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Raffi Hovasapian

Raffi Hovasapian

Equilibrium: Reaction Quotient

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 (21)

1 answer

Last reply by: Jeffrey McNeary
Mon Jul 4, 2016 4:25 PM

Post by Jeffrey McNeary on July 4, 2016

for example 1 part A, you found Q using pressure. You then compared Q to K. My question is this: Why did you not convert Q into a equilibrium constant that used molarity by using the equation: K=Kp(RT)^-n? If you had done this, Q and K would both be in terms of molarity. However, in the problem, it seems like you are comparing pressure to molarity. To me this seems like you are comparing oranges to apples instead of apples to apples.

1 answer

Last reply by: Professor Hovasapian
Tue Dec 2, 2014 2:34 AM

Post by Shih-Kuan Chen on November 30, 2014

Dear Professor,

Just a curious question that has nothing to do with this lecture: Does the AP Chemistry test cover Organic Chemistry?

2 answers

Last reply by: Professor Hovasapian
Sat Aug 2, 2014 3:09 AM

Post by David Restrepo on July 30, 2014

In example 3, if the Q > K then would the +x be on the left side of the equation?
Thank You!

1 answer

Last reply by: Professor Hovasapian
Sun Jul 27, 2014 4:19 AM

Post by Jessica Lee on July 26, 2014

Why can't you just write 1 instead of 1.000, do you need that zeros? Example 3

2 answers

Last reply by: Tim Zhang
Mon Mar 17, 2014 8:41 PM

Post by Tim Zhang on March 16, 2014

I don't understant why a change of reactants are always negative and of products are positive, in such a reaction that can proceed reversibly.

1 answer

Last reply by: Professor Hovasapian
Sun Mar 16, 2014 11:20 PM

Post by Tim Zhang on March 16, 2014

In equailibrium, to calculate the concentation of a reaction the inital values start in Molarity. Does this mean the changes(C) and finial value (F) should all be in the unit of Molarity?
When a question tells you the changes in moles, Do you have to convert to molarity ?

2 answers

Last reply by: joebert binalinbing
Wed Feb 5, 2014 1:15 AM

Post by joebert binalinbing on February 3, 2014

for example 2 what happen if the ratio is 2 to  2 to 1. please help  thanks

3 answers

Last reply by: Professor Hovasapian
Tue Sep 3, 2013 2:29 AM

Post by Marian Iskandar on September 2, 2013

For example 2, 8.7x10^-3 - 2.00x10^3 = 0.0067 (6.7x1^-3), not 0.067. Slight error, but thought I'd point it out.

Equilibrium: Reaction Quotient

  • The Equilibrium Constant is a numerical measure of far a reaction has proceeded before it “stops”
  • The Reaction Quotient, Q, is the same expression as the Equilibrium Expression, but contains concentrations at any given point.
  • Comparing Q with K, we can decide in which direction the reaction has yet to proceed – forward or reverse.

Equilibrium: Reaction Quotient

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 0:57
    • Reaction Quotient
    • If Q > K
    • If Q < K
    • If Q = K
    • Example 1: Part A
    • Example 1: Part B
    • Example 2: Question
    • Example 2: Answer
    • Example 3: Question
    • Example 3: Answer
    • Steps in Solving Equilibrium Problems

Transcription: Equilibrium: Reaction Quotient

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

We are going to continue our discussion of equilibrium, and today we are going to introduce something called the reaction quotient.0005

As you will see in a minute, the reaction quotient is just like the equilibrium expression, except it's used at any given time during the reaction.0011

It actually will tell us in which direction the reaction must go in order to reach equilibrium.0021

Sometimes it's too far to the left; it wants to go to the right; sometimes the reaction is too far to the right; it wants to go to the left.0027

Sometimes, it is exactly at equilibrium.0032

So, this reaction quotient is going to be one of the fundamental things that we use, and we will actually see it over and over again as we continue to decide where our reaction is and in what direction it needs to proceed.0035

And, as you will see when we start doing the problems, it is going to describe the mathematics--what are we adding? What are we subtracting?--things like that.0045

So, anyway, let's go ahead and get started.0055

OK, so let's do a quick review; so let's do this in blue.0059

We'll take our equation that we have been dealing with, which is nitrogen gas, plus 3 moles of hydrogen gas, in equilibrium to form 2 moles of NH3 gas--ammonia.0065

