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Raffi Hovasapian

Raffi Hovasapian

Thermochemistry Example Problems

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Table of Contents

Section 1: Classical Thermodynamics Preliminaries
The Ideal Gas Law

46m 5s

Intro
0:00
Course Overview
0:16
Thermodynamics & Classical Thermodynamics
0:17
Structure of the Course
1:30
The Ideal Gas Law
3:06
Ideal Gas Law: PV=nRT
3:07
Units of Pressure
4:51
Manipulating Units
5:52
Atmosphere : atm
8:15
Millimeter of Mercury: mm Hg
8:48
SI Unit of Volume
9:32
SI Unit of Temperature
10:32
Value of R (Gas Constant): Pv = nRT
10:51
Extensive and Intensive Variables (Properties)
15:23
Intensive Property
15:52
Extensive Property
16:30
Example: Extensive and Intensive Variables
18:20
Ideal Gas Law
19:24
Ideal Gas Law with Intensive Variables
19:25
Graphing Equations
23:51
Hold T Constant & Graph P vs. V
23:52
Hold P Constant & Graph V vs. T
31:08
Hold V Constant & Graph P vs. T
34:38
Isochores or Isometrics
37:08
More on the V vs. T Graph
39:46
More on the P vs. V Graph
42:06
Ideal Gas Law at Low Pressure & High Temperature
44:26
Ideal Gas Law at High Pressure & Low Temperature
45:16
Math Lesson 1: Partial Differentiation

46m 2s

Intro
0:00
Math Lesson 1: Partial Differentiation
0:38
Overview
0:39
Example I
3:00
Example II
6:33
Example III
9:52
Example IV
17:26
Differential & Derivative
21:44
What Does It Mean?
21:45
Total Differential (or Total Derivative)
30:16
Net Change in Pressure (P)
33:58
General Equation for Total Differential
38:12
Example 5: Total Differential
39:28
Section 2: Energy
Energy & the First Law I

1h 6m 45s

Intro
0:00
Properties of Thermodynamic State
1:38
Big Picture: 3 Properties of Thermodynamic State
1:39
Enthalpy & Free Energy
3:30
Associated Law
4:40
Energy & the First Law of Thermodynamics
7:13
System & Its Surrounding Separated by a Boundary
7:14
In Other Cases the Boundary is Less Clear
10:47
State of a System
12:37
State of a System
12:38
Change in State
14:00
Path for a Change in State
14:57
Example: State of a System
15:46
Open, Close, and Isolated System
18:26
Open System
18:27
Closed System
19:02
Isolated System
19:22
Important Questions
20:38
Important Questions
20:39
Work & Heat
22:50
Definition of Work
23:33
Properties of Work
25:34
Definition of Heat
32:16
Properties of Heat
34:49
Experiment #1
42:23
Experiment #2
47:00
More on Work & Heat
54:50
More on Work & Heat
54:51
Conventions for Heat & Work
1:00:50
Convention for Heat
1:02:40
Convention for Work
1:04:24
Schematic Representation
1:05:00
Energy & the First Law II

1h 6m 33s

Intro
0:00
The First Law of Thermodynamics
0:53
The First Law of Thermodynamics
0:54
Example 1: What is the Change in Energy of the System & Surroundings?
8:53
Energy and The First Law II, cont.
11:55
The Energy of a System Changes in Two Ways
11:56
Systems Possess Energy, Not Heat or Work
12:45
Scenario 1
16:00
Scenario 2
16:46
State Property, Path Properties, and Path Functions
18:10
Pressure-Volume Work
22:36
When a System Changes
22:37
Gas Expands
24:06
Gas is Compressed
25:13
Pressure Volume Diagram: Analyzing Expansion
27:17
What if We do the Same Expansion in Two Stages?
35:22
Multistage Expansion
43:58
General Expression for the Pressure-Volume Work
46:59
Upper Limit of Isothermal Expansion
50:00
Expression for the Work Done in an Isothermal Expansion
52:45
Example 2: Find an Expression for the Maximum Work Done by an Ideal Gas upon Isothermal Expansion
56:18
Example 3: Calculate the External Pressure and Work Done
58:50
Energy & the First Law III

1h 2m 17s

Intro
0:00
Compression
0:20
Compression Overview
0:34
Single-stage compression vs. 2-stage Compression
2:16
Multi-stage Compression
8:40
Example I: Compression
14:47
Example 1: Single-stage Compression
14:47
Example 1: 2-stage Compression
20:07
Example 1: Absolute Minimum
26:37
More on Compression
32:55
Isothermal Expansion & Compression
32:56
External & Internal Pressure of the System
35:18
Reversible & Irreversible Processes
37:32
Process 1: Overview
38:57
Process 2: Overview
39:36
Process 1: Analysis
40:42
Process 2: Analysis
45:29
Reversible Process
50:03
Isothermal Expansion and Compression
54:31
Example II: Reversible Isothermal Compression of a Van der Waals Gas
58:10
Example 2: Reversible Isothermal Compression of a Van der Waals Gas
58:11
Changes in Energy & State: Constant Volume

1h 4m 39s

Intro
0:00
Recall
0:37
State Function & Path Function
0:38
First Law
2:11
Exact & Inexact Differential
2:12
Where Does (∆U = Q - W) or dU = dQ - dU Come from?
8:54
Cyclic Integrals of Path and State Functions
8:55
Our Empirical Experience of the First Law
12:31
∆U = Q - W
18:42
Relations between Changes in Properties and Energy
22:24
Relations between Changes in Properties and Energy
22:25
Rate of Change of Energy per Unit Change in Temperature
29:54
Rate of Change of Energy per Unit Change in Volume at Constant Temperature
32:39
Total Differential Equation
34:38
Constant Volume
41:08
If Volume Remains Constant, then dV = 0
41:09
Constant Volume Heat Capacity
45:22
Constant Volume Integrated
48:14
Increase & Decrease in Energy of the System
54:19
Example 1: ∆U and Qv
57:43
Important Equations
1:02:06
Joule's Experiment

16m 50s

Intro
0:00
Joule's Experiment
0:09
Joule's Experiment
1:20
Interpretation of the Result
4:42
The Gas Expands Against No External Pressure
4:43
Temperature of the Surrounding Does Not Change
6:20
System & Surrounding
7:04
Joule's Law
10:44
More on Joule's Experiment
11:08
Later Experiment
12:38
Dealing with the 2nd Law & Its Mathematical Consequences
13:52
Changes in Energy & State: Constant Pressure

43m 40s

Intro
0:00
Changes in Energy & State: Constant Pressure
0:20
Integrating with Constant Pressure
0:35
Defining the New State Function
6:24
Heat & Enthalpy of the System at Constant Pressure
8:54
Finding ∆U
12:10
dH
15:28
Constant Pressure Heat Capacity
18:08
Important Equations
25:44
Important Equations
25:45
Important Equations at Constant Pressure
27:32
Example I: Change in Enthalpy (∆H)
28:53
Example II: Change in Internal Energy (∆U)
34:19
The Relationship Between Cp & Cv

32m 23s

Intro
0:00
The Relationship Between Cp & Cv
0:21
For a Constant Volume Process No Work is Done
0:22
For a Constant Pressure Process ∆V ≠ 0, so Work is Done
1:16
The Relationship Between Cp & Cv: For an Ideal Gas
3:26
The Relationship Between Cp & Cv: In Terms of Molar heat Capacities
5:44
Heat Capacity Can Have an Infinite # of Values
7:14
The Relationship Between Cp & Cv
11:20
When Cp is Greater than Cv
17:13
2nd Term
18:10
1st Term
19:20
Constant P Process: 3 Parts
22:36
Part 1
23:45
Part 2
24:10
Part 3
24:46
Define : γ = (Cp/Cv)
28:06
For Gases
28:36
For Liquids
29:04
For an Ideal Gas
30:46
The Joule Thompson Experiment

39m 15s

Intro
0:00
General Equations
0:13
Recall
0:14
How Does Enthalpy of a System Change Upon a Unit Change in Pressure?
2:58
For Liquids & Solids
12:11
For Ideal Gases
14:08
For Real Gases
16:58
The Joule Thompson Experiment
18:37
The Joule Thompson Experiment Setup
18:38
The Flow in 2 Stages
22:54
Work Equation for the Joule Thompson Experiment
24:14
Insulated Pipe
26:33
Joule-Thompson Coefficient
29:50
Changing Temperature & Pressure in Such a Way that Enthalpy Remains Constant
31:44
Joule Thompson Inversion Temperature
36:26
Positive & Negative Joule-Thompson Coefficient
36:27
Joule Thompson Inversion Temperature
37:22
Inversion Temperature of Hydrogen Gas
37:59
Adiabatic Changes of State

