Integrated Rate Laws — Use integrated rate laws for 0th, 1st, and 2nd order reactions - Chemistry 2

SECTION 1

Historical Context & Motivation

For centuries, chemists knew that some reactions happen quickly while others proceed at a crawl, but they lacked a mathematical vocabulary to describe how fast a reactant disappears. The quest to quantify reaction speed gave birth to chemical kinetics, the branch of chemistry that studies reaction rates. Early breakthroughs came from studying simple processes—acid-catalyzed reactions and radioactive decay—where concentration measurements over time revealed elegant patterns.

1850 • Wilhelmy's Sucrose Experiment

Ludwig Wilhelmy measured the rate of sucrose inversion by an acid, showing that the rate of disappearance was proportional to the concentration of sucrose—one of the first quantitative rate studies.

1864 • Guldberg & Waage's Law of Mass Action

Cato Guldberg and Peter Waage proposed that the rate of a reaction depends on the concentrations of the reacting substances raised to specific powers, formalizing the concept of a rate law.

1903 • Rutherford & Soddy's Decay Law

Ernest Rutherford and Frederick Soddy described radioactive decay using an exponential equation that mirrors the integrated first-order rate law, linking kinetics to nuclear science.

1913 • Bodenstein & the Steady-State Concept

Max Bodenstein developed methods for analyzing multi-step reaction mechanisms, showing that integrated rate laws could reveal the order of complex reactions when combined with experimental data.

A differential rate law tells you how the rate depends on concentration at a single instant. That's useful, but it doesn't answer a practical question every chemist asks: "If I start with a known amount of reactant, how much is left after a certain time?" The integrated rate law answers exactly that question by relating concentration directly to elapsed time.

SECTION 2

Core Principles & Definitions

Before diving into equations, you need a solid grasp of a few foundational ideas. The reaction order describes how the rate depends on the concentration of a particular reactant. It is determined experimentally—you cannot simply look at a balanced equation and guess the order. Once the order is known, the differential rate law can be transformed via calculus (don't worry, the final result is what matters) into an integrated rate law that directly connects concentration [A] to time t.

Reaction Order

The exponent to which a reactant's concentration is raised in the rate law. Common orders are 0, 1, and 2. The overall order is the sum of all individual exponents.

Rate Constant (k)

A proportionality constant unique to each reaction at a given temperature. Its units change depending on the reaction order, which is actually a handy way to confirm order from data.

Initial Concentration [A]₀

The concentration of reactant A at time t = 0. Integrated rate laws use [A]₀ as the starting point to predict how [A] evolves over time.

Half-Life (t₁/₂)

The time required for the reactant concentration to drop to half of its initial value. Each reaction order has a distinct half-life expression, which is a powerful diagnostic tool.

Linear Plots

Each integrated rate law can be rearranged into a straight-line form (y = mx + b). Plotting experimental data in these forms reveals the reaction order from whichever plot is linear.

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KEY TAKEAWAY

Think of the integrated rate law like a GPS tracker for your reactant. The differential rate law is the speedometer—it tells you how fast you're going right now. The integrated rate law is the full trip log—it tells you where you are (concentration) after any amount of time has passed since departure.

SECTION 3

Visual Explanation — Concentration vs. Time

The most immediate way to distinguish reaction orders is to look at how concentration decreases over time. Each order produces a characteristic curve shape. The diagram below plots [A] versus time for all three orders starting from the same initial concentration.

All three curves start at the same initial concentration [A]₀ = 1.0 M. The zero-order (purple) curve is a straight line dropping to zero. The first-order (cyan) curve is an exponential decay that approaches zero asymptotically. The second-order (pink) curve also decays but more gradually at higher concentrations and more steeply at low concentrations.

Notice how the zero-order curve hits zero concentration at a finite time—the reaction simply stops when the reactant is consumed. In contrast, the first- and second-order curves technically never reach zero; the concentration just gets infinitely close to it. This visual difference is one reason each order has a distinct half-life behavior, which we'll explore in the math section.

SECTION 4

Mathematical Framework

Each reaction order has its own integrated rate law, its own linear plot form, and its own half-life expression. Below are the three equations you need to memorize and understand. In every case, [A] is the concentration of the reactant at time t, [A]₀ is the initial concentration, and k is the rate constant.

