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Raffi Hovasapian

Raffi Hovasapian

Second Order & Zero-Order Rate Laws

Slide Duration:

Table of Contents

I. Review
Naming Compounds

41m 24s

Intro
0:00
Periodic Table of Elements
0:15
Naming Compounds
3:13
Definition and Examples of Ions
3:14
Ionic (Symbol to Name): NaCl
5:23
Ionic (Name to Symbol): Calcium Oxide
7:58
Ionic - Polyatoms Anions: Examples
12:45
Ionic - Polyatoms Anions (Symbol to Name): KClO
14:50
Ionic - Polyatoms Anions (Name to Symbol): Potassium Phosphate
15:49
Ionic Compounds Involving Transition Metals (Symbol to Name): Co₂(CO₃)₃
20:48
Ionic Compounds Involving Transition Metals (Name to Symbol): Palladium 2 Acetate
22:44
Naming Covalent Compounds (Symbol to Name): CO
26:21
Naming Covalent Compounds (Name to Symbol): Nitrogen Trifluoride
27:34
Naming Covalent Compounds (Name to Symbol): Dichlorine Monoxide
27:57
Naming Acids Introduction
28:11
Naming Acids (Name to Symbol): Chlorous Acid
35:08
% Composition by Mass Example
37:38
Stoichiometry

37m 19s

Intro
0:00
Stoichiometry
0:25
Introduction to Stoichiometry
0:26
Example 1
5:03
Example 2
10:17
Example 3
15:09
Example 4
24:02
Example 5: Questions
28:11
Example 5: Part A - Limiting Reactant
30:30
Example 5: Part B
32:27
Example 5: Part C
35:00
II. Aqueous Reactions & Stoichiometry
Precipitation Reactions

31m 14s

Intro
0:00
Precipitation Reactions
0:53
Dissociation of ionic Compounds
0:54
Solubility Guidelines for ionic Compounds: Soluble Ionic Compounds
8:15
Solubility Guidelines for ionic Compounds: Insoluble ionic Compounds
12:56
Precipitation Reactions
14:08
Example 1: Mixing a Solution of BaCl₂ & K₂SO₄
21:21
Example 2: Mixing a Solution of Mg(NO₃)₂ & KI
26:10
Acid-Base Reactions

43m 21s

Intro
0:00
Acid-Base Reactions
1:00
Introduction to Acid: Monoprotic Acid and Polyprotic Acid
1:01
Introduction to Base
8:28
Neutralization
11:45
Example 1
16:17
Example 2
21:55
Molarity
24:50
Example 3
26:50
Example 4
30:01
Example 4: Limiting Reactant
37:51
Example 4: Reaction Part
40:01
Oxidation Reduction Reactions

47m 58s

Intro
0:00
Oxidation Reduction Reactions
0:26
Oxidation and Reduction Overview
0:27
How Can One Tell Whether Oxidation-Reduction has Taken Place?
7:13
Rules for Assigning Oxidation State: Number 1
11:22
Rules for Assigning Oxidation State: Number 2
12:46
Rules for Assigning Oxidation State: Number 3
13:25
Rules for Assigning Oxidation State: Number 4
14:50
Rules for Assigning Oxidation State: Number 5
15:41
Rules for Assigning Oxidation State: Number 6
17:00
Example 1: Determine the Oxidation State of Sulfur in the Following Compounds
18:20
Activity Series and Reduction Properties
25:32
Activity Series and Reduction Properties
25:33
Example 2: Write the Balance Molecular, Total Ionic, and Net Ionic Equations for Al + HCl
31:37
Example 3
34:25
Example 4
37:55
Stoichiometry Examples

31m 50s

Intro
0:00
Stoichiometry Example 1
0:36
Example 1: Question and Answer
0:37
Stoichiometry Example 2
6:57
Example 2: Questions
6:58
Example 2: Part A Solution
12:16
Example 2: Part B Solution
13:05
Example 2: Part C Solution
14:00
Example 2: Part D Solution
14:38
Stoichiometry Example 3
17:56
Example 3: Questions
17:57
Example 3: Part A Solution
19:51
Example 3: Part B Solution
21:43
Example 3: Part C Solution
26:46
III. Gases
Pressure, Gas Laws, & The Ideal Gas Equation

49m 40s

Intro
0:00
Pressure
0:22
Pressure Overview
0:23
Torricelli: Barometer
4:35
Measuring Gas Pressure in a Container
7:49
Boyle's Law
12:40
Example 1
16:56
Gas Laws
21:18
Gas Laws
21:19
Avogadro's Law
26:16
Example 2
31:47
Ideal Gas Equation
38:20
Standard Temperature and Pressure (STP)
38:21
Example 3
40:43
Partial Pressure, Mol Fraction, & Vapor Pressure

32m

Intro
0:00
Gases
0:27
Gases
0:28
Mole Fractions
5:52
Vapor Pressure
8:22
Example 1
13:25
Example 2
22:45
Kinetic Molecular Theory and Real Gases

31m 58s

Intro
0:00
Kinetic Molecular Theory and Real Gases
0:45
Kinetic Molecular Theory 1
0:46
Kinetic Molecular Theory 2
4:23
Kinetic Molecular Theory 3
5:42
Kinetic Molecular Theory 4
6:27
Equations
7:52
Effusion
11:15
Diffusion
13:30
Example 1
19:54
Example 2
23:23
Example 3
26:45
AP Practice for Gases

25m 34s

Intro
0:00
Example 1
0:34
Example 1
0:35
Example 2
6:15
Example 2: Part A
6:16
Example 2: Part B
8:46
Example 2: Part C
10:30
Example 2: Part D
11:15
Example 2: Part E
12:20
Example 2: Part F
13:22
Example 3
14:45
Example 3
14:46
Example 4
18:16
Example 4
18:17
Example 5
21:04
Example 5
21:05
IV. Thermochemistry
Energy, Heat, and Work

37m 32s

Intro
0:00
Thermochemistry
0:25
Temperature and Heat
0:26
Work
3:07
System, Surroundings, Exothermic Process, and Endothermic Process
8:19
Work & Gas: Expansion and Compression
16:30
Example 1
24:41
Example 2
27:47
Example 3
31:58
Enthalpy & Hess's Law

32m 34s

Intro
0:00
Thermochemistry
1:43
Defining Enthalpy & Hess's Law
1:44
Example 1
6:48
State Function
13:11
Example 2
17:15
Example 3
24:09
Standard Enthalpies of Formation

23m 9s

Intro
0:00
Thermochemistry
1:04
Standard Enthalpy of Formation: Definition & Equation
1:05
∆H of Formation
10:00
Example 1
11:22
Example 2
19:00
Calorimetry