35m 52s

Intro
0:00
Adiabatic Changes of State
0:10
Adiabatic Changes of State
0:18
Work & Energy in an Adiabatic Process
3:44
Pressure-Volume Work
7:43
Adiabatic Changes for an Ideal Gas
9:23
Adiabatic Changes for an Ideal Gas
9:24
Equation for a Fixed Change in Volume
11:20
Maximum & Minimum Values of Temperature
14:20
Adiabatic Path
18:08
Adiabatic Path Diagram
18:09
Reversible Adiabatic Expansion
21:54
Reversible Adiabatic Compression
22:34
Fundamental Relationship Equation for an Ideal Gas Under Adiabatic Expansion
25:00
More on the Equation
28:20
Important Equations
32:16
Important Adiabatic Equation
32:17
Reversible Adiabatic Change of State Equation
33:02
Section 3: Energy Example Problems
1st Law Example Problems I

42m 40s

Intro
0:00
Fundamental Equations
0:56
Work
2:40
Energy (1st Law)
3:10
Definition of Enthalpy
3:44
Heat capacity Definitions
4:06
The Mathematics
6:35
Fundamental Concepts
8:13
Isothermal
8:20
Adiabatic
8:54
Isobaric
9:25
Isometric
9:48
Ideal Gases
10:14
Example I
12:08
Example I: Conventions
12:44
Example I: Part A
15:30
Example I: Part B
18:24
Example I: Part C
19:53
Example II: What is the Heat Capacity of the System?
21:49
Example III: Find Q, W, ∆U & ∆H for this Change of State
24:15
Example IV: Find Q, W, ∆U & ∆H
31:37
Example V: Find Q, W, ∆U & ∆H
38:20
1st Law Example Problems II

1h 23s

Intro
0:00
Example I
0:11
Example I: Finding ∆U
1:49
Example I: Finding W
6:22
Example I: Finding Q
11:23
Example I: Finding ∆H
16:09
Example I: Summary
17:07
Example II
21:16
Example II: Finding W
22:42
Example II: Finding ∆H
27:48
Example II: Finding Q
30:58
Example II: Finding ∆U
31:30
Example III
33:33
Example III: Finding ∆U, Q & W
33:34
Example III: Finding ∆H
38:07
Example IV
41:50
Example IV: Finding ∆U
41:51
Example IV: Finding ∆H
45:42
Example V
49:31
Example V: Finding W
49:32
Example V: Finding ∆U
55:26
Example V: Finding Q
56:26
Example V: Finding ∆H
56:55
1st Law Example Problems III

44m 34s

Intro
0:00
Example I
0:15
Example I: Finding the Final Temperature
3:40
Example I: Finding Q
8:04
Example I: Finding ∆U
8:25
Example I: Finding W
9:08
Example I: Finding ∆H
9:51
Example II
11:27
Example II: Finding the Final Temperature
11:28
Example II: Finding ∆U
21:25
Example II: Finding W & Q
22:14
Example II: Finding ∆H
23:03
Example III
24:38
Example III: Finding the Final Temperature
24:39
Example III: Finding W, ∆U, and Q
27:43
Example III: Finding ∆H
28:04
Example IV
29:23
Example IV: Finding ∆U, W, and Q
25:36
Example IV: Finding ∆H
31:33
Example V
32:24
Example V: Finding the Final Temperature
33:32
Example V: Finding ∆U
39:31
Example V: Finding W
40:17
Example V: First Way of Finding ∆H
41:10
Example V: Second Way of Finding ∆H
42:10
Thermochemistry Example Problems

59m 7s

Intro
0:00
Example I: Find ∆H° for the Following Reaction
0:42
Example II: Calculate the ∆U° for the Reaction in Example I
5:33
Example III: Calculate the Heat of Formation of NH₃ at 298 K
14:23
Example IV
32:15
Part A: Calculate the Heat of Vaporization of Water at 25°C
33:49
Part B: Calculate the Work Done in Vaporizing 2 Mols of Water at 25°C Under a Constant Pressure of 1 atm
35:26
Part C: Find ∆U for the Vaporization of Water at 25°C
41:00
Part D: Find the Enthalpy of Vaporization of Water at 100°C
43:12
Example V
49:24
Part A: Constant Temperature & Increasing Pressure
50:25
Part B: Increasing temperature & Constant Pressure
56:20
Section 4: Entropy
Entropy

49m 16s

Intro
0:00
Entropy, Part 1
0:16
Coefficient of Thermal Expansion (Isobaric)
0:38
Coefficient of Compressibility (Isothermal)
1:25
Relative Increase & Relative Decrease
2:16
More on α
4:40
More on κ
8:38
Entropy, Part 2
11:04
Definition of Entropy
12:54
Differential Change in Entropy & the Reversible Path
20:08
State Property of the System
28:26
Entropy Changes Under Isothermal Conditions
35:00
Recall: Heating Curve
41:05
Some Phase Changes Take Place Under Constant Pressure
44:07
Example I: Finding ∆S for a Phase Change
46:05
Math Lesson II

33m 59s

Intro
0:00
Math Lesson II
0:46
Let F(x,y) = x²y³
0:47
Total Differential
3:34
Total Differential Expression
6:06
Example 1
9:24
More on Math Expression
13:26
Exact Total Differential Expression
13:27
Exact Differentials
19:50
Inexact Differentials
20:20
The Cyclic Rule
21:06
The Cyclic Rule
21:07
Example 2
27:58
Entropy As a Function of Temperature & Volume

54m 37s

Intro
0:00
Entropy As a Function of Temperature & Volume
0:14
Fundamental Equation of Thermodynamics
1:16
Things to Notice
9:10
Entropy As a Function of Temperature & Volume
14:47
Temperature-dependence of Entropy
24:00
Example I
26:19
Entropy As a Function of Temperature & Volume, Cont.
31:55
Volume-dependence of Entropy at Constant Temperature
31:56
Differentiate with Respect to Temperature, Holding Volume Constant
36:16
Recall the Cyclic Rule
45:15
Summary & Recap
46:47
Fundamental Equation of Thermodynamics
46:48
For Entropy as a Function of Temperature & Volume
47:18
The Volume-dependence of Entropy for Liquids & Solids
52:52
Entropy as a Function of Temperature & Pressure

31m 18s

Intro
0:00
Entropy as a Function of Temperature & Pressure
0:17
Entropy as a Function of Temperature & Pressure
0:18
Rewrite the Total Differential
5:54
Temperature-dependence
7:08
Pressure-dependence
9:04
Differentiate with Respect to Pressure & Holding Temperature Constant
9:54
Differentiate with Respect to Temperature & Holding Pressure Constant
11:28
Pressure-Dependence of Entropy for Liquids & Solids
18:45
Pressure-Dependence of Entropy for Liquids & Solids
18:46
Example I: ∆S of Transformation
26:20
Summary of Entropy So Far

23m 6s

Intro
0:00
Summary of Entropy So Far
0:43
Defining dS
1:04
Fundamental Equation of Thermodynamics
3:51
Temperature & Volume
6:04
Temperature & Pressure
9:10
Two Important Equations for How Entropy Behaves
13:38
State of a System & Heat Capacity
15:34
Temperature-dependence of Entropy
19:49
Entropy Changes for an Ideal Gas

25m 42s

Intro
0:00
Entropy Changes for an Ideal Gas
1:10
General Equation
1:22
The Fundamental Theorem of Thermodynamics
2:37
Recall the Basic Total Differential Expression for S = S (T,V)
5:36
For a Finite Change in State
7:58
If Cv is Constant Over the Particular Temperature Range
9:05
Change in Entropy of an Ideal Gas as a Function of Temperature & Pressure
11:35
Change in Entropy of an Ideal Gas as a Function of Temperature & Pressure
11:36
Recall the Basic Total Differential expression for S = S (T, P)
15:13
For a Finite Change
18:06
Example 1: Calculate the ∆S of Transformation
22:02
Section 5: Entropy Example Problems
Entropy Example Problems I

43m 39s

Intro
0:00
Entropy Example Problems I
0:24
Fundamental Equation of Thermodynamics
1:10
Entropy as a Function of Temperature & Volume
2:04
Entropy as a Function of Temperature & Pressure
2:59
Entropy For Phase Changes
4:47
Entropy For an Ideal Gas
6:14
Third Law Entropies
8:25
Statement of the Third Law
9:17
Entropy of the Liquid State of a Substance Above Its Melting Point
10:23
Entropy For the Gas Above Its Boiling Temperature
13:02
Entropy Changes in Chemical Reactions
15:26
Entropy Change at a Temperature Other than 25°C
16:32
Example I
19:31
Part A: Calculate ∆S for the Transformation Under Constant Volume
20:34
Part B: Calculate ∆S for the Transformation Under Constant Pressure
25:04
Example II: Calculate ∆S fir the Transformation Under Isobaric Conditions
27:53
Example III
30:14
Part A: Calculate ∆S if 1 Mol of Aluminum is taken from 25°C to 255°C
31:14
Part B: If S°₂₉₈ = 28.4 J/mol-K, Calculate S° for Aluminum at 498 K
33:23
Example IV: Calculate Entropy Change of Vaporization for CCl₄
34:19
Example V
35:41
Part A: Calculate ∆S of Transformation
37:36
Part B: Calculate ∆S of Transformation
39:10
Entropy Example Problems II

