Skip to content

Integrated Rate Laws and Half-Life

Master integrated rate laws for zero, first, and second order reactions, calculate concentrations over time, and understand half-life concepts.

Written and reviewed by the Study Mondo Education TeamLast updated
🎯⭐ INTERACTIVE LESSON

Try the Interactive Version!

Learn step-by-step with practice exercises built right in.

Start Interactive Lesson →

Integrated Rate Laws and Half-Life

Overview

Integrated rate laws relate concentration to time directly, allowing us to:

  • Predict reactant concentration at any time
  • Determine reaction order from experimental data
  • Calculate rate constants from concentration-time data

These equations are derived from differential rate laws using calculus.

Zero Order Reactions

Rate law: Rate = k

Integrated form:

[A]t=[A]0−kt[A]_t = [A]_0 - kt

Key characteristics:

  • Concentration decreases linearly with time
  • Plot [A] vs t gives straight line with slope = -k
  • Half-life depends on initial concentration

Half-life formula:

t1/2=[A]02kt_{1/2} = \frac{[A]_0}{2k}

Example: Surface-catalyzed reactions where catalyst surface is saturated


First Order Reactions

Rate law: Rate = k[A]

Integrated form:

ln⁡[A]t=ln⁡[A]0−kt\ln[A]_t = \ln[A]_0 - kt

Alternative form:

[A]t=[A]0e−kt[A]_t = [A]_0 e^{-kt}

Key characteristics:

  • Concentration decreases exponentially
  • Plot ln[A] vs t gives straight line with slope = -k
  • Half-life is constant (independent of concentration)

Half-life formula:

t1/2=0.693k=ln⁡2kt_{1/2} = \frac{0.693}{k} = \frac{\ln 2}{k}

Common examples: Radioactive decay, many decomposition reactions


Second Order Reactions

Rate law: Rate = k[A]²

Integrated form:

1[A]t=1[A]0+kt\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt

Key characteristics:

  • Plot 1/[A] vs t gives straight line with slope = +k
  • Half-life increases as reaction proceeds

Half-life formula:

t1/2=1k[A]0t_{1/2} = \frac{1}{k[A]_0}

Example: Gas-phase dimerization reactions (2A → A₂)


Comparison at a Glance (Mobile‑Friendly)

Rather than a cramped table, here’s a stacked, easy‑scan summary for each key property:

Rate law

  • Zero order: Rate = k
  • First order: Rate = k[A]
  • Second order: Rate = k[A]²

Integrated law

  • Zero order: [A]t=[A]0−kt[A]_t = [A]_0 - kt
  • First order: ln⁡[A]t=ln⁡[A]0−kt\ln[A]_t = \ln[A]_0 - kt (equivalently, [A]t=[A]0e−kt[A]_t = [A]_0 e^{-kt})
  • Second order: 1[A]t=1[A]0+kt\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt

Linear plot for straight line

  • Zero order: [A] vs t (slope = −k)
  • First order: ln[A] vs t (slope = −k)
  • Second order: 1/[A] vs t (slope = +k)

Half‑life

  • Zero order: t1/2=[A]02kt_{1/2} = \dfrac{[A]_0}{2k} (depends on [A]₀)
  • First order: t1/2=0.693kt_{1/2} = \dfrac{0.693}{k} (constant)
  • Second order: t1/2=1k[A]0t_{1/2} = \dfrac{1}{k[A]_0} (depends on [A]₀)

Units of k

  • Zero order: M·s⁻¹
  • First order: s⁻¹
  • Second order: M⁻¹·s⁻¹

Determining Reaction Order Graphically

Method:

  1. Collect data: Measure [A] at various times
  2. Make three plots:
    • [A] vs t
    • ln[A] vs t
    • 1/[A] vs t
  3. Identify the linear plot - this reveals the order:
    • [A] vs t is linear → Zero order
    • ln[A] vs t is linear → First order
    • 1/[A] vs t is linear → Second order

Extracting the rate constant:

  • Zero/First order: k = -slope (note negative!)
  • Second order: k = +slope (positive)

Half-Life Concepts

Definition: Time required for concentration to decrease to half its initial value

After n half-lives:

[A]=[A]02n[A] = \frac{[A]_0}{2^n}

Examples:

  • After 1 half-life: [A] = [A]₀/2 = 50% remaining
  • After 2 half-lives: [A] = [A]₀/4 = 25% remaining
  • After 3 half-lives: [A] = [A]₀/8 = 12.5% remaining

Key insight: Only first-order reactions have a constant half-life!


