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.
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]ยฒ
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 โ Aโ)
Comparison at a Glance (MobileโFriendly)
Rather than a cramped table, hereโs a stacked, easyโscan summary for each key property:
Rate law
- Zero order: Rate = k
- First order: Rate = k[A]
- Second order: Rate = k[A]ยฒ
Integrated law
- Zero order:
- First order: (equivalently, )
- Second order:
Linear plot for straight line
- Zero order: [A] vs t (slope = โk)
- First order: ln[A] vs t (slope = โk)
- Second order: 1/[A] vs t (slope = +k)
Halfโlife
- Zero order: (depends on [A]โ)
- First order: (constant)
- Second order: (depends on [A]โ)
Units of k
- Zero order: Mยทsโปยน
- First order: sโปยน
- Second order: Mโปยนยทsโปยน
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 โ Zero order
- ln[A] vs t is linear โ First order
- 1/[A] vs t is linear โ 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]โ/2 = 50% remaining
- After 2 half-lives: [A] = [A]โ/4 = 25% remaining
- After 3 half-lives: [A] = [A]โ/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] (Mโปยน) | |----------|---------|-------|-------------| | 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 โ Linear! (slope โ -0.01)
Conclusion: First order reaction with k = 0.01 sโปยน
Half-life: tโ/โ = 0.693/0.01 = 69.3 seconds
Real-World Applications
Radioactive Dating:
- Carbon-14 dating: tโ/โ = 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 tโ/โ โ 2-3 hours
Environmental Chemistry:
- Pollutant degradation rates
- Predicting cleanup timelines
Chemical Engineering:
- Reactor design and optimization
- Determining optimal reaction times
๐ Practice Problems
1Problem 1easy
โ Question:
The decomposition of NโOโ is first order with k = 5.0 ร 10โปโด sโปยน. If [NโOโ ]โ = 0.200 M, find: (a) [NโOโ ] after 100 seconds, (b) time to reach 0.050 M, (c) half-life.
๐ก Show Solution
Given:
- First order: k = 5.0 ร 10โปโด sโปยน
- [NโOโ ]โ = 0.200 M
First order integrated law: ln[A]_t = ln[A]โ - kt
(a) Find [NโOโ ] after t = 100 s
ln[NโOโ ]โโโ = ln(0.200) - (5.0 ร 10โปโด)(100) ln[NโOโ ]โโโ = -1.609 - 0.050 = -1.659
[NโOโ ]โโโ = eโปยนยทโถโตโน = 0.190 M
Answer: 0.190 M
(b) Time to reach 0.050 M
Rearrange: t = (1/k)ln([A]โ/[A]_t)
t = (1/(5.0 ร 10โปโด))ln(0.200/0.050) t = 2000 ร ln(4) = 2000 ร 1.386 t = 2772 s = 46.2 min
Answer: 2772 s or 46 min
(c) Half-life
For first order: tโ/โ = 0.693/k
tโ/โ = 0.693/(5.0 ร 10โปโด) = 1386 s = 23.1 min
Answer: 1386 s or 23 min
Check: In part (b), 0.200 M โ 0.050 M is a factor of 4 = 2ยฒ So time = 2 half-lives = 2(1386) = 2772 s โ
2Problem 2medium
โ 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
๐ก Show Solution
Data: | t (s) | [A] (M) | ln[A] | 1/[A] (Mโปยน) | |-------|---------|-------|-------------| | 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: ฮ(ln[A])/ฮt โ -0.031 per second (constant!)
- Linear! โ 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 = ฮ(ln[A])/ฮt = (-1.238 - 0.000)/(40 - 0) = -0.031 sโปยน
For first order: slope = -k k = 0.031 sโปยน
Verification: Calculate tโ/โ = 0.693/k = 0.693/0.031 = 22.4 s Check data: At t โ 22 s, [A] should be โ 0.50 M (half of 1.00) Interpolating between t=20 (0.46 M) and t=30 (0.36 M) gives โ 0.48 M โ
Answer: First order, k = 0.031 sโปยน
3Problem 3hard
โ Question:
A second-order reaction has k = 0.54 Mโปยนยทsโปยน and [A]โ = 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?
