Reaction Rates - AP Chemistry

SECTION 1

Historical Context & Motivation

Long before chemists could write balanced equations or identify molecular structures, practitioners of alchemy and early chemistry recognized that some transformations occur almost instantaneously—explosions, precipitations—while others, like the rusting of iron or the fermentation of wine, unfold over hours, days, or even years. The systematic study of reaction rates, known as chemical kinetics, began in earnest during the nineteenth century when chemists sought not merely to catalogue what reactions produce, but to understand how quickly they do so and why. This question has enormous practical significance: controlling reaction speed determines the yield of an industrial process, the shelf life of a pharmaceutical, and the efficiency of a catalytic converter.

1850 • Wilhelmy's Sucrose Hydrolysis

Ludwig Wilhelmy performed the first quantitative kinetics experiment, tracking the rate of sucrose inversion with a polarimeter and establishing that the rate was proportional to the concentration of sucrose remaining.

1864 • Guldberg & Waage — Law of Mass Action

Cato Guldberg and Peter Waage formulated the law of mass action, connecting the rate of a reaction to the concentrations of reactants raised to experimentally determined powers—laying the algebraic foundation of rate laws.

1889 • Arrhenius Equation

Svante Arrhenius proposed the exponential relationship between the rate constant and temperature, introducing the concept of activation energy (Eₐ) and the pre-exponential factor A.

1913 • Bodenstein's Steady-State Approximation

Max Bodenstein introduced the steady-state approximation for reactive intermediates, enabling the derivation of rate laws for complex multi-step mechanisms from elementary step kinetics.

1935 • Eyring's Transition State Theory

Henry Eyring and colleagues developed transition state theory, providing a statistical-mechanical foundation for the Arrhenius equation and connecting reaction rates to the geometry and energy of activated complexes.

The central question of kinetics remains deceptively simple: How fast does a given reaction proceed under specified conditions, and what molecular-level events dictate that speed? Answering it requires measuring changes in concentration over time, formulating mathematical rate laws, and connecting those macroscopic observations to the microscopic collisions and rearrangements of atoms. This lesson develops the tools you need to do exactly that.

SECTION 2

Core Principles & Definitions

Before diving into equations and mechanisms, it is essential to establish a precise vocabulary. The rate of a reaction is defined as the change in concentration of a reactant or product per unit time, expressed in units of mol·L⁻¹·s⁻¹ (or M/s). Because reactant concentrations decrease while product concentrations increase, convention dictates that rates are always reported as positive values. For a generic reaction aA + bB → cC + dD, the rate is related to the disappearance of reactants and the appearance of products through stoichiometric coefficients, ensuring a single, unambiguous rate regardless of which species is monitored.

Average vs. Instantaneous Rate

The average rate is Δ[species]/Δt over a finite interval. The instantaneous rate is the slope of the concentration-vs-time curve at a single moment—mathematically, the derivative d[species]/dt.

Rate Law & Rate Constant

The rate law expresses rate as a function of reactant concentrations: rate = k[A]ᵐ[B]ⁿ. The rate constant k encapsulates temperature and catalyst effects; its units depend on overall reaction order.

Reaction Order

The exponent on each reactant concentration in the rate law is its individual order; the sum of all exponents is the overall order. Orders are determined experimentally—they are NOT deduced from stoichiometric coefficients.

Factors Affecting Rate

Four principal factors influence reaction rate: concentration of reactants, temperature, the presence of a catalyst, and surface area (for heterogeneous reactions). Each acts by changing either collision frequency or the fraction of collisions with sufficient energy.

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KEY TAKEAWAY

Think of a chemical reaction like rush-hour traffic merging onto a highway. The rate at which cars clear the on-ramp depends on how many cars are waiting (concentration), how aggressively drivers accelerate (temperature/energy), and whether a traffic officer is directing flow (catalyst). Just as a traffic engineer can speed things up without building new roads, a catalyst accelerates a reaction without being consumed. Crucially, the rate law is the mathematical speed limit—it must be measured from real data, not guessed from the road map (balanced equation).

