Raffi Hovasapian

Raffi Hovasapian

Calorimetry

Slide Duration:

Table of Contents

Section 1: 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
Section 2: 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
Section 3: 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
Section 4: 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
Section 5: 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
Section 6: 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
Section 7: 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
Section 8: 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
Section 9: 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
Section 10: 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
Section 11: 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
Section 12: 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
Section 13: 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
Section 14: 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
Section 15: 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
Section 16: 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 (18)

0 answers

Post by Julia Zhu on April 23, 2017

Hi Prof. Raffi,

In my AP Chem class, we say that qlost=-qgained, which means that c for example three would be positive. The link to the Chemistry textbook we use for this is http://www.nonsibihighschool.org/advancedch13.php. Why would this not be correct? Thank you so much!

Best,
Julia

1 answer

Last reply by: Professor Hovasapian
Thu Nov 3, 2016 9:44 PM

Post by Warwick Shaw on October 31, 2016

In my AP Chem class we use the term calorimeter constant, is this the same as C sub m?

Also, my class is asked to calculate the q of solution as the -(q of aq + q of cal) when using a calorimeter. So I am asked to use your constant volume equation to find q of cal and plug it into this equation. Is that the purpose of your constant volume equation, to find the heat lost or gained by the calorimeter?

1 answer

Last reply by: Professor Hovasapian
Sat Aug 8, 2015 9:58 PM

Post by Jim Tang on August 8, 2015

Are we gonna have to know how to calculate C_m values?

2 answers

Last reply by: Jim Tang
Sun Aug 9, 2015 2:31 AM

Post by Jim Tang on August 8, 2015

Did you copy wrong? Where does 2884.2 come from?

1 answer

Last reply by: Professor Hovasapian
Fri Dec 26, 2014 3:17 AM

Post by Bryan M on December 25, 2014

In Example 1, why would 100 ml of the solution be equivalent to 100 ml of water.  The 100 ml of solution contains HCL and NaOH, does that not effect the conversion to 100 gram of water because the solution is not 100 percent water?  Thanks for the help and clarification.

1 answer

Last reply by: Professor Hovasapian
Wed Nov 19, 2014 5:44 AM

Post by Burhan Akram on November 17, 2014

Hello Prof. Raffi,

Are you planning on teaching anything related to Physics anytime soon?

Regards,

1 answer

Last reply by: Professor Hovasapian
Thu Oct 30, 2014 1:15 AM

Post by Delaney Kranz on October 27, 2014

Hello...in the first example, when you were looking at molar heat, why did you only work with the HOH? What about the molar heat required for the chemicals?

1 answer

Last reply by: Professor Hovasapian
Sun Jul 15, 2012 9:25 PM

Post by Ciara Flynn on April 9, 2012

How do you know which substance in a reaction will be "absorbing" the heat? I.e., which substance's mass do you calculate? And will they always give you the heat capacity of the substance, or should we memorize the common ones?

1 answer

Last reply by: Professor Hovasapian
Sun Jul 15, 2012 11:14 PM

Post by NGAWANG TSERING on March 8, 2012

hi...for constant pressure calorimetry, u told that molar heat capacity = j/ degree celsius x mol but example 1... to find molar heat you just divide 2884.2/ 0.050 mol ...why u didn't put degree celsius

Calorimetry

  • Specific Heat: Amount of heat(Energy) required to raise the temp of 1 g of a substance by 1°C. Unit is J/g°C.
  • Heat lost/gained by a substance in a process is
    q=mCΔT
    m is mass in grams
    C is Heat Capacity of the substance
    ΔT is the Temperature Change experienced by the substance.
  • Constant Pressure Calorimetry: Heat lost/gained by one substance equals heat gained/lost by other substance. One of the substances is usually water:
    q1 = −q2
  • Constant Volume Calorimetry: Mass and Heat Capacity are combined into a single Heat Capacity for the apparatus, Cm, so the equation becomes:
    q = CmΔT

Calorimetry

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
  • Thermochemistry 0:21
    • Heat Capacity
    • Molar Heat Capacity
    • Constant Pressure Calorimetry
    • Example 1
    • Constant Volume Calorimetry
    • Example 2
    • Example 3

