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Raffi Hovasapian

Raffi Hovasapian

Equilibrium, Part 2

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 (15)

1 answer

Last reply by: Professor Hovasapian
Wed Nov 26, 2014 12:32 AM

Post by Shih-Kuan Chen on November 26, 2014

Dear Professor,

Is it right to say that the ICE method discusses the partial pressures of each type of gas?
I, standing for "Initial Pressure," C, standing for "Change in Pressure," and E, standing for "Equilibrium Pressure?"

1 answer

Last reply by: Professor Hovasapian
Sun Jul 27, 2014 4:25 AM

Post by Jessica Lee on July 25, 2014

For 7:36 is it suppose to be a brackets or is it Parentheses ?




1 answer

Last reply by: Hong Zhao
Mon May 5, 2014 7:14 PM

Post by Hong Zhao on May 5, 2014

Why the initial concentration of PCL5 is 0.5? Thank you so much.

0 answers

Post by Winnie Hu on January 2, 2014

can somebody help my answers

0 answers

Post by Winnie Hu on January 2, 2014

the reaction 2NO2(g)=4NO(G)
has an equilipbrium constant of 4.5X10^3 at a certain temperature.
what is the equilibrium constant of 2N204(g)=4NO2(g)

0 answers

Post by Marian Iskandar on September 2, 2013

I love the ICE method...makes it so much easier to break down the problem, and work towards a solution. Great teaching, Professor!

1 answer

Last reply by: Professor Hovasapian
Wed Mar 20, 2013 2:46 AM

Post by Joseph Grosse on March 19, 2013

In example 3, we ignored the existence of the solid ammonium chloride in the flask when it came time to consider pressure.

I'm having trouble understanding that. According to the ideal gas law, pressure is dependent on the volume of the flask. So even if the solids are not directly contributing to the pressure of the gas wouldn't the space they are occupying restrict the volume available to the gases and so, affect the pressure?

Thank you.

Joseph

2 answers

Last reply by: okechukwu okigbo
Tue Jan 14, 2014 4:58 AM

Post by Gabriel Fuentes on March 11, 2013

For question 2 at 32:36 why is there a -1 delta n?

0 answers

Post by Abdihakim Mohamed on March 4, 2013

Thank u very much, you explain so great and to the point its not boring lessons, and I love it. The way we go straight to the point, and lot of samples, thank you again

Equilibrium, Part 2

  • The equilibrium expression can be expressed in terms of pressure instead of concentration. A mathematical expression exists relating the two constants.
  • A chart known as an ICE chart is used to solve most equilibrium problems. ICE stands for Initial, Change, Equilibrium.

Equilibrium, Part 2

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 1:31
    • Equilibriums Involving Gases
    • General Equation
    • Example 1: Question
    • Example 1: Answer
    • Example 2: Question
    • Example 2: Answer
    • Example 3: Question
    • Example 3: Answer

Transcription: Equilibrium, Part 2

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

We are going to continue our discussion of equilibrium--in my opinion, absolutely the single most important concept, not just in chemistry, but in all of science.0004

Equilibrium is the thing that all systems tend to--this notion of balance; they don't like being off balance (too much of this, too much of that).0013

All systems tend toward equilibrium, and chemical systems are no different.0023

Last lesson, we introduced the notion of an equilibrium expression, this Keq where, beginning with any sort of concentrations of a particular reaction, once the system has come to equilibrium, we measure those concentrations at equilibrium.0027

We put it into the equilibrium expression, and it ends up being the constant.0043

Different equilibrium conditions=different values; but the relationship among those values is a constant.0047

Today, we are going to continue that discussion, and we are going to introduce a variation of the expression for reactions that involve a gas.0054

As it turns out, when you have a reaction that involves a gas, you can either work in moles per liter (like we did last time--you are certainly welcome to do so by taking the number of moles in a flask and the volume of the flask, and dividing to get your concentration in moles per liter) or, as it turns out, pressure--you can work with pressures, because pressure is actually a measure of concentration.0062

We will show you how, mathematically; it is actually very, very nice.0087

Let's get started.0090

OK, so now, let's just go ahead and write, "For equilibriums involving gases, the K can be expressed in terms of P, pressure."0112