56m 44s

Intro
0:00
Example I
0:09
Example I: Calculate ∆U
1:28
Example I: Calculate Q
3:29
Example I: Calculate Cp
4:54
Example I: Calculate ∆S
6:14
Example II
7:13
Example II: Calculate W
8:14
Example II: Calculate ∆U
8:56
Example II: Calculate Q
10:18
Example II: Calculate ∆H
11:00
Example II: Calculate ∆S
12:36
Example III
18:47
Example III: Calculate ∆H
19:38
Example III: Calculate Q
21:14
Example III: Calculate ∆U
21:44
Example III: Calculate W
23:59
Example III: Calculate ∆S
24:55
Example IV
27:57
Example IV: Diagram
29:32
Example IV: Calculate W
32:27
Example IV: Calculate ∆U
36:36
Example IV: Calculate Q
38:32
Example IV: Calculate ∆H
39:00
Example IV: Calculate ∆S
40:27
Example IV: Summary
43:41
Example V
48:25
Example V: Diagram
49:05
Example V: Calculate W
50:58
Example V: Calculate ∆U
53:29
Example V: Calculate Q
53:44
Example V: Calculate ∆H
54:34
Example V: Calculate ∆S
55:01
Entropy Example Problems III

57m 6s

Intro
0:00
Example I: Isothermal Expansion
0:09
Example I: Calculate W
1:19
Example I: Calculate ∆U
1:48
Example I: Calculate Q
2:06
Example I: Calculate ∆H
2:26
Example I: Calculate ∆S
3:02
Example II: Adiabatic and Reversible Expansion
6:10
Example II: Calculate Q
6:48
Example II: Basic Equation for the Reversible Adiabatic Expansion of an Ideal Gas
8:12
Example II: Finding Volume
12:40
Example II: Finding Temperature
17:58
Example II: Calculate ∆U
19:53
Example II: Calculate W
20:59
Example II: Calculate ∆H
21:42
Example II: Calculate ∆S
23:42
Example III: Calculate the Entropy of Water Vapor
25:20
Example IV: Calculate the Molar ∆S for the Transformation
34:32
Example V
44:19
Part A: Calculate the Standard Entropy of Liquid Lead at 525°C
46:17
Part B: Calculate ∆H for the Transformation of Solid Lead from 25°C to Liquid Lead at 525°C
52:23
Section 6: Entropy and Probability
Entropy & Probability I

54m 35s

Intro
0:00
Entropy & Probability
0:11
Structural Model
3:05
Recall the Fundamental Equation of Thermodynamics
9:11
Two Independent Ways of Affecting the Entropy of a System
10:05
Boltzmann Definition
12:10
Omega
16:24
Definition of Omega
16:25
Energy Distribution
19:43
The Energy Distribution
19:44
In How Many Ways can N Particles be Distributed According to the Energy Distribution
23:05
Example I: In How Many Ways can the Following Distribution be Achieved
32:51
Example II: In How Many Ways can the Following Distribution be Achieved
33:51
Example III: In How Many Ways can the Following Distribution be Achieved
34:45
Example IV: In How Many Ways can the Following Distribution be Achieved
38:50
Entropy & Probability, cont.
40:57
More on Distribution
40:58
Example I Summary
41:43
Example II Summary
42:12
Distribution that Maximizes Omega
42:26
If Omega is Large, then S is Large
44:22
Two Constraints for a System to Achieve the Highest Entropy Possible
47:07
What Happened When the Energy of a System is Increased?
49:00
Entropy & Probability II

35m 5s

Intro
0:00
Volume Distribution
0:08
Distributing 2 Balls in 3 Spaces
1:43
Distributing 2 Balls in 4 Spaces
3:44
Distributing 3 Balls in 10 Spaces
5:30
Number of Ways to Distribute P Particles over N Spaces
6:05
When N is Much Larger than the Number of Particles P
7:56
Energy Distribution
25:04
Volume Distribution
25:58
Entropy, Total Entropy, & Total Omega Equations
27:34
Entropy, Total Entropy, & Total Omega Equations
27:35
Section 7: Spontaneity, Equilibrium, and the Fundamental Equations
Spontaneity & Equilibrium I

28m 42s

Intro
0:00
Reversible & Irreversible
0:24
Reversible vs. Irreversible
0:58
Defining Equation for Equilibrium
2:11
Defining Equation for Irreversibility (Spontaneity)
3:11
TdS ≥ dQ
5:15
Transformation in an Isolated System
11:22
Transformation in an Isolated System
11:29
Transformation at Constant Temperature
14:50
Transformation at Constant Temperature
14:51
Helmholtz Free Energy
17:26
Define: A = U - TS
17:27
Spontaneous Isothermal Process & Helmholtz Energy
20:20
Pressure-volume Work
22:02
Spontaneity & Equilibrium II

34m 38s

Intro
0:00
Transformation under Constant Temperature & Pressure
0:08
Transformation under Constant Temperature & Pressure
0:36
Define: G = U + PV - TS
3:32
Gibbs Energy
5:14
What Does This Say?
6:44
Spontaneous Process & a Decrease in G
14:12
Computing ∆G
18:54
Summary of Conditions
21:32
Constraint & Condition for Spontaneity
21:36
Constraint & Condition for Equilibrium
24:54
A Few Words About the Word Spontaneous
26:24
Spontaneous Does Not Mean Fast
26:25
Putting Hydrogen & Oxygen Together in a Flask
26:59
Spontaneous Vs. Not Spontaneous
28:14
Thermodynamically Favorable
29:03
Example: Making a Process Thermodynamically Favorable
29:34
Driving Forces for Spontaneity
31:35
Equation: ∆G = ∆H - T∆S
31:36
Always Spontaneous Process
32:39
Never Spontaneous Process
33:06
A Process That is Endothermic Can Still be Spontaneous
34:00
The Fundamental Equations of Thermodynamics

30m 50s

Intro
0:00
The Fundamental Equations of Thermodynamics
0:44
Mechanical Properties of a System
0:45
Fundamental Properties of a System
1:16
Composite Properties of a System
1:44
General Condition of Equilibrium
3:16
Composite Functions & Their Differentiations
6:11
dH = TdS + VdP
7:53
dA = -SdT - PdV
9:26
dG = -SdT + VdP
10:22
Summary of Equations
12:10
Equation #1
14:33
Equation #2
15:15
Equation #3
15:58
Equation #4
16:42
Maxwell's Relations
20:20
Maxwell's Relations
20:21
Isothermal Volume-Dependence of Entropy & Isothermal Pressure-Dependence of Entropy
26:21
The General Thermodynamic Equations of State

34m 6s

Intro
0:00
The General Thermodynamic Equations of State
0:10
Equations of State for Liquids & Solids
0:52
More General Condition for Equilibrium
4:02
General Conditions: Equation that Relates P to Functions of T & V
6:20
The Second Fundamental Equation of Thermodynamics
11:10
Equation 1
17:34
Equation 2
21:58
Recall the General Expression for Cp - Cv
28:11
For the Joule-Thomson Coefficient
30:44
Joule-Thomson Inversion Temperature
32:12
Properties of the Helmholtz & Gibbs Energies

39m 18s

Intro
0:00
Properties of the Helmholtz & Gibbs Energies
0:10
Equating the Differential Coefficients
1:34
An Increase in T; a Decrease in A
3:25
An Increase in V; a Decrease in A
6:04
We Do the Same Thing for G
8:33
Increase in T; Decrease in G
10:50
Increase in P; Decrease in G
11:36
Gibbs Energy of a Pure Substance at a Constant Temperature from 1 atm to any Other Pressure.
14:12
If the Substance is a Liquid or a Solid, then Volume can be Treated as a Constant
18:57
For an Ideal Gas
22:18
Special Note
24:56
Temperature Dependence of Gibbs Energy
27:02
Temperature Dependence of Gibbs Energy #1
27:52
Temperature Dependence of Gibbs Energy #2
29:01
Temperature Dependence of Gibbs Energy #3
29:50
Temperature Dependence of Gibbs Energy #4
34:50
The Entropy of the Universe & the Surroundings

