]₁ / [A]₂)ᵐ

When [B] is held constant between trials 1 and 2, dividing the two rate law expressions cancels k and [B]ⁿ, isolating the order m with respect to A. Solving for m: m = ln(Rate₁/Rate₂) / ln([A]₁/[A]₂).

UNITS OF THE RATE CONSTANT

k units = M¹⁻⁽ᵐ⁺ⁿ⁾ · s⁻¹

For overall order 0: M·s⁻¹; for order 1: s⁻¹; for order 2: M⁻¹·s⁻¹; for order 3: M⁻²·s⁻¹. The units of k must always compensate so that the right side simplifies to M·s⁻¹.

💡

AP Exam Tip

The College Board frequently tests whether students can distinguish between the stoichiometric coefficients in a balanced equation and the reaction orders in the rate law. Never assume orders equal coefficients unless you are told the reaction is elementary. Free-response rubrics award explicit statements that orders are experimentally determined.

SECTION 5

Reaction Orders — Zero, First, and Second

Different reaction orders produce dramatically different kinetic behavior. A zero-order reaction proceeds at a constant rate regardless of reactant concentration; the rate is set entirely by k. A first-order reaction has a rate directly proportional to reactant concentration—double the concentration and the rate doubles. A second-order reaction shows a rate proportional to the square of the concentration—double [A] and the rate quadruples. Understanding these patterns is essential for interpreting experimental data and predicting how a reaction behaves over time.

A plot of instantaneous rate versus [A] for the three most common orders. The zero-order line is horizontal (rate is constant). The first-order line is linear through the origin. The second-order curve is parabolic, rising steeply at higher concentrations.

Property	Zero Order	First Order	Second Order
Rate law	Rate = k	Rate = k[A]	Rate = k[A]²
Units of k	M·s⁻¹	s⁻¹	M⁻¹·s⁻¹
Effect of doubling [A]	Rate unchanged	Rate doubles	Rate quadruples
Linear plot	[A] vs. t	ln[A] vs. t	1/[A] vs. t

Summary of common reaction orders and their diagnostic features.

SECTION 6

Worked Example — Determining a Rate Law from Data

Consider the reaction 2 NO(g) + Cl₂(g) → 2 NOCl(g). Three initial-rate experiments are performed at the same temperature.

Trial	[NO]₀ (M)	[Cl₂]₀ (M)	Initial Rate (M·s⁻¹)
1	0.10	0.10	1.2 × 10⁻³
2	0.20	0.10	4.8 × 10⁻³
3	0.10	0.20	2.4 × 10⁻³

Initial-rate data for 2 NO + Cl₂ → 2 NOCl

Deriving the Rate Law from Initial-Rate Data

1
Step 1 — Write the general rate law

Assume Rate = k [NO]^m [Cl₂]ⁿ, where m and n are to be determined.
2
Step 2 — Find the order with respect to NO (compare Trials 1 & 2)

[Cl₂] is constant. Rate₂/Rate₁ = (4.8 × 10⁻³)/(1.2 × 10⁻³) = 4.0. [NO]₂/[NO]₁ = 0.20/0.10 = 2.0. So 4.0 = 2.0^m, which gives m = 2.

m = 2 (second order in NO)
3
Step 3 — Find the order with respect to Cl₂ (compare Trials 1 & 3)

[NO] is constant. Rate₃/Rate₁ = (2.4 × 10⁻³)/(1.2 × 10⁻³) = 2.0. [Cl₂]₃/[Cl₂]₁ = 0.20/0.10 = 2.0. So 2.0 = 2.0ⁿ, which gives n = 1.

n = 1 (first order in Cl₂)
4
Step 4 — Write the rate law and calculate k

Rate = k [NO]² [Cl₂]. Overall order = 2 + 1 = 3. Using Trial 1: k = Rate / ([NO]²[Cl₂]) = (1.2 × 10⁻³) / ((0.10)²(0.10)) = (1.2 × 10⁻³) / (1.0 × 10⁻³) = 1.2 M⁻²·s⁻¹.

Rate = 1.2 M⁻²·s⁻¹ × [NO]² [Cl₂]

⚠

Common Mistake

Students sometimes set the order with respect to NO equal to 2 because of the coefficient 2 in front of NO in the balanced equation. In this case the orders happen to match, but that is coincidence—the rate law was derived from experimental data, not from stoichiometry.