OK, now we said that the equilibrium expression, K...we are no longer writing Keq, remember--if we just do K, that will mean that we are talking about moles per liter, as opposed to KP, which means we are definitely dealing in partial pressures, either in torr or atmosphere, as long as the units are consistent.0079

OK, so K is products over reactants, so it is going to be the concentration of NH3 squared, over the concentration of N2 (and again, I am using parentheses for concentration instead of the standard brackets, simply to make this a little cleaner, because my brackets tend to be a bit messy) and H23.0098

So, this is our equilibrium expression; and now, just as a review for what this equilibrium expression tells us: depending on what this number is--if it's a really, really big number, well, notice: if it's a big number, that means the numerator is a lot bigger than the denominator.0121

That means that there is more product than there is reactant.0137

That means that the equation favors the products; that means that, at equilibrium, most of what is in that flask is going to be products.0142

We say it is really far to the right.0151

If this is a really small number, this ratio, that means that the numerator is small and the denominator is big.0153

That means that the reactants actually are favored in the equilibrium; so at equilibrium, you are going to find most of the stuff on the left-hand side; it's going to be mostly nitrogen and hydrogen, instead of the ammonia.0162

That is what this equilibrium constant is a measure of; it is a measure of the extent to which a reaction moves forward or doesn't move forward.0173

That is all it is; it is just a numerical value expressing that.0181

So, we can speak about it qualitatively--"It's far to the right; it's far to the left"--or we can be precise and quantitative--"It is far to the right, because the Keq is 500," or "The Keq is .001."0185

This tells us specifically how far to the left or right; so that is all that that is.0197

OK, now, as we said, there is also something called the reaction quotient, which we can use to tell us in which direction a reaction must go in order to reach equilibrium--in other words, where it is at that given moment.0202

OK, so let's go ahead and define our Q; this is called the reaction quotient.0216

It is the same expression as the equilibrium constant; so, in other words, it's the same thing as this (products over reactants, raised to their stoichiometric coefficients).0229

It's the same expression as Keq, but concentrations/pressures are taken at any given moment.0241

So you know that this constant--equilibrium constant--it is the ratio of these concentrations, once the system has come to equilibrium.0275

Well, we can measure the concentration any time we want (of the NH3, the N2, the H2, or the products, or the reactants), and we can put those into this expression, and we can see how far away from the equilibrium constant it is, and that will tell us whether it is too far to the left or too far to the right.0283

That is all we are doing here; so, for a general reaction, aA + bB going to cC + dD, we have that the reaction quotient, Q, is equal to the concentration of C raised to the c power, the concentration of D raised to the d power, over the concentration of A raised to the a power, the concentration of B raised to the b power.0304

It is exactly the same as the equilibrium expression, except these concentrations are at any given moment.0330

That is all; now, here are the criteria by which we decide where a reaction is--how far from equilibrium.0337

If Q is bigger than K, then the reaction will proceed (let's say...yes, you know, I think "proceed" is a good word; I wanted to use "shift," but I think "proceed" is better, or "move") to the left to reach equilibrium.0346

In other words, if Q is bigger than K, that means this is bigger than that, that means it has gone too far to the right; or it has not gone too far to the right--it is too far to the right.0381

In order for it to reach its equilibrium point (which, as we said, is a fingerprint for that particular reaction at a given temperature), it needs to move to the left; that means it needs to decompose product to form more reactant.0396

It needs to decompose this to form more this; it needs to move to the left--that is what that means.0408

So now, if Q is less than K, well, it's just the opposite: then, the reaction will proceed to the right to reach equilibrium.0413

At any given moment, if we take a bunch of concentrations of products and reactants and we stick it into this expression, we solve it, and it ends up being less than the K, that means there is too much of this and it needs to move in this direction to reach equilibrium.0438

That means this denominator is too big; Q is too small.0452

It needs to go this way, so it's going to proceed to the right to reach equilibrium.0456

So, in this case, reactants are depleting; products are forming.0461

And of course, last but not least, if Q equals K, well, you know the answer to this one: then the reaction is at equilibrium (here we go again with the stray lines; OK, that is nice).0467

That is it--nice and simple; a mathematical way to see where a reaction is and to see where a reaction is going.0487