SECTION 3

Concentration vs. Time — A Visual Tour

The most fundamental kinetics experiment involves measuring the concentration of a reactant or product at successive time intervals and plotting the data. The shape of the resulting curve reveals the reaction order. The diagram below illustrates concentration-versus-time profiles for zero-, first-, and second-order reactions, each starting at the same initial concentration. Notice how the curvature becomes more pronounced as the order increases: a zero-order reaction depletes linearly, a first-order reaction follows an exponential decay, and a second-order reaction curves even more steeply at early times yet lingers at low concentrations longer.

All three curves begin at [A]₀ = 1.00 M. The zero-order curve (yellow) is linear, the first-order curve (cyan) is an exponential decay, and the second-order curve (violet) decays steeply at first but approaches zero more slowly.

The shape of the raw [A]-vs-time curve already suggests the order, but a more rigorous diagnostic is to plot linearized forms. For a zero-order reaction, [A] versus t is linear. For first order, ln[A] versus t is linear. For second order, 1/[A] versus t is linear. The slope of the appropriate linear plot yields the rate constant k (with the correct sign convention), and the y-intercept gives the initial concentration in the corresponding transformed variable.

Gas collection with a movable syringe A gas syringe collects gas from a reaction, allowing its volume to be followed over time.

SECTION 4

Mathematical Framework

The mathematical heart of kinetics is the differential rate law, which expresses the instantaneous rate in terms of concentrations. By integrating this expression, we obtain the integrated rate law, which directly relates concentration to elapsed time. Both forms appear on the AP Chemistry exam and serve complementary purposes: the differential form is best for initial-rate data, while the integrated form is best for concentration-time data.

General Differential Rate Law

DIFFERENTIAL RATE LAW

Rate = k [A]ᵐ [B]ⁿ

k = rate constant; [A], [B] = molar concentrations of reactants; m, n = reaction orders with respect to A and B (determined experimentally). Overall order = m + n.

Integrated Rate Laws

ZERO ORDER (n = 0)

[A]ₜ = [A]₀ − kt

Plot [A] vs. t → straight line with slope = −k and y-intercept = [A]₀. Half-life: t₁/₂ = [A]₀ / (2k).

FIRST ORDER (n = 1)

ln[A]ₜ = ln[A]₀ − kt

Plot ln[A] vs. t → straight line with slope = −k. Half-life: t₁/₂ = 0.693 / k (constant, independent of [A]₀).

SECOND ORDER (n = 2)

1/[A]ₜ = 1/[A]₀ + kt

Plot 1/[A] vs. t → straight line with slope = +k. Half-life: t₁/₂ = 1 / (k[A]₀), which increases as [A] decreases.

Arrhenius Equation

ARRHENIUS EQUATION

k = A e^(−Eₐ / RT)

A = pre-exponential (frequency) factor; Eₐ = activation energy (J·mol⁻¹); R = 8.314 J·mol⁻¹·K⁻¹; T = absolute temperature (K). The linearized form is ln k = −Eₐ/R × (1/T) + ln A.

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AP Exam Tip

When using the method of initial rates, compare two trials in which only one reactant concentration changes. The ratio of rates equals the ratio of concentrations raised to the unknown order: Rate₂/Rate₁ = ([A]₂/[A]₁)ᵐ. Solve for m by inspection or logarithms.

SECTION 5

Determining Reaction Order — Linearized Plots

One of the most important skills on the AP Chemistry exam is identifying the order of a reaction from experimental data. Two complementary strategies exist: the method of initial rates and the graphical (integrated rate law) method. The method of initial rates involves performing multiple trials in which the initial concentration of only one reactant is varied while others are held constant; comparing the resulting initial rates reveals the order with respect to each reactant. The graphical method involves collecting concentration-time data from a single trial and testing which linearized plot—[A] vs. t, ln[A] vs. t, or 1/[A] vs. t—yields a straight line.

Top row: Each panel shows the linearized plot for one reaction order. Only the correct transformation yields a straight line. Bottom panel: Half-life expressions and their dependence on initial concentration for each order.

On the AP exam, you may be presented with concentration-time data and asked to determine the order graphically. The diagnostic strategy is straightforward: compute [A], ln[A], and 1/[A] at each time point, plot each against t, and identify which yields the best straight line. Alternatively, you might be given a graph and asked to read k directly from its slope. Remember that for zero- and first-order plots the slope is −k, whereas for the second-order plot the slope is +k.