19m 40s

Intro
0:00
Entropy of the Universe & the Surroundings
0:08
Equation: ∆G = ∆H - T∆S
0:20
Conditions of Constant Temperature & Pressure
1:14
Reversible Process
3:14
Spontaneous Process & the Entropy of the Universe
5:20
Tips for Remembering Everything
12:40
Verify Using Known Spontaneous Process
14:51
Section 8: Free Energy Example Problems
Free Energy Example Problems I

54m 16s

Intro
0:00
Example I
0:11
Example I: Deriving a Function for Entropy (S)
2:06
Example I: Deriving a Function for V
5:55
Example I: Deriving a Function for H
8:06
Example I: Deriving a Function for U
12:06
Example II
15:18
Example III
21:52
Example IV
26:12
Example IV: Part A
26:55
Example IV: Part B
28:30
Example IV: Part C
30:25
Example V
33:45
Example VI
40:46
Example VII
43:43
Example VII: Part A
44:46
Example VII: Part B
50:52
Example VII: Part C
51:56
Free Energy Example Problems II

31m 17s

Intro
0:00
Example I
0:09
Example II
5:18
Example III
8:22
Example IV
12:32
Example V
17:14
Example VI
20:34
Example VI: Part A
21:04
Example VI: Part B
23:56
Example VI: Part C
27:56
Free Energy Example Problems III

45m

Intro
0:00
Example I
0:10
Example II
15:03
Example III
21:47
Example IV
28:37
Example IV: Part A
29:33
Example IV: Part B
36:09
Example IV: Part C
40:34
Three Miscellaneous Example Problems

58m 5s

Intro
0:00
Example I
0:41
Part A: Calculating ∆H
3:55
Part B: Calculating ∆S
15:13
Example II
24:39
Part A: Final Temperature of the System
26:25
Part B: Calculating ∆S
36:57
Example III
46:49
Section 9: Equation Review for Thermodynamics
Looking Back Over Everything: All the Equations in One Place

25m 20s

Intro
0:00
Work, Heat, and Energy
0:18
Definition of Work, Energy, Enthalpy, and Heat Capacities
0:23
Heat Capacities for an Ideal Gas
3:40
Path Property & State Property
3:56
Energy Differential
5:04
Enthalpy Differential
5:40
Joule's Law & Joule-Thomson Coefficient
6:23
Coefficient of Thermal Expansion & Coefficient of Compressibility
7:01
Enthalpy of a Substance at Any Other Temperature
7:29
Enthalpy of a Reaction at Any Other Temperature
8:01
Entropy
8:53
Definition of Entropy
8:54
Clausius Inequality
9:11
Entropy Changes in Isothermal Systems
9:44
The Fundamental Equation of Thermodynamics
10:12
Expressing Entropy Changes in Terms of Properties of the System
10:42
Entropy Changes in the Ideal Gas
11:22
Third Law Entropies
11:38
Entropy Changes in Chemical Reactions
14:02
Statistical Definition of Entropy
14:34
Omega for the Spatial & Energy Distribution
14:47
Spontaneity and Equilibrium
15:43
Helmholtz Energy & Gibbs Energy
15:44
Condition for Spontaneity & Equilibrium
16:24
Condition for Spontaneity with Respect to Entropy
17:58
The Fundamental Equations
18:30
Maxwell's Relations
19:04
The Thermodynamic Equations of State
20:07
Energy & Enthalpy Differentials
21:08
Joule's Law & Joule-Thomson Coefficient
21:59
Relationship Between Constant Pressure & Constant Volume Heat Capacities
23:14
One Final Equation - Just for Fun
24:04
Section 10: Quantum Mechanics Preliminaries
Complex Numbers

34m 25s

Intro
0:00
Complex Numbers
0:11
Representing Complex Numbers in the 2-Dimmensional Plane
0:56
Addition of Complex Numbers
2:35
Subtraction of Complex Numbers
3:17
Multiplication of Complex Numbers
3:47
Division of Complex Numbers
6:04
r & θ
8:04
Euler's Formula
11:00
Polar Exponential Representation of the Complex Numbers
11:22
Example I
14:25
Example II
15:21
Example III
16:58
Example IV
18:35
Example V
20:40
Example VI
21:32
Example VII
25:22
Probability & Statistics

59m 57s

Intro
0:00
Probability & Statistics
1:51
Normalization Condition
1:52
Define the Mean or Average of x
11:04
Example I: Calculate the Mean of x
14:57
Example II: Calculate the Second Moment of the Data in Example I
22:39
Define the Second Central Moment or Variance
25:26
Define the Second Central Moment or Variance
25:27
1st Term
32:16
2nd Term
32:40
3rd Term
34:07
Continuous Distributions
35:47
Continuous Distributions
35:48
Probability Density
39:30
Probability Density
39:31
Normalization Condition
46:51
Example III
50:13
Part A - Show that P(x) is Normalized
51:40
Part B - Calculate the Average Position of the Particle Along the Interval
54:31
Important Things to Remember
58:24
Schrӧdinger Equation & Operators

42m 5s

Intro
0:00
Schrӧdinger Equation & Operators
0:16
Relation Between a Photon's Momentum & Its Wavelength
0:17
Louis de Broglie: Wavelength for Matter
0:39
Schrӧdinger Equation
1:19
Definition of Ψ(x)
3:31
Quantum Mechanics
5:02
Operators
7:51
Example I
10:10
Example II
11:53
Example III
14:24
Example IV
17:35
Example V
19:59
Example VI
22:39
Operators Can Be Linear or Non Linear
27:58
Operators Can Be Linear or Non Linear
28:34
Example VII
32:47
Example VIII
36:55
Example IX
39:29
Schrӧdinger Equation as an Eigenvalue Problem

30m 26s

Intro
0:00
Schrӧdinger Equation as an Eigenvalue Problem
0:10
Operator: Multiplying the Original Function by Some Scalar
0:11
Operator, Eigenfunction, & Eigenvalue
4:42
Example: Eigenvalue Problem
8:00
Schrӧdinger Equation as an Eigenvalue Problem
9:24
Hamiltonian Operator
15:09
Quantum Mechanical Operators
16:46
Kinetic Energy Operator
19:16
Potential Energy Operator
20:02
Total Energy Operator
21:12
Classical Point of View
21:48
Linear Momentum Operator
24:02
Example I
26:01
The Plausibility of the Schrӧdinger Equation

21m 34s

Intro
0:00
The Plausibility of the Schrӧdinger Equation
1:16
The Plausibility of the Schrӧdinger Equation, Part 1
1:17
The Plausibility of the Schrӧdinger Equation, Part 2
8:24
The Plausibility of the Schrӧdinger Equation, Part 3
13:45
Section 11: The Particle in a Box
The Particle in a Box Part I

56m 22s

Intro
0:00
Free Particle in a Box
0:28
Definition of a Free Particle in a Box
0:29
Amplitude of the Matter Wave
6:22
Intensity of the Wave
6:53
Probability Density
9:39
Probability that the Particle is Located Between x & dx
10:54
Probability that the Particle will be Found Between o & a
12:35
Wave Function & the Particle
14:59
Boundary Conditions
19:22
What Happened When There is No Constraint on the Particle
27:54
Diagrams
34:12
More on Probability Density
40:53
The Correspondence Principle
46:45
The Correspondence Principle
46:46
Normalizing the Wave Function
47:46
Normalizing the Wave Function
47:47
Normalized Wave Function & Normalization Constant
52:24
The Particle in a Box Part II

45m 24s

Intro
0:00
Free Particle in a Box
0:08
Free Particle in a 1-dimensional Box
0:09
For a Particle in a Box
3:57
Calculating Average Values & Standard Deviations
5:42
Average Value for the Position of a Particle
6:32
Standard Deviations for the Position of a Particle
10:51
Recall: Energy & Momentum are Represented by Operators
13:33
Recall: Schrӧdinger Equation in Operator Form
15:57
Average Value of a Physical Quantity that is Associated with an Operator
18:16
Average Momentum of a Free Particle in a Box
20:48
The Uncertainty Principle
24:42
Finding the Standard Deviation of the Momentum
25:08
Expression for the Uncertainty Principle
35:02
Summary of the Uncertainty Principle
41:28
The Particle in a Box Part III

48m 43s

Intro
0:00
2-Dimension
0:12
Dimension 2
0:31
Boundary Conditions
1:52
Partial Derivatives
4:27
Example I
6:08
The Particle in a Box, cont.
11:28
Operator Notation
12:04
Symbol for the Laplacian
13:50
The Equation Becomes…
14:30
Boundary Conditions
14:54
Separation of Variables
15:33
Solution to the 1-dimensional Case
16:31
Normalization Constant
22:32
3-Dimension
28:30
Particle in a 3-dimensional Box
28:31
In Del Notation
32:22
The Solutions
34:51
Expressing the State of the System for a Particle in a 3D Box
39:10
Energy Level & Degeneracy
43:35
Section 12: Postulates and Principles of Quantum Mechanics
The Postulates & Principles of Quantum Mechanics, Part I