Zero-Order Reactions

ZERO-ORDER INTEGRATED RATE LAW

[A] = [A]₀ − kt

Linear in [A] vs. t with slope = −k and y-intercept = [A]₀. Units of k: mol·L⁻¹·s⁻¹ (or M/s).

ZERO-ORDER HALF-LIFE

t₁/₂ = [A]₀ / (2k)

The half-life depends on initial concentration. Higher [A]₀ means a longer half-life. Each successive half-life is shorter than the previous one.

First-Order Reactions

FIRST-ORDER INTEGRATED RATE LAW

ln[A] = ln[A]₀ − kt

Linear in ln[A] vs. t with slope = −k and y-intercept = ln[A]₀. Units of k: s⁻¹. Can also be written as [A] = [A]₀ × e^−kt.

FIRST-ORDER HALF-LIFE

t₁/₂ = 0.693 / k

The half-life is constant—it does not depend on [A]₀. This is the hallmark of first-order kinetics and is the same math behind radioactive decay.

Second-Order Reactions

SECOND-ORDER INTEGRATED RATE LAW

1/[A] = 1/[A]₀ + kt

Linear in 1/[A] vs. t with slope = +k and y-intercept = 1/[A]₀. Units of k: L·mol⁻¹·s⁻¹ (or M⁻¹s⁻¹).

SECOND-ORDER HALF-LIFE

t₁/₂ = 1 / (k[A]₀)

The half-life depends on [A]₀ and gets longer as the reaction proceeds—each successive half-life is twice as long as the one before.

SECTION 5

Determining Order from Linear Plots

In the lab, you don't usually know the reaction order in advance. The standard technique is to collect concentration-versus-time data and then make three separate plots: [A] vs. t, ln[A] vs. t, and 1/[A] vs. t. Whichever plot yields a straight line tells you the order.

Each panel shows the linear form of one integrated rate law. For zero order, plot [A] vs. t. For first order, plot ln[A] vs. t. For second order, plot 1/[A] vs. t. Note that the second-order plot has a positive slope.

Property	Zero Order	First Order	Second Order
Rate Law	rate = k	rate = k[A]	rate = k[A]²
Integrated Form	[A] = [A]₀ − kt	ln[A] = ln[A]₀ − kt	1/[A] = 1/[A]₀ + kt
Linear Plot	[A] vs. t	ln[A] vs. t	1/[A] vs. t
Slope	−k	−k	+k
Units of k	M·s⁻¹	s⁻¹	M⁻¹·s⁻¹
Half-Life	[A]₀ / (2k)	0.693 / k	1 / (k[A]₀)

Summary of integrated rate law properties by reaction order

SECTION 6

Worked Example

Let's apply these equations to a concrete problem. We'll work through a first-order example step by step, then do a quick second-order calculation.

First-Order Decomposition of N₂O₅

1
Step 1 — Identify Given Values

Dinitrogen pentoxide (N₂O₅) decomposes in a first-order reaction. The rate constant is k = 6.82 × 10⁻³ s⁻¹ at 45 °C, and the initial concentration is [N₂O₅]₀ = 0.100 M. We want to find [N₂O₅] after 100 seconds.
2
Step 2 — Select the Correct Integrated Rate Law

Since the reaction is first order, we use: ln[A] = ln[A]₀ − kt. We can also use the equivalent exponential form: [A] = [A]₀ × e^−kt.
3
Step 3 — Substitute Values

[N₂O₅] = 0.100 M × e^{−(6.82 × 10⁻³ s⁻¹)(100 s)} = 0.100 M × e^−0.682
4
Step 4 — Calculate the Exponential

e^−0.682 ≈ 0.5057. Multiply: [N₂O₅] = 0.100 × 0.5057 = 0.0506 M.

[N₂O₅] ≈ 0.0506 M after 100 s
5
Step 5 — Verify with Half-Life

t₁/₂ = 0.693 / (6.82 × 10⁻³) ≈ 101.6 s. After about 102 seconds the concentration should be roughly half of the initial. Our answer of 0.0506 M at 100 s is indeed very close to half of 0.100 M, which confirms our calculation is reasonable.

t₁/₂ ≈ 102 s — consistent!