What that means is that I can either express my equilibrium expression with moles per liter concentrations or with pressure.0135

I'm going to write the two expressions, and then we will see how they are actually related, because there is a relationship between the two, which allows us to go back and forth, depending on what the problem is asking.0142

Because sometimes, it might be easier to deal with pressure--sometimes, easier with concentration; it just depends.0153

So again, we are going to stick with our tried-and-true nitrogen, plus 3 moles of hydrogen gas, going to 2 moles of ammonia.0158

We know that the actual Keq expression in concentrations is: the concentration of ammonia squared, over the concentration of nitrogen times the concentration of hydrogen cubed.0170

Well, as it turns out, I can also express it this way: and now, I put a P down as a subscript to the equilibrium expression--P for pressure.0189

KP--it's the same thing, and it equals the partial pressure of NH3 (partial pressure means the pressure just of that gas in the container, because you have three gases in a container: one gas has one pressure, on has another pressure--we call those the partial pressures), squared (so again, everything is the same; it's products, over reactants, raised to the stoichiometric coefficients; the pressure of NH3, squared...), because this is a 2, divided by the pressure of nitrogen gas, raised to a 1 power, because this stoichiometric coefficient is 1, times the partial pressure of H2 cubed, because its stoichiometric coefficient is 3.0198

So now, let's talk about how these two are related.0240

Well, whenever we talk about gases, the one equation that we talk about is the PV=nRT expression.0244

Let me just erase a little bit of this and give me a little more room.0251

Pressure times volume equals (let me make my V a little more clear) the number of moles, times the gas constant, times the temperature in Kelvin.0254

Let's rearrange this a little bit.0267

I'm going to write P=nRT/V.0269

Now, I'm going to take two of these and combine them: n/V times RT: just a little bit of mathematical manipulation--I can do this.0275

I can take the denominator and put it with one of the numbers in the numerator.0286

Now, notice what I have: what is n/V?--n is the number of moles, and V is volume in liters; so, as it turns out, P and n/V are related by a constant, RT.0289

As it turns out, pressure is an alternative form of representing concentration in moles per liter.0308

n/V is just moles per liter, so this says, if I have something which is a certain number of moles per a certain number of liters, if I multiply that by RT, I actually get the pressure.0314

Therefore, instead of concentration, I can just put the particular pressure.0326

That is all this is: in dealing with gases, it is often difficult to...not difficult to deal in concentrations; it tends to be easier, simply by nature of gases, by measuring pressure.0330

That is all this is; so, whenever we have an equilibrium condition that involves some gases, we can use its pressure, because pressure is just an alternative form of concentration; that is it.0342

Heads, tails--it's just another way of looking at concentration.0353

Now, let's see what the actual relationship is between these two values.0357

OK, so P is equal to n/V(RT); so P is equal to concentration times RT (we will just use C as concentration, instead of writing it as n/V) or, we can write C=P/RT: concentration is equal to pressure over RT.0361

This is just standard mathematical manipulation.0383

So now, this is concentration; we are going to rewrite the expression here, and we are going to put concentration back in to see how it is related here.0386

We will do: Keq is equal to...it's pressure over RT, the concentration: so we get the partial pressure of NH3, over RT, squared.0397

So all I have done is: I have just taken this expression, put it into here, and divided by pressure of N2/RT to the first power, times the pressure of H2/RT (that is concentration squared, concentration raised to the first, concentration cubed, right?)--just basic math.0413

And then, I have partial pressure of NH3, squared, times 1/RT, squared, over the partial pressure of N2 times 1/RT, times the partial pressure of H2, cubed, times 1/RT, cubed--so far, so good.0443

I have: partial pressure of NH3, squared, over partial pressure of N2, partial pressure of H2, cubed, times 1/RT, squared, over 1/RT to the fourth power.0474

All right, and now, I am going to have...this 2 is going to cancel that; I'm going to end up with (let me write it one more time--well, two more times, actually): PNH3, squared, over PO2, times the PH2, cubed, times 1/(1/RT, squared) (because this cancels two of those, leaving that) and 1/(1/RT squared) is RT squared.0497