๐ก Show Solution
Given:
- Second order: k = 0.54 Mโปยนยทsโปยน
- [A]โ = 0.10 M
Second order integrated law: 1/[A]_t = 1/[A]โ + kt
(a) [A] after 2.0 s
1/[A]โ = 1/0.10 + (0.54)(2.0) 1/[A]โ = 10 + 1.08 = 11.08 Mโปยน
[A]โ = 1/11.08 = 0.0903 M
Answer: 0.090 M
(b) First half-life
For second order: tโ/โ = 1/(k[A]โ)
tโ/โ = 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]โโ.โ = 10 + (0.54)(18.5) = 10 + 10 = 20 Mโปยน [A]โโ.โ = 1/20 = 0.050 M โ
(c) Second half-life
Key point: For second order, half-life increases!
After first tโ/โ: [A] = 0.050 M (new starting point)
Second tโ/โ = 1/(k ร 0.050) = 1/((0.54)(0.050)) Second tโ/โ = 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 ร previous
- Zero order: each half-life = 1/2 ร previous
4Problem 4medium
โ Question:
Zero-order decay: A โ products with k = 2.5ร10โปยณ Mยทsโปยน. 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?
๐ก Show Solution
Zero order: [A]_t = [A]_0 โ kt. (a) [A]_t = 0.300 โ (2.5ร10โปยณ)t (M). (b) t = (0.300 โ 0.120)/(2.5ร10โปยณ) = 0.180/0.0025 = 72 s. (c) tโ/โ = [A]_0/(2k) = 0.300/(2ร2.5ร10โปยณ) = 60 s. (d) t = 4.0 min = 240 s โ [A] = 0.300 โ (2.5ร10โปยณ)(240) = 0.300 โ 0.600 = negative โ zero-order model predicts exhaustion at t = [A]_0/k = 120 s, so after 240 s, [A] โ 0 (reaction complete).
5Problem 5easy
โ Question:
First-order: A first-order reaction has k = 0.035 sโปยน. (a) Time to reach 25% of [A]_0? (b) Fraction remaining after 1.0 minute?
๐ก Show Solution
First order: [A]_t = [A]_0 e^{โkt}. (a) Set [A]_t/[A]_0 = 0.25 = e^{โkt} โ t = (1/k)ln(1/0.25) = (1/0.035)ln(4) โ 28.6ร1.386 โ 39.7 s. (b) t = 60 s โ fraction = e^{โ0.035ร60} = e^{โ2.10} โ 0.122 = 12.2% remains.
6Problem 6medium
โ 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.
๐ก Show Solution
Compute ln[A]: 0.80โโ0.223; 0.57โโ0.562; 0.41โโ0.894; 0.30โโ1.204. ฮ(ln[A]) per 50 s โ โ0.339, โ0.332, โ0.310 (nearly constant) โ ln[A] vs t is ~linear โ first-order. Slope โ (โ1.204 โ (โ0.223))/(150 โ 0) = (โ0.981)/150 = โ0.00654 sโปยน โ k โ 6.5ร10โปยณ sโปยน. Check 1/[A] vs t curvature to rule out second order.
7Problem 7hard
โ Question:
Second-order: For 2A โ products with k = 0.22 Mโปยนยทsโปยน and [A]_0 = 0.50 M, how long to reach 0.20 M? What are the first and second half-lives?
๐ก Show Solution
Second order: 1/[A]_t = 1/[A]_0 + kt. Solve for t to [A]_t = 0.20: t = (1/[A]_t โ 1/[A]_0)/k = (1/0.20 โ 1/0.50)/0.22 = (5.00 โ 2.00)/0.22 = 3.00/0.22 โ 13.6 s. Half-lives: tโ/โ = 1/(k[A]_0) = 1/(0.22ร0.50) = 1/0.11 โ 9.09 s; second half-life = 1/(kร0.25) = 1/0.055 โ 18.2 s (double).
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