OK, let's go ahead and do an example.0496

Example 1 (let's see): For the equation H2O gas (you know, that's OK...well, I don't know; I want to sort of skip writing the gas, but I guess it's pretty important) + Cl2O gas (oops, can't have a double arrow going--I need it to go that way and that way) forms 2 HOCl gas at 25 degrees Celsius, the equilibrium constant equals 0.0900.0510

OK, so: for the reaction H2O gas + dichlorine monoxide, that goes to 2 HOCl gas (hydrogen hypochloride gas, OK?--this is not hypochlorous acid; it's in the gaseous state, so it's not aqueous, so it's not the acid--this is the hydrogen hypochloride), the equilibrium constant for this reaction at 25 degrees Celsius is .0900.0561

Notice, there is no unit here; we will get to that in just a minute.0586

Here is what we want to do: For the following concentrations, determine the direction (I don't want to run out of room over there) the reaction must go in, in order to reach equilibrium.0590

OK, so: For the following concentrations, determine the direction the reaction must go in order to reach equilibrium: a standard reaction quotient problem.0637

It is going to be basic; we are going to do it for most equilibrium problems, because we want to know where equilibrium is.0645

OK, so the first one: we have: A partial pressure of H2 O is equal to 200 torr, and the partial pressure of Cl2O is going to equal 49.8 torr, and the P of HOCl is equal to 21.0 torricelli.0651

Now, our Q is equal to the partial pressure of HOCl squared, over the partial pressure of H2O times the partial pressure of Cl2O.0680

Well, that equals 21.0 squared (that is the HOCl), divided by the partial pressure of H2O, which is 200, and it is 49.8; there we go.0694

Now, in this particular case, this is torr and this is torr squared; so the unit on top is going to be torr squared; this is torr; this is torr; it's going to be torr squared, so it's going to end up without a unit.0715

That is why this equilibrium constant doesn't have a unit.0725

It is because the torricelli, torricelli cancels with torricelli, torricelli, down at the bottom.0729

So, now, we get that Q is equal to 0.0443; so Q is equal to .0443; K is equal to .0900; clearly, Q is less than K, which implies that the reaction will move to the right to reach equilibrium.0733

In other words, reactant will deplete; product will form; it will move to the right to reach equilibrium.0766

It hasn't reached equilibrium yet; it's still moving to the right to reach eq.0774

That is it; that is all you are using the reaction quotient for--to tell you which direction it is going in.0780

OK, all right, let's see: so, let's do another one--another set of conditions.0786

This time, we will do: A 3.0-liter flask contains 0.25 moles of HOCl, 0.0100 mol of Cl2O, and 0.56 mol of H2O.0796

OK, notice: they give us moles, and they give us the actual volume of the flask.0828

Well, again, when we deal with these reaction quotients and equilibrium expressions, they have to be in moles per liter--in concentrations.0832

Let's just take each one and find the concentrations before we put them into our reaction quotient.0841

So, our concentration of HOCl is equal to 0.25 mol, divided by 3.0 liters; that is going to equal 0.0833 Molar (that m with a line over it means molarity; it's an older expression; you are accustomed to seeing it: capital M).0846

The Cl2O: that is equal to 0.0100 moles, again divided by 3 liters, because it is in the flask; so, its concentration is 0.00333 Molar.0875

And finally, our H2O concentration is equal to 0.56 mol, divided by 3.0 liter; and this concentration is 0.1867 Molar.0895

Now that we have the molarities, we can put them into our reaction quotient; the reaction quotient is going to be exactly the same thing.0913

Now, we have: Q is equal to the concentration, as we said, of HOCl squared, over the concentration of H2O, times the concentration of Cl2O, which is equal to (drop down a little bit here) 0.0833 squared, times 0.1867, times 0.0033; and we get 11.16.0922

Now, Q is 11.16; we said that K was equal to 0.0900; so, clearly, Q is much larger than K, which implies that the reaction will move to the right to reach equilibrium.0968

In other words, it is still moving--I'm sorry; not to the right--to the left!0992

The Q is bigger; yes, Q is bigger--it is going to move to the left.0998

Sorry about that; that means that there is too much product at this temperature, given the equilibrium, which is a fingerprint for that reaction; so, there is too much product; the product needs to decompose to form reactant.1004

It is moving to the left to reach equilibrium.1019

That is it; OK, now, again, notice that there are no units for that; there are no units for this particular equilibrium constant.1028

In fact, notice in the question: the question actually said, "Equilibrium constant equals"--it didn't say "K equals" or "KP equals."1040