46m 18s

Intro
0:00
Postulate I
0:31
Probability That The Particle Will Be Found in a Differential Volume Element
0:32
Example I: Normalize This Wave Function
11:30
Postulate II
18:20
Postulate II
18:21
Quantum Mechanical Operators: Position
20:48
Quantum Mechanical Operators: Kinetic Energy
21:57
Quantum Mechanical Operators: Potential Energy
22:42
Quantum Mechanical Operators: Total Energy
22:57
Quantum Mechanical Operators: Momentum
23:22
Quantum Mechanical Operators: Angular Momentum
23:48
More On The Kinetic Energy Operator
24:48
Angular Momentum
28:08
Angular Momentum Overview
28:09
Angular Momentum Operator in Quantum Mechanic
31:34
The Classical Mechanical Observable
32:56
Quantum Mechanical Operator
37:01
Getting the Quantum Mechanical Operator from the Classical Mechanical Observable
40:16
Postulate II, cont.
43:40
Quantum Mechanical Operators are Both Linear & Hermetical
43:41
The Postulates & Principles of Quantum Mechanics, Part II

39m 28s

Intro
0:00
Postulate III
0:09
Postulate III: Part I
0:10
Postulate III: Part II
5:56
Postulate III: Part III
12:43
Postulate III: Part IV
18:28
Postulate IV
23:57
Postulate IV
23:58
Postulate V
27:02
Postulate V
27:03
Average Value
36:38
Average Value
36:39
The Postulates & Principles of Quantum Mechanics, Part III

35m 32s

Intro
0:00
The Postulates & Principles of Quantum Mechanics, Part III
0:10
Equations: Linear & Hermitian
0:11
Introduction to Hermitian Property
3:36
Eigenfunctions are Orthogonal
9:55
The Sequence of Wave Functions for the Particle in a Box forms an Orthonormal Set
14:34
Definition of Orthogonality
16:42
Definition of Hermiticity
17:26
Hermiticity: The Left Integral
23:04
Hermiticity: The Right Integral
28:47
Hermiticity: Summary
34:06
The Postulates & Principles of Quantum Mechanics, Part IV

29m 55s

Intro
0:00
The Postulates & Principles of Quantum Mechanics, Part IV
0:09
Operators can be Applied Sequentially
0:10
Sample Calculation 1
2:41
Sample Calculation 2
5:18
Commutator of Two Operators
8:16
The Uncertainty Principle
19:01
In the Case of Linear Momentum and Position Operator
23:14
When the Commutator of Two Operators Equals to Zero
26:31
Section 13: Postulates and Principles Example Problems, Including Particle in a Box
Example Problems I

54m 25s

Intro
0:00
Example I: Three Dimensional Box & Eigenfunction of The Laplacian Operator
0:37
Example II: Positions of a Particle in a 1-dimensional Box
15:46
Example III: Transition State & Frequency
29:29
Example IV: Finding a Particle in a 1-dimensional Box
35:03
Example V: Degeneracy & Energy Levels of a Particle in a Box
44:59
Example Problems II

46m 58s

Intro
0:00
Review
0:25
Wave Function
0:26
Normalization Condition
2:28
Observable in Classical Mechanics & Linear/Hermitian Operator in Quantum Mechanics
3:36
Hermitian
6:11
Eigenfunctions & Eigenvalue
8:20
Normalized Wave Functions
12:00
Average Value
13:42
If Ψ is Written as a Linear Combination
15:44
Commutator
16:45
Example I: Normalize The Wave Function
19:18
Example II: Probability of Finding of a Particle
22:27
Example III: Orthogonal
26:00
Example IV: Average Value of the Kinetic Energy Operator
30:22
Example V: Evaluate These Commutators
39:02
Example Problems III

44m 11s

Intro
0:00
Example I: Good Candidate for a Wave Function
0:08
Example II: Variance of the Energy
7:00
Example III: Evaluate the Angular Momentum Operators
15:00
Example IV: Real Eigenvalues Imposes the Hermitian Property on Operators
28:44
Example V: A Demonstration of Why the Eigenfunctions of Hermitian Operators are Orthogonal
35:33
Section 14: The Harmonic Oscillator
The Harmonic Oscillator I

35m 33s

Intro
0:00
The Harmonic Oscillator
0:10
Harmonic Motion
0:11
Classical Harmonic Oscillator
4:38
Hooke's Law
8:18
Classical Harmonic Oscillator, cont.
10:33
General Solution for the Differential Equation
15:16
Initial Position & Velocity
16:05
Period & Amplitude
20:42
Potential Energy of the Harmonic Oscillator
23:20
Kinetic Energy of the Harmonic Oscillator
26:37
Total Energy of the Harmonic Oscillator
27:23
Conservative System
34:37
The Harmonic Oscillator II

43m 4s

Intro
0:00
The Harmonic Oscillator II
0:08
Diatomic Molecule
0:10
Notion of Reduced Mass
5:27
Harmonic Oscillator Potential & The Intermolecular Potential of a Vibrating Molecule
7:33
The Schrӧdinger Equation for the 1-dimensional Quantum Mechanic Oscillator
14:14
Quantized Values for the Energy Level
15:46
Ground State & the Zero-Point Energy
21:50
Vibrational Energy Levels
25:18
Transition from One Energy Level to the Next
26:42
Fundamental Vibrational Frequency for Diatomic Molecule
34:57
Example: Calculate k
38:01
The Harmonic Oscillator III

26m 30s

Intro
0:00
The Harmonic Oscillator III
0:09
The Wave Functions Corresponding to the Energies
0:10
Normalization Constant
2:34
Hermite Polynomials
3:22
First Few Hermite Polynomials
4:56
First Few Wave-Functions
6:37
Plotting the Probability Density of the Wave-Functions
8:37
Probability Density for Large Values of r
14:24
Recall: Odd Function & Even Function
19:05
More on the Hermite Polynomials
20:07
Recall: If f(x) is Odd
20:36
Average Value of x
22:31
Average Value of Momentum
23:56
Section 15: The Rigid Rotator
The Rigid Rotator I

41m 10s

Intro
0:00
Possible Confusion from the Previous Discussion
0:07
Possible Confusion from the Previous Discussion
0:08
Rotation of a Single Mass Around a Fixed Center
8:17
Rotation of a Single Mass Around a Fixed Center
8:18
Angular Velocity
12:07
Rotational Inertia
13:24
Rotational Frequency
15:24
Kinetic Energy for a Linear System
16:38
Kinetic Energy for a Rotational System
17:42
Rotating Diatomic Molecule
19:40
Rotating Diatomic Molecule: Part 1
19:41
Rotating Diatomic Molecule: Part 2
24:56
Rotating Diatomic Molecule: Part 3
30:04
Hamiltonian of the Rigid Rotor
36:48
Hamiltonian of the Rigid Rotor
36:49
The Rigid Rotator II

30m 32s

Intro
0:00
The Rigid Rotator II
0:08
Cartesian Coordinates
0:09
Spherical Coordinates
1:55
r
6:15
θ
6:28
φ
7:00
Moving a Distance 'r'
8:17
Moving a Distance 'r' in the Spherical Coordinates
11:49
For a Rigid Rotator, r is Constant
13:57
Hamiltonian Operator
15:09
Square of the Angular Momentum Operator
17:34
Orientation of the Rotation in Space
19:44
Wave Functions for the Rigid Rotator
20:40
The Schrӧdinger Equation for the Quantum Mechanic Rigid Rotator
21:24
Energy Levels for the Rigid Rotator
26:58
The Rigid Rotator III

35m 19s

Intro
0:00
The Rigid Rotator III
0:11
When a Rotator is Subjected to Electromagnetic Radiation
1:24
Selection Rule
2:13
Frequencies at Which Absorption Transitions Occur
6:24
Energy Absorption & Transition
10:54
Energy of the Individual Levels Overview
20:58
Energy of the Individual Levels: Diagram
23:45
Frequency Required to Go from J to J + 1
25:53
Using Separation Between Lines on the Spectrum to Calculate Bond Length
28:02
Example I: Calculating Rotational Inertia & Bond Length
29:18
Example I: Calculating Rotational Inertia
29:19
Example I: Calculating Bond Length
32:56
Section 16: Oscillator and Rotator Example Problems
Example Problems I

