Concentration Changes Over Time - AP Chemistry

SECTION 1

Historical Context & Motivation

The desire to understand how fast chemical reactions proceed — and why some reactions are explosive while others take millennia — has driven scientific inquiry since the birth of modern chemistry. Early chemists observed that the rate of a reaction depended on the amounts of reactants present, but expressing this dependence mathematically required tools from calculus that were only beginning to be applied to chemistry in the nineteenth century. The development of integrated rate laws gave chemists the ability to predict the concentration of a reactant at any future time, transforming kinetics from qualitative description into quantitative science.

1850 • Wilhelmy's Sucrose Hydrolysis

Ludwig Wilhelmy performed one of the first quantitative kinetics experiments, measuring the rate of acid-catalyzed sucrose inversion with a polarimeter and showing it followed a first-order rate law.

1864 • Guldberg & Waage — Law of Mass Action

Cato Guldberg and Peter Waage formalized the relationship between reaction rate and concentration, establishing the foundation for differential rate laws.

1884 • van 't Hoff's Études

Jacobus van 't Hoff published his landmark work classifying reactions by order and integrating the differential rate expressions to yield concentration-versus-time equations.

1889 • Arrhenius Equation

Svante Arrhenius connected the rate constant k to temperature, completing the kinetic framework so that both concentration and temperature effects on rate could be predicted.

Despite these advances, a central question remained for any new reaction encountered in the laboratory: Given a starting concentration, how much reactant will remain after a given time, and how can we determine the reaction order from experimental data? Answering this question requires moving from the differential rate law (rate as a function of instantaneous concentration) to the integrated rate law (concentration as a function of time). That transition is the subject of this lesson.

SECTION 2

Core Principles & Definitions

Before deriving the integrated forms, it is essential to distinguish between a differential rate law and an integrated rate law. The differential rate law expresses the instantaneous rate in terms of concentration (e.g., rate = k[A]ⁿ), while the integrated form solves that differential equation to give [A] as a function of t. Each reaction order yields a distinct integrated expression, a unique linear plot, and a characteristic half-life relationship.

Reaction Order

The exponent n in rate = k[A]ⁿ determines how concentration enters the rate law. Common orders are 0, 1, and 2. The order is determined experimentally, not from stoichiometry.

Integrated Rate Law

The result of integrating the differential rate law from t = 0 to t, yielding [A] (or a function of [A]) as a linear function of time. Each order has its own straight-line form.

Half-Life (t₁/₂)

The time required for [A] to fall to half its initial value. For first-order reactions, t₁/₂ is constant and independent of [A]₀, a defining experimental signature.

Graphical Analysis

Plotting the appropriate function of [A] versus t yields a straight line whose slope contains k. This is the primary method for determining reaction order from experimental data.

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KEY TAKEAWAY

Think of the integrated rate law as the GPS of kinetics: if the differential rate law tells you your speed at any moment, the integrated rate law tells you where you'll be after driving for a given time. Each reaction order (0, 1, 2) corresponds to a different kind of 'traffic pattern' — constant consumption, exponential decay, or inverse growth — and produces its own linear map when plotted correctly.

SECTION 3

Concentration vs. Time — Visual Overview

All three curves begin at the same initial concentration [A]₀ = 1.0 M. The zero-order curve (amber) is a straight line; first-order (cyan) shows exponential decay; and second-order (violet) curves more gradually and never quite reaches zero on the same timescale.

The diagram above captures the essential visual difference among the three most common reaction orders. For a zero-order reaction, concentration decreases linearly with time because the rate is constant regardless of how much reactant remains. The first-order reaction exhibits exponential decay — the rate slows proportionally as [A] drops, producing a curve that always takes the same amount of time to halve. The second-order reaction decays even more slowly at low concentrations because the rate depends on [A]², making complete consumption practically unreachable in finite time. On the AP exam, recognizing these characteristic shapes is often the first step in a kinetics problem.

SECTION 4

Mathematical Framework — Integrated Rate Laws

Each integrated rate law is derived by separating variables in the differential rate law −d[A]/dt = k[A]ⁿ and integrating from t = 0 (where [A] = [A]₀) to an arbitrary time t. Below are the three key results for the AP Chemistry course.

ZERO-ORDER INTEGRATED RATE LAW

[A] = [A]₀ − kt

A plot of [A] vs. t is linear with slope = −k and y-intercept = [A]₀. Half-life: t₁/₂ = [A]₀ / (2k).

FIRST-ORDER INTEGRATED RATE LAW

ln[A] = ln[A]₀ − kt

Equivalently, [A] = [A]₀ e^−kt. A plot of ln[A] vs. t is linear with slope = −k. Half-life: t₁/₂ = ln 2 / k ≈ 0.693 / k (constant, independent of [A]₀).