Transcription: Calorimetry

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

Today, we're going to be talking about calorimetry; we are going to be finishing off our discussion of thermochemistry.0004

We have talked about enthalpy, and a little bit about Hess's law, and things like that; we are going to finish it off with a discussion of calorimetry today.0011

So, let's just jump on in and see what we can do.0018

OK, so let's recall that, at constant pressure, the ΔH, the enthalpy, is actually equal to the heat of the reaction--which is very, very convenient for chemistry.0022

We do things at constant pressure so, when we talk about the heat and the enthalpy, they are the same.0032

Let's just write that down to begin with: recall that, at constant pressure, ΔH is equal to the q.0038

Remember energy: you can transfer energy to and from a system...or the change in energy of a system--the internal energy of a system, ΔE...there are two ways that energy can be changed.0053

It can either transfer as heat in or out of the system, or as work; in other words, you can do work on the system, or work out of the system.0066

So, q was heat; w was work, and it was usually pressure-volume work; and at the constant pressure, as it turns out, the enthalpy is equal to the heat of reaction.0074

We speak about the heat of a particular reaction: 2 H2 + O2 goes to 2 H2O; there is a certain heat that is given off--it is the enthalpy; it's written as ΔH=...negative for exothermic, positive for endothermic, things like that.0085

Now, let us define something called the heat capacity.0100

No, I'm not going to write "definition"; I just said that it is a definition.0106

Heat capacity: the heat capacity is basically the heat absorbed by a material, divided by the increase in temperature.0110

For example, let's say you had absorbed 20 Joules of heat--some material, we don't know what it is--and the rise in temperature was, let's say, 4 degrees Celsius.0134

Well, of course, you know, when you do division, basically what you are doing with all division is taking the denominator to 1; so, this is going to be Joules per 1 degree Celsius.0147

That is the whole idea behind units; so 20 over 4--you get 5 Joules per degree Celsius.0156

That means, for every 5 Joules of energy that you put into this thing, the temperature of that thing is going to rise 1 degree Celsius.0163

That is a general definition of heat capacity: heat absorbed, divided by the increase in temperature.0170

OK, now, temperature is going to be expressed sometimes in degrees Celsius, degree Kelvin--the scales for degree Celsius and degree Kelvin are the same, so the actual gradation is the same; they just differ by 273; so it's not a problem.0176

Sometimes, you will see J/K; sometimes you will see J/ΔC.0193

All right, now different substances have different heat capacities, and this is what is kind of interesting.0198

Now, as a standard, in order for us to have something to discuss...as a standard, we choose 1 gram of a substance.0204

When we choose 1 gram of a particular substance to apply a certain amount of heat to, to see how much the temperature rises, it is called the specific heat--so not just the heat capacity, but the specific heat capacity.0217

So, as a standard, we choose 1 gram of a substance, and call it the specific heat capacity.0228

Now, heat capacity is Joules per degree Celsius; well, when we choose one gram of that, it's going to be Joules per degree Celsius per gram.0252

The unit for specific heat capacity: the unit is Joules per degree Celsius per gram of substance, which is often written as Joules over gram-degree Celsius; that is the unit of heat capacity.0263

OK, there is also molar heat capacity, and it is exactly what you think it is.0283

It is going to be the amount of energy that I need to...the amount of energy that one mole of something absorbs, and rises by 1 degree Celsius.0295

Or, if I want another way of saying it, if I want to raise the temperature of 1 mole of something by 1 degree Celsius, or 1 degree Kelvin, how much heat do I have to put into it?0305

1 gram; 1 mole; we can work with both; so the unit of that is, of course, Joules per degree Celsius per mole, which is equivalent to (the way you will see it is) mole-degree Celsius.0314

And again, this can be a K instead of a C; it just depends on what the problem is asking for.0333

Specific heat capacity/molar heat capacity: the amount of heat required to raise a particular amount of something 1 degree Celsius.0337

