Integrated Rate Laws and Half-Life - Complete Interactive Lesson
Part 1: Zero-Order Reactions
๐ Zero-Order Integrated Rate Law
Part 1 of 7 โ When Rate Doesn't Depend on Concentration
Topics in This Part
Section
๐ Derivation of the Zero-Order Integrated Rate Law
This is the equation of a straight line!
Plot: [A] vs t โ straight line for zero-order
๐ Zero-Order Half-Life
Successive Half-Lives
๐ Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.
What You'll Master in Part 1
Understanding the core concepts covered in Part 1
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic
๐ Derivation of the Zero-Order Integrated Rate Law
For a zero-order reaction: Rate=k
โdtd[A]โ
Zero-Order Graphical Analysis ๐ฏ
๐ Zero-Order Half-Life
The half-life (t1/2โ) is the time for the concentration to drop to half its initial value.
Set [A]=[A]0 in the integrated rate law:
Zero-Order Half-Life Concepts ๐
Zero-Order Calculations ๐งฎ
A zero-order reaction has k=5.0ร10โ3 M/s and [A]0โ M.
๐ง How to Identify Zero-Order from Data
Method: Test Different Plots
Given concentration-time data, make three plots:
Plot
Straight line if...
[A] vs t
Zero-order
ln[A] vs t
First-order
Exit Quiz โ Zero-Order Integrated Rate Law โ
Part 2: First-Order Reactions
๐ First-Order Integrated Rate Law
Part 2 of 7 โ Exponential Decay
Topics in This Part
Section
๐ Derivation
Linear Form: ln[A] vs t
๐ First-Order Half-Life
Connection to Radioactive Decay
๐ Useful Ratio Form
๐ Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.
What You'll Master in Part 2
Understanding the core concepts covered in Part 2
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic
๐ Derivation
For a first-order reaction:
Part 3: Second-Order Reactions
๐ Second-Order Integrated Rate Law
Part 3 of 7 โ Inverse Concentration and Time
Topics in This Part
Section
โฑ๏ธ Derivation (for Rate = k[A]ยฒ)
Linear Form: 1/[A] vs t
๐ Second-Order Half-Life
Successive Half-Lives
โ๏ธ Comparing All Three Orders
๐ Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.
What You'll Master in Part 3
Understanding the core concepts covered in Part 3
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic
โฑ๏ธ Derivation (for Rate = k[A]ยฒ)
Part 4: Half-Life
๐ Graphical Analysis
Part 4 of 7 โ Identifying Reaction Order from Plots
Topics in This Part
Section
๐ฏ The Three-Plot Strategy
How to Check Linearity
๐งช Worked Example: Identifying Order
๐ Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.
What You'll Master in Part 4
Understanding the core concepts covered in Part 4
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic
๐ฏ The Three-Plot Strategy
Given concentration-vs-time data, create:
Plot
If Linear โ
Slope =
y-intercept =
[A] vs
Part 5: Graphical Analysis of Order
โฑ๏ธ Half-Life Problems
Part 5 of 7 โ Calculations for Each Order and Radioactive Decay
Topics in This Part
Section
๐ Half-Life Formulas Summary
๐ Radioactive Decay: Always First-Order
Carbon-14 Dating
Example
๐ Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.
What You'll Master in Part 5
Understanding the core concepts covered in Part 5
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic
๐ Half-Life Formulas Summary
Order
Half-Life Formula
Dependence on [A]
Part 6: Problem-Solving Workshop
๐ง Problem-Solving Workshop
Part 6 of 7 โ Mixed Order Identification and Calculations
Practice Makes Perfect
This workshop features multi-step problems that mirror the AP Chemistry exam format. Each problem requires you to combine concepts from previous parts and show your work clearly.
๐ Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems โ structured practice is the best preparation.
What You'll Master in Part 6
Working through complete multi-step problems from start to finish
Building problem-solving strategies you can apply on the AP exam
Identifying which concepts to apply and in what order
๐ ๏ธ Problem-Solving Flowchart
Step 1: Identify the Order
Method
How It Works
Result
Graphical
Plot [A], ln[A], and 1/[A] vs. t
The linear plot reveals the order
Successive half-lives
Compare values
Part 7: Synthesis & AP Review
๐ Synthesis & AP Review
Part 7 of 7 โ AP-Style Integrated Rate Law Problems
Bringing It All Together
This comprehensive review connects every concept from Parts 1โ6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam โ multi-step, multi-concept, and requiring clear written explanations.