SECOND-ORDER INTEGRATED RATE LAW

1/[A] = 1/[A]₀ + kt

A plot of 1/[A] vs. t is linear with slope = +k and y-intercept = 1/[A]₀. Half-life: t₁/₂ = 1 / (k[A]₀), which increases as [A]₀ decreases.

💡

AP Exam Tip

You will not be asked to perform the calculus integration on the AP exam, but you must be able to identify which plot yields a straight line for each order, extract k from the slope, and compute half-lives. Memorize: [A] vs. t for zero-order, ln[A] vs. t for first-order, 1/[A] vs. t for second-order.

SECTION 5

Graphical Method — Determining Reaction Order

The most reliable experimental method for determining reaction order is the method of graphical analysis. Given concentration-versus-time data, you construct three plots — [A] vs. t, ln[A] vs. t, and 1/[A] vs. t — and determine which gives the best straight line. The linear plot reveals the order, and the slope gives k (with appropriate sign conventions).

For each reaction order, the correct transformation of [A] yields a straight line. Note that the second-order plot has a positive slope (+k), while the zero-order and first-order plots have negative slopes (−k).

Order	Linear Plot	Slope	y-Intercept	Half-Life
0	[A] vs. t	−k	[A]₀	[A]₀ / (2k)
1	ln[A] vs. t	−k	ln[A]₀	0.693 / k
2	1/[A] vs. t	+k	1/[A]₀	1 / (k[A]₀)

Summary of integrated rate law properties for orders 0, 1, and 2.

SECTION 6

Worked Example — First-Order Decomposition

The decomposition of dinitrogen pentoxide, 2 N₂O₅(g) → 4 NO₂(g) + O₂(g), is first-order with k = 5.1 × 10⁻⁴ s⁻¹ at 45 °C. If the initial concentration of N₂O₅ is 0.250 M, find the concentration after 1200 s and the half-life of the reaction.

First-Order Integrated Rate Law Calculation

1
Step 1 — Identify Given Values and Rate Law

The reaction is first-order, so the integrated rate law is ln[A] = ln[A]₀ − kt. We have [A]₀ = 0.250 M, k = 5.1 × 10⁻⁴ s⁻¹, and t = 1200 s.
2
Step 2 — Substitute into the Integrated Rate Law

ln[A] = ln(0.250) − (5.1 × 10⁻⁴ s⁻¹)(1200 s) = −1.386 − 0.612 = −1.998

ln[A] = −1.998
3
Step 3 — Solve for [A]

[A] = e⁻¹·⁹⁹⁸ = 0.136 M. After 1200 seconds, the N₂O₅ concentration has decreased from 0.250 M to 0.136 M.

[N₂O₅] = 0.136 M
4
Step 4 — Calculate the Half-Life

For a first-order reaction, t₁/₂ = 0.693 / k = 0.693 / (5.1 × 10⁻⁴ s⁻¹) = 1359 s ≈ 1.36 × 10³ s. Notice this is independent of [A]₀; it doesn't matter what starting concentration we chose.

t₁/₂ ≈ 1360 s
5
Step 5 — Verify Reasonableness

After 1200 s (which is slightly less than one half-life of 1360 s), we expect [A] to be slightly more than half of 0.250 M = 0.125 M. Our answer of 0.136 M is consistent, confirming the calculation.

SECTION 7

Comparing Reaction Orders — Strengths & Limitations

Integrated rate laws are powerful tools, but they come with assumptions and caveats. Each order model applies only when the reaction genuinely obeys that rate law over the measured time range. Furthermore, the graphical method assumes that one reactant dominates or that pseudo-order conditions (large excess of one reagent) hold. Below is a comparison of key features and common pitfalls for each order.

Feature	Zero-Order	First-Order	Second-Order
Rate depends on [A]?	No — rate is constant	Yes — directly proportional	Yes — proportional to [A]²
Half-life behavior	Decreases as [A]₀ decreases	Constant (independent of [A]₀)	Increases as [A]₀ decreases
Units of k	M s⁻¹ (or mol L⁻¹ s⁻¹)	s⁻¹	M⁻¹ s⁻¹ (or L mol⁻¹ s⁻¹)
Common examples	Enzyme-saturated reactions, surface catalysis at high coverage	Radioactive decay, N₂O₅ decomposition	NO₂ decomposition, some bimolecular gas-phase reactions
Limitation	Predicts negative [A] after t > [A]₀/k (unphysical)	[A] never truly reaches 0; model holds only while mechanism is unchanged	Applies only to single-reactant or pseudo-second-order conditions

Comparison of integrated rate law properties for reaction orders 0, 1, and 2.