If you are given Keq, K, or KP, it will specifically mean that we are talking about pressures, or we are talking about moles per liter; but in this particular case, because there is no unit, that actually means that the KP and the K are the same.1050

You remember the definition of the relationship between KP and K; well, the fact that there is no unit means that there is no...if you look at the equation, Δn=0, so that RT that we had is 1.1069

So, when it says the "equilibrium constant," but it doesn't specifically specify whether it is a K or a KP, well, it's the same thing--they are actually equal to each other, which is why we use the same number, .0900, with molarity and with the one that we just did, which was done in terms of pressure.1085

In both cases, we use the .0900; that comes from the fact that there is no unit that tells us that the K and the KP are the same.1103

These are the little things that you have to watch out for; in other circumstances, when you do have a unit, you have to watch out; you have to actually (if you are dealing with molarity and you are given the KP, you have to) either convert the KP to a K, or you have to convert the molarities to pressures, if you can, depending on what the problem is asking.1112

Again, these are the sort of things; there is a lot that is going to be going on--there is a lot that you have to watch out for.1132

It isn't just "plug and play"--you don't just put numbers into an equation and hope things will fall out.1139

You have to understand what is happening; this is real science, and real science means conversions, units, and strange things.1145

OK, so let's do another example.1155

Here is where we begin to actually explore some of the diversity of these equilibrium problems.1160

What we are going to do is: most of our learning is actually going to come through the problems themselves.1165

That is why we are going to do a fair number of these equilibrium problems; it's very, very important that you have a reasonably solid understanding of how to handle these things, because it's going to be the bread and butter of what you do for the rest of chemistry--certainly for the rest of the AP and the free response questions; the electrochemistry; acid-base; the thermodynamics.1170

It's precisely this kind of reasoning, and equilibrium is fundamental to it all.1191

That is why it comes before everything else does.1196

Equilibrium is chemistry; it is that simple.1198

OK, so let's do Example #2: let's do this one in red--how is that?1203

OK, Example 2: The question is going to be a bit long, but...let's see.1211

At a certain temperature, a 1.0-liter flask contains 0.298 mol of PCl3 and 8.70x10-3 mol of PCl5.1221

OK, now after the system comes to equilibrium (comes to eq), 2.00x10-3 mol of Cl2 gas was formed in the flask.1257

Now, PCl5 decomposes according to the following: PCl5 decomposes into PCl3 + Cl2 gas.1289

What we would like you to do is calculate the equilibrium concentrations of all species and the Keq.1310

OK, so we would like you to calculate the equilibrium concentrations of all the species (in other words, the PCl5, the PCl3, and the Cl2) and we would like you to tell us what the Keq is--what the K is.1337

OK, so let's read this again: At a certain temperature, a 1-liter flask contains .298 moles of PCl3, and 8.70x10-3 moles of PCl5.1349

After the system comes to equilibrium, 2.0x10-3 moles of Cl2 is formed in the flask.1362

Great! So, let's go ahead and start this off; so I'm going to go ahead and move to a new page so I can rewrite the equation.1368

We are going to do our little ICE chart here: Initial, Change, Equilibrium.1378

PCl5 decomposes into PCl3 plus Cl2; our initial concentration, our change, and our equilibrium concentration--which is what we actually want here.1383

It's telling me that PCl5 was 8.7x10-3 initially, right?1396

8.70x10 to the negative...oh, let's do...yes, that's fine; OK; I can just do this down below.1406

Notice: they give us (well, here; let me do this over to the side)--with the PCl5, they gave us 8.70x10-3 moles.1420

Now, that is not a concentration--that is moles, not moles per liter, but they said we have a 1.0-liter flask.1436

Again, this is one of the other things: we have to make sure to actually calculate the concentrations; so, in this case, it's going to be 8.70x10-3 moles, over 1.0 liters.1443

Well, because it's 1 liter--it's a 1-liter flask--the number of moles is equal to the molarity.1457

So, I can just go ahead and put these numbers here.1462

If this were not a 1-liter flask--if it were anything other than a 1-liter flask--I would have to actually calculate the initial concentration, and those are the values that I use in my ICE chart.1465

OK, so in my ICE chart, I'm working with concentrations, not moles.1476

OK, so we have: 8.70x10-3 molarity, and we said the PCl3 was 0.298, and there is no chlorine gas.1482

