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Raffi Hovasapian

Raffi Hovasapian

Equilibrium: Examples

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 (9)

0 answers

Post by Jonathan Garcia on April 27, 2015

I'm just not sure how to utilize the different temperatures and enthalpy of the reaction. Any pointers could help. Thank you

1 answer

Last reply by: Professor Hovasapian
Mon Apr 27, 2015 2:42 AM

Post by Jonathan Garcia on April 27, 2015

Hello Professor,
Thanks for your lectures, they really help me in my AP chem class. I have one equilibrium problem that I'm not sure how to attack. It is stated:
•One of the most important industrial sources of ethanol is the reaction of steam with ethane derived from crude oil:  
C2H4(g) + H2O(g) ⇌ C2H5OH(g) ∆Hrxn = –47.8 kJ  
For this reaction, Kc = 9 x 103 at 330 °C.  Calculate Kc at 180 °C.

2 answers

Last reply by: Professor Hovasapian
Thu Mar 12, 2015 4:27 AM

Post by bob singh on March 11, 2015

Hi Professor,

Can you explain small number approximation? Graphing utilities are not allowed on my exams.

0 answers

Post by Patrick Poplawska on March 23, 2012

Solved by examining Example 3.

0 answers

Post by Patrick Poplawska on March 23, 2012

"Use this Data to calculate the K(EQ) for
H20g+CO <--->2CO2g+2H2g

I apologize for splitting the question into two parts.

0 answers

Post by Patrick Poplawska on March 23, 2012

Today I was given a complex problem, I believe it is testing my comprehensions of manipulating the K(EQ) given a formula.

"At 25C the following reactions have the equilibrium constants

2C0g + 02 ---> 2C02 K = 3.3x10^91
2H29g + 02 ---> 2H20 K=9.1x10^80

Please Advice, and Thank You for the Excellent Instruction.

Equilibrium: Examples

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
  • Equilibrium 1:09
    • Example 1: Question
    • Example 1: Answer
    • Example 2: Question
    • Example 2: Answer
    • Example 3: Question
    • Example 3: Answer

Transcription: Equilibrium: Examples

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

Today, we are going to continue our detailed discussion of equilibrium.0005

We are just going to be doing examples.0009

We have talked about what equilibrium is, the expression; we have actually done a fair number of problems so far, and problems that are reasonably complicated.0012

But, I wanted to do some more that go a little bit more in-depth, as far as--they have some slight variations; they are a little more detailed and require a little bit more care--for the purposes of showing you where things can possibly go wrong, but mostly just to get you really, really accustomed to these equilibrium problems.0020

It has been my experience, from all of my students who have had a solid grounding in equilibrium--the rest of chemistry is a complete breeze for them.0039

They know exactly what to do; they know where to go; I virtually didn't have to teach the class after this.0046

We hammered the equilibrium so much that they knew exactly what was going on; that is why this is important.0051

If you can get your head around this, and be able to handle at least a good 70 or 80 percent of these problems, the AP test should be absolutely a breeze for you, I promise.0057

OK, so let's jump right on in.0066

Our first example is going to be a synthesis of hydrogen fluoride from hydrogen and fluorine gas.0070

Let's go: Example 1: So, in a synthesis of hydrogen fluoride gas...0077

Now notice, I didn't say hydrofluoric acid, because this is hydrogen fluoride gas--it's in gaseous form; remember, an acid is something that is when you take the thing and you drop it in the water, and it dissociates into a hydrogen ion and fluoride ion, or hydrogen ion and any other anion.0093

That is when it is an acid--the acid is the H+; when HF is together, it is not an acid--it won't do any damage.0111

Well, I won't say it won't do any damage, but it won't do any damage as an acid--let's put it that way.0119

In the synthesis of HF from H2 and F2 at a given temperature, 3.000 moles of H2 and 6.000 moles of F2 are mixed (it might be nice if I actually knew how to spell--yes, that would be good) in a 3.0-liter flask.0124

So again, we have moles, and we have a 3-liter flask; this time it is not 1-liter, so we are going to have to actually calculate the concentration.0168

