📈 Zero-Order Integrated Rate Law

Part 1 of 7 — 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.

Derivation of the Zero-Order Integrated Rate Law

For a zero-order reaction: $\text{Rate} = k$

$-\frac{d[A]}{dt} = k$

Integrating from $[A]_0$ to $[A]$ and from $0$ to $t$:

$\int_{[A]_0}^{[A]} d[A] = -\int_0^t k \, dt$

$[A] - [A]_0 = -kt$

$\boxed{[A] = -kt + [A]_0}$

This is the equation of a straight line!

$y = mx + b$

Plot: $[A]$ vs $t$ → straight line for zero-order

• Slope = $-k$
• y-intercept = $[A]_0$
• $[A]$ decreases linearly with time

Zero-Order Graphical Analysis 🎯

Zero-Order Half-Life

The half-life ($t_{1/2}$) is the time for the concentration to drop to half its initial value.

Set $[A] = [A]_0/2$ in the integrated rate law:

$\frac{[A]_0}{2} = -kt_{1/2} + [A]_0$

$kt_{1/2} = [A]_0 - \frac{[A]_0}{2} = \frac{[A]_0}{2}$

$\boxed{t_{1/2} = \frac{[A]_0}{2k}}$

Key Feature

The zero-order half-life depends on $[A]_0$:

• Higher initial concentration → longer half-life
• Each successive half-life is shorter than the previous one
• The reaction reaches [A] = 0 in a finite time: $t_{\text{complete}} = [A]_0/k$

Successive Half-Lives

Each successive half-life is exactly half the duration of the previous one.

Zero-Order Half-Life Concepts 🔍

Zero-Order Calculations 🧮

A zero-order reaction has $k = 5.0 \times 10^{-3}$ M/s and $[A]_0 = 0.60$ M.

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)

How to Identify Zero-Order from Data

Method: Test Different Plots

Given concentration-time data, make three plots:

For zero-order, the $[A]$ vs $t$ plot will be linear with slope $= -k$.

Data Test

Check constant differences: Δ[A] = −0.050 M per 10 s → constant → zero-order ✓

$k = 0.050/10 = 0.005$ M/s

Exit Quiz — Zero-Order Integrated Rate Law ✅

📉 First-Order Integrated Rate Law

Part 2 of 7 — 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.

Derivation

For a first-order reaction: $\text{Rate} = k[A]$

$-\frac{d[A]}{dt} = k[A]$

Separating variables and integrating:

$\int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = -\int_0^t k \, dt$

$\ln[A] - \ln[A]_0 = -kt$

$\boxed{\ln[A] = -kt + \ln[A]_0}$

Or equivalently:

$\boxed{[A] = [A]_0 e^{-kt}}$

Linear Form: $\ln[A]$ vs $t$

Key Feature

A plot of $\ln[A]$ vs $t$ is linear for a first-order reaction. The slope equals $-k$.

First-Order Graphical Analysis 🎯

First-Order Half-Life

Set $[A] = [A]_0/2$:

$\ln\frac{[A]_0/2}{[A]_0} = -kt_{1/2}$

$\ln\frac{1}{2} = -kt_{1/2}$

$-0.693 = -kt_{1/2}$

$\boxed{t_{1/2} = \frac{0.693}{k}}$

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:

$N = N_0 e^{-\lambda t}, \quad t_{1/2} = \frac{0.693}{\lambda}$

where $\lambda$ is the decay constant (equivalent to $k$).

First-Order Calculations 🧮

A first-order reaction has $k = 0.0100$ s⁻¹ and $[A]_0 = 2.00$ 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 🔍

Useful Ratio Form

Often you need to find how much reactant remains at time $t$ without knowing $[A]_0$ explicitly:

$\ln\frac{[A]_t}{[A]_0} = -kt$

$\frac{[A]_t}{[A]_0} = e^{-kt}$

Quick Calculations with Half-Lives

Exit Quiz — First-Order Integrated Rate Law ✅

📊 Second-Order Integrated Rate Law

Part 3 of 7 — 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.

Derivation (for Rate = k[A]²)

$-\frac{d[A]}{dt} = k[A]^2$

Separating variables:

$\frac{d[A]}{[A]^2} = -k \, dt$

Integrating:

$\int_{[A]_0}^{[A]} [A]^{-2} \, d[A] = -\int_0^t k \, dt$

$-\frac{1}{[A]} + \frac{1}{[A]_0} = -kt$

$\boxed{\frac{1}{[A]} = kt + \frac{1}{[A]_0}}$

Linear Form: $1/[A]$ vs $t$

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 🎯

Second-Order Half-Life

Set $[A] = [A]_0/2$:

$\frac{1}{[A]_0/2} = kt_{1/2} + \frac{1}{[A]_0}$

$\frac{2}{[A]_0} - \frac{1}{[A]_0} = kt_{1/2}$

$\frac{1}{[A]_0} = kt_{1/2}$

$\boxed{t_{1/2} = \frac{1}{k[A]_0}}$

Key Feature

The half-life of a second-order reaction is inversely proportional to $[A]_0$:

• Higher $[A]_0$ → shorter half-life
• Each successive half-life is longer (doubles each time!)

Successive Half-Lives

Each successive half-life is twice the previous one. This is a telltale sign of second-order kinetics.

Second-Order Calculations 🧮

A second-order reaction has $k = 0.50$ M⁻¹s⁻¹ and $[A]_0 = 0.80$ 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)

Comparing All Three Orders

Identify the Order 🔍

Exit Quiz — Second-Order Integrated Rate Law ✅

📊 Graphical Analysis

Part 4 of 7 — 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.