๐ Why this matters: AP Chemistry exam questions rarely test one concept in isolation โ success requires connecting ideas across topics.
What You'll Master in Part 7
Solving AP-style questions that integrate multiple concepts from this unit
Writing clear, concise explanations using proper chemistry terminology
Identifying and avoiding common AP exam traps and mistakes
๐ Complete Integrated Rate Laws Summary
๐ Key Concept: Memorize this table โ it is the foundation for all integrated rate law problems on the AP exam.
Zero-Order
First-Order
Second-Order
Differential
Rate = k
Rate =
=
k
Integrating from [A]0โ to [A] and from 0 to t:
โซ[A]0โ[A]โd[A]=โโซ0tโkdt
[A]โ[A]0โ=โkt
[A]=โkt+[A]0โโ
This is the equation of a straight line!
y=mx+b
Variable
Corresponds To
y
[A]
m (slope)
โk
x
t
b (y-intercept)
[A]0โ
Plot: [A] vs t โ straight line for zero-order
๐ Key Concept: If [A] vs t is a straight line, the reaction is zero-order. The slope gives โk.
Slope = โk
y-intercept = [A]0โ
[A] decreases linearly with time
โ
/2
2[A]0โโ=โkt1/2โ+[A]0โ
kt1/2โ=[A]0โโ2[A]0โโ=2[A]0โโ
t1/2โ=2k[A]0โโโ
๐ Key Concept: The zero-order half-life depends on[A]0โ:
Higher initial concentration โ longer half-life
Each successive half-life is shorter than the previous one
The reaction reaches [A] = 0 in a finite time: tcompleteโ=[A]0โ/k
Successive Half-Lives
Half-life
[A] at start
[A] at end
Duration
1st
[A]0โ
[A]0โ/2
[A]0โ/(2k)
2nd
[A]0โ/2
[A]0โ/4
3rd
[A]0โ/4
[A]0โ/8
Each successive half-life is exactly half the duration of the previous one.
โ ๏ธ Warning: Zero-order is the only order where the reaction reaches [A]=0 in finite time. Don't assume all reactions behave this way!
=
0.60
1) What is [A] after 40 s? (in M, 3 significant figures)
2) What is the half-life? (in seconds, whole number)
3) How long until the reaction is complete ([A] = 0)? (in seconds, whole number)
1/[A] vs t
Second-order
For zero-order, the [A] vs t plot will be linear with slope =โk.
Data Test
t (s)
[A] (M)
ln[A]
1/[A] (Mโปยน)
0
0.500
โ0.693
2.00
10
0.450
โ0.799
2.22
20
0.400
โ0.916
2.50
30
0.350
โ1.050
2.86
Check constant differences: ฮ[A] = โ0.050 M per 10 s โ constant โ zero-order โ
k=0.050/10=0.005 M/s
Rate=k[A]
โdtd[A]โ=k[A]
Separating variables and integrating:
โซ[A]0โ[A]โ[A]d[A]โ=โโซ0tโkdt
ln[A]โln[A]0โ=โkt
ln[A]=โkt+ln[A]0โโ
Or equivalently:
[A]=[A]0โeโktโ
Linear Form: ln[A] vs t
Variable
Corresponds To
y
ln[A]
m (slope)
โk
x
t
b (y-intercept)
ln[A]0โ
๐ Key Concept: A plot of ln[A] vs t is linear for a first-order reaction. The slope equals โk.
First-Order Graphical Analysis ๐ฏ
๐ First-Order Half-Life
Set [A]=[A]0โ/2:
ln[A]0โ[A]0โ/2โ=โkt1/2โ
ln21โ=โkt1/2โ
โ0.693=โkt1/2โ
t1/2โ=k0.693โ
๐ Key Concept: The half-life of a first-order reaction is independent of initial concentration.
This means:
Every half-life has the same duration
After 1 half-life: 50% remains
After 2 half-lives: 25% remains
After 3 half-lives: 12.5% remains
After n half-lives: (1/2)n remains
Connection to Radioactive Decay
All radioactive decay follows first-order kinetics:
N=N0โeโฮปt,t
where ฮป is the decay constant (equivalent to k).