OK, the eq constant at this temperature is 1.15x102.0177

What are the equilibrium concentrations of all species?0196

Or I should say, "What is the equilibrium concentration of each species?"0202

Being able to handle these ICE charts is the real key to the majority of the second half of chemistry.0218

The ICE chart itself is always going to be a ubiquitous feature of the problems that we solve.0225

The difference is what the ICE chart looks like: acid-base problems, equilibrium problems, thermodynamic problems, whatever it is...the ICE chart itself is going to change, depending on what the problem is asking.0229

Being able to handle that--that is the real deal; that is when you know you actually know what is going on--when you know how to arrange the ICE chart as necessary; the rest is just math, basic algebra.0242

OK, so in the synthesis of HF from H2 and F2 at a given temperature, 3 moles of hydrogen and 6 moles of fluorine are mixed in a 3-liter flask.0255

The equilibrium constant is 1.15x102, or 115.0267

What is the equilibrium concentration of each species?0272

OK, so let's write our equation; like we said, that is our process: H2 + F2 goes to 2 HF.0275

We want to make sure that the equation is balanced; we want to write our equilibrium expression--it is going to be the concentration of HF squared (stoichiometric coefficient), over the concentration of H2, times the concentration of F2.0283

OK, now let's go ahead and calculate the initial concentration of H2.0298

Well, it says we have 3.00 mol of H2 in a 3.000-liter flask; so we have 1.000 Molar H2.0304

That is what this little 0 here, down at the bottom is: H0; it just means the initial concentration.0319

If you want to put an i, that is fine, too.0324

F2, the initial concentration--it says we have 6 moles of fluorine gas, over (again, it's in the same flask, so it's) 3 liters, so our concentration is 2.00 Molar.0327

Well, Q--the next thing we want to do is, we want to calculate the Q in this case, the reaction quotient, to tell us what direction it is in.0343

There are a couple of ways to do this in this particular problem: we can just plug it in; the concentration of HF to begin with...well, there is no HF, so it's just 0...over 1, times 2, which is 0.0350

That is fine: you can do it that way, or you can just say, "Well, since there is no HF to begin with, I know the reaction is going to move in that direction; there is none of this yet--there is only this and this--so it has to move forward; there is nowhere else for it to go."0364

But, I think it is best to stick with the process, and just go ahead and put the numbers in, and do it that way.0382

OK, so this is 0, which implies that the reaction will move to the right.0388

Moving to the right means it will form product and it will deplete reactant.0397

H2 and F2 concentration will go down; HF concentration will rise.0402

Now, we can do our ICE chart: H2 + F2 (I'll give myself plenty of room here) goes to 2 HF; Initial, Change, Equilibrium.0407

We start off with 1 Molar of hydrogen; 2 Molar of fluorine; 0 Molar of HF.0421

This is going to deplete by x; this is going to deplete by a certain amount, x; this is going to increase by an amount, 2x, because of that 2.0429

That is the whole idea: 1, 1, 2.0440

We get equilibrium concentrations of 1.000-x, 2.000-x, and we get 2x.0443

Now, we plug in these equilibrium concentrations into this expression, and since we already know what that is, we solve the algebraic equation.0454

So, let's write: K is equal to...well, it's equal to the concentration of HF squared, so it's 2x squared, over 1.000-x, times 2.000-x.0472

That is our equation, and we know it is equal to 1.15x102.0494

They gave us the K; now, let's solve for x.0500

Well, this is just a quadratic equation; so you are just going to have to sort of do it.0503

You can...there are several ways you can do this: you can do it by hand, do the quadratic formula; you can go ahead and use your graphic utility, for those of you that have the TI-83s and 84s and 87 calculators.0515

You can handle these really, really easily; that is what I prefer; that is what I use; I have a TI-84--that is how I solve these.0527

Let's just go ahead and at least work out the algebra, and then we'll just write out the answer, presuming that we actually used a graphical utility or something like that to do it.0534

We end up with 4x2=1.15x102 (2.000-3.000x+x2), and we get 4x2=230-345x+115x2.0546

