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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?

Chemical kinetics is the study of how fast reactions occur. While thermodynamics tells us whether a reaction is favorable, kinetics tells us how quickly it reaches products. Some reactions (like explosions) happen in microseconds; others (like rusting) take years.

In this part, we define reaction rate precisely and learn how to measure it.

⚗️ 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:

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

Key Points

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

Example

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

Solution:

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

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.

Rate Definition Quiz 🎯

⏱️ Average Rate vs. Instantaneous Rate

Average Rate

The average rate is calculated over a finite time interval:

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

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?

Not all reactions proceed at the same speed. Some are explosively fast, while others take geological timescales. In this part, we explore the four major factors that determine how quickly a reaction occurs.

📊 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:

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

Part 3: Determining Rate Law from Data

📐 Rate Laws

Part 3 of 7 — The Mathematical Heart of Kinetics

A rate law is a mathematical equation that relates the rate of a reaction to the concentrations of reactants. Rate laws must be determined experimentally — you cannot simply read them off the balanced equation.

📏 The General Rate Law

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

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

Part 4: Method of Initial Rates

📊 Method of Initial Rates

Part 4 of 7 — Systematic Rate Law Determination

The method of initial rates is the gold standard for determining a rate law experimentally. You measure the initial rate of a reaction for several trials, varying concentrations systematically, and use the data to find each reactant's order and the rate constant $k$ .

📋 Step-by-Step Method

Given Data Table Format

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

The Algorithm

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

Step 2: Pick two experiments where only one concentration changes

Part 5: Factors Affecting Rate

📏 Units of the Rate Constant k

Part 5 of 7 — How Units Change with Reaction Order

The rate constant $k$ has different units depending on the overall order of the reaction. Understanding why — and being able to derive or recognize the correct units — is a key AP Chemistry skill.

📌 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

This workshop focuses on building fluency with rate law determination from experimental data. You will work through progressively challenging problems that mirror AP Chemistry exam questions.

🔄 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 $k$ are consistent with the overall order

Tip for the AP Exam

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

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Part 7 of 7 — AP-Style Rate Law Determination

This final part brings together all the concepts from Parts 1–6 with AP exam-level questions. You will work through complete rate law problems including determining orders, calculating k, predicting rates, and interpreting results.

📋 Key Concepts Summary

Rate Expression

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

$Rate = - \frac{1}{a} \frac{Δ [}{}$

4 \times 0.024 = 0.096

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

2 \times 0.024 = 0.048

−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:

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

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

Beer's Law Connection

For colored species, absorbance is directly proportional to concentration:

$A = \varepsilon b c$

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)

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.

Quantitative: The Arrhenius Equation

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

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 ( $E_a$ ).

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

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

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 ✅

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
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.

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

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

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

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

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 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: $m = \frac{\ln(\text{Rate}_2/\text{Rate}_1)}{\ln([A]_2/[A]_1)}$

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).

🧪 Worked Example 1

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

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

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: $4.0 \times 10^{-5} = k(0.100)^2$ $k = \frac{4.0 \times 10^{- 5}}{0.0100} = 4.0 \times 10^{- 3}$

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

For $2\text{NO}(g) + \text{Cl}_2(g) \rightarrow 2\text{NOCl}(g)$ :

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

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

Note: The orders (2 and 1) match the stoichiometric coefficients here, but this is coincidental — it happens because this reaction proceeds via an elementary bimolecular step.

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).

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

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

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	$k = \frac{\text{M/s}}{\text{M}^3} = \text{M}^{-2}\text{s}^{-1}$

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

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

Units of k Quiz 🎯

⚗️ Zero-Order Reactions

$\text{Rate} = k$

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 ( $n = 1$ )

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

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

Second-Order ( $n = 2$ )

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

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 ✅

m = \frac{l o g ( Rate ratio )}{l o g ( 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 ✅

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

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

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

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

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 ✅

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

⚗️ Defining Reaction Rate

Key Points

Example

⏱️ 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

📊 Factor 1: Concentration of Reactants

Why?

Mathematical Connection

Part 3: Determining Rate Law from Data

📐 Rate Laws

📏 The General Rate Law

Part 4: Method of Initial Rates

📊 Method of Initial Rates

📋 Step-by-Step Method

Given Data Table Format

The Algorithm

Part 5: Factors Affecting Rate

📏 Units of the Rate Constant k

📌 Deriving Units of k

Part 6: Problem-Solving Workshop

🔧 Problem-Solving Workshop

🔄 Problem-Solving Strategy Review

Checklist

Tip for the AP Exam

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

📋 Key Concepts Summary

Rate Expression

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

Using Logarithms for Non-Integer Orders

🧪 Worked Example 1

🧪 Worked Example 2: Three Reactants

Summary Table

The Pattern

⚗️ Zero-Order Reactions

When Do Zero-Order Reactions Occur?

Example

⚗️ First-Order and Second-Order Reactions

First-Order (n=1n = 1n=1)

Second-Order (n=2n = 2n=2)

Comparing Half-Lives

Rate Law

Units of k

Key Relationships

First-Order ( $n = 1$ )

Second-Order ( $n = 2$ )