33m 48s

Intro
0:00
Equations Review
0:11
Energy of the Harmonic Oscillator
0:12
Selection Rule
3:02
Observed Frequency of Radiation
3:27
Harmonic Oscillator Wave Functions
5:52
Rigid Rotator
7:26
Selection Rule for Rigid Rotator
9:15
Frequency of Absorption
9:35
Wave Numbers
10:58
Example I: Calculate the Reduced Mass of the Hydrogen Atom
11:44
Example II: Calculate the Fundamental Vibration Frequency & the Zero-Point Energy of This Molecule
13:37
Example III: Show That the Product of Two Even Functions is even
19:35
Example IV: Harmonic Oscillator
24:56
Example Problems II

46m 43s

Intro
0:00
Example I: Harmonic Oscillator
0:12
Example II: Harmonic Oscillator
23:26
Example III: Calculate the RMS Displacement of the Molecules
38:12
Section 17: The Hydrogen Atom
The Hydrogen Atom I

40m

Intro
0:00
The Hydrogen Atom I
1:31
Review of the Rigid Rotator
1:32
Hydrogen Atom & the Coulomb Potential
2:50
Using the Spherical Coordinates
6:33
Applying This Last Expression to Equation 1
10:19
Angular Component & Radial Component
13:26
Angular Equation
15:56
Solution for F(φ)
19:32
Determine The Normalization Constant
20:33
Differential Equation for T(a)
24:44
Legendre Equation
27:20
Legendre Polynomials
31:20
The Legendre Polynomials are Mutually Orthogonal
35:40
Limits
37:17
Coefficients
38:28
The Hydrogen Atom II

35m 58s

Intro
0:00
Associated Legendre Functions
0:07
Associated Legendre Functions
0:08
First Few Associated Legendre Functions
6:39
s, p, & d Orbital
13:24
The Normalization Condition
15:44
Spherical Harmonics
20:03
Equations We Have Found
20:04
Wave Functions for the Angular Component & Rigid Rotator
24:36
Spherical Harmonics Examples
25:40
Angular Momentum
30:09
Angular Momentum
30:10
Square of the Angular Momentum
35:38
Energies of the Rigid Rotator
38:21
The Hydrogen Atom III

36m 18s

Intro
0:00
The Hydrogen Atom III
0:34
Angular Momentum is a Vector Quantity
0:35
The Operators Corresponding to the Three Components of Angular Momentum Operator: In Cartesian Coordinates
1:30
The Operators Corresponding to the Three Components of Angular Momentum Operator: In Spherical Coordinates
3:27
Z Component of the Angular Momentum Operator & the Spherical Harmonic
5:28
Magnitude of the Angular Momentum Vector
20:10
Classical Interpretation of Angular Momentum
25:22
Projection of the Angular Momentum Vector onto the xy-plane
33:24
The Hydrogen Atom IV

33m 55s

Intro
0:00
The Hydrogen Atom IV
0:09
The Equation to Find R( r )
0:10
Relation Between n & l
3:50
The Solutions for the Radial Functions
5:08
Associated Laguerre Polynomials
7:58
1st Few Associated Laguerre Polynomials
8:55
Complete Wave Function for the Atomic Orbitals of the Hydrogen Atom
12:24
The Normalization Condition
15:06
In Cartesian Coordinates
18:10
Working in Polar Coordinates
20:48
Principal Quantum Number
21:58
Angular Momentum Quantum Number
22:35
Magnetic Quantum Number
25:55
Zeeman Effect
30:45
The Hydrogen Atom V: Where We Are

51m 53s

Intro
0:00
The Hydrogen Atom V: Where We Are
0:13
Review
0:14
Let's Write Out ψ₂₁₁
7:32
Angular Momentum of the Electron
14:52
Representation of the Wave Function
19:36
Radial Component
28:02
Example: 1s Orbital
28:34
Probability for Radial Function
33:46
1s Orbital: Plotting Probability Densities vs. r
35:47
2s Orbital: Plotting Probability Densities vs. r
37:46
3s Orbital: Plotting Probability Densities vs. r
38:49
4s Orbital: Plotting Probability Densities vs. r
39:34
2p Orbital: Plotting Probability Densities vs. r
40:12
3p Orbital: Plotting Probability Densities vs. r
41:02
4p Orbital: Plotting Probability Densities vs. r
41:51
3d Orbital: Plotting Probability Densities vs. r
43:18
4d Orbital: Plotting Probability Densities vs. r
43:48
Example I: Probability of Finding an Electron in the 2s Orbital of the Hydrogen
45:40
The Hydrogen Atom VI

51m 53s

Intro
0:00
The Hydrogen Atom VI
0:07
Last Lesson Review
0:08
Spherical Component
1:09
Normalization Condition
2:02
Complete 1s Orbital Wave Function
4:08
1s Orbital Wave Function
4:09
Normalization Condition
6:28
Spherically Symmetric
16:00
Average Value
17:52
Example I: Calculate the Region of Highest Probability for Finding the Electron
21:19
2s Orbital Wave Function
25:32
2s Orbital Wave Function
25:33
Average Value
28:56
General Formula
32:24
The Hydrogen Atom VII

34m 29s

Intro
0:00
The Hydrogen Atom VII
0:12
p Orbitals
1:30
Not Spherically Symmetric
5:10
Recall That the Spherical Harmonics are Eigenfunctions of the Hamiltonian Operator
6:50
Any Linear Combination of These Orbitals Also Has The Same Energy
9:16
Functions of Real Variables
15:53
Solving for Px
16:50
Real Spherical Harmonics
21:56
Number of Nodes
32:56
Section 18: Hydrogen Atom Example Problems
Hydrogen Atom Example Problems I

43m 49s

Intro
0:00
Example I: Angular Momentum & Spherical Harmonics
0:20
Example II: Pair-wise Orthogonal Legendre Polynomials
16:40
Example III: General Normalization Condition for the Legendre Polynomials
25:06
Example IV: Associated Legendre Functions
32:13
The Hydrogen Atom Example Problems II

1h 1m 57s

Intro
0:00
Example I: Normalization & Pair-wise Orthogonal
0:13
Part 1: Normalized
0:43
Part 2: Pair-wise Orthogonal
16:53
Example II: Show Explicitly That the Following Statement is True for Any Integer n
27:10
Example III: Spherical Harmonics
29:26
Angular Momentum Cones
56:37
Angular Momentum Cones
56:38
Physical Interpretation of Orbital Angular Momentum in Quantum mechanics
1:00:16
The Hydrogen Atom Example Problems III

48m 33s

Intro
0:00
Example I: Show That ψ₂₁₁ is Normalized
0:07
Example II: Show That ψ₂₁₁ is Orthogonal to ψ₃₁₀
11:48
Example III: Probability That a 1s Electron Will Be Found Within 1 Bohr Radius of The Nucleus
18:35
Example IV: Radius of a Sphere
26:06
Example V: Calculate <r> for the 2s Orbital of the Hydrogen-like Atom
36:33
The Hydrogen Atom Example Problems IV

48m 33s

Intro
0:00
Example I: Probability Density vs. Radius Plot
0:11
Example II: Hydrogen Atom & The Coulombic Potential
14:16
Example III: Find a Relation Among <K>, <V>, & <E>
25:47
Example IV: Quantum Mechanical Virial Theorem
48:32
Example V: Find the Variance for the 2s Orbital
54:13
The Hydrogen Atom Example Problems V

48m 33s

Intro
0:00
Example I: Derive a Formula for the Degeneracy of a Given Level n
0:11
Example II: Using Linear Combinations to Represent the Spherical Harmonics as Functions of the Real Variables θ & φ
8:30
Example III: Using Linear Combinations to Represent the Spherical Harmonics as Functions of the Real Variables θ & φ
23:01
Example IV: Orbital Functions
31:51
Section 19: Spin Quantum Number and Atomic Term Symbols
Spin Quantum Number: Term Symbols I

59m 18s

Intro
0:00
Quantum Numbers Specify an Orbital
0:24
n
1:10
l
1:20
m
1:35
4th Quantum Number: s
2:02
Spin Orbitals
7:03
Spin Orbitals
7:04
Multi-electron Atoms
11:08
Term Symbols
18:08
Russell-Saunders Coupling & The Atomic Term Symbol
18:09
Example: Configuration for C
27:50
Configuration for C: 1s²2s²2p²
27:51
Drawing Every Possible Arrangement
31:15
Term Symbols
45:24
Microstate
50:54
Spin Quantum Number: Term Symbols II