39m 28s

Intro
0:00
Thermochemistry
0:21
Heat Capacity
0:22
Molar Heat Capacity
4:44
Constant Pressure Calorimetry
5:50
Example 1
12:24
Constant Volume Calorimetry
21:54
Example 2
24:40
Example 3
31:03
V. Kinetics
Reaction Rates and Rate Laws

36m 24s

Intro
0:00
Kinetics
2:18
Rate: 2 NO₂ (g) → 2NO (g) + O₂ (g)
2:19
Reaction Rates Graph
7:25
Time Interval & Average Rate
13:13
Instantaneous Rate
15:13
Rate of Reaction is Proportional to Some Power of the Reactant Concentrations
23:49
Example 1
27:19
Method of Initial Rates

30m 48s

Intro
0:00
Kinetics
0:33
Rate
0:34
Idea
2:24
Example 1: NH₄⁺ + NO₂⁻ → NO₂ (g) + 2 H₂O
5:36
Example 2: BrO₃⁻ + 5 Br⁻ + 6 H⁺ → 3 Br₂ + 3 H₂O
19:29
Integrated Rate Law & Reaction Half-Life

32m 17s

Intro
0:00
Kinetics
0:52
Integrated Rate Law
0:53
Example 1
6:26
Example 2
15:19
Half-life of a Reaction
20:40
Example 3: Part A
25:41
Example 3: Part B
28:01
Second Order & Zero-Order Rate Laws

26m 40s

Intro
0:00
Kinetics
0:22
Second Order
0:23
Example 1
6:08
Zero-Order
16:36
Summary for the Kinetics Associated with the Reaction
21:27
Activation Energy & Arrhenius Equation

40m 59s

Intro
0:00
Kinetics
0:53
Rate Constant
0:54
Collision Model
2:45
Activation Energy
5:11
Arrhenius Proposed
9:54
2 Requirements for a Successful Reaction
15:39
Rate Constant
17:53
Arrhenius Equation
19:51
Example 1
25:00
Activation Energy & the Values of K
32:12
Example 2
36:46
AP Practice for Kinetics

29m 8s

Intro
0:00
Kinetics
0:43
Example 1
0:44
Example 2
6:53
Example 3
8:58
Example 4
11:36
Example 5
16:36
Example 6: Part A
21:00
Example 6: Part B
25:09
VI. Equilibrium
Equilibrium, Part 1

46m

Intro
0:00
Equilibrium
1:32
Introduction to Equilibrium
1:33
Equilibrium Rules
14:00
Example 1: Part A
16:46
Example 1: Part B
18:48
Example 1: Part C
22:13
Example 1: Part D
24:55
Example 2: Part A
27:46
Example 2: Part B
31:22
Example 2: Part C
33:00
Reverse a Reaction
36:04
Example 3
37:24
Equilibrium, Part 2

40m 53s

Intro
0:00
Equilibrium
1:31
Equilibriums Involving Gases
1:32
General Equation
10:11
Example 1: Question
11:55
Example 1: Answer
13:43
Example 2: Question
19:08
Example 2: Answer
21:37
Example 3: Question
33:40
Example 3: Answer
35:24
Equilibrium: Reaction Quotient

45m 53s

Intro
0:00
Equilibrium
0:57
Reaction Quotient
0:58
If Q > K
5:37
If Q < K
6:52
If Q = K
7:45
Example 1: Part A
8:24
Example 1: Part B
13:11
Example 2: Question
20:04
Example 2: Answer
22:15
Example 3: Question
30:54
Example 3: Answer
32:52
Steps in Solving Equilibrium Problems
42:40
Equilibrium: Examples

31m 51s

Intro
0:00
Equilibrium
1:09
Example 1: Question
1:10
Example 1: Answer
4:15
Example 2: Question
13:04
Example 2: Answer
15:20
Example 3: Question
25:03
Example 3: Answer
26:32
Le Chatelier's principle & Equilibrium

40m 52s

Intro
0:00
Le Chatelier
1:05
Le Chatelier Principle
1:06
Concentration: Add 'x'
5:25
Concentration: Subtract 'x'
7:50
Example 1
9:44
Change in Pressure
12:53
Example 2
20:40
Temperature: Exothermic and Endothermic
24:33
Example 3
29:55
Example 4
35:30
VII. Acids & Bases
Acids and Bases

50m 11s

Intro
0:00
Acids and Bases
1:14
Bronsted-Lowry Acid-Base Model
1:28
Reaction of an Acid with Water
4:36
Acid Dissociation
10:51
Acid Strength
13:48
Example 1
21:22
Water as an Acid & a Base
25:25
Example 2: Part A
32:30
Example 2: Part B
34:47
Example 3: Part A
35:58
Example 3: Part B
39:33
pH Scale
41:12
Example 4
43:56
pH of Weak Acid Solutions

43m 52s

Intro
0:00
pH of Weak Acid Solutions
1:12
pH of Weak Acid Solutions
1:13
Example 1
6:26
Example 2
14:25
Example 3
24:23
Example 4
30:38
Percent Dissociation: Strong & Weak Bases

43m 4s

Intro
0:00
Bases
0:33
Percent Dissociation: Strong & Weak Bases
0:45
Example 1
6:23
Strong Base Dissociation
11:24
Example 2
13:02
Weak Acid and General Reaction
17:38
Example: NaOH → Na⁺ + OH⁻
20:30
Strong Base and Weak Base
23:49
Example 4
24:54
Example 5
33:51
Polyprotic Acids

35m 34s

Intro
0:00
Polyprotic Acids
1:04
Acids Dissociation
1:05
Example 1
4:51
Example 2
17:30
Example 3
31:11
Salts and Their Acid-Base Properties

41m 14s

Intro
0:00
Salts and Their Acid-Base Properties
0:11
Salts and Their Acid-Base Properties
0:15
Example 1
7:58
Example 2
14:00
Metal Ion and Acidic Solution
22:00
Example 3
28:35
NH₄F → NH₄⁺ + F⁻
34:05
Example 4
38:03
Common Ion Effect & Buffers

41m 58s

Intro
0:00
Common Ion Effect & Buffers
1:16
Covalent Oxides Produce Acidic Solutions in Water
1:36
Ionic Oxides Produce Basic Solutions in Water
4:15
Practice Example 1
6:10
Practice Example 2
9:00
Definition
12:27
Example 1: Part A
16:49
Example 1: Part B
19:54
Buffer Solution
25:10
Example of Some Buffers: HF and NaF
30:02
Example of Some Buffers: Acetic Acid & Potassium Acetate
31:34
Example of Some Buffers: CH₃NH₂ & CH₃NH₃Cl
33:54
Example 2: Buffer Solution
36:36
Buffer

