Enter your Sign on user name and password.

Forgot password?
Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Use Chrome browser to play professor video
Franklin Ow

Franklin Ow

Energy & Chemical Reactions

Slide Duration:

Table of Contents

I. Basic Concepts & Measurement of Chemistry
Basic Concepts of Chemistry

16m 26s

Intro
0:00
Lesson Overview
0:07
Introduction
0:56
What is Chemistry?
0:57
What is Matter?
1:16
Solids
1:43
General Characteristics
1:44
Particulate-level Drawing of Solids
2:34
Liquids
3:39
General Characteristics of Liquids
3:40
Particulate-level Drawing of Liquids
3:55
Gases
4:23
General Characteristics of Gases
4:24
Particulate-level Drawing Gases
5:05
Classification of Matter
5:27
Classification of Matter
5:26
Pure Substances
5:54
Pure Substances
5:55
Mixtures
7:06
Definition of Mixtures
7:07
Homogeneous Mixtures
7:11
Heterogeneous Mixtures
7:52
Physical and Chemical Changes/Properties
8:18
Physical Changes Retain Chemical Composition
8:19
Chemical Changes Alter Chemical Composition
9:32
Physical and Chemical Changes/Properties, cont'd
10:55
Physical Properties
10:56
Chemical Properties
11:42
Sample Problem 1: Chemical & Physical Change
12:22
Sample Problem 2: Element, Compound, or Mixture?
13:52
Sample Problem 3: Classify Each of the Following Properties as chemical or Physical
15:03
Tools in Quantitative Chemistry

29m 22s

Intro
0:00
Lesson Overview
0:07
Units of Measurement
1:23
The International System of Units (SI): Mass, Length, and Volume
1:39
Percent Error
2:17
Percent Error
2:18
Example: Calculate the Percent Error
2:56
Standard Deviation
3:48
Standard Deviation Formula
3:49
Standard Deviation cont'd
4:42
Example: Calculate Your Standard Deviation
4:43
Precisions vs. Accuracy
6:25
Precision
6:26
Accuracy
7:01
Significant Figures and Uncertainty
7:50
Consider the Following (2) Rulers
7:51
Consider the Following Graduated Cylinder
11:30
Identifying Significant Figures
12:43
The Rules of Sig Figs Overview
12:44
The Rules for Sig Figs: All Nonzero Digits Are Significant
13:21
The Rules for Sig Figs: A Zero is Significant When It is In-Between Nonzero Digits
13:28
The Rules for Sig Figs: A Zero is Significant When at the End of a Decimal Number
14:02
The Rules for Sig Figs: A Zero is not significant When Starting a Decimal Number
14:27
Using Sig Figs in Calculations
15:03
Using Sig Figs for Multiplication and Division
15:04
Using Sig Figs for Addition and Subtraction
15:48
Using Sig Figs for Mixed Operations
16:11
Dimensional Analysis
16:20
Dimensional Analysis Overview
16:21
General Format for Dimensional Analysis
16:39
Example: How Many Miles are in 17 Laps?
17:17
Example: How Many Grams are in 1.22 Pounds?
18:40
Dimensional Analysis cont'd
19:43
Example: How Much is Spent on Diapers in One Week?
19:44
Dimensional Analysis cont'd
21:03
SI Prefixes
21:04
Dimensional Analysis cont'd
22:03
500 mg → ? kg
22:04
34.1 cm → ? um
24:03
Summary
25:11
Sample Problem 1: Dimensional Analysis
26:09
II. Atoms, Molecules, and Ions
Atoms, Molecules, and Ions

52m 18s

Intro
0:00
Lesson Overview
0:08
Introduction to Atomic Structure
1:03
Introduction to Atomic Structure
1:04
Plum Pudding Model
1:26
Introduction to Atomic Structure Cont'd
2:07
John Dalton's Atomic Theory: Number 1
2:22
John Dalton's Atomic Theory: Number 2
2:50
John Dalton's Atomic Theory: Number 3
3:07
John Dalton's Atomic Theory: Number 4
3:30
John Dalton's Atomic Theory: Number 5
3:58
Introduction to Atomic Structure Cont'd
5:21
Ernest Rutherford's Gold Foil Experiment
5:22
Introduction to Atomic Structure Cont'd
7:42
Implications of the Gold Foil Experiment
7:43
Relative Masses and Charges
8:18
Isotopes
9:02
Isotopes
9:03
Introduction to The Periodic Table
12:17
The Periodic Table of the Elements
12:18
Periodic Table, cont'd
13:56
Metals
13:57
Nonmetals
14:25
Semimetals
14:51
Periodic Table, cont'd
15:57
Group I: The Alkali Metals
15:58
Group II: The Alkali Earth Metals
16:25
Group VII: The Halogens
16:40
Group VIII: The Noble Gases
17:08
Ionic Compounds: Formulas, Names, Props.
17:35
Common Polyatomic Ions
17:36
Predicting Ionic Charge for Main Group Elements
18:52
Ionic Compounds: Formulas, Names, Props.
20:36
Naming Ionic Compounds: Rule 1
20:51
Naming Ionic Compounds: Rule 2
21:22
Naming Ionic Compounds: Rule 3
21:50
Naming Ionic Compounds: Rule 4
22:22
Ionic Compounds: Formulas, Names, Props.
22:50
Naming Ionic Compounds Example: Al₂O₃
22:51
Naming Ionic Compounds Example: FeCl₃
23:21
Naming Ionic Compounds Example: CuI₂ 3H₂O
24:00
Naming Ionic Compounds Example: Barium Phosphide
24:40
Naming Ionic Compounds Example: Ammonium Phosphate
25:55
Molecular Compounds: Formulas and Names
26:42
Molecular Compounds: Formulas and Names
26:43
The Mole
28:10
The Mole is 'A Chemist's Dozen'
28:11
It is a Central Unit, Connecting the Following Quantities
30:01
The Mole, cont'd
32:07
Atomic Masses
32:08
Example: How Many Moles are in 25.7 Grams of Sodium?
32:28
Example: How Many Atoms are in 1.2 Moles of Carbon?
33:17
The Mole, cont'd
34:25
Example: What is the Molar Mass of Carbon Dioxide?
34:26
Example: How Many Grams are in 1.2 Moles of Carbon Dioxide?
25:46
Percentage Composition
36:43
Example: How Many Grams of Carbon Contained in 65.1 Grams of Carbon Dioxide?
36:44
Empirical and Molecular Formulas
39:19
Empirical Formulas
39:20
Empirical Formula & Elemental Analysis
40:21
Empirical and Molecular Formulas, cont'd
41:24
Example: Determine Both the Empirical and Molecular Formulas - Step 1
41:25
Example: Determine Both the Empirical and Molecular Formulas - Step 2
43:18
Summary
46:22
Sample Problem 1: Determine the Empirical Formula of Lithium Fluoride
47:10
Sample Problem 2: How Many Atoms of Carbon are Present in 2.67 kg of C₆H₆?
49:21
III. Chemical Reactions
Chemical Reactions

43m 24s

Intro
0:00
Lesson Overview
0:06
The Law of Conservation of Mass and Balancing Chemical Reactions
1:49
The Law of Conservation of Mass
1:50
Balancing Chemical Reactions
2:50
Balancing Chemical Reactions Cont'd
3:40
Balance: N₂ + H₂ → NH₃
3:41
Balance: CH₄ + O₂ → CO₂ + H₂O
7:20
Balancing Chemical Reactions Cont'd
9:49
Balance: C₂H₆ + O₂ → CO₂ + H₂O
9:50
Intro to Chemical Equilibrium
15:32
When an Ionic Compound Full Dissociates
15:33
When an Ionic Compound Incompletely Dissociates
16:14
Dynamic Equilibrium
17:12
Electrolytes and Nonelectrolytes
18:03
Electrolytes
18:04
Strong Electrolytes and Weak Electrolytes
18:55
Nonelectrolytes
19:23
Predicting the Product(s) of an Aqueous Reaction
20:02
Single-replacement
20:03
Example: Li (s) + CuCl₂ (aq) → 2 LiCl (aq) + Cu (s)
21:03
Example: Cu (s) + LiCl (aq) → NR
21:23
Example: Zn (s) + 2HCl (aq) → ZnCl₂ (aq) + H₂ (g)
22:32
Predicting the Product(s) of an Aqueous Reaction
23:37
Double-replacement
23:38
Net-ionic Equation
25:29
Predicting the Product(s) of an Aqueous Reaction
26:12
Solubility Rules for Ionic Compounds
26:13
Predicting the Product(s) of an Aqueous Reaction
28:10
Neutralization Reactions
28:11
Example: HCl (aq) + NaOH (aq) → ?
28:37
Example: H₂SO₄ (aq) + KOH (aq) → ?
29:25
Predicting the Product(s) of an Aqueous Reaction
30:20
Certain Aqueous Reactions can Produce Unstable Compounds
30:21
Example 1
30:52
Example 2
32:16
Example 3
32:54
Summary
33:54
Sample Problem 1
34:55
ZnCO₃ (aq) + H₂SO₄ (aq) → ?
35:09
NH₄Br (aq) + Pb(C₂H₃O₂)₂ (aq) → ?
36:02
KNO₃ (aq) + CuCl₂ (aq) → ?
37:07
Li₂SO₄ (aq) + AgNO₃ (aq) → ?
37:52
Sample Problem 2
39:09
Question 1
39:10
Question 2
40:36
Question 3
41:47
Chemical Reactions II

55m 40s

Intro
0:00
Lesson Overview
0:10
Arrhenius Definition
1:15
Arrhenius Acids
1:16
Arrhenius Bases
3:20
The Bronsted-Lowry Definition
4:48
Acids Dissolve In Water and Donate a Proton to Water: Example 1
4:49
Acids Dissolve In Water and Donate a Proton to Water: Example 2
6:54
Monoprotic Acids & Polyprotic Acids
7:58
Strong Acids
11:30
Bases Dissolve In Water and Accept a Proton From Water
12:41
Strong Bases
16:36
The Autoionization of Water
17:42
Amphiprotic
17:43
Water Reacts With Itself
18:24
Oxides of Metals and Nonmetals
20:08
Oxides of Metals and Nonmetals Overview
20:09
Oxides of Nonmetals: Acidic Oxides
21:23
Oxides of Metals: Basic Oxides
24:08
Oxidation-Reduction (Redox) Reactions
25:34
Redox Reaction Overview
25:35
Oxidizing and Reducing Agents
27:02
Redox Reaction: Transfer of Electrons
27:54
Oxidation-Reduction Reactions Cont'd
29:55
Oxidation Number Overview
29:56
Oxidation Number of Homonuclear Species
31:17
Oxidation Number of Monatomic Ions
32:58
Oxidation Number of Fluorine
33:27
Oxidation Number of Oxygen
34:00
Oxidation Number of Chlorine, Bromine, and Iodine
35:07
Oxidation Number of Hydrogen
35:30
Net Sum of All Oxidation Numbers In a Compound
36:21
Oxidation-Reduction Reactions Cont'd
38:19
Let's Practice Assigning Oxidation Number
38:20
Now Let's Apply This to a Chemical Reaction
41:07
Summary
44:19
Sample Problems
45:29
Sample Problem 1
45:30
Sample Problem 2: Determine the Oxidizing and Reducing Agents
48:48
Sample Problem 3: Determine the Oxidizing and Reducing Agents
50:43
IV. Stoichiometry
Stoichiometry I