OK, now let's talk about calorimetry.0347

There are two types of calorimetry: there is constant-pressure calorimetry, and there is constant-volume calorimetry.0354

We're going to start with constant-pressure calorimetry.0359

Constant-pressure calorimetry measures heats of reaction (you remember that little rxn; it's a shorthand for "reaction"), or the enthalpies of reaction, for reactions occurring in solution--so in aqueous solution or some kind of solution; it doesn't necessarily need to be water as the solvent--it could be anything else--but something that happens in solution.0366

If I have hydrochloric acid, and I mix it with sodium hydroxide, and I stir it all up to...you know that it's an acid-base neutralization reaction...it is going to give off some heat.0409

Well, in calorimetry, what you are doing is: that reaction is taking place in solution, so the heat that that reaction either gives off or absorbs is going to be reflected in the temperature of the solution.0422

We actually use the solution to measure decrease in temperature, and because we know the particular heat capacity of water, it is an indirect way of measuring the actual amount of heat that is being given off, or released, from the particular reaction.0436

It will make more sense when we do a little bit of a problem.0452

Now, recall that energy is conserved; so, if a reaction gives off heat (exothermic), the solution gets warmer.0456

You know that from experience; if a particular reaction is taking place in solution, the solvent that is containing it is like some sort of a container, if you will.0489

The reaction itself gives off heat; well, the water absorbs that heat, so the heat given off is equal to the heat absorbed.0497

That is sort of the fundamental idea behind calorimetry.0504

Calorimetry is a device that measures heats given off or absorbed, using a particular medium that we know the properties of; and that gives us the heat of a particular reaction.0507

OK, so we measure the rise in the temperature or the decline in the temperature, and then we calculate heat absorbed or released.0521

OK, now, energy released equals energy absorbed.0529

If a system is releasing heat, the surroundings are absorbing that heat.0544

If a system is absorbing heat, it is the surroundings that are giving off that heat; it's a trade-off, and it happens at the boundary.0548

That is why we talk about systems and surroundings and universes.0556

OK, now, this is the particular equation that we are interested in: mCΔt.0561

q is the energy or the heat; so energy or heat...heat and energy are interchangeable.0571

This right here--this is the mass of the solution, or the mass of the particular thing that you are discussing--it could be a solid; it could be a liquid--its mass, generally in grams.0582

This C right here is the symbol for heat capacity--specific heat capacity.0594

Heat capacity: I'll say it's heat capacity, and I'll put "specific," because more often than not, it is specific heat capacity.0601

Remember, we are talking about a gram of something; this is why this mass is in grams.0610

This is going to be specific heat capacity, and this ΔT (I'm sorry, this shouldn't be a small t) is a change in temperature.0614

So, that is the change in temperature; now, mind you, it is not the beginning or the ending temperature--it is the actual change in the temperature.0625

It is the final temperature, minus the initial temperature--that is the definition of Δ.0633

It is the change in temperature; so, if a particular problem says the temperature went from 25 to 30, our ΔT in this case is 5.0639

Or, if they say (a problem could say) there was a temperature increase of 4.7 degrees, our ΔT is 4.7; so this is the change.0648

OK, so once again: mass, times heat capacity, times the change in the temperature, is equal to the energy; let's make sure that the units match here.0658

Mass is going to be in grams; the heat capacity is in Joules per gram per degree Celsius; and the change in temperature is going to be in degrees Celsius.0667

Degrees Celsius cancels degrees Celsius; gram cancels gram, leaving you Joules--sure enough, Joules: heat or energy is in Joules.0679

This equation, right here, is our fundamental equation for calorimetry.0688

Anything having to do with heat capacity, work, the heat absorbed or released by a substance (liquid, solid, whatever) is equal to the mass of that substance, times its heat capacity, times the change in temperature that it experiences.0692

It's fantastic: 1, 2, 3, 4: you have 4 variables--given any three of them, you can find the fourth.0711

Let's say you have the heat, the mass, and the change in temperature; you can calculate the heat capacity.0718

Let's say you have the heat capacity, the temperature, and the q; you can calculate the mass.0723