32m 24s

Intro
0:00
Buffers
1:20
Buffer Solution
1:21
Adding Base
5:03
Adding Acid
7:14
Example 1: Question
9:48
Example 1: Recall
12:08
Example 1: Major Species Upon Addition of NaOH
16:10
Example 1: Equilibrium, ICE Chart, and Final Calculation
24:33
Example 1: Comparison
29:19
Buffers, Part II

40m 6s

Intro
0:00
Buffers
1:27
Example 1: Question
1:32
Example 1: ICE Chart
3:15
Example 1: Major Species Upon Addition of OH⁻, But Before Rxn
7:23
Example 1: Equilibrium, ICE Chart, and Final Calculation
12:51
Summary
17:21
Another Look at Buffering & the Henderson-Hasselbalch equation
19:00
Example 2
27:08
Example 3
32:01
Buffers, Part III

38m 43s

Intro
0:00
Buffers
0:25
Buffer Capacity Part 1
0:26
Example 1
4:10
Buffer Capacity Part 2
19:29
Example 2
25:12
Example 3
32:02
Titrations: Strong Acid and Strong Base

42m 42s

Intro
0:00
Titrations: Strong Acid and Strong Base
1:11
Definition of Titration
1:12
Sample Problem
3:33
Definition of Titration Curve or pH Curve
9:46
Scenario 1: Strong Acid- Strong Base Titration
11:00
Question
11:01
Part 1: No NaOH is Added
14:00
Part 2: 10.0 mL of NaOH is Added
15:50
Part 3: Another 10.0 mL of NaOH & 20.0 mL of NaOH are Added
22:19
Part 4: 50.0 mL of NaOH is Added
26:46
Part 5: 100.0 mL (Total) of NaOH is Added
27:26
Part 6: 150.0 mL (Total) of NaOH is Added
32:06
Part 7: 200.0 mL of NaOH is Added
35:07
Titrations Curve for Strong Acid and Strong Base
35:43
Titrations: Weak Acid and Strong Base

42m 3s

Intro
0:00
Titrations: Weak Acid and Strong Base
0:43
Question
0:44
Part 1: No NaOH is Added
1:54
Part 2: 10.0 mL of NaOH is Added
5:17
Part 3: 25.0 mL of NaOH is Added
14:01
Part 4: 40.0 mL of NaOH is Added
21:55
Part 5: 50.0 mL (Total) of NaOH is Added
22:25
Part 6: 60.0 mL (Total) of NaOH is Added
31:36
Part 7: 75.0 mL (Total) of NaOH is Added
35:44
Titration Curve
36:09
Titration Examples & Acid-Base Indicators

52m 3s

Intro
0:00
Examples and Indicators
0:25
Example 1: Question
0:26
Example 1: Solution
2:03
Example 2: Question
12:33
Example 2: Solution
14:52
Example 3: Question
23:45
Example 3: Solution
25:09
Acid/Base Indicator Overview
34:45
Acid/Base Indicator Example
37:40
Acid/Base Indicator General Result
47:11
Choosing Acid/Base Indicator
49:12
VIII. Solubility
Solubility Equilibria

36m 25s

Intro
0:00
Solubility Equilibria
0:48
Solubility Equilibria Overview
0:49
Solubility Product Constant
4:24
Definition of Solubility
9:10
Definition of Solubility Product
11:28
Example 1
14:09
Example 2
20:19
Example 3
27:30
Relative Solubilities
31:04
Solubility Equilibria, Part II

42m 6s

Intro
0:00
Solubility Equilibria
0:46
Common Ion Effect
0:47
Example 1
3:14
pH & Solubility
13:00
Example of pH & Solubility
15:25
Example 2
23:06
Precipitation & Definition of the Ion Product
26:48
If Q > Ksp
29:31
If Q < Ksp
30:27
Example 3
32:58
Solubility Equilibria, Part III

43m 9s

Intro
0:00
Solubility Equilibria
0:55
Example 1: Question
0:56
Example 1: Step 1 - Check to See if Anything Precipitates
2:52
Example 1: Step 2 - Stoichiometry
10:47
Example 1: Step 3 - Equilibrium
16:34
Example 2: Selective Precipitation (Question)
21:02
Example 2: Solution
23:41
Classical Qualitative Analysis
29:44
Groups: 1-5
38:44
IX. Complex Ions
Complex Ion Equilibria

43m 38s

Intro
0:00
Complex Ion Equilibria
0:32
Complex Ion
0:34
Ligan Examples
1:51
Ligand Definition
3:12
Coordination
6:28
Example 1
8:08
Example 2
19:13
Complex Ions & Solubility

31m 30s

Intro
0:00
Complex Ions and Solubility
0:23
Recall: Classical Qualitative Analysis
0:24
Example 1
6:10
Example 2
16:16
Dissolving a Water-Insoluble Ionic Compound: Method 1
23:38
Dissolving a Water-Insoluble Ionic Compound: Method 2
28:13
X. Chemical Thermodynamics
Spontaneity, Entropy, & Free Energy, Part I

56m 28s

Intro
0:00
Spontaneity, Entropy, Free Energy
2:25
Energy Overview
2:26
Equation: ∆E = q + w
4:30
State Function/ State Property
8:35
Equation: w = -P∆V
12:00
Enthalpy: H = E + PV
14:50
Enthalpy is a State Property
17:33
Exothermic and Endothermic Reactions
19:20
First Law of Thermodynamic
22:28
Entropy
25:48
Spontaneous Process
33:53
Second Law of Thermodynamic
36:51
More on Entropy
42:23
Example
43:55
Spontaneity, Entropy, & Free Energy, Part II

39m 55s

Intro
0:00
Spontaneity, Entropy, Free Energy
1:30
∆S of Universe = ∆S of System + ∆S of Surrounding
1:31
Convention
3:32
Examining a System
5:36
Thermodynamic Property: Sign of ∆S
16:52
Thermodynamic Property: Magnitude of ∆S
18:45
Deriving Equation: ∆S of Surrounding = -∆H / T
20:25
Example 1
25:51
Free Energy Equations
29:22
Spontaneity, Entropy, & Free Energy, Part III

30m 10s

Intro
0:00
Spontaneity, Entropy, Free Energy
0:11
Example 1
2:38
Key Concept of Example 1
14:06
Example 2
15:56
Units for ∆H, ∆G, and S
20:56
∆S of Surrounding & ∆S of System
22:00
Reaction Example
24:17
Example 3
26:52
Spontaneity, Entropy, & Free Energy, Part IV

30m 7s

Intro
0:00
Spontaneity, Entropy, Free Energy
0:29
Standard Free Energy of Formation
0:58
Example 1
4:34
Reaction Under Non-standard Conditions
13:23
Example 2
16:26
∆G = Negative
22:12
∆G = 0
24:38
Diagram Example of ∆G
26:43
Spontaneity, Entropy, & Free Energy, Part V