42m 10s

Intro
0:00
Lesson Overview
0:23
Mole to Mole Ratios
1:32
Example 1: In 1 Mole of H₂O, How Many Moles Are There of Each Element?
1:53
Example 2: In 2.6 Moles of Water, How Many Moles Are There of Each Element?
2:24
Mole to Mole Ratios Cont'd
5:13
Balanced Chemical Reaction
5:14
Mole to Mole Ratios Cont'd
7:25
Example 3: How Many Moles of Ammonia Can Form If you Have 3.1 Moles of H₂?
7:26
Example 4: How Many Moles of Hydrogen Gas Are Required to React With 6.4 Moles of Nitrogen Gas?
9:08
Mass to mass Conversion
11:06
Mass to mass Conversion
11:07
Example 5: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂?
12:37
Example 6: How Many Grams of Hydrogen Gas Are Required to React With 6.4 Grams of Nitrogen Gas?
15:34
Example 7: How Man Milligrams of Ammonia Can Form If You Have 1.2 kg of H₂?
17:29
Limiting Reactants, Percent Yields
20:42
Limiting Reactants, Percent Yields
20:43
Example 8: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂ and 3.1 Grams of N₂
22:25
Percent Yield
25:30
Example 9: How Many Grams of The Excess Reactant Remains?
26:37
Summary
29:34
Sample Problem 1: How Many Grams of Carbon Are In 2.2 Kilograms of Carbon Dioxide?
30:47
Sample Problem 2: How Many Milligrams of Carbon Dioxide Can Form From 23.1 Kg of CH₄(g)?
33:06
Sample Problem 3: Part 1
36:10
Sample Problem 3: Part 2 - What Amount Of The Excess Reactant Will Remain?
40:53
Stoichiometry II

42m 38s

Intro
0:00
Lesson Overview
0:10
Molarity
1:14
Solute and Solvent
1:15
Molarity
2:01
Molarity Cont'd
2:59
Example 1: How Many Grams of KBr are Needed to Make 350 mL of a 0.67 M KBr Solution?
3:00
Example 2: How Many Moles of KBr are in 350 mL of a 0.67 M KBr Solution?
5:44
Example 3: What Volume of a 0.67 M KBr Solution Contains 250 mg of KBr?
7:46
Dilutions
10:01
Dilution: M₁V₂=M₁V₂
10:02
Example 5: Explain How to Make 250 mL of a 0.67 M KBr Solution Starting From a 1.2M Stock Solution
12:04
Stoichiometry and Double-Displacement Precipitation Reactions
14:41
Example 6: How Many grams of PbCl₂ Can Form From 250 mL of 0.32 M NaCl?
15:38
Stoichiometry and Double-Displacement Precipitation Reactions
18:05
Example 7: How Many grams of PbCl₂ Can Form When 250 mL of 0.32 M NaCl and 150 mL of 0.45 Pb(NO₃)₂ Mix?
18:06
Stoichiometry and Neutralization Reactions
21:01
Example 8: How Many Grams of NaOh are Required to Neutralize 4.5 Grams of HCl?
21:02
Stoichiometry and Neutralization Reactions
23:03
Example 9: How Many mL of 0.45 M NaOH are Required to Neutralize 250 mL of 0.89 M HCl?
23:04
Stoichiometry and Acid-Base Standardization
25:28
Introduction to Titration & Standardization
25:30
Acid-Base Titration
26:12
The Analyte & Titrant
26:24
The Experimental Setup
26:49
The Experimental Setup
26:50
Stoichiometry and Acid-Base Standardization
28:38
Example 9: Determine the Concentration of the Analyte
28:39
Summary
32:46
Sample Problem 1: Stoichiometry & Neutralization
35:24
Sample Problem 2: Stoichiometry
37:50
V. Thermochemistry
Energy & Chemical Reactions

55m 28s

Intro
0:00
Lesson Overview
0:14
Introduction
1:22
Recall: Chemistry
1:23
Energy Can Be Expressed In Different Units
1:57
The First Law of Thermodynamics
2:43
Internal Energy
2:44
The First Law of Thermodynamics Cont'd
6:14
Ways to Transfer Internal Energy
6:15
Work Energy
8:13
Heat Energy
8:34
∆U = q + w
8:44
Calculating ∆U, Q, and W
8:58
Changes In Both Volume and Temperature of a System
8:59
Calculating ∆U, Q, and W Cont'd
11:01
The Work Equation
11:02
Example 1: Calculate ∆U For The Burning Fuel
11:45
Calculating ∆U, Q, and W Cont'd
14:09
The Heat Equation
14:10
Calculating ∆U, Q, and W Cont'd
16:03
Example 2: Calculate The Final Temperature
16:04
Constant-Volume Calorimetry
18:05
Bomb Calorimeter
18:06
The Effect of Constant Volume On The Equation For Internal Energy
22:11
Example 3: Calculate ∆U
23:12
Constant-Pressure Conditions
26:05
Constant-Pressure Conditions
26:06
Calculating Enthalpy: Phase Changes
27:29
Melting, Vaporization, and Sublimation
27:30
Freezing, Condensation and Deposition
28:25
Enthalpy Values For Phase Changes
28:40
Example 4: How Much Energy In The Form of heat is Required to Melt 1.36 Grams of Ice?
29:40
Calculating Enthalpy: Heats of Reaction
31:22
Example 5: Calculate The Heat In kJ Associated With The Complete Reaction of 155 g NH₃
31:23
Using Standard Enthalpies of Formation
33:53
Standard Enthalpies of Formation
33:54
Using Standard Enthalpies of Formation
36:12
Example 6: Calculate The Standard Enthalpies of Formation For The Following Reaction
36:13
Enthalpy From a Series of Reactions
39:58
Hess's Law
39:59
Coffee-Cup Calorimetry
42:43
Coffee-Cup Calorimetry
42:44
Example 7: Calculate ∆H° of Reaction
45:10
Summary
47:12
Sample Problem 1
48:58
Sample Problem 2
51:24
VI. Quantum Theory of Atoms
Structure of Atoms

42m 33s

Intro
0:00
Lesson Overview
0:07
Introduction
1:01
Rutherford's Gold Foil Experiment
1:02
Electromagnetic Radiation
2:31
Radiation
2:32
Three Parameters: Energy, Frequency, and Wavelength
2:52
Electromagnetic Radiation
5:18
The Electromagnetic Spectrum
5:19
Atomic Spectroscopy and The Bohr Model
7:46
Wavelengths of Light
7:47
Atomic Spectroscopy Cont'd
9:45
The Bohr Model
9:46
Atomic Spectroscopy Cont'd
12:21
The Balmer Series
12:22
Rydberg Equation For Predicting The Wavelengths of Light
13:04
The Wave Nature of Matter
15:11
The Wave Nature of Matter
15:12
The Wave Nature of Matter
19:10
New School of Thought
19:11
Einstein: Energy
19:49
Hertz and Planck: Photoelectric Effect
20:16
de Broglie: Wavelength of a Moving Particle
21:14
Quantum Mechanics and The Atom
22:15
Heisenberg: Uncertainty Principle
22:16
Schrodinger: Wavefunctions
23:08
Quantum Mechanics and The Atom
24:02
Principle Quantum Number
24:03
Angular Momentum Quantum Number
25:06
Magnetic Quantum Number
26:27
Spin Quantum Number
28:42
The Shapes of Atomic Orbitals
29:15
Radial Wave Function
29:16
Probability Distribution Function
32:08
The Shapes of Atomic Orbitals
34:02
3-Dimensional Space of Wavefunctions
34:03
Summary
35:57
Sample Problem 1
37:07
Sample Problem 2
40:23
VII. Electron Configurations and Periodicity
Periodic Trends

38m 50s

Intro
0:00
Lesson Overview
0:09
Introduction
0:36
Electron Configuration of Atoms
1:33
Electron Configuration & Atom's Electrons
1:34
Electron Configuration Format
1:56
Electron Configuration of Atoms Cont'd
3:01
Aufbau Principle
3:02
Electron Configuration of Atoms Cont'd
6:53
Electron Configuration Format 1: Li, O, and Cl
6:56
Electron Configuration Format 2: Li, O, and Cl
9:11
Electron Configuration of Atoms Cont'd
12:48
Orbital Box Diagrams
12:49
Pauli Exclusion Principle
13:11
Hund's Rule
13:36
Electron Configuration of Atoms Cont'd
17:35
Exceptions to The Aufbau Principle: Cr
17:36
Exceptions to The Aufbau Principle: Cu
18:15
Electron Configuration of Atoms Cont'd
20:22
Electron Configuration of Monatomic Ions: Al
20:23
Electron Configuration of Monatomic Ions: Al³⁺
20:46
Electron Configuration of Monatomic Ions: Cl
21:57
Electron Configuration of Monatomic Ions: Cl¹⁻
22:09
Electron Configuration Cont'd
24:31
Paramagnetism
24:32
Diamagnetism
25:00
Atomic Radii
26:08
Atomic Radii
26:09
In a Column of the Periodic Table
26:25
In a Row of the Periodic Table
26:46
Ionic Radii
27:30
Ionic Radii
27:31
Anions
27:42
Cations
27:57
Isoelectronic Species
28:12
Ionization Energy
29:00
Ionization Energy
29:01
Electron Affinity
31:37
Electron Affinity
31:37
Summary
33:43
Sample Problem 1: Ground State Configuration and Orbital Box Diagram
34:21
Fe
34:48
P
35:32
Sample Problem 2
36:38
Which Has The Larger Ionization Energy: Na or Li?
36:39
Which Has The Larger Atomic Size: O or N ?
37:23
Which Has The Larger Atomic Size: O²⁻ or N³⁻ ?
38:00
VIII. Molecular Geometry & Bonding Theory
Bonding & Molecular Structure

