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.
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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:
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:
Example: Surface-catalyzed reactions where catalyst surface is saturated
First Order Reactions
Rate law: Rate = k[A]
Integrated form:
Alternative form:
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:
Common examples: Radioactive decay, many decomposition reactions
Second Order Reactions
Rate law: Rate = k[A]²
Integrated form:
Key characteristics:
- Plot 1/[A] vs t gives straight line with slope = +k
- Half-life increases as reaction proceeds
Half-life formula:
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:
- First order: (equivalently, )
- Second order:
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: (depends on [A]₀)
- First order: (constant)
- Second order: (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:
- Collect data: Measure [A] at various times
- Make three plots:
- [A] vs t
- ln[A] vs t
- 1/[A] vs t
- 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:
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) |
|---|---|
| 0 | 1.00 |
| 50 | 0.61 |
| 100 | 0.37 |
| 150 | 0.22 |
Determine the order and rate constant.
Solution:
Calculate ln[A] and 1/[A]:
| Time (s) | [A] (M) | ln[A] | 1/[A] (M⁻¹) |
|---|---|---|---|
| 0 | 1.00 | 0.00 | 1.00 |
| 50 | 0.61 | -0.49 | 1.64 |
| 100 | 0.37 | -0.99 | 2.70 |
| 150 | 0.22 | -1.51 | 4.55 |
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) |
|---|---|---|---|
| 0 | 1.00 | 0.000 | 1.00 |
| 10 | 0.63 | -0.462 | 1.59 |
| 20 | 0.46 | -0.777 | 2.17 |
| 30 | 0.36 | -1.022 | 2.78 |
| 40 | 0.29 | -1.238 | 3.45 |
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).
📋 AP Chemistry — Exam Format Guide
| Section | Format | Questions | Time | Weight | Calculator |
|---|---|---|---|---|---|
| Multiple Choice | MCQ | 60 | 90 min | 50% | ✅ |
| Free Response (Long) | FRQ | 3 | 69 min | 30% | ✅ |
| Free Response (Short) | FRQ | 4 | 36 min | 20% | ✅ |
📊 Scoring: 1-5
💡 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
mol of reacts with mol of . How many grams of water are produced? Which is the limiting reagent? ()
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