So, it equals the partial pressure of NH3, squared, over the partial pressure of N2 and the partial pressure of H2, cubed, times RT, squared.0542

Well, this is the Keq; that is this expression.0562

This expression right here is the KP (RT)2.0567

So as it turns out, Keq equals KP, in this particular case, times RT squared.0572

There is a relationship between expressing it with concentrations and expressing it with pressures.0584

The relationship is that the Keq equals KP (for this particular reaction--we'll do the general one in a minute), times RT squared.0590

I can go back and forth between the two; so, if I'm working with KP and I want Keq, in concentrations, I can do that.0599

If I have Keq and I want KP, I can do that by multiplying by the square of RT.0605

OK, so now, let's do the general version; I just wanted to show you where the math came from.0612

For the general expression, aA + bB in equilibrium with cC + dD, our relationship is the following.0618

From now on, I'm not going to put the Keq; I'm just going to put K; whenever you see K with no subscript on it, it just means moles per liter, concentration; K equals Keq.0630

So, K is equivalent to Keq; when we speak of P, we will write KP; that means we are dealing with pressures.0641

When we just see a K, without this eq, they are the same--simply to avoid writing the eq over and over again.0651

As it turns out, the K is equal to KP, times RT to the negative Δn, where Δn equals the sum of the coefficients of the products, minus the sum of the stoichiometric coefficients of the reactants.0659

We can write it that way; or, another way that it is written is in terms of KP.0681

KP is equal to K times (RT)Δn; either one of these is fine.0686

If you have K, you can find KP; if you have KP, you can find K by just using this expression; that is it--nothing more than that.0696

Δn is this coefficient plus that coefficient, minus that plus that; that is all.0705

Let's just do an example, and everything will make sense.0711

Example 1: OK, at 427 degrees Celsius, a 2.0-liter flask contains 40.0 moles of H2, 36.0 moles of CO2, 24.0 mol of H2O, and 11.8 mol of CO, carbon monoxide, at equilibrium.0718

So, at equilibrium, we measure 427 degrees; the system has come to equilibrium; we measure in a 2-liter flask; we find that we have this many moles of each of these species.0766

Now, the reaction is as follows: CO2 gas + H2 gas goes to carbon monoxide gas + H2O gas.0777

All of these are gaseous species: that means all of them are involved in the equilibrium expression.0797

Now, our task here is to find K and KP; so, find the equilibrium constant expression, in terms of moles per liter, and find the KP constant, in terms of partial pressure.0804

Well, we have the relationship, so we can find K, and then we can find KP, based on this--no problem.0824

OK, so now, this is our first example of a real equilibrium problem, in the sense that we really have to watch everything that is going on.0833

No two problems are going to read the same way; we can't follow an algorithmic procedure for solving every problem.0842

Physical systems--now, it just depends; different things can happen--you can have different data.0850

It can be worded in a certain way; you have to be able to extract the information that is necessary.0858

Yes, there are certain things that are universal, that you can always count on, but you have to watch every single little thing.0865

Notice here: they are giving you the volume of a flask, and they are giving you the amounts in moles; so you actually have to calculate the moles per liter, before you put it into the equilibrium expression, because the equilibrium expression requires that it be in concentrations and not in moles.0872

You have to sort of watch for that.0888

Let's write the equilibrium expression, K; we will do K first.0892

It is going to equal the products over the reactants, raised to their stoichiometric coefficients.0895

This is a balanced reaction, so everything is 1:1:1:1.0900

We have the concentration of CO, times the concentration of H2O (everything is a gas), over the concentration of CO2, times the concentration of H2 gas.0904

Well, the concentration is the number of moles per volume; our volume is 2 liters (I'm going to go ahead and do this in red), and our moles are 40, 36, 24, and 11.8.0915

So, CO--the concentration of CO, moles per liter of carbon monoxide--is 11.8; so this is going to be 11.8 moles per 2 liters.0932

I take the number of moles and divide by the liters; it has to be concentration.0945