52m 39s

Intro
0:00
Lesson Overview
0:08
Introduction
1:10
Types of Chemical Bonds
1:53
Ionic Bond
1:54
Molecular Bond
2:42
Electronegativity and Bond Polarity
3:26
Electronegativity (EN)
3:27
Periodic Trend
4:36
Electronegativity and Bond Polarity Cont'd
6:04
Bond Polarity: Polar Covalent Bond
6:05
Bond Polarity: Nonpolar Covalent Bond
8:53
Lewis Electron Dot Structure of Atoms
9:48
Lewis Electron Dot Structure of Atoms
9:49
Lewis Structures of Polyatomic Species
12:51
Single Bonds
12:52
Double Bonds
13:28
Nonbonding Electrons
13:59
Lewis Structures of Polyatomic Species Cont'd
14:45
Drawing Lewis Structures: Step 1
14:48
Drawing Lewis Structures: Step 2
15:16
Drawing Lewis Structures: Step 3
15:52
Drawing Lewis Structures: Step 4
17:31
Drawing Lewis Structures: Step 5
19:08
Drawing Lewis Structure Example: Carbonate
19:33
Resonance and Formal Charges (FC)
24:06
Resonance Structures
24:07
Formal Charge
25:20
Resonance and Formal Charges Cont'd
27:46
More On Formal Charge
27:47
Resonance and Formal Charges Cont'd
28:21
Good Resonance Structures
28:22
VSEPR Theory
31:08
VSEPR Theory Continue
31:09
VSEPR Theory Cont'd
32:53
VSEPR Geometries
32:54
Steric Number
33:04
Basic Geometry
33:50
Molecular Geometry
35:50
Molecular Polarity
37:51
Steps In Determining Molecular Polarity
37:52
Example 1: Polar
38:47
Example 2: Nonpolar
39:10
Example 3: Polar
39:36
Example 4: Polar
40:08
Bond Properties: Order, Length, and Energy
40:38
Bond Order
40:39
Bond Length
41:21
Bond Energy
41:55
Summary
43:09
Sample Problem 1
43:42
XeO₃
44:03
I₃⁻
47:02
SF₅
49:16
Advanced Bonding Theories

1h 11m 41s

Intro
0:00
Lesson Overview
0:09
Introduction
0:38
Valence Bond Theory
3:07
Valence Bond Theory
3:08
spᶟ Hybridized Carbon Atom
4:19
Valence Bond Theory Cont'd
6:24
spᶟ Hybridized
6:25
Hybrid Orbitals For Water
7:26
Valence Bond Theory Cont'd (spᶟ)
11:53
Example 1: NH₃
11:54
Valence Bond Theory Cont'd (sp²)
14:48
sp² Hybridization
14:49
Example 2: BF₃
16:44
Valence Bond Theory Cont'd (sp)
22:44
sp Hybridization
22:46
Example 3: HCN
23:38
Valence Bond Theory Cont'd (sp³d and sp³d²)
27:36
Valence Bond Theory: sp³d and sp³d²
27:37
Molecular Orbital Theory
29:10
Valence Bond Theory Doesn't Always Account For a Molecule's Magnetic Behavior
29:11
Molecular Orbital Theory Cont'd
30:37
Molecular Orbital Theory
30:38
Wavefunctions
31:04
How s-orbitals Can Interact
32:23
Bonding Nature of p-orbitals: Head-on
35:34
Bonding Nature of p-orbitals: Parallel
39:04
Interaction Between s and p-orbital
40:45
Molecular Orbital Diagram For Homonuclear Diatomics: H₂
42:21
Molecular Orbital Diagram For Homonuclear Diatomics: He₂
45:23
Molecular Orbital Diagram For Homonuclear Diatomic: Li₂
46:39
Molecular Orbital Diagram For Homonuclear Diatomic: Li₂⁺
47:42
Molecular Orbital Diagram For Homonuclear Diatomic: B₂
48:57
Molecular Orbital Diagram For Homonuclear Diatomic: N₂
54:04
Molecular Orbital Diagram: Molecular Oxygen
55:57
Molecular Orbital Diagram For Heteronuclear Diatomics: Hydrochloric Acid
1:02:16
Sample Problem 1: Determine the Atomic Hybridization
1:07:20
XeO₃
1:07:21
SF₆
1:07:49
I₃⁻
1:08:20
Sample Problem 2
1:09:04
IX. Gases, Solids, & Liquids
Gases

35m 6s

Intro
0:00
Lesson Overview
0:07
The Kinetic Molecular Theory of Gases
1:23
The Kinetic Molecular Theory of Gases
1:24
Parameters To Characterize Gases
3:35
Parameters To Characterize Gases: Pressure
3:37
Interpreting Pressure On a Particulate Level
4:43
Parameters Cont'd
6:08
Units For Expressing Pressure: Psi, Pascal
6:19
Units For Expressing Pressure: mm Hg
6:42
Units For Expressing Pressure: atm
6:58
Units For Expressing Pressure: torr
7:24
Parameters Cont'd
8:09
Parameters To Characterize Gases: Volume
8:10
Common Units of Volume
9:00
Parameters Cont'd
9:11
Parameters To Characterize Gases: Temperature
9:12
Particulate Level
9:36
Parameters To Characterize Gases: Moles
10:24
The Simple Gas Laws
10:43
Gas Laws Are Only Valid For…
10:44
Charles' Law
11:24
The Simple Gas Laws
13:13
Boyle's Law
13:14
The Simple Gas Laws
15:28
Gay-Lussac's Law
15:29
The Simple Gas Laws
17:11
Avogadro's Law
17:12
The Ideal Gas Law
18:43
The Ideal Gas Law: PV = nRT
18:44
Applications of the Ideal Gas Law
20:12
Standard Temperature and Pressure for Gases
20:13
Applications of the Ideal Gas Law
21:43
Ideal Gas Law & Gas Density
21:44
Gas Pressures and Partial Pressures
23:18
Dalton's Law of Partial Pressures
23:19
Gas Stoichiometry
24:15
Stoichiometry Problems Involving Gases
24:16
Using The Ideal Gas Law to Get to Moles
25:16
Using Molar Volume to Get to Moles
25:39
Gas Stoichiometry Cont'd
26:03
Example 1: How Many Liters of O₂ at STP are Needed to Form 10.5 g of Water Vapor?
26:04
Summary
28:33
Sample Problem 1: Calculate the Molar Mass of the Gas
29:28
Sample Problem 2: What Mass of Ag₂O is Required to Form 3888 mL of O₂ Gas When Measured at 734 mm Hg and 25°C?
31:59
Intermolecular Forces & Liquids

33m 47s

Intro
0:00
Lesson Overview
0:10
Introduction
0:46
Intermolecular Forces (IMF)
0:47
Intermolecular Forces of Polar Molecules
1:32
Ion-dipole Forces
1:33
Example: Salt Dissolved in Water
1:50
Coulomb's Law & the Force of Attraction Between Ions and/or Dipoles
3:06
IMF of Polar Molecules cont'd
4:36
Enthalpy of Solvation or Enthalpy of Hydration
4:37
IMF of Polar Molecules cont'd
6:01
Dipole-dipole Forces
6:02
IMF of Polar Molecules cont'd
7:22
Hydrogen Bonding
7:23
Example: Hydrogen Bonding of Water
8:06
IMF of Nonpolar Molecules
9:37
Dipole-induced Dipole Attraction
9:38
IMF of Nonpolar Molecules cont'd
11:34
Induced Dipole Attraction, London Dispersion Forces, or Vand der Waals Forces
11:35
Polarizability
13:46
IMF of Nonpolar Molecules cont'd
14:26
Intermolecular Forces (IMF) and Polarizability
14:31
Properties of Liquids
16:48
Standard Molar Enthalpy of Vaporization
16:49
Trends in Boiling Points of Representative Liquids: H₂O vs. H₂S
17:43
Properties of Liquids cont'd
18:36
Aliphatic Hydrocarbons
18:37
Branched Hydrocarbons
20:52
Properties of Liquids cont'd
22:10
Vapor Pressure
22:11
The Clausius-Clapeyron Equation
24:30
Properties of Liquids cont'd
25:52
Boiling Point
25:53
Properties of Liquids cont'd
27:07
Surface Tension
27:08
Viscosity
28:06
Summary
29:04
Sample Problem 1: Determine Which of the Following Liquids Will Have the Lower Vapor Pressure
30:21
Sample Problem 2: Determine Which of the Following Liquids Will Have the Largest Standard Molar Enthalpy of Vaporization
31:37
The Chemistry of Solids

25m 13s

Intro
0:00
Lesson Overview
0:07
Introduction
0:46
General Characteristics
0:47
Particulate-level Drawing
1:09
The Basic Structure of Solids: Crystal Lattices
1:37
The Unit Cell Defined
1:38
Primitive Cubic
2:50
Crystal Lattices cont'd
3:58
Body-centered Cubic
3:59
Face-centered Cubic
5:02
Lattice Enthalpy and Trends
6:27
Introduction to Lattice Enthalpy
6:28
Equation to Calculate Lattice Enthalpy
7:21
Different Types of Crystalline Solids
9:35
Molecular Solids
9:36
Network Solids
10:25
Phase Changes Involving Solids
11:03
Melting & Thermodynamic Value
11:04
Freezing & Thermodynamic Value
11:49
Phase Changes cont'd
12:40
Sublimation & Thermodynamic Value
12:41
Depositions & Thermodynamic Value
13:13
Phase Diagrams
13:40
Introduction to Phase Diagrams
13:41
Phase Diagram of H₂O: Melting Point
14:12
Phase Diagram of H₂O: Normal Boiling Point
14:50
Phase Diagram of H₂O: Sublimation Point
15:02
Phase Diagram of H₂O: Point C ( Supercritical Point)
15:32
Phase Diagrams cont'd
16:31
Phase Diagram of Dry Ice
16:32
Summary
18:15
Sample Problem 1, Part A: Of the Group I Fluorides, Which Should Have the Highest Lattice Enthalpy?
19:01
Sample Problem 1, Part B: Of the Lithium Halides, Which Should Have the Lowest Lattice Enthalpy?
19:54
Sample Problem 2: How Many Joules of Energy is Required to Melt 546 mg of Ice at Standard Pressure?
20:55
Sample Problem 3: Phase Diagram of Helium
22:42
X. Solutions, Rates of Reaction, & Equilibrium
Solutions & Their Behavior

38m 6s

Intro
0:00
Lesson Overview
0:10
Units of Concentration
1:40
Molarity
1:41
Molality
3:30
Weight Percent
4:26
ppm
5:16
Like Dissolves Like
6:28
Like Dissolves Like
6:29
Factors Affecting Solubility
9:35
The Effect of Pressure: Henry's Law
9:36
The Effect of Temperature on Gas Solubility
12:16
The Effect of Temperature on Solid Solubility
14:28
Colligative Properties
16:48
Colligative Properties
16:49
Changes in Vapor Pressure: Raoult's Law
17:19
Colligative Properties cont'd
19:53
Boiling Point Elevation and Freezing Point Depression
19:54
Colligative Properties cont'd
26:13
Definition of Osmosis
26:14
Osmotic Pressure Example
27:11
Summary
31:11
Sample Problem 1: Calculating Vapor Pressure
32:53
Sample Problem 2: Calculating Molality
36:29
Chemical Kinetics