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KEY TAKEAWAY

The half-life trend is the single most diagnostic feature for identifying reaction order from experimental data. If successive half-lives remain constant, the reaction is first-order. If each successive half-life doubles, the reaction is second-order. If each successive half-life halves, the reaction is zero-order. On the AP exam, this pattern frequently appears in both MCQ and FRQ formats.

SECTION 8

Connection to Advanced Theory

The integrated rate laws presented above apply to elementary or pseudo-elementary processes involving a single concentration variable. In more advanced kinetics — encountered in university-level physical chemistry — several extensions emerge that build directly on this foundation.

AP Chemistry Scope	Advanced Extension
Single-reactant integrated rate laws (orders 0, 1, 2)	Multi-reactant integrated rate laws; mixed-order kinetics; fractional orders
Half-life as a characteristic time	Relaxation times, mean lifetimes, and lifetime distributions in complex decay
Graphical determination of order using linear plots	Nonlinear regression and fitting to coupled ODEs; Bayesian parameter estimation
Rate constant k at a single temperature	Arrhenius and Eyring equations connecting k to activation energy and transition-state thermodynamics

Bridging AP-level integrated rate laws to university physical chemistry.

The Arrhenius equation, k = Ae^−Eₐ/RT, is tested on the AP exam and links directly to integrated rate laws: once you know k at a given temperature, you can plug it into any integrated rate law to predict concentrations. In future coursework, the Eyring equation replaces the empirical Arrhenius expression with transition-state theory, providing a deeper thermodynamic interpretation of the rate constant.

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 trials. Which of the following is the most likely order of the reaction?

A. Zero-order
B. First-order
C. Second-order
D. Third-order

Show answer (B)

B is correct. A constant half-life independent of initial concentration is the defining characteristic of first-order kinetics (t₁/₂ = 0.693/k). A is wrong because zero-order half-life is t₁/₂ = [A]₀/(2k), which depends on [A]₀ — doubling [A]₀ doubles t₁/₂. C is wrong because second-order half-life is t₁/₂ = 1/(k[A]₀), so halving [A]₀ doubles t₁/₂. D is wrong because third-order half-life also depends on [A]₀ (inversely proportional to [A]₀²).

Question 2: BASIC CALCULATION

The decomposition of SO₂Cl₂ is first-order with k = 2.2 × 10⁻⁵ s⁻¹. If [SO₂Cl₂]₀ = 0.100 M, what is [SO₂Cl₂] after 2.50 × 10⁴ s?

A. 0.0423 M
B. 0.0577 M
C. 0.0450 M
D. 0.0950 M

Show answer (B)

B is correct. Using ln[A] = ln[A]₀ − kt: ln[A] = ln(0.100) − (2.2 × 10⁻⁵)(2.50 × 10⁴) = −2.303 − 0.550 = −2.853. Therefore [A] = e⁻²·⁸⁵³ = 0.0577 M. A (0.0423 M) results from using kt = 0.55 but incorrectly squaring the exponential. C (0.0450 M) would come from misidentifying the reaction as zero-order ([A] = 0.100 − 0.055 = 0.045). D (0.0950 M) comes from treating it as zero-order with a factor-of-ten error in kt.

Question 3: INTERMEDIATE

A reaction A → products is studied, and the following data are obtained: t (s): 0 100 200 300 [A] (M): 0.800 0.606 0.459 0.348 Which integrated rate law best fits the data, and what is the approximate value of k?

A. Zero-order; k = 1.5 × 10⁻³ M s⁻¹
B. First-order; k = 2.8 × 10⁻³ s⁻¹
C. Second-order; k = 2.8 × 10⁻³ M⁻¹ s⁻¹
D. First-order; k = 5.6 × 10⁻³ s⁻¹

Show answer (B)

B is correct. Testing first-order: ln(0.800) = −0.223, ln(0.606) = −0.501, ln(0.459) = −0.779, ln(0.348) = −1.056. The differences are constant (≈ −0.278 per 100 s), confirming linearity. Slope = −0.278/100 = −2.78 × 10⁻³ s⁻¹, so k ≈ 2.8 × 10⁻³ s⁻¹. A is wrong because [A] vs. t is not linear (the decrements are 0.194, 0.147, 0.111 — not constant). C is wrong because even though the numerical value matches, the order and units are incorrect for the data pattern. D doubles the correct k value, which would predict much faster decay than observed.

Question 4: APPLIED

4 pts

A pharmaceutical company needs a drug (initial concentration 0.0500 M in solution) to retain at least 90% of its potency over a 2-year shelf life. Degradation of the drug is first-order with k = 1.80 × 10⁻⁹ s⁻¹ at 25 °C. (a) Calculate the concentration of the drug remaining after exactly 2 years (6.31 × 10⁷ s). (b) Determine whether the drug meets the 90% potency requirement. (c) Calculate the half-life of the drug. (d) The company considers storing the drug at 5 °C where k = 3.60 × 10⁻¹⁰ s⁻¹. Calculate the fraction remaining after 2 years at this temperature and comment on the benefit.