First-Order Calculations ๐งฎ
A first-order reaction has k=0.0100 sโปยน and [A]0โ=2.00 M.
1) What is the half-life? (in seconds, 3 significant figures)
2) What is [A] after 100 s? (in M, 3 significant figures)
3) How long until only 10% of A remains? (in seconds, 3 significant figures)
First-Order Properties ๐
๐ Useful Ratio Form
๐ก Tip: The ratio form lets you calculate the fraction remaining without knowing [A]0โ explicitly โ very useful on the AP exam!
ln[A]0โ[A]tโโ=โkt
[A]0โ[A]tโ
Quick Calculations with Half-Lives
Time
Fraction remaining
Percent remaining
0
1
100%
t1/2โ
1/2
50%
Exit Quiz โ First-Order Integrated Rate Law โ
โdtd[A]โ=k[A]2
Separating variables:
[A]2d[A]โ=โkdt
Integrating:
โซ[A]0โ[A]โ[A]โ2d[A]=โโซ0tโkdt
โ[A]1โ+[A]0โ1โ=โkt
[A]1โ=kt+[A]0โ1โโ
Linear Form: 1/[A] vs t
Variable
Corresponds To
y
1/[A]
m (slope)
+k
x
t
b (y-intercept)
1/[A]0โ
๐ Key Concept: A plot of 1/[A] vs t is linear for a second-order reaction. The slope equals +k (positive!).
Second-Order Graphical Analysis ๐ฏ
๐ Second-Order Half-Life
Set [A]=[A]0โ/2:
[A]0โ/21โ=kt1/2โ+[A]0โ1โ
[A]0โ2โโ
[A]0โ1โ=kt
t1/2โ=k[A]
๐ Key Concept: The half-life of a second-order reaction is inversely proportional to [A]0โ:
Higher [A]0โ โ shorter half-life
Each successive half-life is longer (doubles each time!)
Successive Half-Lives
Half-life
Starting [A]
Duration
1st
[A]0โ
1/(k[A]
Each successive half-life is twice the previous one. This is a telltale sign of second-order kinetics.
Second-Order Calculations ๐งฎ
A second-order reaction has k=0.50 Mโปยนsโปยน and [A]0โ=0.80 M.
1) What is the half-life? (in seconds, to 3 significant figures)
2) What is [A] after 5.0 s? (in M, to 3 significant figures)
3) What is the second half-life (starting from [A] = 0.40 M)? (in seconds, to 3 significant figures)
โ๏ธ Comparing All Three Orders
Feature
Zero-Order
First-Order
Second-Order
Rate law
k
k[A]
k[A]2
Integrated law
[A]=โkt+[A]0โ
Linear plot
[A] vs t
ln[A] vs t
1/[A vs
Slope
โk
โk
+k
Half-life
[A]0โ/(2k)
0.693/k
Successive t1/2โ
Shorter
Constant
Longer
Units of k
M/s
sโปยน
Mโปยนsโปยน
๐ก Tip: Only second-order has a positive slope in its linear plot. Zero and first-order both have slope =โk.
Identify the Order ๐
Exit Quiz โ Second-Order Integrated Rate Law โ
t
Zero-order
โk
[A]0โ
ln[A] vs t
First-order
โk
ln[A]0โ
1/[A] vs t
Second-order
+k
1/[A]0โ
How to Check Linearity
Visual inspection: Does it look like a straight line?
Constant slope: Calculate ฮy/ฮx between successive points โ is it constant?
R2 value: In a calculator, the best fit gives R2 closest to 1.
โ ๏ธ Warning: A plot of [A] vs t is always curved for first and second-order โ don't mistake a gentle curve for a straight line!