37m 45s

Intro
0:00
Lesson Overview
0:06
Introduction
1:09
Chemical Kinetics and the Rate of a Reaction
1:10
Factors Influencing Rate
1:19
Introduction cont'd
2:27
How a Reaction Progresses Through Time
2:28
Rate of Change Equation
6:02
Rate Laws
7:06
Definition of Rate Laws
7:07
General Form of Rate Laws
7:37
Rate Laws cont'd
11:07
Rate Orders With Respect to Reactant and Concentration
11:08
Methods of Initial Rates
13:38
Methods of Initial Rates
13:39
Integrated Rate Laws
17:57
Integrated Rate Laws
17:58
Graphically Determine the Rate Constant k
18:52
Reaction Mechanisms
21:05
Step 1: Reversible
21:18
Step 2: Rate-limiting Step
21:44
Rate Law for the Reaction
23:28
Reaction Rates and Temperatures
26:16
Reaction Rates and Temperatures
26:17
The Arrhenius Equation
29:06
Catalysis
30:31
Catalyst
30:32
Summary
32:02
Sample Problem 1: Calculate the Rate Constant and the Time Required for the Reaction to be Completed
32:54
Sample Problem 2: Calculate the Energy of Activation and the Order of the Reaction
35:24
Principles of Chemical Equilibrium

34m 9s

Intro
0:00
Lesson Overview
0:08
Introduction
1:02
The Equilibrium Constant
3:08
The Equilibrium Constant
3:09
The Equilibrium Constant cont'd
5:50
The Equilibrium Concentration and Constant for Solutions
5:51
The Equilibrium Partial Pressure and Constant for Gases
7:01
Relationship of Kc and Kp
7:30
Heterogeneous Equilibria
8:23
Heterogeneous Equilibria
8:24
Manipulating K
9:57
First Way of Manipulating K
9:58
Second Way of Manipulating K
11:48
Manipulating K cont'd
12:31
Third Way of Manipulating K
12:32
The Reaction Quotient Q
14:42
The Reaction Quotient Q
14:43
Q > K
16:16
Q < K
16:30
Q = K
16:43
Le Chatlier's Principle
17:32
Restoring Equilibrium When It is Disturbed
17:33
Disturbing a Chemical System at Equilibrium
18:35
Problem-Solving with ICE Tables
19:05
Determining a Reaction's Equilibrium Constant With ICE Table
19:06
Problem-Solving with ICE Tables cont'd
21:03
Example 1: Calculate O₂(g) at Equilibrium
21:04
Problem-Solving with ICE Tables cont'd
22:53
Example 2: Calculate the Equilibrium Constant
22:54
Summary
25:24
Sample Problem 1: Calculate the Equilibrium Constant
27:59
Sample Problem 2: Calculate The Equilibrium Concentration
30:30
XI. Acids & Bases Chemistry
Acid-Base Chemistry

43m 44s

Intro
0:00
Lesson Overview
0:06
Introduction
0:55
Bronsted-Lowry Acid & Bronsted -Lowry Base
0:56
Water is an Amphiprotic Molecule
2:40
Water Reacting With Itself
2:58
Introduction cont'd
4:04
Strong Acids
4:05
Strong Bases
5:18
Introduction cont'd
6:16
Weak Acids and Bases
6:17
Quantifying Acid-Base Strength
7:35
The pH Scale
7:36
Quantifying Acid-Base Strength cont'd
9:55
The Acid-ionization Constant Ka and pKa
9:56
Quantifying Acid-Base Strength cont'd
12:13
Example: Calculate the pH of a 1.2M Solution of Acetic Acid
12:14
Quantifying Acid-Base Strength
15:06
Calculating the pH of Weak Base Solutions
15:07
Writing Out Acid-Base Equilibria
17:45
Writing Out Acid-Base Equilibria
17:46
Writing Out Acid-Base Equilibria cont'd
19:47
Consider the Following Equilibrium
19:48
Conjugate Base and Conjugate Acid
21:18
Salts Solutions
22:00
Salts That Produce Acidic Aqueous Solutions
22:01
Salts That Produce Basic Aqueous Solutions
23:15
Neutral Salt Solutions
24:05
Diprotic and Polyprotic Acids
24:44
Example: Calculate the pH of a 1.2 M Solution of H₂SO₃
24:43
Diprotic and Polyprotic Acids cont'd
27:18
Calculate the pH of a 1.2 M Solution of Na₂SO₃
27:19
Lewis Acids and Bases
29:13
Lewis Acids
29:14
Lewis Bases
30:10
Example: Lewis Acids and Bases
31:04
Molecular Structure and Acidity
32:03
The Effect of Charge
32:04
Within a Period/Row
33:07
Molecular Structure and Acidity cont'd
34:17
Within a Group/Column
34:18
Oxoacids
35:58
Molecular Structure and Acidity cont'd
37:54
Carboxylic Acids
37:55
Hydrated Metal Cations
39:23
Summary
40:39
Sample Problem 1: Calculate the pH of a 1.2 M Solution of NH₃
41:20
Sample Problem 2: Predict If The Following Slat Solutions are Acidic, Basic, or Neutral
42:37
Applications of Aqueous Equilibria

55m 26s

Intro
0:00
Lesson Overview
0:07
Calculating pH of an Acid-Base Mixture
0:53
Equilibria Involving Direct Reaction With Water
0:54
When a Bronsted-Lowry Acid and Base React
1:12
After Neutralization Occurs
2:05
Calculating pH of an Acid-Base Mixture cont'd
2:51
Example: Calculating pH of an Acid-Base Mixture, Step 1 - Neutralization
2:52
Example: Calculating pH of an Acid-Base Mixture, Step 2 - React With H₂O
5:24
Buffers
7:45
Introduction to Buffers
7:46
When Acid is Added to a Buffer
8:50
When Base is Added to a Buffer
9:54
Buffers cont'd
10:41
Calculating the pH
10:42
Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer
14:03
Buffers cont'd
14:10
Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer: Step 1 -Neutralization
14:11
Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer: Step 2- ICE Table
15:22
Buffer Preparation and Capacity
16:38
Example: Calculating the pH of a Buffer Solution
16:42
Effective Buffer
18:40
Acid-Base Titrations
19:33
Acid-Base Titrations: Basic Setup
19:34
Acid-Base Titrations cont'd
22:12
Example: Calculate the pH at the Equivalence Point When 0.250 L of 0.0350 M HClO is Titrated With 1.00 M KOH
22:13
Acid-Base Titrations cont'd
25:38
Titration Curve
25:39
Solubility Equilibria
33:07
Solubility of Salts
33:08
Solubility Product Constant: Ksp
34:14
Solubility Equilibria cont'd
34:58
Q < Ksp
34:59
Q > Ksp
35:34
Solubility Equilibria cont'd
36:03
Common-ion Effect
36:04
Example: Calculate the Solubility of PbCl₂ in 0.55 M NaCl
36:30
Solubility Equilibria cont'd
39:02
When a Solid Salt Contains the Conjugate of a Weak Acid
39:03
Temperature and Solubility
40:41
Complexation Equilibria
41:10
Complex Ion
41:11
Complex Ion Formation Constant: Kf
42:26
Summary
43:35
Sample Problem 1: Question
44:23
Sample Problem 1: Part a) Calculate the pH at the Beginning of the Titration
45:48
Sample Problem 1: Part b) Calculate the pH at the Midpoint or Half-way Point
48:04
Sample Problem 1: Part c) Calculate the pH at the Equivalence Point
48:32
Sample Problem 1: Part d) Calculate the pH After 27.50 mL of the Acid was Added
53:00
XII. Thermodynamics & Electrochemistry
Entropy & Free Energy

36m 13s

Intro
0:00
Lesson Overview
0:08
Introduction
0:53
Introduction to Entropy
1:37
Introduction to Entropy
1:38
Entropy and Heat Flow
6:31
Recall Thermodynamics
6:32
Entropy is a State Function
6:54
∆S and Heat Flow
7:28
Entropy and Heat Flow cont'd
8:18
Entropy and Heat Flow: Equations
8:19
Endothermic Processes: ∆S > 0
8:44
The Second Law of Thermodynamics
10:04
Total ∆S = ∆S of System + ∆S of Surrounding
10:05
Nature Favors Processes Where The Amount of Entropy Increases
10:22
The Third Law of Thermodynamics
11:55
The Third Law of Thermodynamics & Zero Entropy
11:56
Problem-Solving involving Entropy
12:36
Endothermic Process and ∆S
12:37
Exothermic Process and ∆S
13:19
Problem-Solving cont'd
13:46
Change in Physical States: From Solid to Liquid to Gas
13:47
Change in Physical States: All Gases
15:02
Problem-Solving cont'd
15:56
Calculating the ∆S for the System, Surrounding, and Total
15:57
Example: Calculating the Total ∆S
16:17
Problem-Solving cont'd
18:36
Problems Involving Standard Molar Entropies of Formation
18:37
Introduction to Gibb's Free Energy
20:09
Definition of Free Energy ∆G
20:10
Spontaneous Process and ∆G
20:19
Gibb's Free Energy cont'd
22:28
Standard Molar Free Energies of Formation
22:29
The Free Energies of Formation are Zero for All Compounds in the Standard State
22:42
Gibb's Free Energy cont'd
23:31
∆G° of the System = ∆H° of the System - T∆S° of the System
23:32
Predicting Spontaneous Reaction Based on the Sign of ∆G° of the System
24:24
Gibb's Free Energy cont'd
26:32
Effect of reactant and Product Concentration on the Sign of Free Energy
26:33
∆G° of Reaction = -RT ln K
27:18
Summary
28:12
Sample Problem 1: Calculate ∆S° of Reaction
28:48
Sample Problem 2: Calculate the Temperature at Which the Reaction Becomes Spontaneous
31:18
Sample Problem 3: Calculate Kp
33:47
Electrochemistry

41m 16s

Intro
0:00
Lesson Overview
0:08
Introduction
0:53
Redox Reactions
1:42
Oxidation-Reduction Reaction Overview
1:43
Redox Reactions cont'd
2:37
Which Reactant is Being Oxidized and Which is Being Reduced?
2:38
Redox Reactions cont'd
6:34
Balance Redox Reaction In Neutral Solutions
6:35
Redox Reactions cont'd
10:37
Balance Redox Reaction In Acidic and Basic Solutions: Step 1
10:38
Balance Redox Reaction In Acidic and Basic Solutions: Step 2 - Balance Each Half-Reaction
11:22
Redox Reactions cont'd
12:19
Balance Redox Reaction In Acidic and Basic Solutions: Step 2 - Balance Hydrogen
12:20
Redox Reactions cont'd
14:30
Balance Redox Reaction In Acidic and Basic Solutions: Step 3
14:34
Balance Redox Reaction In Acidic and Basic Solutions: Step 4
15:38
Voltaic Cells
17:01
Voltaic Cell or Galvanic Cell
17:02
Cell Notation
22:03
Electrochemical Potentials
25:22
Electrochemical Potentials
25:23
Electrochemical Potentials cont'd
26:07
Table of Standard Reduction Potentials
26:08
The Nernst Equation
30:41
The Nernst Equation
30:42
It Can Be Shown That At Equilibrium E =0.00
32:15
Gibb's Free Energy and Electrochemistry
32:46
Gibbs Free Energy is Relatively Small if the Potential is Relatively High
32:47
When E° is Very Large
33:39
Charge, Current and Time
33:56
A Battery Has Three Main Parameters
33:57
A Simple Equation Relates All of These Parameters
34:09
Summary
34:50
Sample Problem 1: Redox Reaction
35:26
Sample Problem 2: Battery
38:00
XIII. Transition Elements & Coordination Compounds
The Chemistry of The Transition Metals