Whatever it is--like any equation, you can rearrange it to get what it is that you need.0728

You will run into different problems where you have to solve for each one of those, so it isn't just solving for one or the other; this is just how it is arranged.0733

You will rearrange it accordingly.0742

OK, well, let's just go ahead and do a problem, and I think it will start to come together.0745

Let's do an example: we have: 50 milliliters of a 1 Molar hydrochloric acid solution (and this m with a line on it--that is just an older symbol for molarity; it is just something that I'm used to; you will see the capital M--just as a reminder) is mixed with 50 milliliters of a 1 Molar sodium hydroxide solution, NaOH.0755

OK, temperature rises from 25 degrees to 31.9 degrees (this is Celsius).0794

Calculate the heat released...let's see...and the molar heat released.0811

OK, so 50 milliliters of a 1 Molar hydrochloric acid solution is mixed with 50 milliliters of a 1 Molar sodium hydroxide solution; the temperature rises from 25 to 31.9 (let's make this 31; sorry, it looks like a 37 here); we want you to calculate the heat released in this reaction when I mix these, and the molar heat released (in other words, the heat released per formation of 1 mole of whatever it is that is actually reacting here).0836

Let's just write a reaction; this is chemistry--you always want to start with a reaction, if you can, which...more often than not, you are going to be starting with some reaction.0867

We are putting together hydrochloric acid, and we are putting together sodium hydroxide.0875

We have H and we have OH; this is a neutralization reaction.0880

Any time you have an H and an OH, they are going to seek each other out, and they are going to form liquid water.0885

Of course, what you have left over is the sodium chloride.0892

Once you actually write it out, then you balance it; in this particular case, everything is balanced, so our reaction is HCl + NaOH → H2O, which I am actually going to rewrite a little differently, simply because I like to.0895

It's up to you; you can write H2O; I have always been a big fan of writing it HOH, because I know that this H (let me do this in red)--this H comes from here, and this OH comes from here.0908

I like to keep them separate, especially when I am balancing things that aren't 1 to 1; it makes it a little bit easier to balance.0921

So, HOH is perfectly acceptable--unless your teacher says otherwise, but hopefully, they will be OK with it.0928

50 milliliters of a 1 Molar--all right, so let's see what we have here.0935

Well, they want the heat released, so q=mCΔT.0942

Well, we need to find the mass of the solution; so the heat that is going to be released in this reaction (let me draw a little picture here)--this reaction takes place throughout this solution.0952

The solution is 100 milliliters of solution; well, the heat released into it is going to go into the water.0963

The water is going to rise in temperature; that is what is happening here.0970

The mass of the water--well, 100 milliliters; if you have 100 milliliters, times...water is 1 gram per milliliter, so you have a mass of 100 grams.0976

This mass has to be expressed in grams.0988

We have 100 grams of water; and again, heat released is heat absorbed.0991

Heat released by the reaction is heat absorbed by the water, by the solution.0995

Well, we have 100 grams of water; now, the heat capacity of water happens to be 4.18 Joules per gram per degree Celsius.1000

What that means is that, if I have 1 gram of water, and if I want to raise it by 1 degree Celsius, I have to put in (I have to heat it by) 4.18 Joules.1010

It is a very high heat capacity; that is what makes water the extraordinary substance that it does--one of the things is the fact that it has such a high heat capacity.1019

The change in temperature, now: 25 to 31.9--what is that--6.9 degrees Celsius?1027

OK, so now, Celsius cancels Celsius; gram cancels gram; we do our multiplication, and, if I am not mistaken, we end up with (if I have done my arithmetic right) 2884.2 Joules.1034

Now, you can express this as kilojoules if you want; that is going to be 28.9 kilojoules; but I'm just going to go ahead and leave it as Joules here--it's not a problem.1054

That is how much heat is released, based on everything that they gave us.1062

Rise in temperature: the heat released by the reaction is absorbed by the water; this is what calorimetry is about; usually, water is the particular...because we know the heat capacity of water, we can measure it, so this reaction released this much heat.1069

Now, this was part A; now we want to find the molar heat; OK.1086

This is: now we want to find the amount of Joules per mole of whatever it is we are discussing.1093