Well, they said that, at equilibrium, there is 2.0x10-3 moles per liter of chlorine gas.1496

So, chlorine gas showed up; so the change was: well, for every mole of chlorine gas that shows up, a mole of PCl3 shows up, and a mole of PCl5 decomposes, because the ratio is 1:1, 1:1, 1:1.1506

If 2.00x10-3 moles shows up, that means here, also, 2.00x10-3 moles shows up; here, it is -2.00x10-3 moles.1526

Again, we are using just basic intuition and what we know about the physical system to decide how the math works.1545

This is what chemistry is all about--this is the single biggest problem with chemistry.1553

There is nothing intuitively strange about any concept in chemistry--it's all very, very clear--it's all very, very basic as far as what is happening; there is nothing esoteric; there is nothing metaphysical going on.1557

It is just that...how does one change the physical situation into the math?1571

Well, this is how you do it; you have to know what is going on physically, and then the math should fall out; just trust your instincts.1576

One mole of this shows up; well, the equation says one mole of this shows up.1583

That means, if one mole of this shows up--that means one mole of this was used up; that is it.1587

So now, we do 8.7x10-3, minus 2.0x10-3, and we end up with 6.7x10-3, which I am going to express as a decimal; so, 0.067.1592

And then, we have this one; when we add it together, we end up with 0.300; and here, we have 2.00x10-3.1606

It is probably not a good idea to mix the scientific notation and decimals, but you know what--actually, it is not that big of a deal--it is what science is all about.1619

OK, so now, because we have the equilibrium concentrations, we have solved the first part of the problem.1627

At equilibrium, we are going to have .067 Molar of the PCl5, .300 Molar of the PCl3, and 2.0x10-3 Molar of the Cl2.1635

Well now, let's just go ahead and put it into our equilibrium expression, and find our Keq.1648

That is the easy part.1654

OK, so Keq is equal to...it is going to be...the Cl2 concentration, times the PCl3 concentration, divided by the PCl5 concentration.1656

That equals 2.00x10-3, times 0.300, divided by 0.067.1671

Yes...no; 6.7x10-3...oops, I think I have my numbers wrong here; this is supposed to be...yes, I should have just left it as scientific notation.1688

You know what, I'm just going to go ahead and leave it as scientific notation.1698

I shouldn't mess with things; 6.7x10-3, and we will write this as 6.7x10-3, also.1702

6.7x10-3: I ended up forgetting a 0; it was .0067; OK.1713

And then, when we solve this, we end up with...(let's see, what number did we get?) 8.96x10-2 Molar.1720

So, in this case, it does have a unit; this is a Keq--this is not a concentration.1732

That is why I am not a big fan of units when it comes to equilibrium constants; as far as equilibrium constants are concerned, I think that units should only be used to decide about conversions.1738

Other than that, I think they should be avoided; but you know what, we will just go ahead and leave it there.1753

This is the Keq; Keq equals 8.96x10-2.1756

That is actually a pretty small number; what does a small Keq mean?1761

That means most of this reaction is over here, on the left; there are not a lot of products.1765

Most of it is PCl5; that is what that means.1769

Don't let these numbers say that it is mostly this, because, remember: we started off with a certain amount of the PCl5; we started off with a whole bunch of the PCl3.1773

So, .298--it only went up to .3; that means it only went up .002; that is not very far.1786

So, don't let these equilibrium amounts fool you into thinking that the reaction went forward; it is this equilibrium constant which tells you the relationship between these three numbers under these conditions.1795

But, this is a fingerprint; this reaction at this temperature will not go very far forward.1811

Most of it is still PCl5; that is what is going on--don't let this .3 fool you; it doesn't mean that it has formed that much.1816

You already started at .298; you only formed .002 moles per liter of the PCl3--not very much at all.1824

It is confirmed by the Keq.1833

OK, let's do another example; that is what we are here to do.1835

Let's see what we have.1842

Yes, OK; let me write it out, and then...well, you know what, I am going to actually start a new page for this one, because I would like to see part of the problem while we are reading; OK.1845

This Example 3: Now, carbon monoxide reacts with steam (this is H2O gas) to produce carbon dioxide and hydrogen at 700 Kelvin (we don't need a comma there).1856

At 700 Kelvin, the eq constant (equilibrium constant) is 5.10.1901

Calculate the eq concentrations of all species if 1.000 mol of each component (each component means each species) is mixed in a 1.0-liter flask.1914