Second-Order Reaction: 2 NO₂ → 2 NO + O₂

1
Step 1 — Given Information

The decomposition of NO₂ is second order with k = 0.543 M⁻¹s⁻¹ at 300 °C. [NO₂]₀ = 0.050 M. Find [NO₂] after 20 seconds.
2
Step 2 — Apply Second-Order Integrated Rate Law

1/[NO₂] = 1/[NO₂]₀ + kt = 1/0.050 + (0.543)(20) = 20.0 + 10.86 = 30.86 M⁻¹
3
Step 3 — Solve for [NO₂]

[NO₂] = 1/30.86 ≈ 0.0324 M

[NO₂] ≈ 0.032 M after 20 s

SECTION 7

Comparing the Three Orders

Understanding the differences and similarities among the three orders helps you choose the right equation and interpret results. The table below highlights the practical implications of 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
Curve shape ([A] vs. t)	Straight line (decreasing)	Exponential decay	Concave-up decay
Successive half-lives	Decrease (each is shorter)	Stay constant	Increase (each is longer)
Common real-world example	Enzyme-catalyzed reactions at saturation	Radioactive decay, drug metabolism	NO₂ decomposition at high T
Strength	Simple, predictable time to full consumption	Constant half-life makes predictions easy	Accounts for bimolecular collision dependence
Limitation	Rare—requires a special mechanism	Doesn't apply when two molecules must collide	Half-life changes, complicating dosing predictions

Practical comparison across reaction orders

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KEY TAKEAWAY

Think about draining water from a bathtub. A zero-order process is like a pump removing water at a fixed rate—no matter how full the tub is, the flow stays the same. A first-order process is like the drain at the bottom of the tub: the more water (higher concentration), the faster it drains, and the fraction lost in each time interval is constant. A second-order process is like needing two specific pool toys to bump into each other before they get removed—the chance of collision drops dramatically as fewer toys remain.

SECTION 8

Connection to Advanced Kinetics

The integrated rate laws you've learned apply to simple, single-step (or pseudo-single-step) reactions. In more advanced chemistry, you'll encounter reactions that don't fit neatly into 0th, 1st, or 2nd order models. The table below previews how this topic extends in college-level courses.

This Lesson (High School)	Advanced Kinetics (College)
Single-reactant rate laws (A → products)	Multi-reactant rate laws (A + B → products) with pseudo-order techniques
Integer orders only (0, 1, 2)	Fractional and negative orders arising from complex mechanisms
Rate constant k is given or measured at one temperature	k varies with temperature via the Arrhenius equation: k = $Ae^{−Eₐ/RT}$
Irreversible reactions only	Reversible reactions and equilibrium kinetics connecting to the equilibrium constant K
Graphical method for order determination	Computational curve fitting and numerical integration for complex data

How integrated rate laws expand in college-level kinetics

Even though advanced kinetics involves more complexity, the foundational integrated rate laws you're learning now remain the building blocks. For example, the Arrhenius equation tells you how the rate constant k changes with temperature, but once you know k at a specific temperature, you still plug it into the same integrated rate law equations from this lesson to predict concentrations over time.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student collects concentration-vs-time data for a reaction and finds that a plot of ln[A] versus time gives a straight line, but plots of [A] versus time and 1/[A] versus time are both curved. What is the order of the reaction? Explain how you know.

Show answer

The reaction is first order. Each reaction order produces a straight line on only one type of plot. The first-order integrated rate law, ln[A] = ln[A]₀ − kt, is already in y = mx + b form when you plot ln[A] on the y-axis and t on the x-axis. Since that is the only plot yielding a straight line, the reaction must follow first-order kinetics. The slope of that line equals −k.

Question 2: BASIC CALCULATION

A zero-order reaction has a rate constant k = 0.015 M·s⁻¹. If [A]₀ = 0.600 M, what is [A] after 20 seconds?