Times the concentration of H2O: H2O is 24 moles, so it's 24.0 divided by 2.0950

The concentration of CO2: well, CO2 is 36 moles, 36.0 moles; it is sitting in a 2-liter flask, so its concentration is 36/2, or 18 moles per liter.0959

And now, the H2 is 40 moles in a 2-liter flask.0973

Now, you might think to yourself, "Well, wait a minute; 2, 2, 2, 2: don't the 2's cancel?"0977

Yes, in this case they cancel; that is because all of these coefficients are 1, 1, 1, 1.0981

But, you can't guarantee that all of the coefficients are 1, 1, 1, 1, so you can't just use the mole values in here.0989

The equilibrium expression explicitly requires that you use concentrations, moles per liter, so let's just use moles per liter.0996

Yes, they will cancel, but at least we know what is going on; we won't lose our way.1003

That is what is important: write down everything--don't do anything in your head; don't cut corners; don't take shortcuts.1007

I promise, it will go badly for you.1013

OK, when we calculate this: 0.197; that is what we wanted--we wanted the K.1015

Now, we want the KP, because that is the other thing that we have to find.1026

Well, we know that KP is equal to K times RTΔn.1029

Well, what is Δn?--Δn is adding the product coefficients and subtracting the other coefficients.1036

Well, Δn is (let me go ahead and write it here) equal to 1+1 (is 2), minus 1+1 (2), is equal to 0.1045

So, in this case, Δn is 0.1056

KP is equal to K, times RT, to the Δn, equals K, times RT to the 0 (anything raised to the 0 power is 1); is equal to K.1058

So, in this case, KP is equal to 0.197...in this case; and the only reason it is this way is because everything is in a 1:1:1:1.1079

2 reactants, 2 products: 1+1 is 2; minus 1+1 (is 2) is 0; that is it.1091

Watch what you are doing very, very carefully; do not cut corners on equilibrium--do not cut corners ever, in any problem that you do.1098

Write everything out; it's very, very important.1104

OK, let's see: now, let's do a slightly more complicated problem, and again, this is going to be an example of taking information that is written in a particular problem and reasoning it out.1107

It isn't just the math--the math is actually pretty simple once you know what is going on.1125

That is the biggest problem with chemistry, or physics, or anything else: it's not how to turn it into math; it is, "What is going on, so that I know which math to use?"1130

That is the real issue, and no two problems are the same; no two problems will be worded the same; you cannot count on that, especially at this level.1139

OK, so Example 2 (and again, we are going to do a lot of problems, like I said, from here on in--from equilibrium all the way through at least electrochemistry, because this is the heart and soul of chemistry, not to mention the free-response questions that you are going to face on the AP exam)...1148

OK, so the question is a bit long: A sample of gaseous PCl5 (phosphorus pentachloride) was placed in an evacuated flask so the pressure of pure PCl5 would be 0.5 atmospheres.1172

So, a sample of gaseous phosphorus pentachloride was placed in an evacuated flask so the pressure of the pure PCl5 would be .5 atmospheres.1221

We stuck it in there so that it would be .5 atmospheres.1229

But, PCl5 decomposes according to: PCl5 goes to PCl3 + Cl2.1232

The final total pressure in the flask was 0.84 atmospheres at 250 degrees Celsius.1259

Calculate K at this temperature.1287

OK, so let's make sure we understand exactly what this problem is asking.1298

Problems are going to be very, very specific; do not read into it anything that is there--read exactly what is there--very, very important.1302

Don't cut corners.1308

A sample of gaseous PCl5 was placed in an evacuated flask so the pressure of pure PCl5 would be .5 atmospheres.1310

In other words, they filled it up with gas, and the pressure of the PCl5 gas was .5 atmospheres.1316

The problem is: "But, PCl5 decomposes"--in other words, the PCl5 they put there at .5 atmospheres--all of a sudden, it starts to come apart.1322

It decomposes into PCl3 and Cl2.1331

Well, now there is an equilibrium that exists; now, you not only have PCl5 at .5 atmospheres; now, you have also produced some phosphorus trichloride gas and some chlorine gas, and there is also some PCl5 gas left over.1334

This is an equilibrium expression; so now, it is becoming a little bit more complicated.1347