OK, our reaction is: they said: Carbon monoxide gas plus steam, which is H2O gas, forms carbon dioxide gas, plus hydrogen gas.1953

That is our reaction, and it is balanced; so it is 1:1:1:1--not a problem.1966

OK, so let's read this: Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen.1973

At 700 Kelvin, the equilibrium constant is 5.10.1978

So, we have our (let me do this in blue now) 5.10.1983

Calculate the equilibrium concentrations of all species if 1.0 mol of each of the components (that means 1, 1, 1, 1 mole of each) is mixed in a 1-liter flask.1987

OK, now the first thing we need to do is: we are going to (and again, this is sort of something we are always going to do--this is the procedure) write the equation, which we have.1999

We are going to write the equilibrium expression, and then we are going to check the reaction quotient to see where the reaction is at that moment, to see which direction it is actually going to be moving in.2010

Let's write the equilibrium expression for this, which is going to be the same as the reaction quotient.2019

It is the concentration of H2, times the concentration of CO2, divided by the concentration of CO, times the concentration of water.2026

OK, and that is also equal to the reaction quotient.2037

The first thing we want to do is calculate the reaction quotient to see which direction it is moving in.2041

Now, they say they put in 1.000 mol of each component.2048

Well, it is sitting in a 1.0-liter flask, so basically, I can work with moles: 1 mole, divided by 1 liter, is 1 mole per liter, so the concentration of each species is 1 mole per liter.2054

Well, now let's calculate the Q.2068

The Q equals...well, it is 1 Molar of the H2, 1 Molar of the CO2, divided by 1 Molar of the CO and 1 Molar of the H2O; the Q equals 1.2071

Now, the reaction quotient Q, which is equal to 1, is less than 5.10 (remember, they gave us the 5.10, which is equal to K).2088

When Q is less than K, that means the reaction wants to move forward to produce more product, in order to reach equilibrium.2096

That means it hasn't reached equilibrium yet; it is still moving forward--it is producing more product to reach equilibrium.2105

That means carbon monoxide and H2O gas are being depleted, and for each amount that these are depleted (because the ratio is 1:1), an equal amount of CO2 and H2 are being formed.2113

Now, we can do the actual equilibrium part of this problem.2124

So again, these "1 Molar"--that came from where we started at that moment.2128

At any given moment, I stick in 1 mole of each in a 1-liter flask, and let me find out what this value is.2133

It is 1; it is less than the equilibrium constant; that means it is going to move forward, to the right.2140

OK, so now let's do our equilibrium part, and we do that by doing our ICE chart, so let me rewrite CO + H2O (I tend to rewrite things a lot--sorry about that; I hope it's not a problem--I'm sure you have different ways of doing it yourself--as long as each of these is here...) goes to CO2 + H2.2146

OK, we have an initial concentration; we have the change; and we have our equilibrium concentrations, which is what we are looking for.2173

Our initial concentrations are 1.000, right?--that is how much we started with.2181

We stuck each of those in a flask.2186

Now, this reaction quotient tells us that the reaction is moving in that direction to reach equilibrium.2189

It is moving in that direction; that means CO is disappearing.2196

H2O is also disappearing by an amount x--that means a certain amount is decomposing--a certain amount of CO is being lost, is being converted.2202

Well, since it is moving to the right, and this is 1:1:1:1, that means this is +x; that means CO2 is forming for every 1 mole of this that is disappearing.2214

This is also +x; I hope that makes sense.2226

Our equilibrium concentration (at equilibrium, once everything has stopped, a certain amount of CO has been used up, so our equilibrium concentration) is going to be 1.00-x.2231

A certain amount of H2O has been used up; that is 1.00-x.2242

A certain amount of CO2 has formed, so it is going to be 1+x.2248

A certain amount of H2 has formed: 1+x.2252

These are our equilibrium (oh, wow, that is interesting; look at that--let's get these lines out of the way) expression, but notice: now, we have x, so we need to actually find x.2258

Fortunately, we can do that: we can plug it into the equilibrium expression, right?--because the equilibrium expression is a measure of these concentrations at equilibrium; that is what these values are.2274