Show answer

Use the zero-order integrated rate law: [A] = [A]₀ − kt. Substituting: [A] = 0.600 M − (0.015 M·s⁻¹)(20 s) = 0.600 − 0.300 = 0.300 M. Notice this is exactly half the initial concentration, so 20 s also happens to be the first half-life. You can verify: t₁/₂ = [A]₀ / (2k) = 0.600 / (2 × 0.015) = 20 s. ✓

Question 3: INTERMEDIATE

The first-order decomposition of hydrogen peroxide (H₂O₂) has a rate constant k = 1.06 × 10⁻³ min⁻¹ at a certain temperature. If you begin with [H₂O₂]₀ = 0.200 M, how long will it take for the concentration to drop to 0.050 M?

Show answer

Start with the first-order law: ln[A] = ln[A]₀ − kt. Rearrange for t: t = (ln[A]₀ − ln[A]) / k = ln([A]₀/[A]) / k. Substitute: t = ln(0.200/0.050) / (1.06 × 10⁻³) = ln(4) / (1.06 × 10⁻³) = 1.386 / (1.06 × 10⁻³) ≈ 1308 minutes (about 21.8 hours). Since 0.050 M is one-quarter of 0.200 M, this time corresponds to two half-lives. Check: t₁/₂ = 0.693 / (1.06 × 10⁻³) ≈ 654 min, and 2 × 654 = 1308 min. ✓

Question 4: APPLIED

A pharmaceutical company determines that a drug decomposes in the body following second-order kinetics with k = 0.40 M⁻¹hr⁻¹. A patient receives an injection resulting in an initial plasma concentration of 2.0 × 10⁻³ M. What is the drug concentration after 5 hours? What is the half-life of the drug at this initial concentration?

Show answer

Using 1/[A] = 1/[A]₀ + kt: 1/[A] = 1/(2.0 × 10⁻³) + (0.40)(5) = 500 + 2.0 = 502 M⁻¹. Therefore [A] = 1/502 ≈ 1.99 × 10⁻³ M. The concentration barely changed because kt is very small relative to 1/[A]₀. For the half-life: t₁/₂ = 1 / (k[A]₀) = 1 / (0.40 × 2.0 × 10⁻³) = 1 / (8.0 × 10⁻⁴) = 1250 hours. This very long half-life explains why the concentration barely dropped in just 5 hours.

Question 5: CRITICAL THINKING

A student monitors a reaction and records these half-life data: the first half-life is 60 s, the second half-life is 60 s, and the third half-life is 60 s. Another student monitors a different reaction and finds that the first half-life is 40 s and the second half-life is 80 s. Determine the order of each reaction. For the second reaction, also explain whether you would expect the third half-life to be longer or shorter than 80 s, and why.

Show answer

Reaction 1 has a constant half-life (60 s each time), which is the hallmark of first-order kinetics. The half-life of a first-order reaction is t₁/₂ = 0.693/k and does not depend on concentration. Reaction 2 has increasing half-lives (40 s → 80 s, each doubling), which is characteristic of second-order kinetics. For second order, t₁/₂ = 1/(k[A]₀), and as [A] decreases, the half-life gets longer. Since each half-life halves the concentration, the third half-life should be approximately 160 s—double the second half-life. This doubling pattern arises because after each half-life, [A] is halved, making 1/[A] double, which doubles t₁/₂ for the next interval.

SUMMARY

Lesson Summary

Summary

Integrated rate laws connect a reactant's concentration directly to elapsed time, allowing you to predict how much reactant remains at any point during a reaction. For a zero-order reaction, [A] = [A]₀ − kt, producing a straight line when [A] is plotted against time, with a half-life of [A]₀/(2k) that decreases over time. For a first-order reaction, ln[A] = ln[A]₀ − kt, giving a straight line on a ln[A] vs. t plot, with a constant half-life of 0.693/k. For a second-order reaction, 1/[A] = 1/[A]₀ + kt, yielding a straight line on a 1/[A] vs. t plot, with a half-life of 1/(k[A]₀) that increases as the reaction proceeds.

To determine the reaction order experimentally, plot your data in all three forms and look for the straight line. The slope of that line gives you the rate constant k (watch the sign—it is negative for zero and first order, positive for second order). Remember that the units of k change with order: M·s⁻¹ for zero, s⁻¹ for first, and M⁻¹·s⁻¹ for second. Mastering these three integrated rate laws gives you a powerful toolkit for predicting concentrations, calculating half-lives, and identifying reaction mechanisms.