SECTION 7

Strengths & Limitations of Rate Laws

Rate laws are powerful analytical tools, but their usefulness comes with constraints that any practicing chemist or student must recognize. A rate law accurately describes the initial kinetics of a reaction under specified conditions, but it does not, by itself, reveal the reaction mechanism or predict behavior at extreme concentrations where side reactions may dominate.

Strengths	Limitations
Provides a quantitative, predictive relationship between concentrations and rate.	Must be determined experimentally for every reaction—cannot be deduced from the balanced equation alone.
Enables comparison of reaction speeds across different conditions and reactions.	Valid only for the conditions (temperature, solvent, catalyst) under which it was measured.
Can suggest (but not prove) a mechanism: the orders constrain which mechanisms are plausible.	Does not directly reveal the mechanism; multiple mechanisms can produce the same rate law.
Integration yields concentration-vs.-time equations, enabling predictions about reaction progress.	Assumes constant temperature and often assumes only the forward reaction is significant (not equilibrium).

Strengths and limitations of rate law expressions.

✦

KEY TAKEAWAY

A rate law is like a regression equation fitted to kinetic data—it describes the trend reliably within the measured domain, but extrapolating beyond those conditions or assuming it explains the underlying mechanism is a logical leap. Just as a linear trendline through data doesn't prove the underlying physics is linear, a rate law doesn't prove the mechanism. It constrains it.

SECTION 8

Connection to Integrated Rate Laws & Mechanisms

The differential rate law you have learned describes how rate depends on concentration at any instant. A natural next step is to integrate the rate law—transforming it from a snapshot of speed into a full description of concentration as a function of time. This yields the integrated rate laws that AP Chemistry treats extensively. Additionally, when a reaction proceeds through multiple elementary steps, the rate law for the overall reaction is governed by the rate-determining step (the slowest step). Proposing a mechanism that is consistent with the experimentally observed rate law is a key skill tested in the free-response section.

Concept	Differential Rate Law (This Lesson)	Integrated Rate Law (Next Steps)
What it tells you	How the instantaneous rate changes with concentration	How concentration changes with time
Typical form	Rate = k[A]ᵐ	ln[A] = −kt + ln[A]₀ (1st order)
Primary use	Determine orders and k from initial-rate data	Predict [A] at future time t, determine half-life
Graphical method	Plot rate vs. [A]—shape reveals order	Plot [A], ln[A], or 1/[A] vs. t—linear plot reveals order

Differential vs. integrated rate laws.

Looking further ahead, the Arrhenius equation k = A·e^−Eₐ/(RT) explains why the rate constant increases with temperature by relating k to the activation energy E_a. Catalysts lower E_a, thereby increasing k without appearing in the rate law expression itself. These advanced connections form the complete kinetic picture that Unit 5 of the AP Chemistry curriculum builds piece by piece.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

For the reaction A + 2 B → C, the experimentally determined rate law is Rate = k[A][B]. Which of the following statements is correct regarding this rate law?

A. The reaction must be an elementary bimolecular step because the overall order is 2.
B. The order with respect to B must equal its stoichiometric coefficient of 2.
C. The overall order is 2, and the orders with respect to A and B are each 1, as determined by experiment.
D. The units of k are M·s⁻¹ because the reaction is second order overall.

Show answer (C)

C is correct. The rate law Rate = k[A][B] shows first-order dependence on both A and B, giving an overall order of 1 + 1 = 2. These orders come from experiment, not stoichiometry. Choice A is wrong because an overall second-order rate law can arise from a multi-step mechanism—it does not require a single elementary step. Choice B is wrong because reaction orders do not have to match stoichiometric coefficients for non-elementary reactions. Choice D is wrong because for an overall second-order reaction, the units of k are M⁻¹·s⁻¹, not M·s⁻¹.

Question 2: BASIC CALCULATION

For a reaction with the rate law Rate = k[X]², the initial rate is measured as 3.6 × 10⁻⁴ M·s⁻¹ when [X] = 0.030 M. What is the value of k?