44m 56s

Intro
0:00
Spontaneity, Entropy, Free Energy
0:56
Equations: ∆G of Reaction, ∆G°, and K
0:57
Example 1: Question
6:50
Example 1: Part A
9:49
Example 1: Part B
15:28
Example 2
17:33
Example 3
23:31
lnK = (- ∆H° ÷ R) ( 1 ÷ T) + ( ∆S° ÷ R)
31:36
Maximum Work
35:57
XI. Electrochemistry
Oxidation-Reduction & Balancing

39m 23s

Intro
0:00
Oxidation-Reduction and Balancing
2:06
Definition of Electrochemistry
2:07
Oxidation and Reduction Review
3:05
Example 1: Assigning Oxidation State
10:15
Example 2: Is the Following a Redox Reaction?
18:06
Example 3: Step 1 - Write the Oxidation & Reduction Half Reactions
22:46
Example 3: Step 2 - Balance the Reaction
26:44
Example 3: Step 3 - Multiply
30:11
Example 3: Step 4 - Add
32:07
Example 3: Step 5 - Check
33:29
Galvanic Cells

43m 9s

Intro
0:00
Galvanic Cells
0:39
Example 1: Balance the Following Under Basic Conditions
0:40
Example 1: Steps to Balance Reaction Under Basic Conditions
3:25
Example 1: Solution
5:23
Example 2: Balance the Following Reaction
13:56
Galvanic Cells
18:15
Example 3: Galvanic Cells
28:19
Example 4: Galvanic Cells
35:12
Cell Potential

48m 41s

Intro
0:00
Cell Potential
2:08
Definition of Cell Potential
2:17
Symbol and Unit
5:50
Standard Reduction Potential
10:16
Example Figure 1
13:08
Example Figure 2
19:00
All Reduction Potentials are Written as Reduction
23:10
Cell Potential: Important Fact 1
26:49
Cell Potential: Important Fact 2
27:32
Cell Potential: Important Fact 3
28:54
Cell Potential: Important Fact 4
30:05
Example Problem 1
32:29
Example Problem 2
38:38
Potential, Work, & Free Energy

41m 23s

Intro
0:00
Potential, Work, Free Energy
0:42
Descriptions of Galvanic Cell
0:43
Line Notation
5:33
Example 1
6:26
Example 2
11:15
Example 3
15:18
Equation: Volt
22:20
Equations: Cell Potential, Work, and Charge
28:30
Maximum Cell Potential is Related to the Free Energy of the Cell Reaction
35:09
Example 4
37:42
Cell Potential & Concentration

34m 19s

Intro
0:00
Cell Potential & Concentration
0:29
Example 1: Question
0:30
Example 1: Nernst Equation
4:43
Example 1: Solution
7:01
Cell Potential & Concentration
11:27
Example 2
16:38
Manipulating the Nernst Equation
25:15
Example 3
28:43
Electrolysis

33m 21s

Intro
0:00
Electrolysis
3:16
Electrolysis: Part 1
3:17
Electrolysis: Part 2
5:25
Galvanic Cell Example
7:13
Nickel Cadmium Battery
12:18
Ampere
16:00
Example 1
20:47
Example 2
25:47
XII. Light
Light

44m 45s

Intro
0:00
Light
2:14
Introduction to Light
2:15
Frequency, Speed, and Wavelength of Waves
3:58
Units and Equations
7:37
Electromagnetic Spectrum
12:13
Example 1: Calculate the Frequency
17:41
E = hν
21:30
Example 2: Increment of Energy
25:12
Photon Energy of Light
28:56
Wave and Particle
31:46
Example 3: Wavelength of an Electron
34:46
XIII. Quantum Mechanics
Quantum Mechanics & Electron Orbitals

54m

Intro
0:00
Quantum Mechanics & Electron Orbitals
0:51
Quantum Mechanics & Electron Orbitals Overview
0:52
Electron Orbital and Energy Levels for the Hydrogen Atom
8:47
Example 1
13:41
Quantum Mechanics: Schrodinger Equation
19:19
Quantum Numbers Overview
31:10
Principal Quantum Numbers
33:28
Angular Momentum Numbers
34:55
Magnetic Quantum Numbers
36:35
Spin Quantum Numbers
37:46
Primary Level, Sublevels, and Sub-Sub-Levels
39:42
Example
42:17
Orbital & Quantum Numbers
49:32
Electron Configurations & Diagrams

34m 4s

Intro
0:00
Electron Configurations & Diagrams
1:08
Electronic Structure of Ground State Atom
1:09
Order of Electron Filling
3:50
Electron Configurations & Diagrams: H
8:41
Electron Configurations & Diagrams: He
9:12
Electron Configurations & Diagrams: Li
9:47
Electron Configurations & Diagrams: Be
11:17
Electron Configurations & Diagrams: B
12:05
Electron Configurations & Diagrams: C
13:03
Electron Configurations & Diagrams: N
14:55
Electron Configurations & Diagrams: O
15:24
Electron Configurations & Diagrams: F
16:25
Electron Configurations & Diagrams: Ne
17:00
Electron Configurations & Diagrams: S
18:08
Electron Configurations & Diagrams: Fe
20:08
Introduction to Valence Electrons
23:04
Valence Electrons of Oxygen
23:44
Valence Electrons of Iron
24:02
Valence Electrons of Arsenic
24:30
Valence Electrons: Exceptions
25:36
The Periodic Table
27:52
XIV. Intermolecular Forces
Vapor Pressure & Changes of State

52m 43s

Intro
0:00
Vapor Pressure and Changes of State
2:26
Intermolecular Forces Overview
2:27
Hydrogen Bonding
5:23
Heat of Vaporization
9:58
Vapor Pressure: Definition and Example
11:04
Vapor Pressures is Mostly a Function of Intermolecular Forces
17:41
Vapor Pressure Increases with Temperature
20:52
Vapor Pressure vs. Temperature: Graph and Equation
22:55
Clausius-Clapeyron Equation
31:55
Example 1
32:13
Heating Curve
35:40
Heat of Fusion
41:31
Example 2
43:45
Phase Diagrams & Solutions

31m 17s

Intro
0:00
Phase Diagrams and Solutions
0:22
Definition of a Phase Diagram
0:50
Phase Diagram Part 1: H₂O
1:54
Phase Diagram Part 2: CO₂
9:59
Solutions: Solute & Solvent
16:12
Ways of Discussing Solution Composition: Mass Percent or Weight Percent
18:46
Ways of Discussing Solution Composition: Molarity
20:07
Ways of Discussing Solution Composition: Mole Fraction
20:48
Ways of Discussing Solution Composition: Molality
21:41
Example 1: Question
22:06
Example 1: Mass Percent
24:32
Example 1: Molarity
25:53
Example 1: Mole Fraction
28:09
Example 1: Molality
29:36
Vapor Pressure of Solutions

