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Reaction Rates and Rate Laws

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Reaction Rates and Rate Laws - Complete Interactive Lesson

Part 1: Measuring Reaction Rates

⚗️ Measuring Reaction Rates

Part 1 of 7 — How Fast Does It Go?

Topics in This Part

Section
⚗️ Defining Reaction Rate
Key Points
Example
⏱️ Average Rate vs. Instantaneous Rate
Average Rate

🔑 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:

$Rate = - \frac{1}{a} \frac{Δ [A]}{Δ t} = - \frac{1}{}$

Rate Definition Quiz 🎯

⏱️ Average Rate vs. Instantaneous Rate

Average Rate

The average rate is calculated over a finite time interval:

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

Average vs. Instantaneous Rate 🔍

⏱️ Experimental Methods for Measuring Rates

Monitoring Concentration Over Time

Method	What It Measures	Best For
Spectrophotometry	Absorbance (Beer's Law: $A = \varepsilon bc$ )	Colored solutions
Pressure change	Total gas pressure	Gas-phase reactions
Conductivity	Ion concentration	Reactions producing/consuming ions
Mass loss	Mass of system	Reactions releasing gas
Titration (aliquot method)	Concentration at specific times	Slow reactions

Rate Calculation Drill 🧮

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

Time (s)	[NO₂] (M)

Exit Quiz — Measuring Reaction Rates ✅

Part 2: Rate Laws & Orders

🌡️ Factors Affecting Reaction Rate

Part 2 of 7 — What Makes Reactions Faster?

Topics in This Part

Section
📊 Factor 1: Concentration of Reactants
Why?
Mathematical Connection
Example
📊 Factor 2: Temperature

🔑 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:

$Rate = k [$

Part 3: Determining Rate Law from Data

📐 Rate Laws

Part 3 of 7 — The Mathematical Heart of Kinetics

Topics in This Part

Section
📏 The General Rate Law
Critical Points
What Does Order Mean?
🔍 Determining Order from Experimental Data
The Method of Initial Rates

🔑 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 $aA + bB \rightarrow \text{products}$ :

Part 4: Method of Initial Rates

📊 Method of Initial Rates

Part 4 of 7 — Systematic Rate Law Determination

Topics in This Part

Section
📋 Step-by-Step Method
Given Data Table Format
The Algorithm
Using Logarithms for Non-Integer Orders
🧪 Worked Example 1

🔑 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

Experiment	[A]₀	[B]₀	Initial Rate
1	value	value	value
2	changed	same	value

Part 5: Factors Affecting Rate

📏 Units of the Rate Constant k

Part 5 of 7 — How Units Change with Reaction Order

Topics in This Part

Section
📌 Deriving Units of k
Summary Table
The Pattern
⚗️ Zero-Order Reactions
When Do Zero-Order Reactions Occur?

🔑 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:

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

Part 6: Problem-Solving Workshop

🔧 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

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

Part 7: Synthesis & AP Review

🎓 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 $aA + bB \rightarrow cC + dD$ :

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

Symbol	Meaning
$\Delta[A]$	Change in molar concentration of A
$\Delta t$	Change in time
Negative sign for reactants	Reactants are consumed, so $\Delta[A] < 0$ ; the negative sign makes rate positive
Stoichiometric coefficients	Divide by coefficient to get a single, unique rate

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

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 disappears at M/s.

t2−t1[A]t2−[A]t1

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

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

Feature	Average Rate	Instantaneous Rate
Time interval	Finite ( $\Delta t$ )	Infinitesimally small ( $dt$ )
Graphically	Slope of secant line	Slope of tangent line
Accuracy	Approximation	Exact at that instant
As $\Delta t \rightarrow 0$	Approaches instantaneous 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.

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.

1) What is the average rate of disappearance of NO₂ 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 O₂ over the interval 0–100 s? (in M/s, 3 significant figures)

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

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

Burning steel wool in pure O₂ (100%) is dramatically faster than in air (21% O₂) because the concentration of oxygen molecules is about 5 times higher.

📊 Factor 2: Temperature

Higher temperature → faster rate

Why?