Practice Example

Problem: A reaction has the following data:

Time (s)[A] (M)
01.00
500.61
1000.37
1500.22

Determine the order and rate constant.

Solution:

Calculate ln[A] and 1/[A]:

Time (s)[A] (M)ln[A]1/[A] (M⁻¹)
01.000.001.00
500.61-0.491.64
1000.37-0.992.70
1500.22-1.514.55

Plot ln[A] vs t → Linear! (slope ≈ -0.01)

Conclusion: First order reaction with k = 0.01 s⁻¹

Half-life: t₁/₂ = 0.693/0.01 = 69.3 seconds


Real-World Applications

Radioactive Dating:

  • Carbon-14 dating: t₁/₂ = 5,730 years (first order)
  • Used to date organic materials up to ~50,000 years old

Pharmacology:

  • Drug metabolism follows first-order kinetics
  • Constant half-life allows predictable dosing schedules
  • Example: Aspirin t₁/₂ ≈ 2-3 hours

Environmental Chemistry:

  • Pollutant degradation rates
  • Predicting cleanup timelines

Chemical Engineering:

  • Reactor design and optimization
  • Determining optimal reaction times

📚 Practice Problems

1Problem 1easy

❓ Question:

The decomposition of N₂O₅ is first order with k = 5.0 × 10⁻⁴ s⁻¹. If [N₂O₅]₀ = 0.200 M, find: (a) [N₂O₅] after 100 seconds, (b) time to reach 0.050 M, (c) half-life.

💡 Show Solution

Given:

  • First order: k = 5.0 × 10⁻⁴ s⁻¹
  • [N₂O₅]₀ = 0.200 M

First order integrated law: ln[A]_t = ln[A]₀ - kt


(a) Find [N₂O₅] after t = 100 s

ln[N₂O₅]₁₀₀ = ln(0.200) - (5.0 × 10⁻⁴)(100) ln[N₂O₅]₁₀₀ = -1.609 - 0.050 = -1.659

[N₂O₅]₁₀₀ = e⁻¹·⁶⁵⁹ = 0.190 M

Answer: 0.190 M


(b) Time to reach 0.050 M

Rearrange: t = (1/k)ln([A]₀/[A]_t)

t = (1/(5.0 × 10⁻⁴))ln(0.200/0.050) t = 2000 × ln(4) = 2000 × 1.386 t = 2772 s = 46.2 min

Answer: 2772 s or 46 min


(c) Half-life

For first order: t₁/₂ = 0.693/k

t₁/₂ = 0.693/(5.0 × 10⁻⁴) = 1386 s = 23.1 min

Answer: 1386 s or 23 min

Check: In part (b), 0.200 M → 0.050 M is a factor of 4 = 2² So time = 2 half-lives = 2(1386) = 2772 s ✓

2Problem 2medium

❓ Question:

Given concentration vs time data, determine the order and rate constant: t(s): 0, 10, 20, 30, 40 | A: 1.00, 0.63, 0.46, 0.36, 0.29

💡 Show Solution

Data:

t (s)[A] (M)ln[A]1/[A] (M⁻¹)
01.000.0001.00
100.63-0.4621.59
200.46-0.7772.17
300.36-1.0222.78
400.29-1.2383.45

Test for zero order: Plot [A] vs t

  • Not linear (curved)