37m 23s

Intro
0:00
Vapor Pressure of Solutions
2:07
Vapor Pressure & Raoult's Law
2:08
Example 1
5:21
When Ionic Compounds Dissolve
10:51
Example 2
12:38
Non-Ideal Solutions
17:42
Negative Deviation
24:23
Positive Deviation
29:19
Example 3
31:40
Colligatives Properties

34m 11s

Intro
0:00
Colligative Properties
1:07
Boiling Point Elevation
1:08
Example 1: Question
5:19
Example 1: Solution
6:52
Freezing Point Depression
12:01
Example 2: Question
14:46
Example 2: Solution
16:34
Osmotic Pressure
20:20
Example 3: Question
28:00
Example 3: Solution
30:16
XV. Bonding
Bonding & Lewis Structure

48m 39s

Intro
0:00
Bonding & Lewis Structure
2:23
Covalent Bond
2:24
Single Bond, Double Bond, and Triple Bond
4:11
Bond Length & Intermolecular Distance
5:51
Definition of Electronegativity
8:42
Bond Polarity
11:48
Bond Energy
20:04
Example 1
24:31
Definition of Lewis Structure
31:54
Steps in Forming a Lewis Structure
33:26
Lewis Structure Example: H₂
36:53
Lewis Structure Example: CH₄
37:33
Lewis Structure Example: NO⁺
38:43
Lewis Structure Example: PCl₅
41:12
Lewis Structure Example: ICl₄⁻
43:05
Lewis Structure Example: BeCl₂
45:07
Resonance & Formal Charge

36m 59s

Intro
0:00
Resonance and Formal Charge
0:09
Resonance Structures of NO₃⁻
0:25
Resonance Structures of NO₂⁻
12:28
Resonance Structures of HCO₂⁻
16:28
Formal Charge
19:40
Formal Charge Example: SO₄²⁻
21:32
Formal Charge Example: CO₂
31:33
Formal Charge Example: HCN
32:44
Formal Charge Example: CN⁻
33:34
Formal Charge Example: 0₃
34:43
Shapes of Molecules

41m 21s

Intro
0:00
Shapes of Molecules
0:35
VSEPR
0:36
Steps in Determining Shapes of Molecules
6:18
Linear
11:38
Trigonal Planar
11:55
Tetrahedral
12:45
Trigonal Bipyramidal
13:23
Octahedral
14:29
Table: Shapes of Molecules
15:40
Example: CO₂
21:11
Example: NO₃⁻
24:01
Example: H₂O
27:00
Example: NH₃
29:48
Example: PCl₃⁻
32:18
Example: IF₄⁺
34:38
Example: KrF₄
37:57
Hybrid Orbitals

40m 17s

Intro
0:00
Hybrid Orbitals
0:13
Introduction to Hybrid Orbitals
0:14
Electron Orbitals for CH₄
5:02
sp³ Hybridization
10:52
Example: sp³ Hybridization
12:06
sp² Hybridization
14:21
Example: sp² Hybridization
16:11
σ Bond
19:10
π Bond
20:07
sp Hybridization & Example
22:00
dsp³ Hybridization & Example
27:36
d²sp³ Hybridization & Example
30:36
Example: Predict the Hybridization and Describe the Molecular Geometry of CO
32:31
Example: Predict the Hybridization and Describe the Molecular Geometry of BF₄⁻
35:17
Example: Predict the Hybridization and Describe the Molecular Geometry of XeF₂
37:09
XVI. AP Practice Exam
AP Practice Exam: Multiple Choice, Part I

52m 34s

Intro
0:00
Multiple Choice
1:21
Multiple Choice 1
1:22
Multiple Choice 2
2:23
Multiple Choice 3
3:38
Multiple Choice 4
4:34
Multiple Choice 5
5:16
Multiple Choice 6
5:41
Multiple Choice 7
6:20
Multiple Choice 8
7:03
Multiple Choice 9
7:31
Multiple Choice 10
9:03
Multiple Choice 11
11:52
Multiple Choice 12
13:16
Multiple Choice 13
13:56
Multiple Choice 14
14:52
Multiple Choice 15
15:43
Multiple Choice 16
16:20
Multiple Choice 17
16:55
Multiple Choice 18
17:22
Multiple Choice 19
18:59
Multiple Choice 20
20:24
Multiple Choice 21
22:20
Multiple Choice 22
23:29
Multiple Choice 23
24:30
Multiple Choice 24
25:24
Multiple Choice 25
26:21
Multiple Choice 26
29:06
Multiple Choice 27
30:42
Multiple Choice 28
33:28
Multiple Choice 29
34:38
Multiple Choice 30
35:37
Multiple Choice 31
37:31
Multiple Choice 32
38:28
Multiple Choice 33
39:50
Multiple Choice 34
42:57
Multiple Choice 35
44:18
Multiple Choice 36
45:52
Multiple Choice 37
48:02
Multiple Choice 38
49:25
Multiple Choice 39
49:43
Multiple Choice 40
50:16
Multiple Choice 41
50:49
AP Practice Exam: Multiple Choice, Part II

32m 15s

Intro
0:00
Multiple Choice
0:12
Multiple Choice 42
0:13
Multiple Choice 43
0:33
Multiple Choice 44
1:16
Multiple Choice 45
2:36
Multiple Choice 46
5:22
Multiple Choice 47
6:35
Multiple Choice 48
8:02
Multiple Choice 49
10:05
Multiple Choice 50
10:26
Multiple Choice 51
11:07
Multiple Choice 52
12:01
Multiple Choice 53
12:55
Multiple Choice 54
16:12
Multiple Choice 55
18:11
Multiple Choice 56
19:45
Multiple Choice 57
20:15
Multiple Choice 58
23:28
Multiple Choice 59
24:27
Multiple Choice 60
26:45
Multiple Choice 61
29:15
AP Practice Exam: Multiple Choice, Part III

32m 50s

Intro
0:00
Multiple Choice
0:16
Multiple Choice 62
0:17
Multiple Choice 63
1:57
Multiple Choice 64
6:16
Multiple Choice 65
8:05
Multiple Choice 66
9:18
Multiple Choice 67
10:38
Multiple Choice 68
12:51
Multiple Choice 69
14:32
Multiple Choice 70
17:35
Multiple Choice 71
22:44
Multiple Choice 72
24:27
Multiple Choice 73
27:46
Multiple Choice 74
29:39
Multiple Choice 75
30:23
AP Practice Exam: Free response Part I