We stick it in here; we know what the Keq is--it's 5.10; and we solve for x.2286

That is it; it is just an algebra problem.2291

So, let's go ahead and do that: so K, which is equal to 1.000+x, times 1.000+x (that is the CO2 and H2 concentrations), divided by the CO and H2O concentrations, which at equilibrium are 1.000-x, 1.000-x, and we know that that equals 5.10.2294

OK, so now let's just handle this algebraically.2329

This is (1.000+x) squared, over 1.000 (oops, too many 0's) minus x, squared, equals 5.10.2333

Now, I know that I said earlier (I think a lesson or two ago): when you are writing out the equilibrium expression, don't put the square--don't square it immediately--if two of the things are the same.2353

That is different than what I am doing now; I wrote the expression as each species separately, so that I can see that I actually have four species in my equilibrium expression.2363

Here, now, I am just dealing with the math.2373

Once you have actually written it out, then you can go ahead and write it like this to deal with the math.2375

Now, it's (1+x)2 over (1-x)2.2380

That is fine; but when you initially write the expression, don't cut corners; write down everything.2384

We want to know that the expression actually consists of four terms, not two terms, each squared.2389

OK, and now we just...well, we have a square here and a square here, so we'll just go ahead and take the square root of both sides.2395

We end up with 1+x over 1-x, equals 2.258, and then we multiply through to get 1.000+x=2.258-2.258 times x (I'm hoping that I am doing my math right here).2406

We end up with 3.258x equals 1.258, and x is equal to 0.386 Molar.2440

We found x; x is .386 Molar.2458

Now, if you go back to your ICE chart, it didn't ask for what x was; it asked for the final concentration.2464

Well, the final concentrations were the 1-x and the 1+x for those four species.2471

So, the CO concentration (carbon monoxide concentration), which also equals the H2O concentration, is equal to 1.000-x, 1.000-.386, equals 0.613 Molar.2476

The carbon monoxide and the water are at .613 molarity.2501

Now, the CO2 concentration, which also happens to equal the H2 concentration, is equal to (OK, let's see if we can clean this up a little bit; I'm not going to have these stray lines driving us crazy all day)...CO2 (I just really need to learn to write slower; I know that that is what it is) equals H2 concentration, equals 1.000+x, equals 1.000+0.386, =1.386 molarity.2505

OK, and my friends, we have done it; we have taken a standard equilibrium problem, and here is how we have approached it.2551

1) Write the equation; this is chemistry--chemistry always begins with some equation; don't just go into the math.2564

Look at the equation--the equation gives you all of the information that you need--in fact, it will tell you everything you want.2570

2) Write the equilibrium expression--write the K expression.2576

Write it out explicitly--don't count on the fact that you will know how to do it later on when you are ready to plug things in.2583

Write it out; don't cut corners; doing things quickly is not impressive--doing things correctly is impressive.2589

3) Find Q, the reaction quotient; find Q to decide which direction the reaction is going in--to decide the reaction direction, if any.2598

That is how we knew...in this problem that we just did, the reaction quotient was less than the equilibrium constant--so that means the reaction was actually moving forward.2614

Well, that is how we knew that the reactants get the -x and the products get the +x.2623

If it were the other way around and the reaction were moving to the left, that means the products would get the -x and the reactants would get the +x.2628

You have to do this; you have to do the reaction quotient--very, very important.2637

After the reaction quotient, well, you set up your ICE chart: Initial, Change, Equilibrium concentration.2642

Once you get a value for the equilibrium concentration, you put the eq concentration expressions (in other words, the concentrations that you calculated there--the equations/expressions) into the Keq expression.2652

Then, last but not least, you solve, depending on what they want.2685

If they want x, you stop there; if they want equilibrium concentrations, you take the x value; you plug it back into the equilibrium expressions; and you add and subtract until you get your equilibrium expressions.2689

If they want something else, they will tell you that they want something else.2701

Again, this process is what you are always going to be doing; this is the algorithm, the general, broad-strokes algorithm.2705

Write the equation; write the expression; find the Q; set up your ICE chart; put the equilibrium concentrations in; and then solve it.2714

Within this are the different variations that make up the different number of problems, which seem to be infinite (I understand completely).2723

That is all you are doing here.2731

OK, thank you for joining us here at Educator.com for our continuation of AP Chemistry in equilibrium.2734

In our next lesson, we are actually going to do more equilibrium problems, because again, this is a profoundly important concept; you have to be able to have a good, good, solid, intuitive understanding of what is going on.2740

So, we will do some more practice.2751

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