And then, we end up with 111x2 - 345 x + 230 = 0.0582

And again, when I put this into my graphical utility, and I solve, I end up with (again, this is a quadratic equation, so I have 2 values): the first x-value is 0.968 Molar, and the second root is going to be 2.14 Molar.0592

Now, we can't have both of these be true; there is only one equilibrium condition; one of these has to be true.0611

Here is how you decide: well, take a look at the original concentration--1 Molar of hydrogen gas, I think it was.0617

Well, 1 molarity - 2.14 molarity--that is going to give you a negative 1.14 molarity; you can't have a negative concentration, so that one drops out.0626

Let's do that in red; that one drops out--this is the x-value.0637

So, we have that .968 Molar is the x-value; we plug those back into the equilibrium concentrations in our ICE chart to get the following.0642

Our H2 concentration is 1.000-0.968=3.2x10-2 Molar at equilibrium.0653

Our F2 concentration is 2.000-0.968, is equal to (let's get a better-looking equals sign than that) 1.032 molarity.0669

And our final HF concentration is 2x, so it's 2 times 0.968, is equal to 1.936 molarity (you know what, these numbers are getting strange again...1.936 molarity).0687

And my friends, we have our final solution.0709

Notice, we started off with 1 Molar of H; we end up with 3.2x10-2; that means most of the H is used up.0714

That makes sense, because again, you are looking at a very, very large Keq; the Keq is 115--that means the reaction is very far to the right.0724

It favors the product formation, not reactants.0735

When it has come to equilibrium, virtually no reactants are left over.0740

The only reason that you have 1.32 moles per liter left over of the other reactant: because you started off with 2 moles per liter of that, and the stoichiometry of the H2 to the F2 is 1:1.0744

You have used up half of it; that is the only reason that it looks like this number is so big, compared to this number.0757

The Keq tells you that it is mostly to the right; you started off with no hydrogen fluoride; you ended up virtually 2 moles per liter of hydrogen fluoride.0764

That is confirmed; so, the numbers match up; everything is good.0774

OK, let's see what is next.0780

Example 2: OK, this is going to be an example of an equilibrium problem where the Keq is actually very, very small.0785

And again, depending on the equation, you might run into some rather complicated things, like cubic, quartic, quintal equations, which you can certainly handle with your graphical utility--it's not a problem--that is the great thing about having graphical utilities.0799

But, the method we are going to show is going to be a slightly simplified version, if you just want to do the math quickly.0810

And then, we will give you a way of checking to see whether any simplifications you made were actually viable--whether you could actually get away with it.0816

There might be situations where you might simplify something, and you can't get away with it.0824

We will give you a rule of thumb for doing that.0827

OK, so Example 2: Gaseous NOCl decomposes to form gaseous NO and Cl2.0830

At 40 degrees Celsius, the eq constant is 1.4x10-5; it is very small--that means there is virtually no product at equilibrium.0858

OK, initially, 1.0 mol of NOCl is placed in a 2.0-liter flask.0876

Here we go again: what is the equilibrium concentration of each species?0901

OK, let's write our equation: 2 NOCl decomposes into 2 NO + Cl2.0923

Let's write our equilibrium expression--that is what we always do: it's going to be the concentration of NO squared, times the concentration of Cl2, divided by the concentration of NOCl squared (is that correct?--yes, that is correct).0933

OK, so now we do that, and let's see what else we have.0955

Initial concentrations--we have to do initial concentrations.0962

The NOCl equals 1.0 mol over...it looks like a 2.0-liter flask: is that correct?--yes.0967

Let me circle my numbers: 2 (oops, let me use blue) liters, 1 mole; that is my Keq; OK.0984

Let's go back to red; that is going to equal 0.50 moles per liter, and then I have my initial NO concentration, which is 0, and my initial Cl2 concentration, which is also 0.0997

Therefore, Q is equal to 0 squared times 0, over 0.5, equals 0, which is definitely smaller--the Keq is small, but it is still bigger than (the 1.4x10-5 is still bigger than) 0, which implies that the reaction moves to the right.1015

