Reaction Rates and Rate Laws

⚗️ Measuring Reaction Rates

Part 1 of 7 — How Fast Does It Go?

Topics in This Part

🔑 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

⚗️ Defining Reaction Rate

For a general reaction:

$aA + bB \rightarrow cC + dD$

The rate of reaction is defined as the change in concentration of a reactant or product per unit time:

$\boxed{\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t} = +\frac{1}{c}\frac{\Delta[C]}{\Delta t} = +\frac{1}{d}\frac{\Delta[D]}{\Delta t}}$

Key Points

🔑 Key Concept: Rate is always defined as a positive quantity. The negative sign for reactants ensures this, since reactant concentrations decrease over time ($\Delta[A] < 0$).

Example

Problem: For $2\text{N}_2\text{O}_5 \rightarrow 4\text{NO}_2 + \text{O}_2$:

Solution:

$\boxed{\text{Rate} = -\frac{1}{2}\frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} = +\frac{1}{4}\frac{\Delta[\text{NO}_2]}{\Delta t} = +\frac{\Delta[\text{O}_2]}{\Delta t}}$

If $\text{O}_2$ appears at $0.024$ M/s, then $\text{NO}_2$ appears at $4 \times 0.024 = 0.096$ M/s and $\text{N}_2\text{O}_5$ disappears at $2 \times 0.024 = 0.048$ M/s.

Rate Definition Quiz 🎯

⏱️ Average Rate vs. Instantaneous Rate

Average Rate

The average rate is calculated over a finite time interval:

$\boxed{\text{Average rate} = -\frac{\Delta[A]}{\Delta t} = -\frac{[A]_{t_2} - [A]_{t_1}}{t_2 - t_1}}$

Instantaneous Rate

The instantaneous rate is the rate at a specific moment — the slope of the tangent line to the concentration-vs-time curve:

$\boxed{\text{Instantaneous rate} = -\frac{d[A]}{dt}}$

Key Differences

Initial Rate

The initial rate is the instantaneous rate at $t = 0$, before significant product buildup. It is especially useful because:

• Concentrations are known precisely (the starting concentrations)
• Reverse reactions have not yet become significant
• It is used in the method of initial rates to determine rate laws

🔑 Key Concept: The initial rate is key to determining rate laws because concentrations are precisely known and the reverse reaction hasn't started.

Average vs. Instantaneous Rate 🔍

⏱️ Experimental Methods for Measuring Rates

Monitoring Concentration Over Time

Beer's Law Connection

For colored species, absorbance is directly proportional to concentration:

$\boxed{A = \varepsilon b c}$

where $\varepsilon$ = molar absorptivity, $b$ = path length, $c$ = concentration. By measuring absorbance over time, you can track $[\text{colored species}]$ over time.

Rate Calculation Drill 🧮

Consider the reaction: $2\text{NO}_2 \rightarrow 2\text{NO} + \text{O}_2$

1) What is the average rate of disappearance of $NO_{2}$ over the first 50 s? (in M/s, 3 significant figures)

2) What is the average rate of the reaction over the first 50 s? (divide by stoichiometric coefficient, 3 significant figures)

3) What is the average rate of appearance of $O_{2}$ over the interval 0–100 s? (in M/s, 3 significant figures)

Exit Quiz — Measuring Reaction Rates ✅

🌡️ Factors Affecting Reaction Rate

Part 2 of 7 — What Makes Reactions Faster?

Topics in This Part

🔑 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

📊 Factor 1: Concentration of Reactants

Higher concentration → faster rate (usually)

Why?

More particles per unit volume means more frequent collisions. More collisions per second means more chances for a successful (reactive) collision.

Mathematical Connection

The rate law (which we will derive in later parts) often takes the form:

$\boxed{\text{Rate} = k[A]^m[B]^n}$

Increasing $[A]$ or $[B]$ directly increases the rate (when $m, n > 0$).

Example

Burning steel wool in pure $O_{2}$ (100%) is dramatically faster than in air (21% $O_{2}$) because the concentration of oxygen molecules is about 5 times higher.

📊 Factor 2: Temperature

Higher temperature → faster rate

Why?

At higher temperatures:

1. Molecules move faster → more frequent collisions
2. A greater fraction of molecules have enough energy to overcome the activation energy barrier

Rule of Thumb

For many reactions, a 10°C increase roughly doubles the rate. This is an approximation — the actual factor depends on the activation energy.