Now, you have three things in the flask, when you only introduced one.1351

We measure the final pressure of the flask, and it comes out to .84 atmospheres at 250 degrees Celsius.1354

Calculate the K at this temperature (K, we said, is K equilibrium, moles per liter).1363

They have given this to us in pressures; so, the first thing we have to do is: we have to find KP and then calculate K.1369

So, we want to find (well, if it's OK, I'm not going to write out everything; we know what we are doing)...so notice: they didn't write and say "Find KP."1379

They said, "Find K"; but the problem as written, since we are dealing with atmospheres and gases (PCl5 gas, PCl3 gas, Cl2 gas)...we are going to deal with pressures, and then from pressures, we are going to use the RT and Δn expression that we just worked with to find the K.1387

Make sure that you understand what it is that they are asking for.1404

Don't just stop by getting the KP.1407

OK, let's see how we are going to do this.1411

We are going to introduce something called an ICE chart; and this ICE chart is something that we are going to use for absolutely the rest of the time that we discuss this.1414

All equilibrium problems--acid-base problems, further aspects of acid-base equilibria, solubility-product equilibria, electrochemistry--we are going to deal with these things called ICE charts.1422

This is sort of an introduction to them, and it is a way of dealing with what is going on in the problem.1433

If you understand these, everything should fall out naturally.1438

OK, so let's write the equation.1442

PCl5 is in equilibrium with PCl3 + Cl2.1445

ICE stands for Initial concentration, before anything happens; C stands for Change--what changes take place; and E stands for Equilibrium concentrations.1452

Well, it is these equilibrium concentrations that go into the equilibrium constant expression, right?1464

That is what we said: the Keq: those values that we put in there are concentrations at equilibrium.1469

I means Initial; C means change.1475

Even if you are not sure what to do in a problem, just start, and just write down what you know; write down what is happening; eventually, the solution will fall out.1482

The single biggest mistake that kids make is: they think that they are supposed to just look at a problem and automatically know what is going on.1491

Even I don't just look at a problem and know what is going on!1497

After all of these years of experience, I have to sit there and stare at it, and sometimes just see where I am going.1500

ICE chart is a great place to start with equilibrium problems.1506

It will give you a sense of what is happening; then, you can put the math together.1511

Don't ever feel that you have to just know what is happening; you are extracting information.1514

And...sorry about that--E stands for equilibrium.1520

OK, so our initial concentration of PCl5 was .5 atmospheres.1528

Remember what we said: pressure and concentration--they are the same; they are just different sides of the same coin, so we can deal in pressures the same way we deal with concentrations.1534

We start with 0.5 atmospheres, where, before anything happens--before PCl5 decomposes--there is no PCl3, and there is no Cl2; so this is our initial condition.1544

It's nice; well, a certain amount of PCl5 decomposes.1555

Well, look at our equation; it's 1:1:1.1561

For every 1 mole that decomposes, 1 mole is produced, and 1 mole is produced of the Cl2.1563

We can say that, if -x is the amount that disappears (or, again...concentration and pressure are the same thing)...so, if the pressure drops--PCl5--by a certain amount, that means it has to increase here by that same amount for each.1571

For every 1 atmosphere that drops, that means a certain amount has been used; well, that means that a certain amount has been produced of the PCl3 and the Cl2.1591

That is what the stoichiometry tells me.1598

That is why I have -x, +x, +x; I hope that makes sense--because, again, when this decomposes, this is forming; that is what is going on.1600

So, if 1 mole of this decomposes, 1 mole of this is formed; 1 mole of this is formed.1610

If 5.2 moles of this decomposes, 5.2 moles of PCl3 is formed; 5.2 moles of Cl2 is formed.1614

We don't know how much is formed yet, so that is why we use x: -x here, +x, +x.1621

Now, we add: the initial plus the change gives us the equilibrium condition.1627

At equilibrium, I have 0.50-x.1633

Here, I have 0+x is x; 0+x is x; so at equilibrium, this is how much I have.1637

Now, they want us to find K; well, what other information do they give us?1645

They give us the fact that our total pressure is equal to .84 atmospheres.1651

Well, our total pressure is the pressure at equilibrium: this plus this plus this.1657