47m 22s

Intro
0:00
Free Response
0:15
Free Response 1: Part A
0:16
Free Response 1: Part B
4:15
Free Response 1: Part C
5:47
Free Response 1: Part D
9:20
Free Response 1: Part E. i
10:58
Free Response 1: Part E. ii
16:45
Free Response 1: Part E. iii
26:03
Free Response 2: Part A. i
31:01
Free Response 2: Part A. ii
33:38
Free Response 2: Part A. iii
35:20
Free Response 2: Part B. i
37:38
Free Response 2: Part B. ii
39:30
Free Response 2: Part B. iii
44:44
AP Practice Exam: Free Response Part II

43m 5s

Intro
0:00
Free Response
0:12
Free Response 3: Part A
0:13
Free Response 3: Part B
6:25
Free Response 3: Part C. i
11:33
Free Response 3: Part C. ii
12:02
Free Response 3: Part D
14:30
Free Response 4: Part A
21:03
Free Response 4: Part B
22:59
Free Response 4: Part C
24:33
Free Response 4: Part D
27:22
Free Response 4: Part E
28:43
Free Response 4: Part F
29:35
Free Response 4: Part G
30:15
Free Response 4: Part H
30:48
Free Response 5: Diagram
32:00
Free Response 5: Part A
34:14
Free Response 5: Part B
36:07
Free Response 5: Part C
37:45
Free Response 5: Part D
39:00
Free Response 5: Part E
40:26
AP Practice Exam: Free Response Part III

28m 36s

Intro
0:00
Free Response
0:43
Free Response 6: Part A. i
0:44
Free Response 6: Part A. ii
3:08
Free Response 6: Part A. iii
5:02
Free Response 6: Part B. i
7:11
Free Response 6: Part B. ii
9:40
Free Response 7: Part A
11:14
Free Response 7: Part B
13:45
Free Response 7: Part C
15:43
Free Response 7: Part D
16:54
Free Response 8: Part A. i
19:15
Free Response 8: Part A. ii
21:16
Free Response 8: Part B. i
23:51
Free Response 8: Part B. ii
25:07
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Lecture Comments (13)

0 answers

Post by Soo bin Hwang on February 27, 2017

Dear Professor Hovasapian,

  Hello, I have learned the slope= -k in the "integrated rate law and reaction half-life" lecture. So for the example of second order, would
k= -0.06145 since m= 0.06145? But then t(1/2)= -1627 seconds. Can reaction time have negative value?

Thank you!

1 answer

Last reply by: Professor Hovasapian
Mon Feb 8, 2016 2:11 AM

Post by Gowrish Vaka on January 31, 2016

Greetings Professor Hovasapian,

The explanation for half-life (t1/2) of a second-order reaction on a University website problem said that the the formula for half life was as follows:

t1/2 = 1/(a*k) * 1/[A0]

It said that the variable "a" was the reaction order, in this case 2. I kept inputting my answer as 140 y, but the explanation stated that the answer was 70 y. I followed the problem's explanation exactly, using the given value of k and the given value for [A0], yet my answer was double the "correct value" for half-life since I didn't divide by 2 (a).

Do you know anything about this mysterious "a" variable in their given equation for the half-life of a second-order rxn? Or did the University make a mistake in their explanation?

Thank you

1 answer

Last reply by: Professor Hovasapian
Mon Aug 10, 2015 6:15 AM

Post by Jim Tang on August 10, 2015

without calculating any values, i feel the rate constant does vary (albeit slightly) when you choose diff. points to find your dy/dx. is this a big deal?

2 answers

Last reply by: Professor Hovasapian
Tue Apr 7, 2015 11:02 PM

Post by Lyngage Tan on April 7, 2015

hello professor im using a non programmable calculator scientific on solving this problems. i got the following values for example 1. K= 0.06126 and T1/2 = 1632 can this values also be considered as correct.

2 answers

Last reply by: peter alabi
Thu May 28, 2015 5:56 AM

Post by Datevig Daghlian on February 18, 2015

Dear Professor Hovasapian,

    As always, thank you very much for your lecture! these lectures are so very helpful! I was just wondering if it was possible to determine if a reaction is first order or second order without plotting the data (in other words, is it possible to formulaicly solve this?). Thank you very much and may God bless you!

Thank You,
George Daghlian

1 answer

Last reply by: Professor Hovasapian
Sun Jun 2, 2013 2:48 PM

Post by Kate Bevan on May 31, 2013

I'd just like to say how helpful I've found all these videos. It's amazing what a difference a good teacher can make!

Second Order & Zero-Order Rate Laws

  • Given Time and Concentration data, if a plot of Reciprocal Concentration (y) vs. Time (x) gives a straight line, then the reaction is 2nd order. The slope of this line is the Rate constant (k) for the reaction.
  • Given Time and Concentration data, if a direct plot of Concentration (y) vs. Time (x) gives a straight line, then the reaction is Zero-Order. The slope of this line is the negative of k, the rate constant.

Second Order & Zero-Order Rate Laws

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Kinetics 0:22
    • Second Order
    • Example 1
    • Zero-Order
    • Summary for the Kinetics Associated with the Reaction

Transcription: Second Order & Zero-Order Rate Laws

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

Today, we are going to continue our discussion of integrated rate laws.0004

Last lesson, we introduced the integrated rate law for a first-order reaction, and also the half-life formula for a first-order reaction.0007

Today, we are going to talk about second-order reactions and zero-order rate laws.0015

So, let's just jump right on in.0019

OK, so again, we are talking about (in order to simplify matters for ourselves, just so we get a good sense of the kinetics) a single reactant that decomposes to products.0024

That is what we have been doing this entire time.0035

Our fundamental reaction that we have been working with is the following products.0037

OK, now, the rate law for a second-order is the following--this is the differential rate law; and again, when we say "rate law," they mean differential rate law.0043

When they mean integrated rate law, they will specifically say "integrated rate law."0055

So, the rate is equal to -Delta;A/Δt; this rate symbol, which is the differential part in differential rate law, is equal to some constant, K, times the concentration of A to the second power.0060

That is the differential rate law for a second-order reaction.0079

Now, when we integrate this particular function, we end up with the following.0084

Let me write; this is the differential rate law; now, the integrated, which relates concentration as a function of time.0091

It is going to be 1 over the concentration of A, is equal to Kt, plus 1 over the initial concentration.0112

For a second-order reaction, your integrated rate law says 1 over the concentration of A is equal to the rate constant, times the time, plus 1 over the initial concentration of A.0124

So again, in this particular case, we can set it up as y=mx+b, but this time, the y is 1 over the concentration--the data that we take.0138

That is what is important; and the slope of the line that we get is the rate constant.0151

Notice, this time it is a positive slope--positive K instead of negative K.0158

What we are going to do is: we are presented with some standard raw kinetic data; we are going to find...we have the time; we have concentration; we are going to also change those concentrations to 1/concentration, and we are also going to do logarithm of concentration, because now, we have to check; now, we have 2 different types of equation, 2 different types of reaction orders.0161