💡 Tip: The "rate doubles per 10°C" rule is a rough estimate. The actual factor depends on $E_a$ — reactions with higher activation energies are more sensitive to temperature changes.

Quantitative: The Arrhenius Equation

$\boxed{k = Ae^{-E_a/(RT)}}$

where:

• $k$ = rate constant
• $A$ = frequency factor (collision frequency × orientation factor)
• $E_a$ = activation energy
• $R$ = 8.314 J/(mol·K)
• $T$ = temperature in Kelvin

We will explore this equation in depth in a later topic.

📊 Factor 3: Surface Area

Greater surface area → faster rate (for heterogeneous reactions)

Why?

In heterogeneous reactions (where reactants are in different phases), the reaction occurs at the interface between phases. More exposed surface = more contact area = faster reaction.

Examples

Dust Explosions

Finely powdered combustible materials (flour, coal dust, grain dust) have enormous surface area. If suspended in air, they can ignite and cause devastating explosions. This is why grain elevators have strict safety protocols.

📊 Factor 4: Catalysts

Catalysts speed up reactions without being consumed

How?

A catalyst provides an alternative reaction pathway with a lower activation energy ($E_a$).

$\boxed{E_a(\text{catalyzed}) < E_a(\text{uncatalyzed})}$

Key Properties of Catalysts

• Not consumed — regenerated at the end of the mechanism
• Lower $E_a$ — more molecules have sufficient energy to react
• Do NOT change $\Delta H$ or $\Delta G$ — thermodynamics is unaffected
• Do NOT shift equilibrium — both forward and reverse rates increase equally
• Speed up both directions equally

🔑 Key Concept: A catalyst lowers $E_a$ but does NOT change $\Delta H$, $\Delta G$, or the equilibrium position. It speeds up both forward and reverse reactions equally.

Types

Factors Affecting Rate Quiz 🎯

Match the Factor 🔍

Application Problems 🧮

1) A reaction has a rate of 0.020 M/s at 25°C. Using the rough rule that rate doubles with each 10°C increase, estimate the rate at 45°C. (in M/s)

2) If the concentration of a reactant is tripled, and the reaction is second-order in that reactant, by what factor does the rate increase? (whole number)

3) A catalyzed reaction has $E_a = 50$ kJ/mol. The uncatalyzed reaction has $E_a = 120$ kJ/mol. By how many kJ/mol does the catalyst lower the activation energy?

Round all answers to 3 significant figures.

Exit Quiz — Factors Affecting Rate ✅

📐 Rate Laws

Part 3 of 7 — The Mathematical Heart of Kinetics

Topics in This Part

🔑 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

📏 The General Rate Law

For a reaction $aA + bB \rightarrow \text{products}$:

$\boxed{\text{Rate} = k[A]^m[B]^n}$

Critical Points

• $m$ and $n$ are NOT necessarily the stoichiometric coefficients $a$ and $b$
• Orders must be determined from experimental data

⚠️ Warning: You cannot read rate law exponents off the balanced equation! Orders must be determined from experimental data. They only equal coefficients for elementary (single-step) reactions.

• Orders can be 0, 1, 2, or even fractional
• The rate constant $k$ depends on temperature but NOT on concentration

What Does Order Mean?

Reaction Order Concepts 🎯

🔍 Determining Order from Experimental Data

The Method of Initial Rates

The most common technique: measure the initial rate of reaction for several experiments where you vary one concentration at a time.

🔑 Key Concept: In the method of initial rates, you vary one concentration at a time while keeping others constant, then take ratios to determine each order.

Example Data

Step 1: Find order in A

Compare Exp 1 and 2 ([B] is constant): $\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[A]_2^m[B]^n}{k[A]_1^m[B]^n} = \left(\frac{[A]_2}{[A]_1}\right)^m$

$\frac{0.060}{0.015} = \left(\frac{0.20}{0.10}\right)^m \Rightarrow 4 = 2^m \Rightarrow m = 2$

Step 2: Find order in B

Compare Exp 1 and 3 ([A] is constant): $\frac{0.030}{0.015} = \left(\frac{0.20}{0.10}\right)^n \Rightarrow 2 = 2^n \Rightarrow n = 1$

Step 3: Write the rate law and find k

$\boxed{\text{Rate} = k[A]^2[B]}$

Using Exp 1: $0.015 = k(0.10)^2(0.10) = k(0.001)$

$\boxed{k = \frac{0.015}{0.001} = 15 \; \text{M}^{-2}\text{s}^{-1}}$

Order Determination Practice 🧮

Given:

1) What is the order with respect to X? (integer)

2) What is the order with respect to Y? (integer)

3) What is the value of k? (in appropriate units, give the number only — e.g., if k = 2.0, enter 2.0)

Round all answers to 3 significant figures.