The Three-Plot Strategy

Given concentration-vs-time data, create:

How to Check Linearity

1. Visual inspection: Does it look like a straight line?
2. Constant slope: Calculate $\Delta y / \Delta x$ between successive points — is it constant?
3. $R^2$ value: In a calculator, the best fit gives $R^2$ 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 $\Delta[A]$ is constant per $\Delta t$ → zero-order
• If $\Delta(\ln[A])$ is constant per $\Delta t$ → first-order
• If $\Delta(1/[A])$ is constant per $\Delta t$ → second-order

Worked Example: Identifying Order

Test [A] vs t: Differences in [A]: −0.393, −0.239, −0.145, −0.088 → NOT constant ✗

Test ln[A] vs t: Differences in ln[A]: −0.500, −0.500, −0.500, −0.500 → CONSTANT ✓

Test 1/[A] vs t: Differences: +0.648, +1.070, +1.766, +2.923 → NOT constant ✗

Conclusion: The reaction is first-order.

$k = -\text{slope} = -(-0.500/10) = 0.0500 \text{ s}^{-1}$

Graphical Analysis Practice 🎯

Given this data:

Identify the Order 🧮

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 🔍

AP-Style Problem 🎯

A student collects concentration-time data and plots all three standard graphs. She finds:

• [A] vs t: curved
• ln[A] vs t: curved
• 1/[A] vs t: straight line with slope 0.45

Exit Quiz — Graphical Analysis ✅

⏱️ Half-Life Problems

Part 5 of 7 — 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).

Half-Life Formulas Summary

Pattern of Successive Half-Lives

Zero-Order Half-Life Problems 🧮

A zero-order reaction has $k = 0.0020$ M/s.

1. If [A]₀ = 0.100 M, what is the half-life? (in seconds)

2. If [A]₀ = 0.200 M, what is the half-life? (in seconds)

3. For [A]₀ = 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 🧮

1. A first-order reaction has $k = 0.0462$ s⁻¹. 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 🧮

A second-order reaction has $k = 0.40$ M⁻¹s⁻¹ and $[A]_0 = 0.50$ 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)

Radioactive Decay: Always First-Order

All radioactive decay processes follow first-order kinetics:

$N = N_0 e^{-\lambda t}$

$\ln\frac{N}{N_0} = -\lambda t$

$t_{1/2} = \frac{0.693}{\lambda}$

where $\lambda$ = decay constant (same role as $k$), $N$ = number of atoms remaining.

Carbon-14 Dating

• $^{14}$C has $t_{1/2} = 5{,}730$ years
• Living organisms maintain constant $^{14}$C/$^{12}$C ratio through intake
• When an organism dies, $^{14}$C decays without replacement
• Measuring the remaining $^{14}$C fraction tells us when it died

Example

A fossil has 25% of original $^{14}$C. How old is it?

$\frac{N}{N_0} = 0.25 = \left(\frac{1}{2}\right)^n \Rightarrow n = 2 \text{ half-lives}$

$\text{Age} = 2 \times 5{,}730 = 11{,}460 \text{ years}$

Radioactive Decay Quiz 🎯

Half-Life Review 🔍

Exit Quiz — Half-Life Problems ✅

🔧 Problem-Solving Workshop

Part 6 of 7 — 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.

Problem-Solving Flowchart

Step 1: Identify the Order

Method A — Graphical: Plot [A], ln[A], and 1/[A] vs t. The linear one wins.

Method B — Successive half-lives:

• Half-lives equal → first-order
• Half-lives decreasing → zero-order
• Half-lives increasing (doubling) → second-order

Method C — 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 = $-k$
• First: slope of ln[A] vs t = $-k$
• Second: slope of 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, or half-life.

Problem 1: Order Identification 🧮

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 🎯

Problem 3: Working Backwards 🧮

A first-order reaction has a half-life of 25.0 minutes. The initial concentration is 1.20 M.

1. What is k? (in min⁻¹, 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 🔍

Challenge Problem 🧮

A certain reaction is second-order with $k = 0.10$ M⁻¹s⁻¹ and $[A]_0 = 2.0$ 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 — Problem-Solving Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — 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.

Complete Integrated Rate Laws Summary

AP Problem 1 🎯

The decomposition of SO₂Cl₂ is first-order:

$\text{SO}_2\text{Cl}_2(g) \rightarrow \text{SO}_2(g) + \text{Cl}_2(g)$

At 320°C, $k = 2.20 \times 10^{-5}$ s⁻¹.

AP Problem 2: Order Determination 🧮

The decomposition of a compound was studied. Data:

Verify: $1/[A]$ values are 1.00, 2.00, 3.00, 4.00 (constant Δ of 1.00 per 20 min).

1. What is the order? (enter 0, 1, or 2)

2. What is k? (in M⁻¹min⁻¹, to 3 significant figures)

3. What is [A] at t = 100 min? (in M, to 3 significant figures)

AP Problem 3: Carbon Dating 🎯

A wooden artifact is found to have 12.5% of the $^{14}$C content of living wood. The half-life of $^{14}$C is 5,730 years.

Synthesis: Match the Scenario 🔍

AP Problem 4: Comprehensive 🧮

A first-order reaction has $k = 3.46 \times 10^{-2}$ s⁻¹.

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]₀ = 0.500 M, what is [A] after 30.0 s? (in M, to 3 significant figures)

Final Exit Quiz — Integrated Rate Laws ✅