39m 3s

Intro
0:00
Lesson Overview
0:11
Coordination Compounds
1:20
Coordination Compounds
1:21
Nomenclature of Coordination Compounds
2:48
Rule 1
3:01
Rule 2
3:12
Rule 3
4:07
Nomenclature cont'd
4:58
Rule 4
4:59
Rule 5
5:13
Rule 6
5:35
Rule 7
6:19
Rule 8
6:46
Nomenclature cont'd
7:39
Rule 9
7:40
Rule 10
7:45
Rule 11
8:00
Nomenclature of Coordination Compounds: NH₄[PtCl₃NH₃]
8:11
Nomenclature of Coordination Compounds: [Cr(NH₃)₄(OH)₂]Br
9:31
Structures of Coordination Compounds
10:54
Coordination Number or Steric Number
10:55
Commonly Observed Coordination Numbers and Geometries: 4
11:14
Commonly Observed Coordination Numbers and Geometries: 6
12:00
Isomers of Coordination Compounds
13:13
Isomers of Coordination Compounds
13:14
Geometrical Isomers of CN = 6 Include: ML₄L₂'
13:30
Geometrical Isomers of CN = 6 Include: ML₃L₃'
15:07
Isomers cont'd
17:00
Structural Isomers Overview
17:01
Structural Isomers: Ionization
18:06
Structural Isomers: Hydrate
19:25
Structural Isomers: Linkage
20:11
Structural Isomers: Coordination Isomers
21:05
Electronic Structure
22:25
Crystal Field Theory
22:26
Octahedral and Tetrahedral Field
22:54
Electronic Structure cont'd
25:43
Vanadium (II) Ion in an Octahedral Field
25:44
Chromium(III) Ion in an Octahedral Field
26:37
Electronic Structure cont'd
28:47
Strong-Field Ligands and Weak-Field Ligands
28:48
Implications of Electronic Structure
30:08
Compare the Magnetic Properties of: [Fe(OH₂)₆]²⁺ vs. [Fe(CN)₆]⁴⁻
30:09
Discussion on Color
31:57
Summary
34:41
Sample Problem 1: Name the Following Compound [Fe(OH)(OH₂)₅]Cl₂
35:08
Sample Problem 1: Name the Following Compound [Co(NH₃)₃(OH₂)₃]₂(SO₄)₃
36:24
Sample Problem 2: Change in Magnetic Properties
37:30
XIV. Nuclear Chemistry
Nuclear Chemistry

16m 39s

Intro
0:00
Lesson Overview
0:06
Introduction
0:40
Introduction to Nuclear Reactions
0:41
Types of Radioactive Decay
2:10
Alpha Decay
2:11
Beta Decay
3:27
Gamma Decay
4:40
Other Types of Particles of Varying Energy
5:40
Nuclear Equations
6:47
Nuclear Equations
6:48
Nuclear Decay
9:28
Nuclear Decay and the First-Order Kinetics
9:29
Summary
11:31
Sample Problem 1: Complete the Following Nuclear Equations
12:13
Sample Problem 2: How Old is the Rock?
14:21
Loading...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of General Chemistry
  • Discussion

  • Study Guides

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books & Services

Lecture Comments (25)

0 answers

Post by Abdullah Alqahtani on June 24, 2017

YOU are very bad lecturer

1 answer

Last reply by: Professor Franklin Ow
Wed Jul 20, 2016 11:02 PM

Post by Magic Fu on July 20, 2016

In 15:44. Why Styrofoam has smaller heat capacities, and still need lot of heat? Shouldn't it be the opposite instead?

0 answers

Post by Peter Ke on September 16, 2015

At 45:00, why did you multiply 1.00 g/mL x 100mL because 1 Liter = 1000 mL so I was thinking that you should multiply it by 1000 and get 1000 grams of solution?

Also, when you got -339 kJ shouldn't it be 339 J without the kilo? Because if you were to set up the equation -100g(4.18 J/gC)(24.21 C - 23.40 C), don't the grams and celsius cancel each other out and you would be left with J?

Please correct me if I'm wrong.

0 answers

Post by Peter Ke on September 15, 2015

At 23:12. I understand how you got Qreaction = - Qbomb, because you subtract Qbomb on both side but I don't understand how you got -Cbomb(DeltaT). Please explain. Thx!

0 answers

Post by Peter Ke on September 15, 2015

At 16:04, can you show me how you got 30.7, cause the answer I got was like 495.1 which totally made no sense. So, please show me how you got 30.7. Thx

0 answers

Post by Peter Ke on September 15, 2015

At 8:59, why 488 is negative? Please explain it in a easier way. Thx

1 answer

Last reply by: Professor Franklin Ow
Mon Jun 1, 2015 5:41 PM

Post by Ivan de La Grange on May 31, 2015

For Sample problem 1, shouldn't it be .0327 grams?

2 answers

Last reply by: Denise Bermudez
Sun Mar 8, 2015 10:04 AM

Post by Denise Bermudez on March 2, 2015

hi professor
can you please explain or re-do how you got the answer of 30.7 degrees celsius. I set the equation correctly but i cant figure out how you got your answer with it. The part that confused me was the *4.18 J/ g celsius.

thanks!

0 answers

Post by Saadman Elman on January 31, 2015

Please look at 13:00 - 14:00 min, You did first part of the problem but you didn't do the second part of the problem. The question was asked Calculate delta U for burning of fuel if the system gives off 875 joule. When all is said and done, the answer that you will end up is -998.6  Joule which is the answer for delta you. You skipped that part. Please verify my answer.

0 answers

Post by Saadman Elman on January 9, 2015

Correct me if i am wrong. You said Insulator such as Styrofoam has small heat capacities. Since they retain the heat so i think they have higher heat capacities.

1 answer

Last reply by: Ryan Lozon
Sun Dec 21, 2014 11:34 AM

Post by Ryan Lozon on December 21, 2014

You mean insulators such as styrofoam having larger heat capacities than metals, right? Because they require more energy to raise one gram by 1 Celsius?

3 answers

Last reply by: Professor Franklin Ow
Mon Aug 4, 2014 11:31 PM

Post by brandon joyner on July 1, 2014

Why does 725 have a negative sign?

2 answers

Last reply by: Peidong, He
Sun Mar 16, 2014 11:53 AM

Post by Peidong, He on March 15, 2014

should insulators, such as Styrofoam, have a higher specific heat capacity?  

0 answers

Post by KyungYeop Kim on July 9, 2013

Regarding a specific situation in which a nonspontaneous reaction under standard conditions becomes spontaneous at lower temperature, how can I describe this phenomenon in relation to enthalpy, entropy, and free energy? and how can I explain it in terms of temperature change? I've succeeded so far in determining that ΔG>0 since it's nonspontaneous under standard conditions, but what about ΔH and ΔS?

Given the equation = ΔG = ΔH*TΔS; I think the fact that temperature(T) can go either from positive to negative or negative to negative seems to confuse me. Are we assuming, in saying lowering temperature, that the T goes from negative(-) to negative(-)?

I know it's a complex problem, and I apologize if I'm asking too much, but I would like to know what the answer is and why. Thank you always!

0 answers

Post by KyungYeop Kim on July 9, 2013

I have a question about effective nuclear charge. As you go down a group in the periodic table, why is it that the effective nuclear charge decreases? from what I know, is it true that as the attraction decreases down the group, it somehow counterbalances the increase in nuclear charge? I'm confused.

Energy & Chemical Reactions

  • State functions are path-independent.
  • The first law of thermodynamics states that energy is conserved during a chemical reaction.
  • Energy can be transferred through heat and work.
  • Calorimetry problems involve either constant-volume or constant-pressure scenarios.
  • Enthalpy is the amount of heat transferred under constant pressure.

Energy & Chemical Reactions

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
  • Lesson Overview 0:14
  • Introduction 1:22
    • Recall: Chemistry
    • Energy Can Be Expressed In Different Units
  • The First Law of Thermodynamics 2:43
    • Internal Energy
  • The First Law of Thermodynamics Cont'd 6:14
    • Ways to Transfer Internal Energy
    • Work Energy
    • Heat Energy
    • ∆U = q + w
  • Calculating ∆U, Q, and W 8:58
    • Changes In Both Volume and Temperature of a System
  • Calculating ∆U, Q, and W Cont'd 11:01
    • The Work Equation
    • Example 1: Calculate ∆U For The Burning Fuel
  • Calculating ∆U, Q, and W Cont'd 14:09
    • The Heat Equation
  • Calculating ∆U, Q, and W Cont'd 16:03
    • Example 2: Calculate The Final Temperature
  • Constant-Volume Calorimetry 18:05
    • Bomb Calorimeter
    • The Effect of Constant Volume On The Equation For Internal Energy
    • Example 3: Calculate ∆U
  • Constant-Pressure Conditions 26:05
    • Constant-Pressure Conditions
  • Calculating Enthalpy: Phase Changes 27:29
    • Melting, Vaporization, and Sublimation
    • Freezing, Condensation and Deposition
    • Enthalpy Values For Phase Changes
    • Example 4: How Much Energy In The Form of heat is Required to Melt 1.36 Grams of Ice?
  • Calculating Enthalpy: Heats of Reaction 31:22
    • Example 5: Calculate The Heat In kJ Associated With The Complete Reaction of 155 g NH₃
  • Using Standard Enthalpies of Formation 33:53
    • Standard Enthalpies of Formation
  • Using Standard Enthalpies of Formation 36:12
    • Example 6: Calculate The Standard Enthalpies of Formation For The Following Reaction
  • Enthalpy From a Series of Reactions 39:58
    • Hess's Law
  • Coffee-Cup Calorimetry 42:43
    • Coffee-Cup Calorimetry
    • Example 7: Calculate ∆H° of Reaction
  • Summary 47:12
  • Sample Problem 1 48:58
  • Sample Problem 2 51:24

Transcription: Energy & Chemical Reactions

Hi, welcome back to Educator.com.0000

Today's lesson from general chemistry is going to be our first discussion on thermodynamics.0002

We are going to be discussing energy and chemical reactions.0010

Let's go ahead and go over the lesson overview.0015

We are going to first start off with a brief introduction.0018

Because this is going to be totally different than all of the previous material we have been seeing.0021

We are then going to discuss the very important first law of thermodynamics.0027

Then the rest of the chapter is highly quantitative.0034

Once we have the introduced the different parameters that are at a focus0037

in thermodynamics, we are going to learn how to calculate each of them.0042

There is going to be different situations when we both have a volume and temperature change.0045

There is also going to be a situation when we have no volume change.0051

Then there is going to be a situation when we have no pressure change.0057

That is what pretty much this chapter is about, studying, introducing these very important physical parameters.0061

Then learning how to calculate them in each of their different situations.0069

We are going to finish up the chapter with a discussion of enthalpy.0074

Followed by our summary and then two sample problems.0078

Let's go ahead and get into the lesson now.0083

Remember we introduced the definition of chemistry way long time ago.0088

We said that chemistry is the study of matter and the changes that it undergoes.0094