At higher temperatures:

Molecules move faster → more frequent collisions
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

Form	Surface Area	Rate
Iron block	Low	Rusts slowly over years
Iron filings	Medium	Rusts in days
Iron nanoparticles	Very high	Can ignite spontaneously

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

Type	Description	Example
Homogeneous	Same phase as reactants	Acid catalysis in solution
Heterogeneous	Different phase (usually solid)	Catalytic converter (Pt surface)
Biological	Enzymes	Lactase breaking down lactose

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 ✅

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

Symbol	Meaning
$k$	Rate constant (temperature-dependent)
$[A], [B]$	Molar concentrations of reactants
$m$	Order with respect to A
$n$	Order with respect to B
$m + n$	Overall order of the reaction

$m$ 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?

Order in A ( $m$ )	Effect of Doubling [A]
0	Rate unchanged (×1)
1	Rate doubles (×2)
2	Rate quadruples (×4)
3	Rate increases 8×

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

Experiment	[A] (M)	[B] (M)	Initial Rate (M/s)
1	0.10	0.10	0.015
2	0.20	0.10	0.060
3	0.10	0.20	0.030

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:

Experiment	[X] (M)	[Y] (M)	Initial Rate (M/s)
1	0.10	0.10	0.0020
2	0.30	0.10	0.018
3	0.10	0.30	0.0060

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 ✅

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.

Experiment	[A] (M)	[B] (M)	Initial Rate (M/s)
1	0.100	0.100	4.0 × 10⁻⁵
2	0.200	0.100	16.0 × 10⁻⁵
3	0.100	0.300	4.0 × 10⁻⁵

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: $4.0 \times 10^{-5} = k(0.100)^2$ $k = \frac{4.0 \times 10^{}}{}$

Practice Problem 1 🧮

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

Experiment	[P] (M)	[Q] (M)	Initial Rate (M/s)
1	0.20	0.10	0.0030
2	0.40	0.10	0.0060
3	0.20	0.20	0.012

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.

Experiment	[NO] (M)	[Cl₂] (M)	Initial Rate (M/s)
1	0.10	0.10	0.18
2	0.10	0.20	0.36
3	0.20	0.10	0.72

Solution:

Order in Cl₂ (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, [Cl₂] 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[Cl₂]

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

Exp	[A] (M)	[B] (M)	[C] (M)	Rate (M/s)
1	0.10	0.10	0.10	0.0050
2	0.20	0.10	0.10	0.010
3	0.10	0.20	0.10	0.020
4	0.10	0.10	0.30	0.0050

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 ✅

Rate always has units of M/s (or mol·L⁻¹·s⁻¹). Concentration has units of M (mol/L).

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

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

Overall Order	Units of $k$	Example
0	M·s⁻¹ (or M/s)	$k = \frac{\text{M/s}}{\text{M}^0} = \text{M/s}$
1	s⁻¹	$k = \frac{\text{M/s}}{\text{M}^1} = \text{s}^{-1}$
2	M⁻¹·s⁻¹	$k = \frac{\text{M/s}}{\text{M}^2} = \text{M}^{-1}\text{s}^{-1}$
3	M⁻²·s⁻¹	$k = \frac{\text{M/s}}{\text{M}^3} = \text{M}^{-2}\text{s}^{-1}$

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

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 NH₃ on a hot tungsten surface: Rate = k. The tungsten surface is saturated with NH₃, 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: $\boxed{t_{1/2} = \frac{0.693}{k}}$ ( 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⁻¹s⁻¹ (for both cases)
Half-life: $t_{1/2} = \frac{1}{k[A]_0}$ (depends on concentration)

Comparing Half-Lives

Order	$t_{1/2}$	Dependence on $[A]_0$
0

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 ✅

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

Exp	[A] (M)	[B] (M)	Rate (M/s)
1	0.050	0.050	1.25 × 10⁻⁴
2	0.100	0.050	5.00 × 10⁻⁴
3	0.050	0.100	2.50 × 10⁻⁴

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

Exp	[X] (M)	[Y] (M)	Rate (M/s)
1	0.10	0.10	0.0040
2	0.30	0.10	0.036
3	0.30	0.30	0.036

1) Order in X? (integer)

2) Order in Y? (integer)

3) What is the overall order?