๐ก Tip โ AP Exam Shortcut: If you are given just the raw data, calculate the transformed values and check which set has constant spacing:
If ฮ[A] is constant per ฮt โ zero-order
If ฮ(ln[A]) is constant per ฮt โ first-order
If ฮ(1/[A]) is constant per ฮt โ second-order
๐งช Worked Example: Identifying Order
Problem: Given the following data, determine the reaction order and find k.
t (s)
[A] (M)
ln[A]
1/[A] (Mโปยน)
0
1.000
0.000
1.000
10
0.607
โ0.500
1.648
20
0.368
โ1.000
2.718
30
0.223
โ1.500
4.484
40
0.135
โ2.000
7.407
Solution:
Test [A] vs t: Differences in [A]: โ0.393, โ0.239, โ0.145, โ0.088 โ NOT constant โ
Test ln[A] vs t: Differences in ln[A]: โ0.500, โ0.500, โ0.500, โ0.500 โ CONSTANT โ
Test 1/[A] vs t: Differences: +0.648, +1.070, +1.766, +2.923 โ NOT constant โ
Conclusion: The reaction is first-order.
k=โslope=โ(โ0.500/10)=0.0500ย sโ1
Graphical Analysis Practice ๐ฏ
Given this data:
t (s)
[B] (M)
0
0.400
100
0.300
200
0.200
300
0.100
Identify the Order ๐งฎ
t (min)
[C] (M)
ln[C]
1/[C] (Mโปยน)
0
0.500
โ0.693
2.00
5
0.333
โ1.099
3.00
10
0.250
โ1.386
4.00
15
0.200
โ1.609
5.00
1) What is the order? (enter 0, 1, or 2)
2) What is k? (number only, to 3 significant figures)
3) What is [C] at t = 25 min? (in M, to 3 significant figures)
Slope Interpretation ๐
AP-Style Problem ๐ฏ
A student collects concentration-time data and plots all three standard graphs. She finds:
โ ๏ธ Warning: For zero and second-order, "n half-lives" does NOT mean nรt1/2โ total time, because each successive half-life has a different duration!
๐ Key Concept: The pattern of successive half-lives uniquely identifies reaction order:
Order
1st tโ/โ
2nd tโ/โ
3rd tโ/โ
Pattern
Zero
T
T/2
T/4
Each is half the previous
First
T
T
T
All equal
Second
T
2T
4T
Each is double the previous
Zero-Order Half-Life Problems ๐งฎ
A zero-order reaction has k=0.0020 M/s.
1) If [A]โ = 0.100 M, what is the half-life? (in seconds)
2) If [A]โ = 0.200 M, what is the half-life? (in seconds)
3) For [A]โ = 0.100 M, how long until 75% has reacted (only 25% remains)? Note: this is NOT simply 2 half-lives for zero-order! Use the integrated rate law. (in seconds, to 3 significant figures)
First-Order Half-Life Problems ๐งฎ
1) A first-order reaction has k=0.0462 sโปยน. What is the half-life? (in seconds, to 3 significant figures)
2) If 93.75% of a first-order reactant has decomposed, how many half-lives have passed? (integer)
3) Iodine-131 has a half-life of 8.02 days. What fraction remains after 24.06 days? Express as a fraction with denominator 8 (e.g., enter 3/8).
Second-Order Half-Life Problems ๐งฎ
A second-order reaction has k=0.40 Mโปยนsโปยน and [A]0โ=0.50 M.
1) What is the first half-life? (in seconds, to 3 significant figures)
2) What is the second half-life? (in seconds, to 3 significant figures)
3) What is [A] after 15 s? (in M, to 3 significant figures)
๐ Radioactive Decay: Always First-Order
All radioactive decay processes follow first-order kinetics:
N=N0โeโฮปtโ
lnN0โNโ=โฮปt
t1/2โ=ฮป0.693โ
where ฮป = decay constant (same role as k), N = number of atoms remaining.
๐ Key Concept: All radioactive decay is first-order โ the probability of any one nucleus decaying is constant per unit time, regardless of how many nuclei remain.
Carbon-14 Dating
14C has t1/2โ=5,730 years
Living organisms maintain constant C/C ratio through intake
Example
Problem: A fossil has 25% of original 14C. How old is it?