We have already discussed the three states of matter.0098

We have already discussed the types of physical and chemical changes.0101

But there is always something that is associated with physical and chemical changes.0105

That is a change in energy.0111

When we talk about energy, energy can be measured and expressed in different units.0115

If you have a utility bill, especially the electricity, that is going to be usually expressed in terms of kilowatt hours.0123

If you are going to look at the side of a cereal box or anything0131

or any type of beverage, some nutritional information, nutritional information tends to use calories.0136

But in the physical sciences, we tend to use what is called the Joule.0142

The Joule is going to be abbreviated capital J.0147

The relationship between all three is the following.0150

One kilowatt hour is 3.6 times 106 joules; one calorie is 4.184 joules0153

Now that we know the units of energy, let's get into some more theory.0163

What we are interested in in general chemistry is what is called internal energy.0171

You have probably taken physics by now.0178

You probably have already learned what is called kinetic energy and potential energy.0181

Where potential energy is just related to position and kinetic energy is the energy related to motion.0187

What internal energy is, internal energy is going to be the sum basically.0197

Internal energy is the sum of a system's potential energy and its kinetic energy.0203

Depending on what textbook you use, internal energy will be either symbolized with capital U or with capital E.0209

We are going to now introduce another term, what is called the state function.0219

Internal energy is an example of a state function; a state function is path independent.0223

In thermodynamics, it is pretty easy to always identify what a state function is.0231

A state function name is going to tend to be capitalized.0235

What do I mean by path independent?0240

Let's go ahead and look at the following reaction.0243

A solid plus B solid goes on to form AB solid.0246

This reaction is the direct combination of A in the solid state reacting with0259

B in the solid state to form the compound AB in the solid state.0265

Let's say that this reaction had associated with it a certain change in energy.0269

This is the most direct route; but I can do this in a stepwise motion.0278

For example, I can first go from A solid to A gas.0283

I can then go from B solid to B gas.0290

Once I have A and B in the gaseous state, I can combine the two.0295

Once in the gaseous state, that can then go on to form AB gas.0300

AB vapor can then condense back down to the solid.0308

What a state function means, what path independent means is the following.0314

If I take the direct route or if I do the stepwise route, in the end0320

the change in energy is going to be the same regardless of the path.0330

Δenergy same regardless of path.0334

Again state functions are path independent.0345

We really don't care about the step by step process that we take.0348

We only really care about the comparison between the initial and the final state.0352

Because of that, we are then often interested in not just internal energy but the change in internal energy.0359

What I am going to be using is the symbol ΔU to represent change in internal energy.0367

When we talk about internal energy, there are essentially two ways to transfer internal energy in and out of a system.0377

When I use the word system, I mean maybe a chemical reaction for example.0386

The two ways to transfer internal energy in and out of a system are through work and through heat.0392

Work is going to be symbolized lowercase w.0401

Heat is going to be symbolized with lowercase q.0404

Because they are lowercase, they are not state functions.0407

Remember from the previous slide that state functions tend to be capitalized.0410

What you want to pay attention to are the signs.0418

If a system performs work, then the work of the system is going to be negative.0424

If a system has work performed on it, we say that the sign is going to be positive.0440

Let's now look at the signs for q.0457

If heat enters a system, we say that the sign of q is going to be positive.0460

If heat leaves a system, we say that the sign of q for the system is going be negative.0473

You have to know these signs because when you are going to be given a0485

word problem, you have to be able to derive the sign from the sentences.0489

A lot of this many of you will recognize this comes immediately from physics.0496

We define work energy as the product of applied force over a certain distance.0512

Many of you will recognize the equation, work is equal to force times distance.0513

Heat energy basically, that is more literal.0516

Heat energy basically is going to be essentially thermal energy.0519

The change in internal energy equals the sum of work and heat energies.0525

The equation for the first law of thermodynamics is ΔU is equal to q plus w.0533

A very common application of the first law of thermodynamics is when the system experiences both a change in volume and in temperature.0542

Now we are going to learn to calculate the previous three parameters; ΔU, q, and w.0553

This is a very traditional type of problem.0564

This involves a piston and a cylinder like you have in your automobile engine.0566

The cylinder and piston are warmed by a flame.0571

The gases in the cylinder expand, doing work on the surroundings by pushing the piston outward.0574

If the system absorbs 559 joules of heat and does 488 joules of work, what is ΔU?0579

Basically we have a piston here inside a cylinder.0587

Here we have a lot of a hot gas inside.0594

This hot gas is going to push against the piston.0599

It is going to apply a force against it.0604

You are told that the system absorbs 559 joules of heat and it does 488 joules of work.0608

You see immediately how important it is to pay attention to the sentences because you have to derive the signs.0615

ΔU is equal to q plus w; let's go ahead and look at q.0621

You are told that the system absorbs 559 joules.0628

That means heat is entering the system; that is going to be +559 joules.0632

In addition, w, you are told that the system is doing, it is performing 488 joules of work.0638

That is going to be a negative sign.0646

When all is said and done, we get an answer of 71 joules for ΔU.0650

It is quite straightforward to use this equation that represents the first law of thermodynamics.0657

Let's now introduce the equations for work and for heat because each of them have their own.0666

Let's first introduce the work equation.0673

Work is equal to -Pexternal times ΔV.0677

In this case, Pexternal, this is going to be a constant external pressure.0682

That is the key; the pressure has to be constant or else you cannot you this equation.0690

Of course ΔV, we immediately recognize as change in volume or Vfinal minus Vinitial.0697

When fuel was burned in the cylinder equipped with a piston, the volume expands0708

from 0.255 liters to 1.45 liters against an external pressure of 1.02 atm.0712

During this process, 875 joules of heat is given off.0719

Calculate ΔU for the burning of the fuel.0723

Work is equal to ?Pexternal times ΔV.0730

You are told in the problem that the external pressure is 1.02 atm; times ΔV.0737

That is going to be Vfinal minus Vinitial or 1.45 liters minus 0.255 liters.0749

When all is said, we are going to get an answer of -1.22.0757

Our units are going to be atmospheres times liters together.0762

We always want to try to get to joules.0768

The conversion factor between liter atmosphere and Joules is the following.0772

One liter atmosphere equals 101.3 Joules.0777

We are just going to use dimensional analysis.0782

-1.22 atmosphere times liter times 101.3 joules over 1 liter atmosphere.0785

That is going to be equal to -123 joules.0793

That is going to be our value for w.0797

Let's see if this sign makes sense.0800

Does it make sense that the sign of work here is going to be negative?0803

We have a piston.0808

The piston is exerting a downward pressure of 1.02 atm.0811

If my system is the...0818

If gas is trapped inside the piston, they have to do work to push against this downward pressure.0822

You are told that the piston is going to expand from 0.255 to 1.45.0828

In order for that increase in volume to occur, the system will be performing work to go against the downward pressure.0834

Yes, the negative sign of w does make sense.0843

The system is performing work; very good.0847

Now that we have introduced the work equation, let's now tackle the heat equation.0852

The heat equation for all you pre-meds is mCaΔT.0858

You will recognize that as MCAT; you will hear q being the MCAT equation.0863

M is just your mass, usually in grams.0869

ΔT is your change in temperature; Tfinal minus Tinitial.0872

C is something new; C is what we call the specific heat capacity.0878

The specific heat capacity is in units of joules per gram degree Celsius.0882

What specific heat capacity is, it is formally defined as the following.0887

It is the energy required to raise the temperature of 1 gram of a compound by exactly 1 degree Celsius.0891

What does the specific heat capacity physically mean?0900

If you think of a metal, metals tend to get very hot very very quickly.0905

They are excellent conductors of thermal energy.0910

It doesn't take much to heat a metal.0914

We say that metals are going to have relatively smaller heat capacities.0918

They don't retain the heat as much.0921

The opposite of metals, remember insulators.0925

Insulators such as Styrofoam are going to have smaller heat capacities.0928

It take a lot of energy to heat up Styrofoam.0933

That is what we know and have experienced.0937

One very important compound of course is going to be liquid water.0943

You should just know its specific heat capacity to be 4.18 joules per gram degree Celsius.0948

That is relatively large, especially compared to metals.0953

If you think about it, yes it does require lots of energy to heat up water.0956

So q is equal to mcΔT.0962

Now let's go ahead and look through an example.0966

You have a 0.055 kilogram sample of water and 27.5 degree Celsius.0969

You are told that it absorbs 725 joules of heat.0975

Calculate its final temperature.0978

q is equal to mcΔT.0982

Let's plug in what we know and see if we can solve for what is remaining.0988

You are told that the water absorbs 725 joules of heat.0994

That means heat is entering the system, which means it is going to be a positive sign.0998

q is equal to +725 joules; that is equal to the mass of the water.1003

Remember we want to use mass in units of grams.1011

That is going to be 55 grams.1014

Times the specific heat capacity of water which is going to be 4.18 joules per gram degree Celsius.1018

Finally times ΔT where ΔT is equal to T2 minus T1.1029

T2 is what we are trying to solve for.1034

The initial temperature is 27.5 degrees Celsius.1038

We are in good shape; it looks like we only have one unknown.1044

That is the final temperature which is what we are trying to solve for anyways.1049

T2 is going to be equal to 30.7 degrees Celsius.1053

Always double check; does it make sense that T2 is greater than T1?1058

Yes, absolutely because you are told that the sample of water is absorbing heat.1064

That heat is going into the system, into the water to warm it up.1068

Yes, our final temperature had better be greater than our initial temperature.1073

Again for thermodynamics, it is always a good habit to double check your work and ask yourself, does it physically make sense?1077

There is different ways of calculating the amount of energy that is involved in a process.1088

One way is using a constant volume process.1096

Now we introduce the word calorimetry.1101

All calorimetry is, it is the determination of the amount of energy transferred in a chemical process.1103

We can do this using a constant volume process.1111

This is what we call bond calorimetry.1115

The energy released from a combustion reaction can be determined experimentally in a bomb calorimeter.1118

Let me go ahead and provide a rough sketch of what a bomb calorimeter is.1125

Basically you have a small cup; inside the cup is going to be your reactants.1131

Here is the key; this cup is going to tend to be made of stainless steel.1142

We know stainless steel to be very very hard and very very rigid.1154

Because it is quite rigid, its volume is not going to change, at least on the macroscopic scale.1158

That is how we guarantee a zero change in volume.1167

Stainless steel maintains volume, maintains a fixed volume during the reaction.1172

This little cup of stainless steel, that contains your reactants.1183

This is essentially hooked up to a fuse.1187

This is why we call it a bomb calorimeter.1190

Basically you are going to ignite the fuse.1192

The fuse is going to provide a spark, a source of energy.1195

You are going to have basically a combustion reaction occur inside the stainless steel apparatus.1199

On the outside of this stainless steel apparatus is a vessel.1206

This is going to be some type of insulative material.1213

This vessel then is going to have a thermometer embedded in it.1221

The thermometer is going to allow us to determine what the temperature change is during the reaction.1227