34m 54s

Intro
0:00
Microstates
0:25
We Started With 21 Possible Microstates
0:26
³P State
2:05
Microstates in ³P Level
5:10
¹D State
13:16
³P State
16:10
²P₂ State
17:34
³P₁ State
18:34
³P₀ State
19:12
9 Microstates in ³P are Subdivided
19:40
¹S State
21:44
Quicker Way to Find the Different Values of J for a Given Basic Term Symbol
22:22
Ground State
26:27
Hund's Empirical Rules for Specifying the Term Symbol for the Ground Electronic State
27:29
Hund's Empirical Rules: 1
28:24
Hund's Empirical Rules: 2
29:22
Hund's Empirical Rules: 3 - Part A
30:22
Hund's Empirical Rules: 3 - Part B
31:18
Example: 1s²2s²2p²
31:54
Spin Quantum Number: Term Symbols III

38m 3s

Intro
0:00
Spin Quantum Number: Term Symbols III
0:14
Deriving the Term Symbols for the p² Configuration
0:15
Table: MS vs. ML
3:57
¹D State
16:21
³P State
21:13
¹S State
24:48
J Value
25:32
Degeneracy of the Level
27:28
When Given r Electrons to Assign to n Equivalent Spin Orbitals
30:18
p² Configuration
32:51
Complementary Configurations
35:12
Term Symbols & Atomic Spectra

57m 49s

Intro
0:00
Lyman Series
0:09
Spectroscopic Term Symbols
0:10
Lyman Series
3:04
Hydrogen Levels
8:21
Hydrogen Levels
8:22
Term Symbols & Atomic Spectra
14:17
Spin-Orbit Coupling
14:18
Selection Rules for Atomic Spectra
21:31
Selection Rules for Possible Transitions
23:56
Wave Numbers for The Transitions
28:04
Example I: Calculate the Frequencies of the Allowed Transitions from (4d) ²D →(2p) ²P
32:23
Helium Levels
49:50
Energy Levels for Helium
49:51
Transitions & Spin Multiplicity
52:27
Transitions & Spin Multiplicity
52:28
Section 20: Term Symbols Example Problems
Example Problems I

1h 1m 20s

Intro
0:00
Example I: What are the Term Symbols for the np¹ Configuration?
0:10
Example II: What are the Term Symbols for the np² Configuration?
20:38
Example III: What are the Term Symbols for the np³ Configuration?
40:46
Example Problems II

56m 34s

Intro
0:00
Example I: Find the Term Symbols for the nd² Configuration
0:11
Example II: Find the Term Symbols for the 1s¹2p¹ Configuration
27:02
Example III: Calculate the Separation Between the Doublets in the Lyman Series for Atomic Hydrogen
41:41
Example IV: Calculate the Frequencies of the Lines for the (4d) ²D → (3p) ²P Transition
48:53
Section 21: Equation Review for Quantum Mechanics
Quantum Mechanics: All the Equations in One Place

18m 24s

Intro
0:00
Quantum Mechanics Equations
0:37
De Broglie Relation
0:38
Statistical Relations
1:00
The Schrӧdinger Equation
1:50
The Particle in a 1-Dimensional Box of Length a
3:09
The Particle in a 2-Dimensional Box of Area a x b
3:48
The Particle in a 3-Dimensional Box of Area a x b x c
4:22
The Schrӧdinger Equation Postulates
4:51
The Normalization Condition
5:40
The Probability Density
6:51
Linear
7:47
Hermitian
8:31
Eigenvalues & Eigenfunctions
8:55
The Average Value
9:29
Eigenfunctions of Quantum Mechanics Operators are Orthogonal
10:53
Commutator of Two Operators
10:56
The Uncertainty Principle
11:41
The Harmonic Oscillator
13:18
The Rigid Rotator
13:52
Energy of the Hydrogen Atom
14:30
Wavefunctions, Radial Component, and Associated Laguerre Polynomial
14:44
Angular Component or Spherical Harmonic
15:16
Associated Legendre Function
15:31
Principal Quantum Number
15:43
Angular Momentum Quantum Number
15:50
Magnetic Quantum Number
16:21
z-component of the Angular Momentum of the Electron
16:53
Atomic Spectroscopy: Term Symbols
17:14
Atomic Spectroscopy: Selection Rules
18:03
Section 22: Molecular Spectroscopy
Spectroscopic Overview: Which Equation Do I Use & Why

50m 2s

Intro
0:00
Spectroscopic Overview: Which Equation Do I Use & Why
1:02
Lesson Overview
1:03
Rotational & Vibrational Spectroscopy
4:01
Frequency of Absorption/Emission
6:04
Wavenumbers in Spectroscopy
8:10
Starting State vs. Excited State
10:10
Total Energy of a Molecule (Leaving out the Electronic Energy)
14:02
Energy of Rotation: Rigid Rotor
15:55
Energy of Vibration: Harmonic Oscillator
19:08
Equation of the Spectral Lines
23:22
Harmonic Oscillator-Rigid Rotor Approximation (Making Corrections)
28:37
Harmonic Oscillator-Rigid Rotor Approximation (Making Corrections)
28:38
Vibration-Rotation Interaction
33:46
Centrifugal Distortion
36:27
Anharmonicity
38:28
Correcting for All Three Simultaneously
41:03
Spectroscopic Parameters
44:26
Summary
47:32
Harmonic Oscillator-Rigid Rotor Approximation
47:33
Vibration-Rotation Interaction
48:14
Centrifugal Distortion
48:20
Anharmonicity
48:28
Correcting for All Three Simultaneously
48:44
Vibration-Rotation

59m 47s

Intro
0:00
Vibration-Rotation
0:37
What is Molecular Spectroscopy?
0:38
Microwave, Infrared Radiation, Visible & Ultraviolet
1:53
Equation for the Frequency of the Absorbed Radiation
4:54
Wavenumbers
6:15
Diatomic Molecules: Energy of the Harmonic Oscillator
8:32
Selection Rules for Vibrational Transitions
10:35
Energy of the Rigid Rotator
16:29
Angular Momentum of the Rotator
21:38
Rotational Term F(J)
26:30
Selection Rules for Rotational Transition
29:30
Vibration Level & Rotational States
33:20
Selection Rules for Vibration-Rotation
37:42
Frequency of Absorption
39:32
Diagram: Energy Transition
45:55
Vibration-Rotation Spectrum: HCl
51:27
Vibration-Rotation Spectrum: Carbon Monoxide
54:30
Vibration-Rotation Interaction

46m 22s

Intro
0:00
Vibration-Rotation Interaction
0:13
Vibration-Rotation Spectrum: HCl
0:14
Bond Length & Vibrational State
4:23
Vibration Rotation Interaction
10:18
Case 1
12:06
Case 2
17:17
Example I: HCl Vibration-Rotation Spectrum
22:58
Rotational Constant for the 0 & 1 Vibrational State
26:30
Equilibrium Bond Length for the 1 Vibrational State
39:42
Equilibrium Bond Length for the 0 Vibrational State
42:13
Bₑ & αₑ
44:54
The Non-Rigid Rotator

29m 24s

Intro
0:00
The Non-Rigid Rotator
0:09
Pure Rotational Spectrum
0:54
The Selection Rules for Rotation
3:09
Spacing in the Spectrum
5:04
Centrifugal Distortion Constant
9:00
Fundamental Vibration Frequency
11:46
Observed Frequencies of Absorption
14:14
Difference between the Rigid Rotator & the Adjusted Rigid Rotator
16:51
Adjusted Rigid Rotator
21:31
Observed Frequencies of Absorption
26:26
The Anharmonic Oscillator

30m 53s

Intro
0:00
The Anharmonic Oscillator
0:09
Vibration-Rotation Interaction & Centrifugal Distortion
0:10
Making Corrections to the Harmonic Oscillator
4:50
Selection Rule for the Harmonic Oscillator
7:50
Overtones
8:40
True Oscillator
11:46
Harmonic Oscillator Energies
13:16
Anharmonic Oscillator Energies
13:33
Observed Frequencies of the Overtones
15:09
True Potential
17:22
HCl Vibrational Frequencies: Fundamental & First Few Overtones
21:10
Example I: Vibrational States & Overtones of the Vibrational Spectrum
22:42
Example I: Part A - First 4 Vibrational States
23:44
Example I: Part B - Fundamental & First 3 Overtones
25:31
Important Equations
27:45
Energy of the Q State
29:14
The Difference in Energy between 2 Successive States
29:23
Difference in Energy between 2 Spectral Lines
29:40
Electronic Transitions

1h 1m 33s

Intro
0:00
Electronic Transitions
0:16
Electronic State & Transition
0:17
Total Energy of the Diatomic Molecule
3:34
Vibronic Transitions
4:30
Selection Rule for Vibronic Transitions
9:11
More on Vibronic Transitions
10:08
Frequencies in the Spectrum
16:46
Difference of the Minima of the 2 Potential Curves
24:48
Anharmonic Zero-point Vibrational Energies of the 2 States
26:24
Frequency of the 0 → 0 Vibronic Transition
27:54
Making the Equation More Compact
29:34
Spectroscopic Parameters
32:11
Franck-Condon Principle
34:32
Example I: Find the Values of the Spectroscopic Parameters for the Upper Excited State
47:27
Table of Electronic States and Parameters
56:41
Section 23: Molecular Spectroscopy Example Problems
Example Problems I