Solution:
N0โNโ=0.25=
Age=2ร5,730=11,460ย years
Radioactive Decay Quiz ๐ฏ
Half-Life Review ๐
Exit Quiz โ Half-Life Problems โ
t1/2โ
Equal = 1st; Decreasing = 0th; Doubling = 2nd
Initial rates
Compare experiments (covered earlier)
Change [A], observe rate change
Step 2: Find k from the Linear Plot
Order
Linear Plot
Slope
Zero
[A] vs. t
โk
First
ln[A] vs. t
โk
Second
1/[A] vs. t
+k
Step 3: Solve the Problem
Use the appropriate integrated rate law to find:
Concentration at any time
Time to reach a certain concentration
Half-life
โ ๏ธ Always check the units of k to confirm the order you identified. Wrong units = wrong order!
Problem 1: Order Identification ๐งฎ
t (min)
[A] (M)
0
0.800
10
0.400
20
0.200
30
0.100
1) What is the order of the reaction? (enter 0, 1, or 2)
2) What is k? (to 3 significant figures)
3) What is [A] at t = 50 min? (in M, to 3 significant figures)
Problem 2: Data Table Analysis ๐ฏ
t (s)
[B] (M)
1/[B]
0
0.500
2.00
50
0.333
3.00
100
0.250
4.00
150
0.200
5.00
Problem 3: Working Backwards ๐งฎ
A first-order reaction has a half-life of 25.0 minutes. The initial concentration is 1.20 M.
1) What is k? (in minโปยน, to 3 significant figures)
2) What is [A] after 75.0 minutes? (in M, to 3 significant figures)
3) How long until [A] = 0.10 M? (in minutes, to 3 significant figures)
Problem 4: Conceptual Matching ๐
Challenge Problem ๐งฎ
A certain reaction is second-order with k=0.10 Mโปยนsโปยน and [A]0โ=2.0 M.
1) What is the first half-life? (in seconds, to 3 significant figures)
2) How long total until 87.5% of A has reacted? (in seconds, to 3 significant figures)
3) What is [A] at t = 35 s? (in M, to 3 significant figures)
Exit Quiz โ Problem-Solving Workshop โ
k[A]
Rate = k[A]2
Integrated
[A]=โkt+[A]0โ
ln[A]=โkt+ln[A]0โ
[A]1โ=kt+[
Linear plot
[A] vs t
ln[A] vs t
[A]1โ vs t
Slope
โk
โk
+k
t1/2โ
2k[A]0โโ
k0.693โ
k[A]0โ1โ
Units of k
M/s
sโปยน
Mโปยนsโปยน
Successive t1/2โ
Decrease (halve)
Constant
Increase (double)
โ ๏ธ Warning: On the AP exam, always write the correct units of k with your answer. M/s (zero), sโปยน (first), Mโปยนsโปยน (second) โ getting units wrong can cost points!
AP Problem 1 ๐ฏ
The decomposition of SOโClโ is first-order:
SO2โCl2โ(g)โSO2โ(g)+Cl2โ(g)
At 320ยฐC, k=2.20ร10โ5 sโปยน.
AP Problem 2: Order Determination ๐งฎ
The decomposition of a compound was studied. Data:
t (min)
[A] (M)
0
1.00
20
0.50
40
0.33
60
0.25
Verify: 1/[A] values are 1.00, 2.00, 3.00, 4.00 (constant ฮ of 1.00 per 20 min).
1) What is the order? (enter 0, 1, or 2)
2) What is k? (in Mโปยนminโปยน, to 3 significant figures)
3) What is [A] at t = 100 min? (in M, to 3 significant figures)
AP Problem 3: Carbon Dating ๐ฏ
A wooden artifact is found to have 12.5% of the 14C content of living wood. The half-life of 14C is 5,730 years.
Synthesis: Match the Scenario ๐
AP Problem 4: Comprehensive ๐งฎ
A first-order reaction has k=3.46ร10โ2 sโปยน.
1) What is the half-life? (in seconds, to 3 significant figures)
2) How long until 90% has reacted? (in seconds, to 3 significant figures)
3) If [A]โ = 0.500 M, what is [A] after 30.0 s? (in M, to 3 significant figures)