📌 Common Mistakes to Avoid

❌ Mistake 1: Using stoichiometric coefficients as orders

For $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$, the rate law is experimentally found to be:

$\text{Rate} = k[\text{NO}]^2[\text{O}_2]$

The orders happen to match the coefficients here, but this is coincidence — it only occurs when the reaction happens in a single elementary step.

❌ Mistake 2: Forgetting to hold one variable constant

When comparing experiments to find the order in A, you must choose experiments where [B] is the same. If both change, you cannot isolate the effect of one.

❌ Mistake 3: Confusing rate with rate constant

• Rate changes as concentrations change during a reaction
• Rate constant $k$ is fixed at a given temperature

Rate Law Concepts 🔍

Exit Quiz — Rate Laws ✅

📊 Method of Initial Rates

Part 4 of 7 — Systematic Rate Law Determination

Topics in This Part

🔑 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

📋 Step-by-Step Method

Given Data Table Format

The Algorithm

Step 1: Assume Rate = k[A]^m[B]^n

Step 2: Pick two experiments where only one concentration changes

Step 3: Take the ratio: $\frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[A]_2}{[A]_1}\right)^m$

Step 4: Solve for $m$ using logarithms if needed: $\boxed{m = \frac{\ln(\text{Rate}_2/\text{Rate}_1)}{\ln([A]_2/[A]_1)}}$

Step 5: Repeat for each reactant

Step 6: Substitute back into any experiment to solve for $k$

Using Logarithms for Non-Integer Orders

If $\frac{\text{Rate}_2}{\text{Rate}_1} = 2.83$ and $\frac{[A]_2}{[A]_1} = 2$:

$m = \frac{\ln(2.83)}{\ln(2)} = \frac{1.04}{0.693} = 1.5$

So the order is $\frac{3}{2}$ (fractional order).

💡 Tip: When the rate ratio isn't a clean power of the concentration ratio, use logarithms to find the order. Fractional orders are common for multi-step mechanisms.

🧪 Worked Example 1

Problem: For the reaction $\text{A} + \text{B} \rightarrow \text{C}$, determine the rate law.

Solution:

Finding order in A (compare Exp 1 & 2, [B] constant): $\frac{16.0 \times 10^{-5}}{4.0 \times 10^{-5}} = \left(\frac{0.200}{0.100}\right)^m \Rightarrow 4 = 2^m \Rightarrow m = 2$

Finding order in B (compare Exp 1 & 3, [A] constant): $\frac{4.0 \times 10^{-5}}{4.0 \times 10^{-5}} = \left(\frac{0.300}{0.100}\right)^n \Rightarrow 1 = 3^n \Rightarrow n = 0$

Rate law: Rate = k[A]${}^{2}$ (zero-order in B!)

Finding k: Using Exp 1: $4.0 \times 10^{-5} = k(0.100)^2$ $\boxed{k = \frac{4.0 \times 10^{-5}}{0.0100} = 4.0 \times 10^{-3} \; \text{M}^{-1}\text{s}^{-1}}$

Practice Problem 1 🧮

For the reaction $\text{P} + \text{Q} \rightarrow \text{R}$:

1) What is the order with respect to P? (integer)

2) What is the order with respect to Q? (integer)

3) What is the value of the rate constant k? (give the number; e.g., enter 7.5 for 7.5)

Round all answers to 3 significant figures.

🧪 Worked Example 2: Three Reactants

Problem: For $2\text{NO}(g) + \text{Cl}_2(g) \rightarrow 2\text{NOCl}(g)$, determine the rate law.