In this particular case, this particular reaction is a neutralization reaction.1100

The reaction that is taking place is: H+ is getting together with OH- to form H2O.1104

What we want to find is the amount of heat that is going to be released per mole of water formed.1112

That is what this is--as is: with 50 milliliters of a 1 Molar HCl, 50 milliliters of a 1 Molar NaOH, what we have is 28.9 kilojoules.1122

Now, we want to know what that is per mole of water formed.1131

OK, here is where we have to calculate how much is actually formed.1135

Well, 50 milliliters is 0.050 liters; times 1 mole per liter--that gives me 0.050 moles of HCl, which gives me 0.050 moles of H+, right?1141

One molecule of HCl releases 1 hydrogen ion.1159

Now, we do the same thing for the hydroxide.1164

We have 0.050 liters of the sodium hydroxide, and it is also 1 Molar; liters cancel; that gives me 0.050 mol of OH-, because 1 mole of sodium hydroxide releases 1 mole of hydroxide.1167

Well, this is nice: the equation is 1 to 1, so there is nothing limiting here; so every H is going to join with an OH--nothing is going to be left over, and we have 1:1:1.1184

We are not going to have .1 here; we don't add these; it says 1 mole of H+ and 1 mole of OH- goes to 1 mole of water.1197

We're going to form 0.050 moles of H2O.1204

OK, so now, we can go ahead and finish the problem.1213

We said that we ended up releasing 2884.2 Joules in this reaction.1219

Well, the amount of water formed was 0.050 moles; so again, when you divide, you are taking the denominator, and you are turning it to 1.1229

So, this becomes 57684 Joules per mole, which is equivalent to 57.7 kilojoules per mole(when the numbers start to get this large, generally they will express it in kilojoules)--per mole of water formed.1244

So, a particular set of conditions gives you the amount of heat, just in Joules.1270

Then, if you adjust that by putting the number of moles of the thing that you actually created, or that you are talking about, in the denominator, that gives you the molar heat released.1276

There you have it; OK.1289

Now, that was constant-pressure calorimetry; and again, constant-pressure calorimetry takes place in solution, because you are not really changing the pressure of the environment when you are pouring--when you are working in a solution.1295

The pressure doesn't change; the pressure is the atmospheric pressure on that solution, so everything works out really nicely.1307

Well, there is also something called constant-volume calorimetry, and so let's talk about that a little bit.1314

Constant-volume calorimetry: this takes place in something called a bomb calorimeter.1324

So basically, what a bomb calorimeter is (I'll give you a quick...): it is this container, and it is full of water, and inside the container, there is another container.1336

That is connected to the outside; and what we do is: we put some compound in there; we ignite it; and we burn it--we combust it completely--we blow it up, basically.1353

Now, because it is constant volume--the volume doesn't change--we're keeping a constant volume, the pressure rises; the heat rises; the heat escapes into the surrounding medium.1365

Well, what we have done with a standard: we have actually already calculated the heat capacity for the entire calorimeter, which involves the metal, any parts, the solution, the water--everything in there.1382

So, for constant-volume calorimetry, it isn't q=mCΔT; it is actually CmΔT; this Cm actually includes (so I'll put it over here--q=mCΔT)...this Cm is the heat capacity of the entire bomb--of the calorimeter itself.1396

It includes--it has already taken into account--the heat capacity of the water, the mass of the water, the heat capacity of the metal, the mass of the metal...everything is taken into account; so this one is actually going to be Joules per degree Celsius, or Joules per degree Kelvin.1422

This is slightly different; so, for constant-volume calorimetry, this is our equation.1441

The mass is accounted for, which is why they ended up putting it as a subscript on the heat capacity; so now, it's just a symbol reminding you that we are talking about constant-volume calorimetry.1448

The principle is the same: so you are just taking something--in this case, you are blowing it up; you are completely combusting it, seeing how much heat it releases or...well, yes, it's going to release heat.1460

The heat passes through the bomb into the solution; the temperature of the solution rises; we measure that temperature change, and then, we just use this.1470