Now, we have to check to see whether it is first-order or second-order.0185

We have to have 2 columns: one with the logarithm of A, one with the reciprocal of the concentration of A.0188

And again, you will see that in just a minute, when we do the example.0195

So, now, this is the integrated rate law; let's go ahead and work out the half-life for a second-order reaction.0199

And again, remember: half-life means that the concentration of A is equal to the initial concentration, over 2; it is when half of it is gone.0209

We will go ahead and write; that implies using this equation with those values: 1 over A0 over 2 equals Kt, plus 1 over A0; this ends up being 2 over A0, minus 1 over A0.0219

I flip this and move this over; it equals Kt; 2 over A0 minus 1 over A0 is 1 over A0, equals Kt.0247

Now, I will divide through by K, and I get that t of 1/2 is equal to 1/K, times the concentration (the initial concentration).0257

This is the formula for the half-life of a second-order reaction.0269

Notice, in this particular (you know what, this is not too clear; let me write this out; I need to erase this, and let me write it bigger over to the side; I think it will be a little more clear; let me get rid of this thing; A0...so, we get t is equal to 1/K, times the initial concentration)...0274

Now notice: in this case, the half-life of a second-order reaction not only depends on K, but it also depends on the initial concentration.0302

That was not the case with the first-order reaction; a first-order reaction does not depend on the concentration--it just depends on K; it's constant.0310

This is not constant; so the half-life--if you start with a certain amount, half of it is going to take a certain amount of time to get rid of.0317

Well, now that you have gotten through half of it, now that is a new initial amount.0325

In order to release half of that, that half-life is going to change; it is actually going to end up being longer.0328

So now, it depends on two things: the rate constant and the concentration.0334

There you go: we have that equation, which is the integrated rate law; we have this equation, which is the half-life; and, of course, we have this equation, which is the differential rate law.0339

These are the things that matter in kinetics: the differential rate law, the integrated rate law, and the half-life formula.0354

Pretty much everything else can be worked out from this information, and all of this, of course, comes from raw kinetic data: time, concentration.0360

OK, now, let's go ahead and do an example, because that is the best way that these things work.0368

Example: we have butadiene; butadiene dimerizes (and dimerization means that two molecules of something stick together); butadiene dimerization was studied, and the following kinetic data were obtained.0377

OK, so let's go ahead and write out the formula.0417

2 C4H6 turns into C8H12.0420

OK, so now, we have our time value; I'm going to go ahead and write it as one big plot.0429

Our time value; we have our concentration of C4H6, which is our reactant.0437

Now, what we want to do here is: we want to find the rate constant; we want to find the order of the reaction; and we want to find the half-life (I'm sorry about that; I should have actually told you what it is that we are actually going to be doing here).0446

So, the first thing we want to do is: we want to find the order of this particular reaction, based on the data that we are given.0459

Then, we want to find out what the rate constant is; and then, we want to find out what the half-life of the reaction is, based on this data.0467

So, t (let me go ahead and write this down) is 0, 1000 seconds, 1800, 2800, 3600, 4400, 5200, and 6200.0475

And what we have is 0.0100, 0.0625, 0.0476...no, that is not right; that is 0.0100; so this is going to be 0.00625--the butadiene is diminishing--0.00476, 0.00370, 0.00313, 0.00270, and we are almost done--no worries--0.00241; and 0.00208.0498

OK, given this raw data, find the order of the reaction; find the rate constant; and find the half-life of this reaction, based on the fact that we started with .0100 moles per liter of this butadiene.0552

All right, so we need to check: is it first-order or second-order?0569

We need to plot both the logarithm of C4H6 versus time, and the reciprocal of C4H6 versus time, to see which one of these gives us a straight line.0574

If this gives us a straight line--the logarithm--it is first-order; if this one gives us a straight line, it's second-order.0591

That is how we do it.0597

OK, so here is what the data looks like.0599

Let's see; let's do logarithm first.0602

We get -4.605, -5.075, -5.348, -5.599, -5.767, -5.915, -6.028, and our last one is -6.175; reciprocals are a lot easier.0604

What we have is 100; 160; 210; 270; 320; 370; 415; and 481.0639

OK, so now we are going to plot (I'll do it in blue): we are going to plot this versus time, and we are going to plot this versus time, to see which one gives us a straight line.0654

Let's go ahead and take a look at what these plots look like.0669

I'm going to go ahead and put them next to each other; so, I'm not going to give you too much detail on these graphs--you can actually go ahead and plot them yourself on a piece of graph paper, or using a software like Excel or something like that.0671

What you are going to end up with, as it turns out--let's see here, this is going to be 200...yes; so let's do the logarithm of the butadiene versus time here; and let's do the 1 over the reciprocal of butadiene (which I'll just call A) over time here.0687

When I do this, the logarithm graph is something like this--definitely not a straight line.0714

When I do the reciprocal, I get a straight line; so I will have you confirm this, but this is exactly what happens.0721

So, because it is the reciprocal that gives us a straight line, it is a second-order reaction.0729

Second-order: so, we can go ahead and write the differential rate law: C4H6/Δt equals the constant times C4H6 to the second power.0737

We know it now; it is a second power; we derived it from the graph now, so we have an order of 2.0754

Now, the next thing we want to know is: what is the rate constant?0763

OK, well, the rate constant--the best thing to do is to go ahead and (let's see, how shall we do the rate constant)...let's just go ahead and use the integrated rate law.0767

We know that the second order is: 1, over the concentration of A, equals -K (actually, you know what?--what am I saying?--let me just...write it out, so we have it)...equals positive K times t, plus 1 over A0.0780

This is y=mx+b.0808

We just graphed the reciprocal over that; we got a straight line, so now, what we want to do is: we want to pick two points on that straight line, take the Δy over the Δx of those two points...0811

And again, the points need to be on the line; if the data points that we have in the original data, when we calculated it, are on the line, that is fine--you can use those data points.0825

But, if they are not, make sure that you are using points that are on the line.0835

When we calculate Δy/Δx from this graph, based on the kinetic data that we derived, we end up with the following.0838

Let's see: let's use a couple of points, actually; so let's use Δy over Δx; it equals...as it turns out, a couple of the data points do fall on the line.0848

I pick the first and the last; so it's going to be 481 minus 100, over 6200 minus 0, which gives me 0.06145, and the unit is going to be liters per mole-second.0862

I know the units for the rate constants don't really make sense; they tend to change, depending on what order it is; it is actually the number that matters.0882

This is our rate constant; and now, we want to find t1/2.0889

OK, well, t1/2 of a second-order reaction is 1/K, times the initial concentration.0897

It is equal to 1 over 0.06145, and the initial concentration was 0.0100; so, when we do that, we get 1627 seconds.0909