Solution:

Order in $Cl_{2}$ (Exp 1 vs 2, [NO] constant): $\frac{0.36}{0.18} = \left(\frac{0.20}{0.10}\right)^n \Rightarrow 2 = 2^n \Rightarrow n = 1$

Order in NO (Exp 1 vs 3, [$Cl_{2}$] constant): $\frac{0.72}{0.18} = \left(\frac{0.20}{0.10}\right)^m \Rightarrow 4 = 2^m \Rightarrow m = 2$

Rate law: Rate = k[NO]${}^{2}$[$Cl_{2}$]

Finding k: $0.18 = k(0.10)^2(0.10) = k(0.001)$ $\boxed{k = 180 \; \text{M}^{-2}\text{s}^{-1}}$

⚠️ Warning: The orders (2 and 1) happen to match the stoichiometric coefficients here, but this is coincidental — it only works because the rate-determining step happens to be bimolecular with these exact stoichiometries.

Method of Initial Rates Quiz 🎯

Challenge: Complete Rate Law Determination 🧮

For $\text{A} + \text{B} + \text{C} \rightarrow \text{Products}$:

1) What is the overall order of the reaction? (integer)

2) What is the rate constant k? (number only)

3) Predict the rate (in M/s) when [A] = 0.30, [B] = 0.20, [C] = 0.50.

Round all answers to 3 significant figures.

Exit Quiz — Method of Initial Rates ✅

📏 Units of the Rate Constant k

Part 5 of 7 — How Units Change with Reaction Order

Topics in This Part

🔑 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

📌 Deriving Units of k

Start from the general rate law:

$\text{Rate} = k[A]^m[B]^n$

Rate always has units of M/s $(or mol\cdot L^{-1}\cdot s^{-1})$. Concentration has units of M (mol/L).

Solving for $k$:

$k = \frac{\text{Rate}}{[A]^m[B]^n}$

$\boxed{\text{Units of } k = \frac{\text{M/s}}{\text{M}^{m+n}} = \text{M}^{1-(m+n)} \cdot \text{s}^{-1}}$

Summary Table

The Pattern

$\boxed{\text{Units of } k = \text{M}^{1-n}\text{s}^{-1}}$

where $n$ = overall order. As order increases by 1, the power of M decreases by 1.

💡 Tip: Units of $k$ vary with reaction order! You can identify the overall order from the units of $k$, and vice versa. This is a common AP exam shortcut.

Units of k Quiz 🎯

⚗️ Zero-Order Reactions

$\boxed{\text{Rate} = k}$

• Rate is constant — independent of concentration
• Units of $k$: M/s
• Half-life: $t_{1/2} = \frac{[A]_0}{2k}$ (depends on initial concentration)

When Do Zero-Order Reactions Occur?

Zero-order kinetics often occur when:

• A catalyst or enzyme is saturated — every active site is occupied
• A reaction occurs on a surface that is fully covered
• Photochemical reactions where rate depends on light intensity, not concentration

Example

Decomposition of $NH_{3}$ on a hot tungsten surface: Rate = k. The tungsten surface is saturated with $NH_{3}$, so adding more does not increase the rate.

⚗️ First-Order and Second-Order Reactions

First-Order ($n = 1$)

$\boxed{\text{Rate} = k[A]}$

• Units of $k$: $s^{-1}$
• Half-life: $\boxed{t_{1/2} = \frac{0.693}{k}}$ (independent of concentration!)
• Examples: Radioactive decay, many decomposition reactions

💡 Tip: First-order half-life is constant — it doesn't depend on how much reactant you start with. This makes it uniquely useful for radioactive decay and pharmacokinetics.

Second-Order ($n = 2$)

$\boxed{\text{Rate} = k[A]^2 \quad \text{or} \quad \text{Rate} = k[A][B]}$

• Units of $k$: $M^{-1}s^{-1}$ (for both cases)
• Half-life: $t_{1/2} = \frac{1}{k[A]_0}$ (depends on concentration)
• Example: $2\text{NO}_2 \rightarrow 2\text{NO} + \text{O}_2$

Comparing Half-Lives

Match the Order to Its Properties 🔍

Units and k Calculations 🧮

1) A third-order reaction has Rate = k[A][B][C]. What is the power of M in the units of k? (e.g., for $M^{-1}$, enter −1)

2) A first-order reaction has k = 0.0250 $s^{-1}$. What is the half-life in seconds? (3 significant figures)

3) A zero-order reaction has k = 0.0040 M/s and [A]${}_{0}$ = 0.80 M. What is the half-life in seconds? (whole number)

Exit Quiz — Units of k ✅

🔧 Problem-Solving Workshop

Part 6 of 7 — Rate Law Problems from Data Tables

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.