Moving to the right means we are forming product.1042

We are depleting reactant.1048

OK, so let's go ahead and set up our ICE chart.1053

We write our equation: 2 NOCl goes to 2 NO + Cl2; Initial, Change, Equilibrium.1056

We start off with 0.50 Molar and none of those; we are going to deplete this by 2x; we are going to form 2x, and we are going to form x.1069

Therefore, the equilibrium concentrations...just add them straight down: what you started with, what you lost, and what you end up with.1086

0.50 minus 2x; 2x; and x; now, let's go ahead and put it into our expression.1093

Our K is equal to 2x squared (that is the squared part--2x squared) times x to the 1 power, divided by that squared, (0.50-2x)2.1105

That equals 1.4x10-5; OK.1132

Well, all right, see now: it is getting a little complicated.1141

You are going to end up with 4x2 times x; it is going to be 4x3; I don't know...you are certainly welcome to go ahead and solve this, just because you have a graphical utility; it's not a problem--you can do that.1144

But, let me give you an alternate procedure, which actually makes things at least a little bit more tractable mathematically.1156

Because we notice that the Keq is very small, which means that the reaction is over here, mostly; that means not a lot of product; this is small, because that is small, meaning that there is not a lot of product--in fact, there is virtually no product at all...it is mostly NOCl still.1169

Well, because of that--because it is mostly NOCl still--that means that NOCl hasn't lost very much.1189

Very little of it has actually decomposed.1197

Well, very little of it has decomposed...well, we started with .5; that means x is probably really, really, really small compared to .5.1199

Because it is so small compared to .5, it is possible to take this term and just leave it out--ignore it.1212

Solve the problem, and then check to see whether it is actually valid or not that we did what we did.1224

So, this is how we do it: so again, to our first approximation, we are going to presume because this is small, it means that most of it is here; that means very little of this is going to decompose; in other words, very little of this is going to form.1233

Because this is small, we are saying it is so small compared to .5 that it is probable that the .5 minus the 2x is going to go unnoticed, so let me just ignore it and solve the easier problem.1246

OK, well, let's see what we can do.1256

We have: It's going to be 2x2 times x, over 0.50 squared, equals 1.4x10-5.1258

Well, what we end up with here is (what we are going to end up with, once we do all the multiplication): we are going to end up with: 4x3 is equal to 3.5x10-6; x3 is equal to 8.75x10-7; x is going to equal 9.6x10-3.1273

We found a value of x, 9.6x10-3; now, we want to check to see whether we were actually justified in ignoring it.1300

Well, if I take (yes, that is fine) 9.6x10-3, and if I divide by the 0.50, and I multiply by 100, I am trying to see what percentage of the .5 this 9.6x10-3 really is.1311

That is what I'm doing; I am trying to see how much of the .5 this is.1341

Well, as it turns out, it actually equals about 1.91%.1346

Now, we did, actually, 2x: so 2x would put us roughly at twice that; but, as you can see, the 1.91% is actually really, really small.1352

As it turns out, if you make an approximation of this nature, and the value that you get ends up being anywhere from about...anywhere less than 5 percent of your total value, it is actually (as a good rule of thumb) a valid approximation.1363

That means, virtually, the mathematics is not going to notice that you actually eliminated that.1379

So, because of the 1.91 percent, the x compared to the .5 (or, in this case, it's going to be 2x, but again, you are still going to be below 5%), you are actually pretty good.1384

And this allows us to sort of keep this value of x, as opposed to having to solve this entire equation, this cubic equation.1394

We just did it in a nice, simple way to get an answer, instead of having to use a graphical utility, and we got this value for x.1403

Given that value of x, we can go ahead and use it now to find our equilibrium concentrations.1411

Our NOCl concentration, final, is equal to 0.50, minus the 9.6...oops, this is going to be minus two times the 9.6x10-3.1418

Our final NO concentration (I will go ahead and let you finish off the arithmetic here) is going to equal 2 times 9.6x10-3.1442