Test for first order: Plot ln[A] vs t

  • Points: (0, 0.000), (10, -0.462), (20, -0.777), (30, -1.022), (40, -1.238)
  • Check linearity: Δ(ln[A])/Δt ≈ -0.031 per second (constant!)
  • Linear! → First order

Test for second order: Plot 1/[A] vs t

  • Not as linear as ln[A] plot

Conclusion: First order

Calculate k from slope:

slope = Δ(ln[A])/Δt = (-1.238 - 0.000)/(40 - 0) = -0.031 s⁻¹

For first order: slope = -k k = 0.031 s⁻¹

Verification: Calculate t₁/₂ = 0.693/k = 0.693/0.031 = 22.4 s Check data: At t ≈ 22 s, [A] should be ≈ 0.50 M (half of 1.00) Interpolating between t=20 (0.46 M) and t=30 (0.36 M) gives ≈ 0.48 M ✓

Answer: First order, k = 0.031 s⁻¹

3Problem 3hard

❓ Question:

A second-order reaction has k = 0.54 M⁻¹·s⁻¹ and [A]₀ = 0.10 M. (a) What is [A] after 2.0 s? (b) What is the first half-life? (c) What is the second half-life?

💡 Show Solution

Given:

  • Second order: k = 0.54 M⁻¹·s⁻¹
  • [A]₀ = 0.10 M

Second order integrated law: 1/[A]_t = 1/[A]₀ + kt


(a) [A] after 2.0 s

1/[A]₂ = 1/0.10 + (0.54)(2.0) 1/[A]₂ = 10 + 1.08 = 11.08 M⁻¹

[A]₂ = 1/11.08 = 0.0903 M

Answer: 0.090 M


(b) First half-life

For second order: t₁/₂ = 1/(k[A]₀)

t₁/₂ = 1/((0.54)(0.10)) = 1/0.054 = 18.5 s

Answer: 18.5 s

Verification: After 18.5 s, [A] should be 0.050 M

1/[A]₁₈.₅ = 10 + (0.54)(18.5) = 10 + 10 = 20 M⁻¹ [A]₁₈.₅ = 1/20 = 0.050 M ✓


(c) Second half-life

Key point: For second order, half-life increases!

After first t₁/₂: [A] = 0.050 M (new starting point)

Second t₁/₂ = 1/(k × 0.050) = 1/((0.54)(0.050)) Second t₁/₂ = 1/0.027 = 37.0 s

Answer: 37.0 s

Important: Second half-life (37 s) is twice the first (18.5 s)

  • For second order: each successive half-life doubles
  • Total time for 2 half-lives: 18.5 + 37.0 = 55.5 s
  • After 55.5 s: [A] = 0.025 M (1/4 of original)

Contrast with first order:

  • First order: all half-lives equal
  • Second order: each half-life = 2 × previous
  • Zero order: each half-life = 1/2 × previous

4Problem 4medium

❓ Question:

Zero-order decay: A → products with k = 2.5×10⁻³ M·s⁻¹. If [A]_0 = 0.300 M, (a) write [A]_t, (b) time to 0.120 M, (c) half-life, (d) how much remains after 4.0 minutes?

💡 Show Solution

Zero order: [A]_t = [A]_0 − kt. (a) [A]_t = 0.300 − (2.5×10⁻³)t (M). (b) t = (0.300 − 0.120)/(2.5×10⁻³) = 0.180/0.0025 = 72 s. (c) t₁/₂ = [A]_0/(2k) = 0.300/(2×2.5×10⁻³) = 60 s. (d) t = 4.0 min = 240 s ⇒ [A] = 0.300 − (2.5×10⁻³)(240) = 0.300 − 0.600 = negative → zero-order model predicts exhaustion at t = [A]_0/k = 120 s, so after 240 s, [A] ≈ 0 (reaction complete).

5Problem 5easy

❓ Question:

First-order: A first-order reaction has k = 0.035 s⁻¹. (a) Time to reach 25% of [A]_0? (b) Fraction remaining after 1.0 minute?