🔑 Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems — 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

🔄 Problem-Solving Strategy Review

Checklist

1. ✅ Write the general rate law: Rate = k[A]^m[B]^n...
2. ✅ Identify pairs of experiments differing in only ONE concentration
3. ✅ Take ratios to find each order
4. ✅ Write the complete rate law with orders
5. ✅ Plug in any experiment to solve for $k$
6. ✅ Verify $k$ using a second experiment
7. ✅ Check that units of $k$ are consistent with the overall order

🔑 Key Concept: Always verify your rate constant by plugging it into a different experiment than the one you used to calculate it. If the predicted rate matches, your rate law is correct.

Tip for the AP Exam

💡 Tip: When the rate ratio is not a clean power, use logarithms:

$\boxed{m = \frac{\log(\text{Rate ratio})}{\log(\text{Concentration ratio})}}$

Problem 1: Two-Reactant System 🧮

For $\text{A} + \text{B} \rightarrow \text{Products}$:

1) Order in A? (integer)

2) Order in B? (integer)

3) Value of k? (number only, no units)

Problem 2: Three Experiments, Non-Obvious Ratios 🧮

For $\text{X} + \text{Y} \rightarrow \text{Z}$:

1) Order in X? (integer)

2) Order in Y? (integer)

3) What is the overall order?

Problem 3: AP-Style Question 🎯

For the reaction $2\text{A} + \text{B} \rightarrow \text{C} + 3\text{D}$:

Problem 4: Finding k and Predicting 🧮

Given Rate = k[M]${}^{2}$[N] and the following data point: when [M] = 0.25 M and [N] = 0.40 M, the rate is 0.050 M/s.

1) Calculate k (number only, to 3 significant figures)

2) What is the rate when [M] = 0.50 M and [N] = 0.20 M? (in M/s, to 3 significant figures)

3) By what factor does the rate change if [M] is halved and [N] is doubled? (give the number)

Quick Concept Check 🔍

Exit Quiz — Workshop Problems ✅

🎓 Synthesis & AP Review

Part 7 of 7 — AP-Style Rate Law Determination

Bringing It All Together

This comprehensive review connects every concept from Parts 1–6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam — multi-step, multi-concept, and requiring clear written explanations.

🔑 Why this matters: AP Chemistry exam questions rarely test one concept in isolation — 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

📋 Key Concepts Summary

Rate Expression

For $aA + bB \rightarrow cC + dD$:

$\boxed{\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = +\frac{1}{c}\frac{\Delta[C]}{\Delta t}}$

Rate Law

$\boxed{\text{Rate} = k[A]^m[B]^n}$

• Determined experimentally (not from coefficients)
• $k$ depends on temperature only
• Overall order = $m + n$

⚠️ Warning: Rate law exponents are determined experimentally — they do NOT come from balanced equation coefficients (except for elementary steps).

Units of k

$\boxed{\text{Units} = \text{M}^{1-(m+n)}\text{s}^{-1}}$

Key Relationships

AP Problem 1 🎯

The decomposition of $\text{N}_2\text{O}_5$ was studied at 45°C:

$2\text{N}_2\text{O}_5(g) \rightarrow 4\text{NO}_2(g) + \text{O}_2(g)$

AP Problem 2: Complete Analysis 🧮

The reaction $\text{A} + 2\text{B} \rightarrow \text{C}$ was studied:

1) What is the order with respect to A? (integer)

2) What is the order with respect to B? (integer)

3) What is the value of k? (number only)

Round all answers to 3 significant figures.

AP Problem 3: Conceptual Questions 🎯

Synthesis Review 🔍

AP Problem 4: Rate Prediction 🧮

A reaction has rate law Rate = k[A]${}^{2}$[B] with k = 0.50 $M^{-2}s^{-1}$.

1) Calculate the rate when [A] = 0.40 M and [B] = 0.60 M. (in M/s, 3 significant figures)

2) If [A] is tripled while [B] is halved, by what factor does the rate change? (to 3 significant figures)

3) What are the units of the rate constant for a reaction that is first-order overall? (enter just the exponent of s: e.g., for $s^{-1}$ enter −1)

Final Exit Quiz — Rate Laws Complete Review ✅