SECTION 6

Worked Example — Method of Initial Rates

Consider the reaction: 2 NO(g) + O₂(g) → 2 NO₂(g). Three experiments yield the following initial-rate data:

Trial	[NO]₀ (M)	[O₂]₀ (M)	Initial Rate (M/s)
1	0.010	0.010	2.5 × 10⁻⁵
2	0.020	0.010	1.0 × 10⁻⁴
3	0.010	0.020	5.0 × 10⁻⁵

Initial rate data for 2 NO + O₂ → 2 NO₂

Determine the Rate Law and Rate Constant

1
Step 1 — Find the order with respect to NO

Compare Trials 1 and 2, where [O₂] is held constant at 0.010 M. [NO] doubles from 0.010 M to 0.020 M (factor of 2). The rate increases from 2.5 × 10⁻⁵ to 1.0 × 10⁻⁴ (factor of 4). Since 2ᵐ = 4, the order m = 2. The reaction is second order in NO.

m = 2
2
Step 2 — Find the order with respect to O₂

Compare Trials 1 and 3, where [NO] is held constant at 0.010 M. [O₂] doubles from 0.010 M to 0.020 M (factor of 2). The rate doubles from 2.5 × 10⁻⁵ to 5.0 × 10⁻⁵ (factor of 2). Since 2ⁿ = 2, the order n = 1. The reaction is first order in O₂.

n = 1
3
Step 3 — Write the rate law

Combining the results: Rate = k[NO]²[O₂]¹. The overall reaction order is 2 + 1 = 3 (third order overall).

Rate = k[NO]²[O₂]
4
Step 4 — Calculate k

Substitute data from any trial (using Trial 1): 2.5 × 10⁻⁵ = k (0.010)²(0.010) = k (1.0 × 10⁻⁶). Therefore k = 2.5 × 10⁻⁵ / 1.0 × 10⁻⁶ = 25. The units for a third-order rate constant are M⁻²·s⁻¹.

k = 25 M⁻²·s⁻¹

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Verification

Always verify your k value by substituting into a different trial. Using Trial 2: Rate = 25 × (0.020)² × (0.010) = 25 × 4.0 × 10⁻⁶ = 1.0 × 10⁻⁴ M/s. This matches the measured rate, confirming the result.

SECTION 7

Factors Affecting Rates — Strengths & Limitations of Models

Two theoretical models underpin our understanding of why reaction rates depend on temperature and molecular identity: collision theory and transition state theory. Collision theory posits that reactant molecules must collide with sufficient kinetic energy (≥ Eₐ) and proper spatial orientation to react. Transition state theory refines this picture by introducing the concept of an activated complex—a transient, high-energy arrangement of atoms at the top of the energy barrier. Both models predict the Arrhenius relationship between k and temperature, but transition state theory provides deeper insight into the role of molecular geometry and entropy.

Factor	Effect on Rate	Molecular Explanation
↑ Concentration	Rate generally increases	More molecules per unit volume → higher collision frequency
↑ Temperature	Rate increases (often ~doubles per 10 K)	Greater average kinetic energy → larger fraction of molecules exceed Eₐ; slightly higher collision frequency
Catalyst present	Rate increases (sometimes by orders of magnitude)	Provides an alternative pathway with lower Eₐ; does not change ΔG or ΔH of the overall reaction
↑ Surface area	Rate increases (heterogeneous reactions)	More exposed reactant sites available for contact; applies to solids reacting with liquids or gases
Nature of reactants	Varies	Ionic reactions in solution are often fast (no bonds broken); reactions requiring multiple bond rearrangements are slower

Summary of factors affecting reaction rate

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KEY TAKEAWAY

Collision theory is like a model of a billiard table: it correctly predicts that more balls (higher concentration) and faster shots (higher temperature) produce more collisions, but it oversimplifies by treating molecules as structureless spheres. Transition state theory adds the nuance of molecular shape—like recognizing that a key must be oriented correctly to enter a lock, not just thrown at it with enough force. The Arrhenius equation emerges from both frameworks, but only transition state theory explains why the pre-exponential factor A varies between reactions.