Of course the temperature is going to go up because it is a combustion reaction.1234

Sometimes the bomb calorimeter is going to have water too.1239

This is going to be with or without water.1244

Basically the water is going to provide a very nice heat sink for us1247

so that we can dissipate the thermal energy coming off of the combustion reaction.1251

What we are assuming is the following.1261

That all of the heat energy that is given off by this combustion reaction is going to be essentially completely transferred to the surroundings.1263

Basically the surroundings of this experiment are going to be the bomb calorimeter itself.1274

If there is the water, it is also the water.1282

Anything that is capable of absorbing the energy.1285

All of this is equal to zero.1292

q of reaction plus q of bomb plus q of water is equal to zero.1294

There is no heat lost to the outside environment.1298

All of it is going from the reaction inside the bomb calorimeter to the bomb calorimeter itself.1301

If there is water, to the water itself.1308

The equation for the heat of the bomb calorimeter is simply equal to the heat capacity of the calorimeter times ΔT.1312

Please make a note that this is going to be independent of its mass; mass independent.1321

Something very interesting happens.1333

What we want to look at is now the effect of constant volume on our equation for internal energy.1335

We know that ΔU is equal to q plus w, which equals to q minus PΔV.1341

If the volume is not changing, ΔV and this whole term essentially goes to zero.1350

We very nicely have the relationship of ΔU is strictly equal to q.1356

That is, in a constant volume process, all of the energy that is1362

transferred is in the form of heat, not in the form of work.1368

All energy transferred as heat.1374

Again constant volume process, ΔU equals to q.1383

Let's go ahead and now look at a typical problem.1393

You have 1.550 grams of C6H14; it is undergoing combustion in a bomb calorimeter.1400

Right away, bomb calorimeter, immediately I would underline that because that tells me that ΔV is zero.1407

So ΔU is q; q of the reaction of course.1416

You are told that the temperature changes from 25.87 to 28.13.1422

Remember that is what is expected; the temperature is going to increase.1426

Remember this is a combustion reaction after all.1429

They are always going to be giving off energy in the form of heat.1431

Calculate ΔU in kilojoules per mole.1436

You are told that the specific heat capacity of the calorimeter is 5.73 kilojoules per degree Celsius.1439

Let's go ahead and solve the problem; you are asked to calculate ΔU.1446

But because this is bomb calorimetry, ΔU is equal to q of the reaction.1450

In addition we know that q of the reaction plus q of the bomb calorimeter1457

plus q of water is equal to zero from the previous slide.1464

But do you see water being mentioned in this example?1468

There is none; this term is not applicable.1472

We have the relationship that q of the reaction is equal to ?q of the bomb.1479

This is equal to ?C of the bomb times ΔT.1486

Let's go ahead and plug in everything we know.1494

That is going to be equal to -12.95 kilojoules.1496

The question asks us for ΔU in units of kilojoules per mole.1508

You see here that we have the mass of the reactant.1513

All we have to do is get that to moles; then divide our kilojoules over the moles.1517

That is going to be 1.550 grams times 1 mole over its molar mass which is approximately 86 grams.1523

That is going to give us x moles of the reactant, C6H14.1536

When all is said and done and we take the kilojoules over the moles, we are going to get -718.5 kJ per mole.1544

When you do these problems, always pay attention to the units that are requested for ΔU.1554

That is how we do a typical bomb calorimetry problem.1561

In addition to having a calorimetry performed under a constant volume process, we can also do a constant pressure process.1567

This is going to be a slight mathematical derivation here.1576

We know that ΔU is equal to q plus w, which is equal to q minus PΔV.1580

From thermodynamics, there is going to be a new state function that is going to introduced right now.1586

This state function is what we call enthalpy; enthalpy is going to be symbolized ΔH.1592

By definition from thermodynamics, ΔH is equal to q plus PΔV.1597

We will call that equation two; let's combine equations one and two right now.1602

ΔH is equal to U plus ΔV, which is equal to q minus PΔV plus PΔV, which is equal to q.1608

In other words, at constant pressure conditions, enthalpy and heat are identical.1619

They are the same thing.1626

In addition, if the volume change is negligible, you see from the equation that ΔH and U are approximately the same value.1629

This is how we define enthalpy.1640

Enthalpy is essentially heat energy under the constraint of constant pressure.1642

The remainder of this lecture deals with calculating enthalpy under various conditions.1652

There is numerous ones; there is about four to five.1662

Which just shows you how important enthalpy is to this chapter.1664

The first way of calculating enthalpy involves phase changes.1669

Every phase change which we know to be physical changes of course is going to be associated with a change of energy.1674

Melting, vaporization, and sublimation; think about it.1682

If you are going to melt something, you have to put energy into it.1686

They all require energy.1688

Those processes that require energy are what we call endothermic processes.1692

Endothermic processes tells us the following.1697

That the sign of ΔH is going to be a positive value.1700

The opposite directions, freezing, condensation, and deposition, all give off energy and are exothermic processes.1706

For exothermic processes, the sign of ΔH is going to be a negative value.1714

Enthalpy values for phase changes are commonly given in units of kilojoules per mole under standard conditions.1721

What do we mean by standard conditions?1733

Basically standard conditions are provided for us to level the playing field1735

because there is so many different parameters we can perform experiments at.1739

It is very for us to come up with a universal accepted condition.1744

That is going to make everything a lot easier.1755

Standard conditions are defined as the following.1757

If we are dealing with a gas, it is going to be 1 atmosphere of pressure.1760

If we are dealing with a solution, it is going to be a 1 molar concentration.1765

The temperature is going to be 25 degrees Celsius.1772

Again 1 atm, 1 molar, and 25 degrees Celsius.1776

Let's go ahead and look at the following question.1784

How much energy in the form of heat is required to melt 1.36 grams of ice?1787

You are given the ΔH of these phase changes to be 6.00 kilojoules per mole.1792

Anytime you see the letters f-u-s, that means a fusion.1798

That is going to be for the melting process; melting process.1802

Anytime you are dealing with a phase change, it is quite simple.1809

q is equal to n times ΔH of the change.1812

Very straightforward equation where n is going to be moles of your compound.1821

ΔH of the phase change is going to be like we said in kilojoules per mole.1827

It is pretty straightforward.1833

We have to get the 1.36 grams of the ice to moles.1834

1.36 grams times 1 mole over 18.016 grams.1837

Then we want to multiply this by kilojoules per mole.1845

That is going to give us 0.453 kJ for our final answer.1851

Does it make sense that the sign of ΔH here is positive?1857

Remember this is a melting process so absolutely this is going to be an endothermic process.1861

Basically you want to add ΔH as a new conversion factor, kilojoules per mole; very good.1867

That is the first way of calculating enthalpy we are going to talk about via a phase change.1877

The next way of calculating enthalpy is what we call the heat of a reaction.1884

Remember that we said that every chemical process is going to have a change in energy associated with it.1888

This change in energy is what we call ΔH of the reaction.1895

ΔH of the reaction is going to be symbolized ΔHr.1898

It is usually given to the right of a chemical reaction.1903

Basically this is telling me that if I am going to perform this reaction exactly one time,1908

and I got the perfect amounts, mole to mole ratios of reactants,1916

it is going to release exactly 906 kilojoules of energy in the form of heat.1919

Calculate the heat in kiloJoules associated with the complete reaction of 155 grams of ammonia.1927

Anytime you are given a ΔH of a reaction problem, all you are going to do is1934

you are going to using ΔH of reaction as a conversion factor; again new conversion factor.1938

Basically the conversion factors are the following.1947

You can have -906 kilojoules for every 4 moles of ammonia.1950

You can have -906 kilojoules for every 5 moles of O2, etc.1959

We have seen this before where we are using the stoichiometric coefficients directly into a conversion factor.1966

Again I want to remind you, this is all the same material.1973

We are just using a slight twist.1976

But it is completely within your realm of what you know how to do already.1979

Let's go ahead and solve the problem; 155 grams of ammonia.1985

Again this is stoichiometry because we are asked to relate any two items in the chemical reaction.1990

One of those items is energy.1995

Times... we have to get everything into moles.1998

Remember that is always our first step.2000

Divided by roughly 17 grams of ammonia.2002

Then I am going to use my conversion factor.2005

That is going to be -906 kilojoules divided by 4 moles of ammonia.2009

When all is said and done, I am left with -2065 kilojoules as our final answer.2016

Again this is how you deal with what is called heats of reaction.2025

They are typically given to the far right of a chemical reaction.2027

The next scenario with calculating enthalpy is what we call standard enthalpies of formation.2035

Standard enthalpies of formation is going to be symbolized as ΔHf.2042

The formal definition is the energy change associated with the formation of a compound2047

from its constituent elements in their standard states at 1 atm in 25 degrees Celsius.2053

Let's take apart this definition; let's talk about water vapor.2060

The elements that are involved are of course hydrogen and oxygen.2067

We have already discussed these.2072

Hydrogen exists as a diatomic gas under standard conditions.2074

Oxygen exists as a diatomic gas under standard conditions.2078

The reaction representing the formation of water vapor under standard conditions is H2 gas plus 1/2O2 gas goes on to form water vapor.2084

This reaction is -241.8 kilojoules for the heat of formation.2097

Every textbook always has an appendix.2104

You are going to find in the appendix a very comprehensive table of heats of formation for many many compounds.2108

You probably don't have to memorize them.2118

You will either be given the information or be able to look it up in the appendix when you do problems.2120

There are two notes that you want to be aware of now.2127

ΔH of formation and all ΔH values, ΔH reaction, ΔH formation, etc, they are all affected by the reaction, the coefficients.2132

ΔH is proportional to the coefficients.2145

If I take the same reaction and I double it, I multiply it through by 2,2148

the ΔH value is going to get increased proportionally.2154

If I take the reverse reaction though, the sign is also going to flip.2160

For the reverse reaction, it is the same magnitude but opposite sign.2166

How do we go ahead and use ΔH of formation to solve ΔH of the reaction?2174

A typical problem is represented as the following.2183

Calculate ΔH of the reaction for the following given the heats of formation.2185

Again if you are not given it, you need to be able to look it up in the appendix.2190

This is going to be a very easy equation to remember.2196

ΔH of the reaction equals the following.2202

Summation n, ΔH of formation of all products, minus summation n ΔH of formation of all reactants,2206

where n is going to be the stoichiometric coefficient from the balanced chemical equation.2212

That is why again it is so important to make sure your chemical equation is balanced.2223

Or else this problem is going to be trouble.2227

We are going to take the products first.2232

You notice that only Al2O3 and Fe2O3 are given to you.2235

What is not given is iron solid and aluminum solid.2241

The reason why is because they are the elements in their standard state--aluminum solid monatomic and iron solid monatomic.2244

What you want to remember is the following.2254

That for elements in their standard states, ΔH of formation is formally going to be equal to zero.2256

That is what you will always want to remember.2273

That requires you to know those elements in their standard state.2277

Again most of them are monatomic, except the following.2281

H2 gas, N2 gas, F2 gas, Cl2 gas, Br2 liquid, I2 solid.2286

Also of course remember mercury is going to be a liquid under standard conditions.2301