Show answer

(a) [1 point] Using ln[A] = ln[A]₀ − kt: ln[A] = ln(0.0500) − (1.80 × 10⁻⁹)(6.31 × 10⁷) = −2.996 − 0.1136 = −3.110. [A] = e⁻³·¹¹⁰ = 0.0446 M. (b) [1 point] Fraction remaining = 0.0446/0.0500 = 0.893, which is 89.3%. This is below 90%, so the drug does NOT meet the potency requirement at 25 °C. (c) [1 point] t₁/₂ = 0.693/(1.80 × 10⁻⁹ s⁻¹) = 3.85 × 10⁸ s ≈ 12.2 years. (d) [1 point] At 5 °C: ln[A] = ln(0.0500) − (3.60 × 10⁻¹⁰)(6.31 × 10⁷) = −2.996 − 0.02272 = −3.019. [A] = 0.0489 M. Fraction remaining = 0.0489/0.0500 = 0.978 = 97.8%. This greatly exceeds the 90% threshold, demonstrating that lowering temperature (which reduces k per the Arrhenius equation) is an effective strategy to extend shelf life.

Question 5: CRITICAL THINKING

5 pts

A student studies the reaction 2 NO₂(g) → 2 NO(g) + O₂(g) at 300 °C and collects the following data: t (s): 0 50 100 200 400 [NO₂] (M): 0.0100 0.00714 0.00556 0.00385 0.00256 (a) Determine the reaction order with respect to NO₂ by testing each integrated rate law. Show calculations for at least two orders. (b) Determine the rate constant k with correct units. (c) Calculate the time required for [NO₂] to reach 0.00200 M. (d) A student claims that because the stoichiometric coefficient of NO₂ is 2, the reaction must be second-order. Evaluate this claim.

Show answer

(a) [2 points — 1 for testing at least two orders with calculations, 1 for correct identification] Testing zero-order: Δ[A]/Δt for successive intervals = (0.00286/50), (0.00158/50), (0.00171/100), (0.00129/200) — not constant, so zero-order is ruled out. Testing first-order: ln values are −4.605, −4.942, −5.193, −5.560, −5.967. Differences per unit time: (−0.337/50), (−0.251/50), (−0.367/100), (−0.407/200) = −6.74×10⁻³, −5.02×10⁻³, −3.67×10⁻³, −2.04×10⁻³ — not constant, so first-order is ruled out. Testing second-order: 1/[A] values are 100, 140, 180, 260, 390.6. Differences per unit time: (40/50), (40/50), (80/100), (130.6/200) = 0.80, 0.80, 0.80, 0.65. These are approximately constant (minor deviation at end due to experimental error), confirming second-order. (b) [1 point] The slope of the 1/[A] vs. t plot equals k. Using the first two points: k = (140 − 100)/(50 − 0) = 0.80 M⁻¹ s⁻¹. Units: M⁻¹ s⁻¹ (or L mol⁻¹ s⁻¹). (c) [1 point] Using 1/[A] = 1/[A]₀ + kt: 1/0.00200 = 1/0.0100 + (0.80)t → 500 = 100 + 0.80t → t = 400/0.80 = 500 s. (d) [1 point] The claim is flawed. Reaction order is an experimentally determined quantity and is NOT deducible from stoichiometric coefficients. The coefficient of 2 for NO₂ reflects the balanced equation, not the rate law. It is coincidental that this reaction happens to be second-order. Many reactions with a coefficient of 2 are first-order or zero-order. Order is determined only by the mechanism (specifically the rate-determining step) or by experiment.

SUMMARY

Lesson Summary

Summary

Integrated rate laws convert the differential rate expression into a concentration-versus-time equation. For a zero-order reaction, [A] = [A]₀ − kt and the plot of [A] vs. t is linear. For a first-order reaction, ln[A] = ln[A]₀ − kt, the plot of ln[A] vs. t is linear, and the half-life is constant (t₁/₂ = 0.693/k). For a second-order reaction, 1/[A] = 1/[A]₀ + kt and the plot of 1/[A] vs. t is linear with a positive slope equal to k.

To determine reaction order experimentally, plot the data in all three forms and identify which gives a straight line. Alternatively, examine successive half-lives: constant half-lives indicate first-order, increasing half-lives suggest second-order, and decreasing half-lives suggest zero-order. Remember that reaction order is always determined experimentally — never assumed from stoichiometric coefficients.