Accessible surfaces of a fragmented solid Breaking a solid into fragments exposes more surface to surrounding reactants.

SECTION 8

Connection to Reaction Mechanisms

Reaction rates provide the experimental bridge to understanding reaction mechanisms—the step-by-step molecular-level sequences through which reactants transform into products. A proposed mechanism must satisfy two criteria: the elementary steps must sum to give the overall balanced equation, and the rate law derived from the mechanism must match the experimentally determined rate law. The rate-determining step (RDS) is typically the slowest elementary step, and its molecularity dictates the form of the predicted rate law. Understanding mechanisms is the natural extension of kinetics and forms a core topic in AP Chemistry Unit 5.

Concept	Reaction Rates (This Lesson)	Reaction Mechanisms (Advanced)
Focus	How fast the overall reaction proceeds	The sequence of elementary steps that constitute the pathway
Rate law	Determined experimentally from data	Derived theoretically from proposed elementary steps
Order	Measured from initial rates or integrated plots	Equals molecularity of the rate-determining step (for elementary steps)
Intermediates	Not explicitly considered	Species produced in one step and consumed in a subsequent step; may appear in derived rate law
Energy profile	Single activation energy Eₐ from Arrhenius	Multi-step energy diagram with intermediates in energy wells between transition states

Reaction rates vs. reaction mechanisms

As you advance to the mechanism-focused portions of AP Chemistry, keep in mind that the experimental rate law is the ultimate arbiter of any proposed mechanism. A mechanism is only as credible as its agreement with measured kinetics data. The tools you have learned here—initial rate comparisons, integrated rate law plots, and Arrhenius analysis—provide the empirical foundation on which mechanistic hypotheses are tested.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student measures the half-life of a reaction at two different initial concentrations and finds that the half-life is the same in both experiments. Which statement best explains this observation?

A. The reaction is zero order, and the half-life is always independent of initial concentration.
B. The reaction is first order, and its half-life equals 0.693/k, which does not depend on [A]₀.
C. The reaction is second order, and doubling [A]₀ halves the half-life, keeping the measured value constant.
D. The reaction order cannot be determined from half-life data alone.

Show answer (B)

The correct answer is B. For a first-order reaction, t₁/₂ = 0.693/k, which depends only on the rate constant and not on [A]₀. Option A is incorrect because the zero-order half-life is t₁/₂ = [A]₀/(2k), which does depend on initial concentration. Option C is incorrect because for a second-order reaction, t₁/₂ = 1/(k[A]₀), which also depends on [A]₀; doubling [A]₀ would halve t₁/₂, not keep it the same. Option D is incorrect because a constant half-life is, in fact, a diagnostic fingerprint uniquely associated with first-order kinetics.

Question 2: BASIC CALCULATION

The decomposition of N₂O₅ is first order with a rate constant of 4.80 × 10⁻⁴ s⁻¹ at 45 °C. If the initial concentration is 0.200 M, what is the concentration after 1500 s?

A. 0.0560 M
B. 0.0975 M
C. 0.144 M
D. 0.200 M

Show answer (B)

The correct answer is B. Using the first-order integrated rate law: ln[A]ₜ = ln[A]₀ − kt = ln(0.200) − (4.80 × 10⁻⁴)(1500) = −1.609 − 0.720 = −2.329. Therefore [A]ₜ = e⁻²·³²⁹ = 0.0974 M ≈ 0.0975 M. Option A (0.0560 M) would result from an incorrectly large kt product. Option C (0.144 M) corresponds to using the zero-order equation [A] = [A]₀ − kt = 0.200 − 0.720 which does not apply here. Option D (0.200 M) implies no reaction has occurred and ignores the passage of time entirely.