The total pressure is the sum of the individual pressures; so, we write: 0.50-x+x+x (write it out; don't do it in your head first) =0.84 atm.1665

Well, this -x cancels with that x, and I'm left with 0.50+x=0.84; this is simple arithmetic--there is nothing hard about this.1682

x=0.84-0.5; it equals 0.34 atmospheres; look at that.1693

I have just solved for x, x, x; I did it; that is nice.1701

I found the value of x: .34 atmospheres of PCl3 show up; .34 atmospheres of Cl2 show up; and .34 atmospheres of PCl5 is lost.1710

So now, we can actually find our K.1724

That is the best part, which is ultimately what we want; so we are solving for KP here.1727

So now, I have partial pressures: I have the partial pressure of chlorine gas, is equal to x, which is 0.34 atm; I have the partial pressure of the PCl3 gas, which, again, is x, which is 0.34 atm; and I have the partial pressure of the PCl5 gas, which is 0.50-x (that is the equilibrium) minus 0.34, which is 0.16 atm.1732

Now, I can put these values into my equilibrium expression.1767

I know what my equilibrium expression is; my equilibrium expression is KP equals the partial pressure of Cl2, times the partial pressure of PCl3, divided by the partial pressure of PCl5, each raised to the first power, because the stoichiometric coefficients are 1.1772

Well, that equals (let's go down here) 0.34, times 0.34, divided by 0.16; I do the multiplication, and I get 0.72; this is my KP; KP equals 0.72.1792

They didn't ask for KP; they asked for K, which means they asked for Keq.1815

Well, the relationship between K and KP is the following.1822

K is equal to KP times (RT)Δn, if that is correct; let me double-check; KP equals...-Δn.1826

OK, so we have...let's see...yes, OK; so now, what is Δn, first of all?1847

Δn is equal to...well, we take the...again, let's write out the equation, so we have it on this page.1862

We have PCl5 in equilibrium with PCl3 plus Cl2; Δn is equal to 1+1-1; 2-1=1.1871

So, K is equal to KP (which is 0.72), times RT...OK, so here is where we have to be careful; R is not...well, the R that we are going to be using here is 0.08206; that is liter-atmosphere/mole-Kelvin; times temperature in Kelvin; it has to be in Kelvin, because the unit of this is liter-atmosphere...1887

You know, let me write out the units here, so you can see it.1920

This is liter-atmosphere per mole-Kelvin; so the temperature has to be in Kelvin.1925

At this particular temperature, add 273; you get 523 Kelvin to the -1 power (negative Δn; that was the relationship).1935

You put in KP; R is .08206; temperature in Kelvin is 523; negative Δn...Δn was 1; negative 1.1948

We do the math, and we end up with 0.017.1957

There you go; in this particular problem, there was a lot going on; we handled it by just sort of stopping, taking a look, and making sure we wrote everything out.1966

We introduced this thing called an ICE chart; we write out the equation on top; underneath, we write the initial concentrations, we discuss the changes that take place, and we add to get our equilibrium.1976

From there, we take a look at what the problem is asking.1988

Sometimes, they might give us a K, and they might ask for a particular concentration.1991

I take those equilibrium concentrations, and I put them into my expression, and I solve that way.1996

In this case, they gave me a total pressure so I could find x.2001

I used that to find the K; it just depends on what they are asking.2005

This is why we are going to do a lot of different types of problems for these equilibrium and so on...at least through electrochemistry.2009

OK, so let's do another example.2018

Example 3: We have: Solid ammonium chloride, solid NH4Cl, was placed in an evacuated chamber, then heated; it decomposed according to: NH4Cl, solid, decomposes into ammonia, NH3, gas, plus hydrogen chloride gas.2027

It's not hydrochloric acid; it's hydrogen chloride gas.2086

Now, after heating, the total pressure was found to be 4.4 atmospheres.2090

Our task is to calculate the KP, the equilibrium constant with respect to pressures.2119

OK, well, let's write out the equilibrium expression first; it's always a great thing to do--write everything out.2125

