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
Integrated rate laws connect concentration to time directly. While the differential rate law tells us how rate depends on concentration, the integrated form lets us calculate concentrations at any future time, determine half-lives, and identify reaction order from graphical data.
We begin with the simplest case: zero-order reactions.
๐ Derivation of the Zero-Order Integrated Rate Law
For a zero-order reaction: Rate=k
โdtd[A]โ=k
Integrating from [A]0โ to [A] and from 0 to t:
โซ[A]0โ[A]โd[A]
[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
Plot: [A] vs t โ straight line for zero-order
Slope = โk
y-intercept = [A]0โ
[A] decreases linearly with time
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
First-order reactions are the most common type in chemistry. Radioactive decay, many decomposition reactions, and most biochemical processes follow first-order kinetics. The math involves logarithms, and the behavior is exponential decay.
๐ Derivation
For a first-order reaction: Rate=k[A]
โ
Part 3: Second-Order Reactions
๐ Second-Order Integrated Rate Law
Part 3 of 7 โ Inverse Concentration and Time
Second-order reactions complete our trio of integrated rate laws. The mathematics involves the reciprocal of concentration, and the behavior is distinctly different from both zero and first-order kinetics.
โฑ๏ธ Derivation (for Rate = k[A]ยฒ)
โdtd[A]โ=k
Part 4: Half-Life
๐ Graphical Analysis
Part 4 of 7 โ Identifying Reaction Order from Plots
On the AP exam, you may be given data or graphs and asked to determine the reaction order. The key technique: make three plots and see which one is linear. This part gives you systematic practice with this approach.
๐ฏ The Three-Plot Strategy
Given concentration-vs-time data, create:
Plot
If Linear โ
Slope =
y-intercept =
[A] vs t
Zero-order
โk
Part 5: Graphical Analysis of Order
โฑ๏ธ Half-Life Problems
Part 5 of 7 โ Calculations for Each Order and Radioactive Decay
Half-life is one of the most-tested topics on the AP Chemistry exam. This part provides intensive practice with half-life calculations for all three orders, plus applications to radioactive decay (which is always first-order).
๐ Half-Life Formulas Summary
Order
Half-Life Formula
Dependence on [A]0โ
Zero
Part 6: Problem-Solving Workshop
๐ง Problem-Solving Workshop
Part 6 of 7 โ Mixed Order Identification and Calculations
This workshop combines all three integrated rate laws in problems that require you to first identify the order, then perform calculations. These mirror the multi-step problems found on the AP exam.
๐ ๏ธ Problem-Solving Flowchart
Step 1: Identify the Order
Method A โ Graphical: Plot [A], ln[A], and 1/[A] vs t. The linear one wins.
Method B โ Successive half-lives:
Half-lives equal โ first-order
Half-lives decreasing โ zero-order
Half-lives increasing (doubling) โ second-order
Method C โ Initial rates: Compare experiments (covered in earlier parts).
Step 2: Find k
Use the slope of the appropriate linear plot:
Zero: slope of [A] vs t = โk
First: slope of ln[A] vs t = โk
Second: slope of 1/[A] vs t =
Part 7: Synthesis & AP Review
๐ Synthesis & AP Review
Part 7 of 7 โ AP-Style Integrated Rate Law Problems
This final part challenges you with comprehensive, exam-level problems that combine order identification, integrated rate law calculations, half-life analysis, and graphical interpretation.
๐ Complete Integrated Rate Laws Summary
Zero-Order
First-Order
Second-Order
Differential
Rate = k
Rate = k[A]
Rate =
=
โโซ0tโkdt
x
t
b (y-intercept)
[A]0โ
โ
/2
2[A]0โโ=โkt1/2โ+[A]0โ
kt1/2โ=[A]0โโ2[A]0โโ=2[A]0โโ
t1/2โ=2k[A]0โโโ
Key Feature
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.
=
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
dt
d[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
Key Feature
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
The Most Important Feature
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
Often you need to find how much reactant remains at time t without knowing [A]0โ explicitly:
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
[
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
Key Feature
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]
Key Feature
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
Identify the Order ๐
Exit Quiz โ Second-Order Integrated Rate Law โ
[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.
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
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
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:
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=N0โeโฮปt
lnN0โNโ=โฮปt
t1/2โ=ฮป0.693โ
where ฮป = decay constant (same role as k), N = number of atoms remaining.
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
+k
Step 3: Solve the Problem
Use the appropriate integrated rate law to find concentration at any time, time to reach a certain concentration, or half-life.
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]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)
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)
Final Exit Quiz โ Integrated Rate Laws โ
[A]0โ/(4k)
[A]0โ/(8k)
โ
1/2
โ
=
ฮป0.693โ
โ
=
eโkt
2
t1/2โ
1/4
25%
3t1/2โ
1/8
12.5%
4t1/2โ
1/16
6.25%
10t1/2โ
1/1024
~0.1%
[A]0โ1โ
=
kt1/2โ
1/2
โ
0
โ
1
โ
โ
0
โ
)
2nd
[A]0โ/2
2/(k[A]0โ)
3rd
[A]0โ/4
4/(k[A]0โ)
ln
[
A
]
=
โkt+
ln[A]0โ
1/[A]=kt+1/[A]0โ
]
t
1/
(
k
[
A
]0โ
)
14
12
When an organism dies, 14C decays without replacement
Measuring the remaining 14C fraction tells us when it died