Same thing; so, let's do an example.1480

OK, let's do this one in red.1484

Example (let's see): 0.1964 grams of quinone (which is C6H4O2) is placed in a bomb calorimeter; the bomb calorimeter is known to have a heat capacity of 1.56 (let me actually write this over here, so I can include the units)...Cm = 1.56 (not 7) kilojoules per degree Celsius.1488

So here, they are actually expressing the specific heat in kilojoules per degree Celsius.1553

Now, the rise in temperature is (I don't know where these lines are coming from; OK) 3.2 degrees Celsius.1559

Now, the rise in temperature--this is the ΔT, right here.1575

Calculate (whoa, that was interesting; we don't want that; OK) the heat of combustion per gram and per mole of quinone.1584

We want to calculate the heat of combustion; we want to calculate the enthalpy of this combustion reaction.1619

Quinone--so, just to let you know, we have this thing quinone--I'm just going to write out the reaction with names.1625

Combustion: Quinone + O2 → CO2 + H2O.1632

All combustion reactions are the same; O2 is the other reactant; they produce carbon dioxide, and they produce water; this is a standard combustion reaction.1638

We want to find out what the heat is per gram of quinone, and per mole of quinone.1647

OK, so we use our constant-volume equation, which is CmΔT.1652

Well, we know what Cm is; it is 1.56 kilojoules per degree Celsius.1665

We want to multiply that by 3.2 degrees Celsius.1673

What you end up with, when you multiply this out: you get 4.99 kilojoules.1678

OK, 4.99 kilojoules is released; now, A: per gram--so 4.99 kilojoules was released, and we said that there were .1964 grams of this quinone, so we take the 4.99 kilojoules, and we divide it by the mass, 0.1964 grams.1686

We want Joules per gram; when you divide by something, you are making the denominator 1.1718

So, you end up with 25.4 kilojoules per gram; that is how much heat is released.1725

If you have a gram of quinone--not a lot--and you combust it, it releases--the bonds in quinone release--25.4 kilojoules of heat; that is a lot of heat.1734

Now, per mole: well, we just have to now change this to moles.1748

Well, we know how to do that: we use the molar mass, grams per mole.1752

We go to 25.4 kilojoules per gram, times grams per mole, the molar mass; and now, we look up quinone--the molecular formula.1755

It ends up being 108 grams per mole; we end up doing the multiplication; we get 2743 kilojoules per mole.1771

108 grams of quinone--that is 1 mole--it releases 2,743 kilojoules (in other words, 2 million, 743 thousand Joules) of heat.1783

That is a very, very, very powerful molecule; there is a lot of energy stored in those bonds.1796

So again, basic calorimetry: constant-pressure calorimetry--let's just see--constant P: you have that the heat comes from mass, times the heat capacity, times the change in temperature.1802

And you have constant volume--Constant V--which is q=CmΔT.1819

The mass has been accounted for; it is just one calorimeter; you have the bomb; everything is accounted for.1828

Now, this equation could be used for anything; as far as calorimetry is concerned, the heat capacity--notice, we have been using this for water, 4.18 Joules per gram-degree Celsius.1834

But, water doesn't have to be the thing that I am heating up; this equation is perfectly valid for anything that you heat up, whether it is iron, concrete...anything at all; water, mercury...1845

Each different substance has its own heat capacity, so there you have it.1857

OK, let's close off with one more problem, and it will sort of tie everything together.1864

OK, a 28.2-gram sample of nickel at 99.8 degrees Celsius is placed in--is dropped in, if you will--150.0 grams of water at 23.5 degrees Celsius.1872

So, the nickel is at 99.8 degrees Celsius; we drop it in water that is at 23.5 degrees Celsius.1909

That is what this says.1915

After thermal equilibrium is reached (in other words, after we allow everything to settle down, where now no more heat is being transferred between the metal and the water)--after thermal eq... (oh, let me write it out, actually--after thermal equilibrium is reached), the temperature of the solution is 25 degrees Celsius.1919