Write out the equilibrium expression; write out the ICE chart; and then, just see where you go from there.2132

So, the equilibrium expression here, based on this equation--well, we have a gas; we have a gas; and we have a solid.2137

Solids don't show up in the equilibrium expression, so in this case, it is just these two.2143

The coefficients are 1, so we have: the KP is equal to the partial pressure of NH3 gas, times the partial pressure of HCl gas.2148

That is it; if we find the partial pressures, we are done--we plug them in, we multiply them, and we are done.2157

OK, now let's do our ICE chart.2163

Do our ICE chart: always do it this way.2165

NH4Cl (write out the equation, and do it underneath; don't do it separately--you want to be able to keep everything straight) goes to NH3 + HCl.2169

Our initial concentration; our change; and our equilibrium concentration...2183

OK, solid NH4Cl--we don't care; it doesn't even matter, so we just put lines there; it doesn't show up in the equilibrium expression--it doesn't matter.2188

Our initial NH3 and HCl concentration--well, we started off with solid NH4Cl; so, initially, there is none of these.2198

The change--well, a certain amount shows up; that certain amount is what we want.2208

So, +x+x, right?--so again, you have to be able to see what the question is saying.2214

It is telling you that you start off with NH4Cl; it decomposes--when something decomposes, that means it is going away; the products are showing up.2220

That is why you have a +x and a +x here.2229

This + goes here; it's not that +.2235

0+x is +x; 0+x is +x; well, they are telling me that the total pressure in this flask is 4.4 atmospheres.2239

Well, which gases are in the flask?2251

Well, the solid is a solid; that doesn't matter--that doesn't do anything for the gas.2256

The gases in here are NH3 gas and HCl gas.2261

So, I basically have: x+x=4.4; they are telling me that the total pressure in there is 4.4 atmospheres; that has to be made up of the amount of NH3 gas and the amount of HCl gas.2265

Well, that is even, because they are forming a 1:1 ratio.2283

So, 2x is equal to 4.4; x is equal to 2.2; well, 2.2--there you go; that is the partial pressure of HCl and the partial pressure of NH3.2287

I have 2.2 and 2.2 (2.2, not 2.4...oh, numbers and arithmetic!); these numbers are the one that I put back in here.2307

So, KP is equal to 2.2, times...OK, now watch this: even though 2.2 and 2.2 is the same, please don't write 2.22.2319

They are different species; I promise you, if you write 2.22, and if somewhere along the way you get lost and you have to come back, you will spend five minutes trying to figure out what happened, because again, stoichiometric coefficients--they show up in the equation, so write them separately; write 2.2 times 2.2.2332

Even though they are the same, they represent different species.2352

Don't mix them up.2355

We multiply that; we get 4.84; 4.84--that is the KP, and we double-check: "Calculate KP"--that is what we wanted; we're done; that is it.2357

So, they gave us a certain amount of information; they gave us an equation to work with; we wrote down the KP expression; we wrote down the equation.2370

We wrote an ICE chart (Initial, Change, Equilibrium--ultimately, it is the equilibrium that we are concerned with, because the system has come to equilibrium).2380

We followed it; we get x and x.2388

They tell us that the total pressure is 4.4; well, the total pressure is the sum of the individual pressures.2392

There are only 2 gases in here (the NH3 and the HCl); each one is x.2397

2x=4.4; x=2.2; at equilibrium, 2.2 atmospheres is hydrogen chloride gas; 2.2 atmospheres is ammonia gas.2401

You plug that into the equilibrium expression; we multiply, and we get 4.84: a standard equilibrium problem.2413

OK, this sort of gets us going with the types of problems we are going to be dealing with with equilibrium.2419

We are going to systematize this, and this whole idea of the ICE chart--our problems are going to start to become a little bit more complex, but this whole idea of using an ICE chart, writing the equilibrium expression, and then seeing what they want, based on what they give us--the ICE chart itself is going to be different.2425

That is what is going to change.2443

The approach does not change.2445

OK, so thank you for joining us here at Educator.com for AP Chemistry and equilibrium.2448

We will see you next time; goodbye.2452

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