💡 Show Solution

First order: [A]_t = [A]_0 e^{−kt}. (a) Set [A]_t/[A]_0 = 0.25 = e^{−kt} ⇒ t = (1/k)ln(1/0.25) = (1/0.035)ln(4) ≈ 28.6×1.386 ≈ 39.7 s. (b) t = 60 s ⇒ fraction = e^{−0.035×60} = e^{−2.10} ≈ 0.122 = 12.2% remains.

6Problem 6medium

❓ Question:

Order identification from data: t (s) = 0, 50, 100, 150; [A] (M) = 0.80, 0.57, 0.41, 0.30. Determine order and k.

💡 Show Solution

Compute ln[A]: 0.80→−0.223; 0.57→−0.562; 0.41→−0.894; 0.30→−1.204. Δ(ln[A]) per 50 s ≈ −0.339, −0.332, −0.310 (nearly constant) ⇒ ln[A] vs t is ~linear ⇒ first-order. Slope ≈ (−1.204 − (−0.223))/(150 − 0) = (−0.981)/150 = −0.00654 s⁻¹ ⇒ k ≈ 6.5×10⁻³ s⁻¹. Check 1/[A] vs t curvature to rule out second order.

7Problem 7hard

❓ Question:

Second-order: For 2A → products with k = 0.22 M⁻¹·s⁻¹ and [A]_0 = 0.50 M, how long to reach 0.20 M? What are the first and second half-lives?

💡 Show Solution

Second order: 1/[A]_t = 1/[A]_0 + kt. Solve for t to [A]_t = 0.20: t = (1/[A]_t − 1/[A]_0)/k = (1/0.20 − 1/0.50)/0.22 = (5.00 − 2.00)/0.22 = 3.00/0.22 ≈ 13.6 s. Half-lives: t₁/₂ = 1/(k[A]_0) = 1/(0.22×0.50) = 1/0.11 ≈ 9.09 s; second half-life = 1/(k×0.25) = 1/0.055 ≈ 18.2 s (double).

Explain using:

📋 AP Chemistry — Exam Format Guide

⏱ 3 hours 15 minutes📝 67 questions📊 3 sections
SectionFormatQuestionsTimeWeightCalculator
Multiple ChoiceMCQ6090 min50%✅
Free Response (Long)FRQ369 min30%✅
Free Response (Short)FRQ436 min20%✅

📊 Scoring: 1-5

5
Extremely Qualified
~12%
4
Well Qualified
~16%
3
Qualified
~24%
2
Possibly Qualified
~24%
1
No Recommendation
~24%

💡 Key Test-Day Tips

  • ✓Memorize common polyatomic ions
  • ✓Practice dimensional analysis
  • ✓Know your gas laws

⚠️ Common Mistakes: Integrated Rate Laws and Half-Life

Avoid these 3 frequent errors

🌍 Real-World Applications: Integrated Rate Laws and Half-Life

See how this math is used in the real world

📝 Worked Example: Stoichiometry — Limiting Reagent

Problem:

22 mol of H2H_2 reacts with 11 mol of O2O_2. How many grams of water are produced? Which is the limiting reagent? (2H2+O2→2H2O2H_2 + O_2 \to 2H_2O)

2Determine the limiting reagent
3Calculate moles of product
4Convert moles to grams

📌 Related Topics in Kinetics

❓ Frequently Asked Questions

What is Integrated Rate Laws and Half-Life?▾
Master integrated rate laws for zero, first, and second order reactions, calculate concentrations over time, and understand half-life concepts.
How can I study Integrated Rate Laws and Half-Life effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 7 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Integrated Rate Laws and Half-Life study guide free?▾
Yes — all study notes, flashcards, and practice problems for Integrated Rate Laws and Half-Life on Study Mondo are free to access. No account is needed.
What course covers Integrated Rate Laws and Half-Life?▾
Integrated Rate Laws and Half-Life is part of the AP Chemistry course on Study Mondo, specifically in the Kinetics section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Integrated Rate Laws and Half-Life?▾
Yes, this page includes 7 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.