A. 0.012 M⁻¹·s⁻¹
B. 0.40 M⁻¹·s⁻¹
C. 4.0 × 10² M⁻¹·s⁻¹
D. 1.2 × 10⁻² s⁻¹

Show answer (B)

B is correct. Rearranging Rate = k[X]² gives k = Rate / [X]² = (3.6 × 10⁻⁴) / (0.030)² = (3.6 × 10⁻⁴) / (9.0 × 10⁻⁴) = 0.40 M⁻¹·s⁻¹. Choice A results from dividing rate by [X] instead of [X]². Choice C likely comes from an arithmetic error with powers of ten. Choice D has incorrect units (s⁻¹ corresponds to a first-order reaction, not second-order).

Question 3: INTERMEDIATE

For the reaction 2 A + B → products, initial-rate data show: (Trial 1) [A] = 0.10 M, [B] = 0.10 M, Rate = 5.0 × 10⁻³ M·s⁻¹; (Trial 2) [A] = 0.20 M, [B] = 0.10 M, Rate = 2.0 × 10⁻² M·s⁻¹; (Trial 3) [A] = 0.10 M, [B] = 0.30 M, Rate = 5.0 × 10⁻³ M·s⁻¹. What is the rate law?

A. Rate = k[A]²[B]
B. Rate = k[A][B]²
C. Rate = k[A]²
D. Rate = k[A][B]

Show answer (C)

C is correct. Comparing Trials 1 and 2 ([B] constant): Rate₂/Rate₁ = (2.0 × 10⁻²)/(5.0 × 10⁻³) = 4 and [A]₂/[A]₁ = 2, so 4 = 2ᵐ gives m = 2. Comparing Trials 1 and 3 ([A] constant): Rate₃/Rate₁ = (5.0 × 10⁻³)/(5.0 × 10⁻³) = 1 and [B]₃/[B]₁ = 3, so 1 = 3ⁿ gives n = 0. The rate law is Rate = k[A]². Choice A incorrectly includes [B] at first order. Choice B reverses the orders. Choice D assumes first order in both, which contradicts the four-fold rate increase when [A] doubles.

Question 4: APPLIED

4 pts

The decomposition of dinitrogen pentoxide follows the equation: 2 N₂O₅(g) → 4 NO₂(g) + O₂(g). In a series of experiments at 45 °C, the following initial-rate data were collected: Trial 1: [N₂O₅]₀ = 0.015 M, Initial Rate = 4.5 × 10⁻⁴ M·s⁻¹ Trial 2: [N₂O₅]₀ = 0.030 M, Initial Rate = 9.0 × 10⁻⁴ M·s⁻¹ Trial 3: [N₂O₅]₀ = 0.060 M, Initial Rate = 1.8 × 10⁻³ M·s⁻¹ (a) Determine the rate law for this reaction. Justify your answer with calculations. (b) Calculate the value of the rate constant k, including proper units. (c) Explain why the exponent in the rate law differs from the stoichiometric coefficient of N₂O₅ in the balanced equation. (d) If the temperature were increased to 55 °C, state and justify what would happen to the value of k and the reaction order.

Show answer

Rubric (4 points): (a) 1 point — Comparing Trials 1 and 2: Rate₂/Rate₁ = (9.0 × 10⁻⁴)/(4.5 × 10⁻⁴) = 2.0; [N₂O₅]₂/[N₂O₅]₁ = 0.030/0.015 = 2.0. Since 2.0 = 2.0ᵐ, m = 1. The rate law is Rate = k[N₂O₅]. (Verification with Trials 2 and 3: ratio of rates = 2.0, ratio of concentrations = 2.0, confirming m = 1.) (b) 1 point — Using Trial 1: k = Rate / [N₂O₅] = (4.5 × 10⁻⁴ M·s⁻¹) / (0.015 M) = 0.030 s⁻¹. Units are s⁻¹, consistent with first-order kinetics. (c) 1 point — The stoichiometric coefficient of N₂O₅ is 2, but the reaction order is 1. This occurs because the overall reaction is not elementary—it proceeds through a multi-step mechanism. The rate law reflects the kinetics of the rate-determining step, not the overall stoichiometry. Only for an elementary step do the orders necessarily match the molecularity. (d) 1 point — Increasing the temperature would increase the value of k because, according to the Arrhenius equation (k = Ae⁻ᴱᵃ/ᴿᵀ), a higher T decreases the magnitude of Eₐ/RT, increasing the exponential term. The reaction order would remain 1 because reaction order is a property of the mechanism, which does not change with temperature (at least within a moderate temperature range where the mechanism is unchanged).