Again these are the ones that you want to remember.2309

These are going to be zero ΔH of formation.2312

The rest are typically going to be monatomic solids.2316

Let's go ahead and do this; we are going to look at the products.2324

That is going to be aluminum oxide.2327

That is going to be -1675.7 kilojoules per mole.2329

My coefficient for aluminum oxide here is just 1.2336

The other product, iron, is just 0.2342

Remember summation tells me I am going to add everything together.2347

That takes care of my products.2350

I am now going to subtract this with the same equation for the reactants.2351

This is going to be for iron oxide.2360

Minus 1 mole of iron oxide times -824.2 kilojoules per mole.2362

Plus 0 for aluminum solid; summation products minus summation reactants.2371

We are going to get -852 kilojoules for the ΔH of the reaction.2381

Again it is a very easy equation to know; it really sticks out.2387

This is a summation equation again for ΔH of formation values under standard conditions.2392

Yet another way to calculate enthalpy is from a series of reactions.2400

We introduce now a very important law from thermodynamics is what we call Hess's law.2406

Hess's law tells us basically the following.2412

That the whole is equal to the sum of the parts.2414

That is, the change in enthalpy for an overall reaction equals to the sum of the enthalpy changes for each individual step.2417

Again the whole is equal to the sum of the parts.2425

Imagine we have a reaction through a three step process.2432

If I am to combine all these reactions, all we are going to do is combine reactions.2437

We are going to add up everything.2442

We are going to sum all reactants on one side.2444

We are going to sum all products on one side.2449

We are basically adding chemical reactions together.2453

We go ahead and do that; let's see what we get.2457

2NO gas plus O2 gas plus N2 gas plus O2 gas plus 2N2O gas goes on to form all of the products.2460

2NO2 gas plus 2NO gas plus 2N2 gas plus O2 gas.2477

You want to treat reactions basically as any algebraic equation.2490

Now we can go ahead and cancel like terms on opposite sides of the equation.2497

Let's see what we can get rid of.2505

One O2 here is gone; that means one O2 here is gone.2510

This N2 is gone which means we only have one N2 remaining right here.2514

2NO on the left is going to cancel with the 2NOs on the right.2519

We have a net balanced combined reaction of 2N2O gas plus O2 gas going on to form 2NO2 gas plus N2 gas.2523

This is our overall net equation.2539

Basically what Hess's law is telling me is that when I combine these three reactions,2543

ΔH of the overall reaction is simply equal to the sum of the individual ΔHs.2549

ΔH1 plus ΔH2 plus ΔH3; again this is Hess's law; very good.2556

The final way to calculate enthalpy is what we call a coffee cup calorimeter.2566

It is called coffee cup calorimetry because the calorimeter is good old Styrofoam.2578

Styrofoam, we know to be a very inexpensive readily available insulator.2584

This type of calorimetry is going to more or less guarantee constant pressure conditions2590

because it is going to be whatever the atmospheric pressure is at the time.2596

q of the reaction is going to be equal to ?q of the solution.2603

Basically for coffee cup calorimetry, you have a coffee cup with a aqueous solution usually with your reactant in here.2608

All it is is the thermometer inserted and usually some type of lid is here used.2619

You see that this reaction is pretty much going to always be at whatever the atmospheric pressure is going to be.2627

Coffee cup calorimetry is not used for combustion reactions but instead for aqueous reactions.2632

That is a big difference right there; aqueous reactions, not combustion.2641

The system is equal to q of the reaction.2647

If the reaction is giving off energy, where is it going to go?2650

It is basically going to go to the aqueous environment, to q of the solution.2654

We are basically assuming that the Styrofoam material, because it is such a good insulator,2661

it is not involved in the transfer of energy; assuming Styrofoam not involved.2667

That is why you don't see any term for the Styrofoam material.2675

Basically all of the energy that is given off by the reaction or absorbed by the reaction2681

is going to come from or go to the aqueous solution that the reaction is placed into.2686

q of reaction is equal to ?q of solution.2693

In other words, q of the system equals ?q of the surroundings.2696

We know the equation for q; we have introduced it before.2703

It is the MCAT equation; that is equal to ?mcΔT.2706

Let's go ahead and look at the following reaction.2711

Silver nitrate is reacting with hydrochloric acid to form silver chloride and nitric acid.2714

You have 50 milliliters of 0.1 molar silver nitrate and 50 milliliters of 0.1 molar HCl.2720

You mix it in a coffee cup calorimeter.2727

The temperature rises from 23.4 to 24.21; calculate ΔH of the reaction.2730

You are told that the density of the solution is 1 gram per milliliter.2738

You are given the heat capacity of the solution.2742

We are assuming it to be that of water.2745

We are combining 50 milliliters of one reactant with 50 milliliters of another.2748

Both of them are aqueous.2753

We are assuming that the density of solution is roughly 1.2755

We can get the mass of the solution.2757

1.00 gram per milliliter times the 100 milliliters of the solution is going to give me 100 grams of the solution.2761

We know that q of the reaction is equal ΔH of the reaction2773

because this is constant pressure after all; coffee cup calorimetry.2781

That is going to be equal to ?q of the solution.2785

That is equal to -100 grams times the heat capacity which is given.2790

Times the ΔT which is going to be 24.21 minus 23.40 degrees Celsius.2796

We are going to get an answer of -339 kilojoules.2804

Once again ask yourself, does it make sense that my sign of ΔH here is negative?2809

You are told that the temperature increases, which means it is warm to the touch.2815

Which means it is going to be giving off energy; this is an exothermic process.2819

Yes, our sign of ΔH had better be negative; this answer does absolutely make sense.2823

That is how we do coffee cup calorimetry.2830

Let's now take a moment and summarize the key points from this opening introductory chapter on thermodynamics.2834

We learn the first law of thermodynamics is basically a restatement for the law of conservation of energy.2841

That energy can neither be created nor destroyed.2846

Everything you have is not going to be lost at all; it is conserved.2849

However we introduced the mathematical form of the first law, where ΔU is equal to q plus w.2857

This is going to be true for any closed isolated system.2865

Basically after our brief introduction, we spent a lot of time doing calculations, calculating these different parameters of thermodynamics.2869

To calculate these parameters, you must consider the conditions.2877

Do we have volume constant?--do we have pressure constant?2880

We saw that the equations are different for each of the situations.2885

We learned a very important experimental technique called calorimetry.2890

Calorimetry is used to determine the energy change of a chemical reaction under either constant volume or constant pressure conditions.2894

Constant volume conditions, we did this using a bomb calorimeter.2905

Constant pressure conditions, we did this using a coffee cup calorimeter.2915

Finally we spent a great deal of time on enthalpy.2920

We learned many many ways of calculating enthalpy.2924

Remember enthalpy is essentially heat transferred under the constraint of constant pressure.2927

Let's now spend some time on a pair of sample problems.2935

Here we are given a chemical reaction; you are given a ΔH of reaction.2941

Right away I know that this is going to be a type of a stoichiometry problem.2946

I am going to be incorporating the ΔH of the reaction as a conversion factor.2951

Again we are going to be incorporating this as a conversion factor with of course the coefficients.2961

What mass of C4H10 is required to produce 1500 joules of heat?2972

Produce, what does produce mean?2980

Produce means it is going to be giving off heat.2982

What is my sign?--is it positive or negative?2987

It is going to be negative; -1500 joules.2990

We are going to be using ΔH of the reaction as a conversion factor.2997

The reactant we are interested in is a C4H10; this equation is already balanced.3000

1 mole of C4H10 over -26 of 58000 joules.3006

I am then going to go ahead and multiply this by the molar mass of the compound, 58 grams roughly over 1 mole.3018

When we get this, we get 32.7 grams of C4H10 required.3029

Once again this was a nice stoichiometry problem where we used ΔH of the reaction as a conversion factor with the stoichiometric coefficients.3039

Pay attention to this.3051

If we had forgotten the negative sign, we would get then a negative mass, which physically doesn't exist.3054

Right away again always check if your signs are correct.3067

Again if we had not placed this as -1500 joules, we would have gotten a negative mass, which is not going to make sense.3072

Let's now move on to the final sample problem two.3086

Consider the following series of chemical reactions.3090

Right away you see three reactions; ΔH 1, 2, and 3.3092

You pretty much know this is a Hess's law type of problem.3097

The question asks us to calculate ΔH of the reaction for N2O gas plus NO2 gas going to 3NO gas.3100

If I try to add up these reactions right away, I do not get the balanced overall reaction.3113

This means that I must manipulate my previous three reactions.3118

Must manipulate; must change at least one reaction.3123

Remember we went over the two ways to manipulate a chemical reaction.3134

We can either multiply through by some type of factor; multiply through.3139

Or we can reverse the reaction.3148

Remember if you multiply through all of the coefficients by a factor, ΔH gets multiplied by the same factor.3154

If you reverse a reaction, remember the sign of ΔH also changes.3170

Those are the two warnings that you must remember whenever manipulating a chemical reaction.3179

I see that NO2 here is on the reactant side.3188

But the only place where NO2 appears is on the product side here.3191

For the first reaction, I am going to go ahead and I am going to reverse it.3195

2NO2 goes to 2NO plus O2.3199

In addition, I only see one NO2 in the whole problem, but I have two here.3206

Which means I am going to have to multiply through by 1-1/2 to get rid of that 2, to reduce it to 1.3214

That means I have -1-1/2 ΔH1.3221

ΔH2 is just N2 plus O2 going to 2NO.3230

That is going to remain exactly the same.3236

Finally the third reaction is 2N2O going to N2 plus O2.3238

That looks good right now because you see that N2O is on the reactant side in the overall reaction.3249

Here it is on the reactant side in this reaction.3254

But what is the difference?--it is the factor of 2 again.3257

Now we are going to multiply through by 1-1/2 again to get rid of the 2.3260

That going to be 1-1/2 ΔH3.3271

When all is said and done, when I add up all three of these equations that we have changed,3278

that we have made changes to some, we get the overall reaction, N2O plus NO2 going to 3NO.3286

Hess's law tells me that my ΔH of the reaction is simply going to be equal to the sum of the individual ΔHs.3295

-1-1/2 ΔH1 plus ΔH2 plus 1-1/2 ΔH3.3304

This problem is a great illustration of Hess's law.3314

That was our first lecture on thermodynamics.3319

Thank you for using Educator.com; I will see you all next time.3324

Educator®

Please sign in for full access to this lesson.

Sign-InORCreate Account

Enter your Sign-on user name and password.

Forgot password?

Start Learning Now

Our free lessons will get you started (Adobe Flash® required).
Get immediate access to our entire library.

Sign up for Educator.com

Membership Overview

  • Unlimited access to our entire library of courses.
  • Search and jump to exactly what you want to learn.
  • *Ask questions and get answers from the community and our teachers!
  • Practice questions with step-by-step solutions.
  • Download lesson files for programming and software training practice.
  • Track your course viewing progress.
  • Download lecture slides for taking notes.