What is C--what is the heat capacity for nickel?1952

So here, we are taking hot nickel and dropping it in water; the heat is released into the water, so the water rises in temperature.1959

Now, from that, we want to be able to measure the heat capacity of the nickel.1969

OK, let's draw a little schematic of what is actually going on.1974

OK, I have the nickel here at a temperature of 99.8; this is actually a good way to do this, because you can actually see it pictorially--what is going on.1978

We have the water, which is at 23.5 degrees Celsius; this is degrees Celsius; now, this one drops in temperature; this one rises in temperature; and at some point, when thermal equilibrium is reached, they are both going to be at 25 degrees Celsius.1988

That is where we ended up; that is the thermal equilibrium.2005

OK, now, energy that is lost by the nickel is energy that is absorbed by the water.2010

So, in other words, the heat that the nickel gives off is equal to the heat that the water absorbs.2023

They are equal to each other; OK.2037

I'm going to run this calculation one way, and then I'm going to do something slightly different--at the end, I'm going to talk about this little negative sign that is going to show up out of nowhere.2041

So, q of nickel is mCΔT, and q of H2O is mCΔT; so now, let me see; let's go ahead and put our values in.2049

We have 28.2 grams of nickel; we are looking for C--that is what we want--and ΔT is 25-99.8 (I don't know if I should do it this way or not; OK, that is fine); 25 minus 99.8--that is equal to the mass of the water, which was 150, and the specific heat of water is 4.18, and the change in temperature is 25 minus 23.5.2077

25 minus 23.5: that is the ΔT for the water--it goes from 23.5 to 25, and this one goes from (the final minus initial) 25 from the 99.8; that is why we have set it up this way.2124

OK, so let's do 28.2C, times (25 minus 99.8), equals 940.5 Joules; when we run through this math, we multiply this by this, divide through; we should end up with .446.2141

OK, this is the heat capacity for the nickel; this is in Joules per gram per degree Celsius.2177

OK, we said that the (so let's stop and take a look at this) heat capacity is not a negative number; this is just a mathematical artifact here, based on how we did the problem.2187

When you are doing this particular kind of problem, what you end up having to do is--because the heat absorbed is equal to the heat released, well, as it turns out, because one is releasing heat, it is losing, and the other one is gaining that heat, so it's positive.2200

So, as it turns out, the sum of those is actually equal to 0; that is the whole idea.2227

From one perspective, the loss is equal to the gain; therefore, to actually make them equal, you really have to stick a negative sign in front of it.2234

Or, you can just run this particular calculation as-is, just leaving everything: mCΔT, mCΔT.2245

Because one is dropping in temperature and one is rising in temperature, the Δ (because it's defined as final minus initial)--one of these is going to be a negative number.2254

That is where this little negative sign shows up; this is an artifact of the mathematics, based on how you did it.2264

If you want to run this same problem and not have to worry about this sign at the end, you can just say the q of one equals negative (the q of the other), because that is what it is.2273

One of them is a positive quantity--it is being absorbed; the other one is a negative quantity--it is being released.2287

It just depends on your perspective.2294

As far as the mathematics is concerned, one is the negative of the other.2296

So, you can go ahead and put the negative sign in here, and then run this calculation; or you can just run this calculation, doing final minus initial for nickel, final minus initial for water--leaving everything as it is, based on the definition of Δ.2300

And then, if you end up with a negative sign, just go ahead and drop that negative sign; the negative sign is a physical feature of the problem--it doesn't necessarily reflect the actual mathematics of the problem.2317

Heat capacity is a positive capacity, so we will just take the .446.2329

As long as you keep that in mind with problems of this nature, where something is dropping in temperature and something is rising in temperature, the final is going to be--one of them is going to be--negative (the ΔTs).2334

We have to account for that negative; you can account for it here, in the beginning, or you can account for it at the end.2345

I hope that that makes sense; we may actually be returning to a particular problem like this, maybe, to talk about it a little bit more, because sometimes there is a little bit of confusion regarding this.2353

OK, thank you for joining us for AP Chemistry, and thank you for coming to see us here at Educator.com.2362

Take good care; goodbye.2367

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