There you go: raw data: time and concentration; we calculated the logarithm of concentration; we calculated the reciprocal of the concentration; we plotted ln of concentration versus time for one graph; we plotted reciprocal of concentration versus time for a second graph.0926

We saw which one of those graphs gave us a straight line; in this particular case, it was the reciprocal versus time that gave us a straight line: second-order.0947

Because it is second order, I go ahead, and I can write the differential rate law, if I need it.0955

We need the rate constant; so I just found a couple of points on that straight line; I took the slope, y2-y1 over x2-x1.0961

I ended up with the rate constant; and then, because I have the half-life formula, and I have the rate constant which I found, and I have the initial concentration (which is the initial concentration from raw data), I was able to calculate the half-life of this reaction--very, very straightforward.0969

Nothing altogether too complicated: again, differential rate law, integrated rate law, half-life--those three things are what is important, as far as the kinetics of a reaction are concerned.0985

OK, that was second-order reaction.0996

Let's talk about a zero-order reaction.0999

All right, so a zero-order reaction is exactly what you think it is; it says that the rate (well, let me actually write it in the top here)...1001

Zero-order reaction: well, the rate (which is equal to -Δ of some reactant/Δt) equals KA to the 0 power, which is K times 1, which is K.1011

In other words, the rate itself is constant; the rate doesn't change.1032

The integrated rate law: when we integrate this particular function, you end up with the following.1039

You end up with the concentration of A (you know what, I am not even going to write these anymore; they are driving me crazy), equals -Kt, plus A0.1045

The concentration at any given time is equal to minus the rate constant, times time, plus the initial concentration.1060

Or, it equals the initial concentration, minus rate constant, times time; it's diminishment.1067

Again, y=mx+b.1072

So now, if you had some raw kinetic data, and you just plotted the concentration versus the time, and you got a straight line, that is a zero-order reaction; there you go.1077

Now, half-life: A=A0/2; so let's put it in here.1088

A0/2=-K times t, plus A0.1097

A0/2 minus A0=-Kt.1111

-A0/2=-Kt.1117

Therefore, t1/2 equals A0 (these lines--they always show up in the most inopportune moments), divided by 2K.1125

Let's do it in red.1143

We have this; we have this; and we have this.1145

For a zero-order reaction, the differential rate law says that the rate is equal to K, the rate constant; it is a constant.1155

The integrated rate law says that the concentration at any time A is equal to -Kt, plus A0.1163

The half-life formula says that the initial concentration, divided by twice the rate constant, will give you the amount of time it takes for half of whatever you started with to disappear.1169

Notice, in this case: again, it depends on the initial concentration, and it depends on K.1181

Now, you might be wondering what a zero-order reaction is; where would you run across a zero-order reaction?1187

When would you have a reaction that doesn't actually depend on the concentration?1192

Zero-order reactions show up in places where...usually in things that involve catalysis; and, extended to the biological realm, that means enzymes.1197

So, for example, if you have--let's say just 10 enzymes, catalyzing a reaction; well, if all of those 10 enzymes are busy (meaning if they are full, doing what they are supposed to do), it doesn't matter how much more reactant you actually put in there.1210

It is called a substrate--the thing that the enzyme grabs onto and fills with; so it doesn't matter--once those 10 enzymes are full, you can't make the reaction go any faster, because it doesn't matter how much concentration--how much more of the reactant--you actually put in; the reaction rate is controlled by how many enzymes are actually doing the work and how fast they are doing it.1229

So, under catalysis conditions, the reaction rate actually depends on the catalyst, not so much the concentration of the reactant.1252

Now, this isn't always the case; we're just saying that, when you run across a zero-order reaction, that more often than not, it is going to be in a catalysis situation.1260

That is just a qualitative bit of information that you should know, if it happens to come up.1268

OK, so now, let us take a global view of what it is that we have done with zero-, first-, and second-order reactions.1273

Let's summarize what we have, and it will give us a good, nice one-page summary of reaction kinetics and how to deal with them.1280

OK, so this is going to be a summary for the kinetics associated with the reaction aA going to products.1288

So again, this entire time we have been talking about a single reactant decomposing into products.1314

It diminishes, and it forms something.1320

How can we figure out the rate?1322

All right, so here, let's go ahead and make ourselves a little table; and we will go ahead and put a line here, and we will say first-order, second-order, zero-order reaction.1324

OK, the differential rate law for a first-order reaction is: the rate is equal to K, times the concentration of A to the first power.1345

The rate is equal to K, times the concentration of A to the second power.1360

And, the rate equals just plain old K.1366

That is the differential rate law.1369

OK, our integrated rate law: we have, for a first-order reaction--we have: ln of A is equal to -Kt, plus (oops, I always do an h for ln; I don't know why) ln of A0; again, concentrations, kinetics--we are talking about concentrations.1371

Now, the rate law for a second-order is reciprocal: 1 over A is equal to Kt, plus 1 over A0.1395

And, this one is: A is equal to -Kt, plus A0.1408

These are the integrated rate laws.1414

OK, a plot giving a straight line is going to be ln of A versus t, y versus x.1416

So, the plot of ln(A) for a first-order reaction--the plot of ln(A) versus t gives a straight line.1440

Here, it is 1/A versus t that gives a straight line.1445

Here, it is A versus t that gives a straight line.1450

And again, it is just the left-hand side of the integrated rate law: ln(A), 1/A...you check the data.1456

Slope: well, the slope is -K, K, -K of the particular line, if you happen to get one.1464

In other words, the slope is that; so you can find K; that is how you find K.1478

And, half-life: the t1/2 is equal to ln(2) over K; notice, for a first-order reaction, it is kind of interesting; it does not depend on concentration at all; it is constant.1483

The half-life for a second-order reaction is equal to 1/K times the concentration of A0.1503

It depends on the initial concentration.1512

And, t1/2 of this one equals the initial concentration over 2K.1515

So, this is everything that we have done: first-order, second-order, zero-order reaction rates.1521

When we are given kinetic data, which involves time and concentration, we have to plot the logarithm of the concentration versus time, the concentration versus time, and 1 over the concentration versus time.1528

Whichever one of these graphs gives us a straight line, that is the order of the reaction.1542

When we have the order of the reaction (let's say, for example--boom!--we come up with second-order), we know the differential rate law; we can write the integrated rate law; we can find K from the slope; and we have the half-life formula.1547

We can tell you what the concentration is at any time; we can tell you what the concentration is going to be at any percentage that we want along the way, instead of the time--all of this information right here.1561

This is a summary of standard kinetic operating procedure.1573

We'll go ahead and leave it at that; next time, we will talk about temperature dependence of reaction rates, and talk about something called the Arrhenius equation, and activation energy, and things like that.1581

But until then, thank you for joining us for AP Chemistry and kinetics, and thank you for joining us at Educator.com.1592

Take care; goodbye.1599

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