Our final Cl2 concentration is going to equal, well, just x: which is 9.6x10-3.1461

The rest is just arithmetic, which I will leave to you--nice and straightforward.1473

Hopefully you are getting a sense of the general procedure and of the things that you have to sort of watch out for.1480

OK, so let's see: let's go ahead and close this off with an interesting type of problem; it should go pretty quickly, actually.1487

Let's do...yes, that is fine; we can start it on this page.1499

OK, Example 3: Calculate the value of the equilibrium constant for the reaction O2 gas + oxygen gas goes to ozone gas, given reaction 1 (this is going to be sort of a Hess's Law kind of problem): NO2 in equilibrium with NO + O, and K1 is equal to 6.8x10-49 (wow, that is really, really small) and that one, which is O3 + NO goes to NO2 + O2; K2 equals 5.8x10-34.1508

Let's see what they are asking: they are asking you to calculate the value of the equilibrium constant for this reaction (let's do this in blue), given these two reactions and their corresponding equilibrium constants.1593

Well, we know from Hess's Law that if we want to find a final reaction that involves some of the reactants that we have, we have to rearrange them by either flipping them or multiplying them by constants.1607

In doing so, if we add all of the equations together, and then get our final equation--well, when we did ΔHs, when we did enthalpies, we just added the enthalpy; with Ks, with equilibrium constants, it is actually different.1621

When we add equations to get a final equation, what we do to equilibrium constants is: we actually multiply them.1635

So, let's go ahead and do this one.1643

In order to actually come up with this, I am going to flip Equation 1.1646

I am going to flip Equation 1, and that will give me: NO + O goes to NO2.1654

Now, when I flip an equation, I take the reciprocal of the equilibrium constant, right?--because you are just flipping products and reactants.1664

That equilibrium constant, now, is 1.47x1048--huge!1678

I am also going to flip the second reaction; I am going to make the reactants the products and the products the reactants; when I do that, I end up with NO2 + O2 in equilibrium with NO + O3.1686

Well, this equilibrium constant--again, I flipped it, so I take the reciprocal of that, and I get 1.72x1033.1702

Now I add these two equations; NO2 cancels NO2; NO cancels NO; I am left with O + O2 goes to O3; this was the equation that we wanted.1714

Now, in order to get the final equilibrium constant, I have to...I don't add these; I multiply them; it becomes K1' times K2'.1735

It equals 1.47x1048, times 1.72x1033, and I end up with getting some huge number, if I am not mistaken.1749

2.53x1088: that is huge!--that means that any time oxygen gas and free oxygen atom come together, you will not find them separately!1766

This huge number tells me that the reaction is way to the right; there is no equilibrium here--not really.1782

Any time you have a bunch of equations, if you add them together to come up with a final equation, you multiply the equilibrium constants for all of the individual equations.1789

So, when we add equations to get a final net equation, our final K is equal to K1 times K2 times...all the way to Kn, if we had n equations; and that is it.1803

OK, so we have gone ahead and dealt with a fair number of problems in equilibrium.1837

Next time, we actually are going to continue our discussion of equilibrium; we are going to talk about Le Chatelier's Principle, and then we are going to do some more problems involving Le Chatelier's Principle.1843

And Le Chatelier's Principle, just to give you a little bit of a preamble, is basically just: If I have a system at equilibrium, if I stress that system out somehow--if I put pressure on it, meaning if I add this or heat it up or cool it down, what happens to that equilibrium.1854

So, I can shift the equilibrium; well, we know (well, I can tell you) that a system will always seek out equilibrium.1869

When I apply stress to a particular system, the system is going to respond by doing whatever is necessary to relieve that stress.1878

You know this from just being a human being: any time, any system that you apply stress to, the response of that system will be doing the things that relieve that stress.1885

Well, a chemical system behaves in exactly the same way, and it is a very deep, fundamental thing called Le Chatelier's Principle.1896

It is actually quite beautiful; so we look forward to seeing you next time.1903

Thank you for joining us at Educator.com and AP Chemistry.1907

We'll see you next time; goodbye.1910

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