Question 3: INTERMEDIATE

The rate constant for a reaction is 0.0140 s⁻¹ at 400 K and 0.110 s⁻¹ at 450 K. What is the activation energy of this reaction? (R = 8.314 J·mol⁻¹·K⁻¹)

A. 6.17 × 10⁴ J/mol
B. 3.09 × 10⁴ J/mol
C. 1.23 × 10⁵ J/mol
D. 7.86 × 10³ J/mol

Show answer (A)

The correct answer is A. Using the two-point form of the Arrhenius equation: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂). First, compute the ratio of rate constants: k₂/k₁ = 0.110/0.0140 = 7.857, so ln(7.857) = 2.061. Next, compute the temperature term: 1/T₁ − 1/T₂ = 1/400 − 1/450 = 0.002500 − 0.002222 = 2.778 × 10⁻⁴ K⁻¹. Solving for Eₐ: Eₐ = ln(k₂/k₁) × R / (1/T₁ − 1/T₂) = 2.061 × 8.314 / 2.778 × 10⁻⁴ = 17.13 / 2.778 × 10⁻⁴ = 6.17 × 10⁴ J/mol. Option B halves the correct value, which could result from an algebraic error such as omitting a factor of 2 in the temperature term. Option C doubles the correct value, possibly from reversing the temperature subtraction sign and compensating incorrectly. Option D is far too small and would imply a negligible temperature dependence of the rate constant.

Question 4: APPLIED

5 pts

The decomposition of hydrogen peroxide, 2 H₂O₂(aq) → 2 H₂O(l) + O₂(g), is studied at 25 °C. The following data are collected: Time (s): 0, 120, 300, 600, 1200 [H₂O₂] (M): 1.000, 0.910, 0.780, 0.590, 0.370 (a) Using the integrated rate law method, determine whether this reaction is zero, first, or second order with respect to H₂O₂. Show your work clearly. (b) Calculate the rate constant k, including proper units. (c) Calculate the half-life of this reaction. (d) Predict [H₂O₂] at t = 1800 s. (e) If a catalyst (MnO₂) is added, explain qualitatively how the rate constant, activation energy, and half-life would change.

Show answer

Point-by-point rubric: (a) 1 point — Determine the reaction order by testing linearized plots. Calculate ln[H₂O₂] at each time: 0, −0.0943, −0.2485, −0.5276, −0.9943. A plot of ln[H₂O₂] vs. t yields evenly spaced data consistent with a straight line, confirming FIRST ORDER. (Award 1 point for correct order with supporting calculation or graph description.) (b) 1 point — Using any two data points in the first-order integrated rate law: k = (ln[A]₀ − ln[A]ₜ)/t. Using t = 1200 s: k = (0 − (−0.9943))/1200 = 0.9943/1200 = 8.29 × 10⁻⁴ s⁻¹. (Award 1 point for correct k value with proper units of s⁻¹.) (c) 1 point — t₁/₂ = 0.693/k = 0.693/(8.29 × 10⁻⁴) ≈ 836 s. (Award 1 point for correct half-life using the first-order formula.) (d) 1 point — [H₂O₂] at 1800 s: ln[A]₁₈₀₀ = 0 − (8.29 × 10⁻⁴)(1800) = −1.492. [A]₁₈₀₀ = e⁻¹·⁴⁹² = 0.224 M. (Award 1 point for correct prediction using the integrated rate law.) (e) 1 point — A catalyst lowers the activation energy Eₐ by providing an alternative reaction pathway. This increases the rate constant k (since k = Ae⁻ᴱᵃ/ᴿᵀ, lowering Eₐ increases the exponential term). Because k increases and t₁/₂ = 0.693/k, the half-life decreases. The catalyst does not change the overall thermodynamics (ΔH or ΔG). (Award 1 point for correctly explaining all three changes: Eₐ decreases, k increases, t₁/₂ decreases.)

Question 5: CRITICAL THINKING

5 pts

A research team studies the reaction A → Products at five temperatures. They report the following data: T (K): 280, 300, 320, 340, 360 k (s⁻¹): 1.20 × 10⁻⁵, 7.50 × 10⁻⁵, 3.90 × 10⁻⁴, 1.70 × 10⁻³, 6.40 × 10⁻³ (a) Describe how you would construct an Arrhenius plot from these data. What quantities would you plot on each axis? (b) Calculate 1/T and ln k for each data point. (c) Using the data from T = 280 K and T = 360 K, calculate the activation energy Eₐ. (d) The researchers claim that a catalyst reduces Eₐ to 40 kJ/mol. Predict the new rate constant at 300 K, assuming the pre-exponential factor A remains the same. (e) Evaluate whether the "rule of thumb" that reaction rate approximately doubles for every 10 K increase in temperature holds for this reaction. Use specific data from the table to support your answer.

