General Chemistry II · Chemical Kinetics

Integrated Rate Laws

6 min read

On this page 8 sections

In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Study tools
Sources & references

In 30 seconds

The differential rate law tells you how rate depends on concentration. The integrated rate law goes one step further and tells you how concentration depends on time. It is obtained by integrating the differential equation and is the workhorse for prediction: given a starting concentration and k, you can compute [A] at any later time, or find the time needed to reach a target concentration.

Why this matters

Integrated rate laws are how kinetic data becomes prediction. Radiocarbon dating, drug half-life and dosing schedules, pollutant decay, and food-spoilage estimates all use the first-order equation. The linear-plot method is the practical way a chemist decides what order a reaction obeys from a table of concentration–time measurements.

The college version

Core Concept

Key Ideas

Each reaction order has a characteristic integrated equation and a characteristic linear plot.
The half-life (t½) — the time for [A] to fall to half its initial value — behaves differently for each order.
Fitting data to a straight line is the standard way to determine order experimentally.
First-order kinetics (like radioactive decay) has a half-life independent of concentration.

Equations and Variables

Zero order (Rate = k):

$[A] t = [A] 0 - kt t 1/2 = [A] 0 2k$

Linear plot: [A] vs t (slope = −k).

First order (Rate = k[A]):

$ln [A] t = ln [A] 0 - kt t 1/2 = ln 2 k = 0.693 k$

Linear plot: ln[A] vs t (slope = −k).

Second order (Rate = k[A]²):

$1 [A] t = 1 [A] 0 + kt t 1/2 = 1 k[A] 0$

Linear plot: 1/[A] vs t (slope = +k).

[A]₀ = initial concentration (M)
[A]ₜ = concentration at time t (M)
k = rate constant (units depend on order)
t½ = half-life (s)

How It Works

Each integrated law is a straight-line equation in disguise. If you plot [A] against time and get a straight line, the reaction is zero order and the slope equals −k. If ln[A] vs t is linear, it's first order with slope −k. If 1/[A] vs t is linear, it's second order with slope +k. This "linearization" trick turns a messy curve into a diagnostic: whichever plot gives the best straight line reveals the order.

Half-lives behave very differently. For a first-order reaction, t½ = 0.693/k is constant — every half-life takes the same time regardless of how much is left (this is why carbon-14 dating works). For a second-order reaction, t½ = 1/(k[A]₀) grows as the concentration drops, so each successive half-life takes longer. For zero order, t½ = [A]₀/2k shrinks as [A]₀ falls.

Worked Example

Dinitrogen pentoxide decomposes by a first-order process: 2 N₂O₅ → 4 NO₂ + O₂, with k = 6.2 × 10⁻⁴ s⁻¹ at 45 °C.

(a) Half-life.

$t 1/2 = 0.693 6.2 \times 10 -4 s -1 = 1.1 \times 103 s (\approx 19 min)$

(b) Concentration after 2.00 × 10³ s if [N₂O₅]₀ = 0.100 M.

Use the first-order integrated law:

$ln [A] t = ln (0.100) - (6.2 \times 10 -4 s -1)(2.00 \times 103 s)$

$ln [A] t = -2.303 - 1.24 = -3.54$

$[A] t = e -3.54 = 0.029 M$

(c) Time for [N₂O₅] to fall from 0.100 M to 0.0250 M.

Each half-life halves the concentration: 0.100 → 0.050 → 0.025 requires two half-lives, so t = 2 × 1.1 × 10³ s = 2.2 × 10³ s. (Equivalently, 0.0250/0.100 = ¼ = (½)².)

How it works

Common confusions

Mixing up the signs. First order subtracts kt (decay); second order adds kt (1/[A] grows). Don't blindly copy a sign.
Forgetting the concentration dependence of t½. Only first-order half-lives are constant.
Using the wrong plot. Check all three linearizations; the best straight line identifies the order.
Unit errors on k. First-order k is s⁻¹; second-order k is M⁻¹·s⁻¹.

Quick review

Write the integrated law for a first-order reaction and identify the linear plot.
What is the half-life formula for first, second, and zero order?
Which order has a concentration-independent half-life?
How do you determine reaction order from a table of [A] vs t data?
If a first-order reaction has k = 0.0231 min⁻¹, what is its half-life? (Answer: 30.0 min)

Eli explains

The same idea, in plain words

Explain it like I’m 10

The differential rate law is the "speed"; the integrated rate law is the "mileage log" — it tells you where you are at any point on the trip. For a first-order reaction, the trip is special: no matter where you are, it always takes the same amount of time to go halfway to empty. That's why "half-life" is so convenient for drugs and radioisotopes — it never changes. For other orders, the half-life depends on how much you started with, so the "halfway" markers move. To find out which type of trip you're on, plot your data three ways and see which one makes a straight line.

Worked example

Worked Example

Dinitrogen pentoxide decomposes by a first-order process: 2 N₂O₅ → 4 NO₂ + O₂, with k = 6.2 × 10⁻⁴ s⁻¹ at 45 °C.

(a) Half-life.

$t 1/2 = 0.693 6.2 \times 10 -4 s -1 = 1.1 \times 103 s (\approx 19 min)$

(b) Concentration after 2.00 × 10³ s if [N₂O₅]₀ = 0.100 M.

Use the first-order integrated law:

$ln [A] t = ln (0.100) - (6.2 \times 10 -4 s -1)(2.00 \times 103 s)$

$ln [A] t = -2.303 - 1.24 = -3.54$

$[A] t = e -3.54 = 0.029 M$

(c) Time for [N₂O₅] to fall from 0.100 M to 0.0250 M.

Each half-life halves the concentration: 0.100 → 0.050 → 0.025 requires two half-lives, so t = 2 × 1.1 × 10³ s = 2.2 × 10³ s. (Equivalently, 0.0250/0.100 = ¼ = (½)².)

Key takeaways

### High-Yield Facts
Zero order: [A]ₜ = [A]₀ − kt; linear in [A]; t½ = [A]₀/2k.
First order: ln[A]ₜ = ln[A]₀ − kt; linear in ln[A]; t½ = 0.693/k (constant).
Second order: 1/[A]ₜ = 1/[A]₀ + kt; linear in 1/[A]; t½ = 1/(k[A]₀).
The slope of the correct linear plot gives k (magnitude).
Only the first-order half-life is independent of concentration.

Keep learning

Ready to build on this? Continue to the next lesson.

Practice General Chemistry II

This lesson has no separate scored set. Practice draws from the subject’s question bank.

Study tools & related lessonsYou’ll learn to · Related

You’ll learn to

Write the integrated rate laws for zero-, first-, and second-order reactions.
Use the linearized plots ([A] vs t, ln[A] vs t, 1/[A] vs t) to identify reaction order from data.
Calculate and interpret half-lives for each order.
Apply the appropriate equation to compute concentration at a given time.

Sources & references

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.

In 30 seconds

Why this matters

The college version

Core Concept

Key Ideas

Equations and Variables

How It Works

Worked Example

How it works

Common confusions

Quick review

The same idea, in plain words

Worked example

Worked Example

Key takeaways

Keep learning

You’ll learn to

Related lessons

Sources & references