Question 5: CRITICAL THINKING

4 pts

A student investigates the kinetics of the reaction: BrO₃⁻(aq) + 5 Br⁻(aq) + 6 H⁺(aq) → 3 Br₂(aq) + 3 H₂O(l). The student collects the following initial-rate data: Trial 1: [BrO₃⁻] = 0.10 M, [Br⁻] = 0.10 M, [H⁺] = 0.10 M, Rate = 8.0 × 10⁻⁴ M·s⁻¹ Trial 2: [BrO₃⁻] = 0.20 M, [Br⁻] = 0.10 M, [H⁺] = 0.10 M, Rate = 1.6 × 10⁻³ M·s⁻¹ Trial 3: [BrO₃⁻] = 0.10 M, [Br⁻] = 0.20 M, [H⁺] = 0.10 M, Rate = 1.6 × 10⁻³ M·s⁻¹ Trial 4: [BrO₃⁻] = 0.10 M, [Br⁻] = 0.10 M, [H⁺] = 0.20 M, Rate = 3.2 × 10⁻³ M·s⁻¹ (a) Determine the full rate law, showing your work for each reactant's order. (b) Calculate the rate constant k, with correct units. (c) A classmate claims the overall reaction order must be 11 because the sum of stoichiometric coefficients of the reactants is 1 + 5 + 6 = 12 and 'you subtract 1 for the overall order.' Evaluate this claim using the data. (d) Predict the initial rate if [BrO₃⁻] = 0.15 M, [Br⁻] = 0.25 M, and [H⁺] = 0.15 M.

Show answer

Rubric (4 points): (a) 1 point — Order w.r.t. BrO₃⁻ (Trials 1 & 2): Rate₂/Rate₁ = (1.6 × 10⁻³)/(8.0 × 10⁻⁴) = 2.0; [BrO₃⁻] ratio = 2.0; 2.0 = 2.0ᵃ → a = 1. Order w.r.t. Br⁻ (Trials 1 & 3): Rate₃/Rate₁ = 2.0; [Br⁻] ratio = 2.0; 2.0 = 2.0ᵇ → b = 1. Order w.r.t. H⁺ (Trials 1 & 4): Rate₄/Rate₁ = (3.2 × 10⁻³)/(8.0 × 10⁻⁴) = 4.0; [H⁺] ratio = 2.0; 4.0 = 2.0ᶜ → c = 2. Rate law: Rate = k[BrO₃⁻][Br⁻][H⁺]². (b) 1 point — Overall order = 1 + 1 + 2 = 4. Using Trial 1: k = Rate / ([BrO₃⁻][Br⁻][H⁺]²) = (8.0 × 10⁻⁴) / ((0.10)(0.10)(0.10)²) = (8.0 × 10⁻⁴) / (1.0 × 10⁻⁴) = 8.0 M⁻³·s⁻¹. (c) 1 point — The classmate's claim is incorrect. Reaction orders are experimentally determined, not derived from stoichiometric coefficients. The data clearly show the overall order is 4, not 11. The stoichiometric coefficients reflect the mole ratios of consumption, while the rate law reflects the kinetics of the rate-determining step in a multi-step mechanism. (d) 1 point — Rate = (8.0 M⁻³·s⁻¹)(0.15)(0.25)(0.15)² = (8.0)(0.15)(0.25)(0.0225) = (8.0)(8.4375 × 10⁻⁴) = 6.75 × 10⁻³ M·s⁻¹ ≈ 6.8 × 10⁻³ M·s⁻¹.

SUMMARY

Lesson Summary

Summary

The rate law is the central equation of chemical kinetics, expressed as Rate = k[A]ᵐ[B]ⁿ, where the reaction orders m and n are determined experimentally using the initial-rate method. The rate constant k encodes temperature dependence and has units that adjust with the overall order so that Rate always carries units of M·s⁻¹.

Key takeaways: orders do not necessarily equal stoichiometric coefficients; zero-order reactions have constant rates, first-order rates scale linearly with concentration, and second-order rates scale with the square. The differential rate law provides the foundation for integrated rate laws, half-life expressions, and mechanistic analysis that you will encounter in subsequent lessons.