Show answer

Point-by-point rubric: (a) 1 point — An Arrhenius plot graphs ln k (y-axis) vs. 1/T (x-axis). The result should be a straight line with slope = −Eₐ/R and y-intercept = ln A. (Award 1 point for correctly identifying both axes and the meaning of the slope.) (b) 1 point — Calculated values: 1/T (K⁻¹): 3.571 × 10⁻³, 3.333 × 10⁻³, 3.125 × 10⁻³, 2.941 × 10⁻³, 2.778 × 10⁻³ ln k: −11.33, −9.50, −7.85, −6.38, −5.05 (Award 1 point for at least 4 of 5 pairs calculated correctly.) (c) 1 point — Using T₁ = 280 K and T₂ = 360 K: ln(k₂/k₁) = ln(6.40 × 10⁻³ / 1.20 × 10⁻⁵) = ln(533.3) = 6.28. 1/T₁ − 1/T₂ = 3.571 × 10⁻³ − 2.778 × 10⁻³ = 7.94 × 10⁻⁴ K⁻¹. Eₐ = 6.28 × 8.314 / 7.94 × 10⁻⁴ = 52.2 / 7.94 × 10⁻⁴ = 6.57 × 10⁴ J/mol ≈ 65.7 kJ/mol. (Award 1 point for a correct Eₐ value between 64 and 68 kJ/mol.) (d) 1 point — First find A using the uncatalyzed data at 300 K: ln A = ln k + Eₐ/(RT) = −9.50 + 65700/(8.314 × 300) = −9.50 + 26.35 = 16.85, so A = e¹⁶·⁸⁵ = 2.08 × 10⁷ s⁻¹. With Eₐ,cat = 40000 J/mol: ln k_cat = ln A − Eₐ,cat/(RT) = 16.85 − 40000/(8.314 × 300) = 16.85 − 16.04 = 0.81. k_cat = e⁰·⁸¹ = 2.25 s⁻¹. (Award 1 point for correctly calculating the new k at 300 K, showing that the catalyst increases the rate constant by roughly 5 orders of magnitude.) (e) 1 point — Test the rule using adjacent 20 K intervals: k₃₂₀/k₃₀₀ = 3.90 × 10⁻⁴ / 7.50 × 10⁻⁵ = 5.2 for a 20 K jump. If the rate doubled every 10 K, we would expect a factor of 2² = 4 per 20 K. Similarly, k₃₆₀/k₃₄₀ = 6.40 × 10⁻³ / 1.70 × 10⁻³ = 3.76 for a 20 K jump. The actual ratios (3.8–5.2 per 20 K) are somewhat larger than the expected factor of 4, but in the right ballpark. The rule of thumb is a rough approximation—it works better for reactions with Eₐ ≈ 50 kJ/mol and near room temperature, and this reaction (Eₐ ≈ 66 kJ/mol) shows slightly more sensitivity to temperature. (Award 1 point for using specific data to evaluate the claim and providing a nuanced conclusion.)

SUMMARY

Reaction Rates — Key Concepts at a Glance

Summary

The rate of a reaction is the change in concentration per unit time and is expressed in M/s. The rate law (Rate = k[A]ᵐ[B]ⁿ) is determined experimentally—never from stoichiometric coefficients. The method of initial rates reveals the order with respect to each reactant by comparing trials where only one concentration changes. Integrated rate laws relate concentration to time and are used to construct linearized diagnostic plots: [A] vs. t for zero order, ln[A] vs. t for first order, and 1/[A] vs. t for second order.

The four major factors that influence rate are concentration, temperature, catalysts, and surface area. The Arrhenius equation (k = Ae⁻ᴱᵃ/ᴿᵀ) quantifies the temperature dependence of the rate constant and provides a pathway to determining the activation energy. A unique diagnostic of first-order kinetics is a constant half-life (t₁/₂ = 0.693/k), independent of initial concentration. These tools form the empirical foundation for proposing and testing reaction mechanisms.