Problem 3: AP-Style Question 🎯

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

Exp	[A] (M)	[B] (M)	Rate (M/s)
1	0.10	0.20	0.010
2	0.20	0.20	0.020
3	0.20	0.40	0.080

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 ✅

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

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

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).

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

If doubling [A] causes rate to...	Order in A
Stay the same	0
Double	1
Quadruple	2
Increase 8×	3

AP Problem 1 🎯

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

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

Exp	[N₂O₅]₀ (M)	Initial Rate (M/s)
1	0.020	4.8 × 10⁻⁶
2	0.040	9.6 × 10⁻⁶
3	0.060	14.4 × 10⁻⁶

AP Problem 2: Complete Analysis 🧮

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

Exp	[A] (M)	[B] (M)	Rate (M/s)
1	0.10	0.10	3.0 × 10⁻³
2	0.20	0.10	1.2 × 10⁻²
3	0.10	0.30	3.0 × 10⁻³

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⁻²s⁻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 ✅

4 \times 0.024 = 0.096

\text{N}_2\text{O}_5

2 \times 0.024 = 0.048

k = \frac{4.0 \times 1 0 ^{- 5}}{0.0100} = 4.0 \times 1 0^{- 3} M^{- 1} s^{- 1}

k = 180 M^{- 2} s^{- 1}

Example:

2\text{NO}_2 \rightarrow 2\text{NO} + \text{O}_2

Reaction Rates and Rate Laws

Reaction Rates and Rate Laws - Complete Interactive Lesson

Part 1: Measuring Reaction Rates

⚗️ Measuring Reaction Rates

Topics in This Part

What You'll Master in Part 1

⚗️ Defining Reaction Rate

⏱️ Average Rate vs. Instantaneous Rate

Average Rate

⏱️ Experimental Methods for Measuring Rates

Monitoring Concentration Over Time

Part 2: Rate Laws & Orders

🌡️ Factors Affecting Reaction Rate

Topics in This Part

What You'll Master in Part 2

📊 Factor 1: Concentration of Reactants

Why?

Mathematical Connection

Part 3: Determining Rate Law from Data

📐 Rate Laws

Topics in This Part

What You'll Master in Part 3

📏 The General Rate Law

Part 4: Method of Initial Rates

📊 Method of Initial Rates

Topics in This Part

What You'll Master in Part 4

📋 Step-by-Step Method

Given Data Table Format

Part 5: Factors Affecting Rate

📏 Units of the Rate Constant k

Topics in This Part

What You'll Master in Part 5

📌 Deriving Units of k

Part 6: Problem-Solving Workshop

🔧 Problem-Solving Workshop

Practice Makes Perfect

What You'll Master in Part 6

🔄 Problem-Solving Strategy Review

Checklist

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Bringing It All Together

What You'll Master in Part 7

📋 Key Concepts Summary

Rate Expression

Key Points

Example

Instantaneous Rate

Key Differences

Initial Rate

Beer's Law Connection

Example

📊 Factor 2: Temperature

Why?

Rule of Thumb

Quantitative: The Arrhenius Equation

📊 Factor 3: Surface Area

Why?

Examples

Dust Explosions

📊 Factor 4: Catalysts

How?

Key Properties of Catalysts

Types

Critical Points

What Does Order Mean?

🔍 Determining Order from Experimental Data

The Method of Initial Rates

Example Data

Step 1: Find order in A

Step 2: Find order in B

Step 3: Write the rate law and find k

📌 Common Mistakes to Avoid

❌ Mistake 1: Using stoichiometric coefficients as orders

❌ Mistake 2: Forgetting to hold one variable constant

❌ Mistake 3: Confusing rate with rate constant

The Algorithm

Using Logarithms for Non-Integer Orders

🧪 Worked Example 1

First-Order ( $n = 1$ )

Second-Order ( $n = 2$ )