33m 47s

Intro
0:00
Example I: Calculate the Bond Length
0:10
Example II: Calculate the Rotational Constant
7:39
Example III: Calculate the Number of Rotations
10:54
Example IV: What is the Force Constant & Period of Vibration?
16:31
Example V: Part A - Calculate the Fundamental Vibration Frequency
21:42
Example V: Part B - Calculate the Energies of the First Three Vibrational Levels
24:12
Example VI: Calculate the Frequencies of the First 2 Lines of the R & P Branches of the Vib-Rot Spectrum of HBr
26:28
Example Problems II

1h 1m 5s

Intro
0:00
Example I: Calculate the Frequencies of the Transitions
0:09
Example II: Specify Which Transitions are Allowed & Calculate the Frequencies of These Transitions
22:07
Example III: Calculate the Vibrational State & Equilibrium Bond Length
34:31
Example IV: Frequencies of the Overtones
49:28
Example V: Vib-Rot Interaction, Centrifugal Distortion, & Anharmonicity
54:47
Example Problems III

33m 31s

Intro
0:00
Example I: Part A - Derive an Expression for ∆G( r )
0:10
Example I: Part B - Maximum Vibrational Quantum Number
6:10
Example II: Part A - Derive an Expression for the Dissociation Energy of the Molecule
8:29
Example II: Part B - Equation for ∆G( r )
14:00
Example III: How Many Vibrational States are There for Br₂ before the Molecule Dissociates
18:16
Example IV: Find the Difference between the Two Minima of the Potential Energy Curves
20:57
Example V: Rotational Spectrum
30:51
Section 24: Statistical Thermodynamics
Statistical Thermodynamics: The Big Picture

1h 1m 15s

Intro
0:00
Statistical Thermodynamics: The Big Picture
0:10
Our Big Picture Goal
0:11
Partition Function (Q)
2:42
The Molecular Partition Function (q)
4:00
Consider a System of N Particles
6:54
Ensemble
13:22
Energy Distribution Table
15:36
Probability of Finding a System with Energy
16:51
The Partition Function
21:10
Microstate
28:10
Entropy of the Ensemble
30:34
Entropy of the System
31:48
Expressing the Thermodynamic Functions in Terms of The Partition Function
39:21
The Partition Function
39:22
Pi & U
41:20
Entropy of the System
44:14
Helmholtz Energy
48:15
Pressure of the System
49:32
Enthalpy of the System
51:46
Gibbs Free Energy
52:56
Heat Capacity
54:30
Expressing Q in Terms of the Molecular Partition Function (q)
59:31
Indistinguishable Particles
1:02:16
N is the Number of Particles in the System
1:03:27
The Molecular Partition Function
1:05:06
Quantum States & Degeneracy
1:07:46
Thermo Property in Terms of ln Q
1:10:09
Example: Thermo Property in Terms of ln Q
1:13:23
Statistical Thermodynamics: The Various Partition Functions I

47m 23s

Intro
0:00
Lesson Overview
0:19
Monatomic Ideal Gases
6:40
Monatomic Ideal Gases Overview
6:42
Finding the Parition Function of Translation
8:17
Finding the Parition Function of Electronics
13:29
Example: Na
17:42
Example: F
23:12
Energy Difference between the Ground State & the 1st Excited State
29:27
The Various Partition Functions for Monatomic Ideal Gases
32:20
Finding P
43:16
Going Back to U = (3/2) RT
46:20
Statistical Thermodynamics: The Various Partition Functions II

54m 9s

Intro
0:00
Diatomic Gases
0:16
Diatomic Gases
0:17
Zero-Energy Mark for Rotation
2:26
Zero-Energy Mark for Vibration
3:21
Zero-Energy Mark for Electronic
5:54
Vibration Partition Function
9:48
When Temperature is Very Low
14:00
When Temperature is Very High
15:22
Vibrational Component
18:48
Fraction of Molecules in the r Vibration State
21:00
Example: Fraction of Molecules in the r Vib. State
23:29
Rotation Partition Function
26:06
Heteronuclear & Homonuclear Diatomics
33:13
Energy & Heat Capacity
36:01
Fraction of Molecules in the J Rotational Level
39:20
Example: Fraction of Molecules in the J Rotational Level
40:32
Finding the Most Populated Level
44:07
Putting It All Together
46:06
Putting It All Together
46:07
Energy of Translation
51:51
Energy of Rotation
52:19
Energy of Vibration
52:42
Electronic Energy
53:35
Section 25: Statistical Thermodynamics Example Problems
Example Problems I

48m 32s

Intro
0:00
Example I: Calculate the Fraction of Potassium Atoms in the First Excited Electronic State
0:10
Example II: Show That Each Translational Degree of Freedom Contributes R/2 to the Molar Heat Capacity
14:46
Example III: Calculate the Dissociation Energy
21:23
Example IV: Calculate the Vibrational Contribution to the Molar heat Capacity of Oxygen Gas at 500 K
25:46
Example V: Upper & Lower Quantum State
32:55
Example VI: Calculate the Relative Populations of the J=2 and J=1 Rotational States of the CO Molecule at 25°C
42:21
Example Problems II

57m 30s

Intro
0:00
Example I: Make a Plot of the Fraction of CO Molecules in Various Rotational Levels
0:10
Example II: Calculate the Ratio of the Translational Partition Function for Cl₂ and Br₂ at Equal Volume & Temperature
8:05
Example III: Vibrational Degree of Freedom & Vibrational Molar Heat Capacity
11:59
Example IV: Calculate the Characteristic Vibrational & Rotational temperatures for Each DOF
45:03
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Lecture Comments (4)

1 answer

Last reply by: Professor Hovasapian
Fri Jan 16, 2015 9:55 PM

Post by tiffany yang on January 16, 2015

In example four, part B asked about the work, so we need to find change in volume. However, can delta V really be V of gas - V of liquid?
I thought that delta V would just be V of the gas....because no gas presents initially...something I learned by looking at the solution of my hw problem

Thanks Raffi!

1 answer

Last reply by: Professor Hovasapian
Sun Dec 21, 2014 8:45 PM

Post by David Llewellyn on December 18, 2014

In example II the delta(H)o 1000 of the reaction is given as -123.8 kJ/mol but at the end of the problem you divided the delta(H)o 298 obtained by delta(H)o 1000 - Int[delta(Cp)dt]1000 298 by 2 to give the delta(H)o form.  The answer you obtained is close to the tabulated value so shouldn't delta(H)o 1000 just be the heat of reaction in Joules rather than Joules/mol?

Thermochemistry Example Problems

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
  • Example I: Find ∆H° for the Following Reaction 0:42
  • Example II: Calculate the ∆U° for the Reaction in Example I 5:33
  • Example III: Calculate the Heat of Formation of NH₃ at 298 K 14:23
  • Example IV 32:15
    • Part A: Calculate the Heat of Vaporization of Water at 25°C
    • Part B: Calculate the Work Done in Vaporizing 2 Mols of Water at 25°C Under a Constant Pressure of 1 atm
    • Part C: Find ∆U for the Vaporization of Water at 25°C
    • Part D: Find the Enthalpy of Vaporization of Water at 100°C
  • Example V 49:24
    • Part A: Constant Temperature & Increasing Pressure
    • Part B: Increasing temperature & Constant Pressure

Transcription: Thermochemistry Example Problems

Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.0000

Today, we are going to talk about Thermal Chemistry.0005

We just finished our discussion with several examples of Thermodynamics.0007

Discussing energy and heat and work and things like that.0011

I thought it would be best instead of discussing Thermal Chemistry theoretically and then doing problems, I think it is best just do the problems.0015

And then within the problems itself, discuss anything that needs to be discussed theoretically which is actually very little.0021

Much of this is stuff that you already know from General Chemistry.0031

Just a couple of the problems are a little bit more sophisticated than what you are you are used to, except for the first which is going to be very basic.0034

Let us jump right on in.0041

This one was nice and easy, it says find δ H for the following reaction.0044

We have titanium dioxide + chlorine gas goes to titanium chloride and oxygen gas.0049

Notice we have solid gas, solid gas.0059

Finding the δ H for the reaction is really simple.0063

Again we know this from general chemistry.0066

Let us see what color I’m going to use, I think I’m going to do blue.0068

We know that the δ H is equal to what would you say the δ H of the reaction and0074

we remember that this little degree sign up above there represents standard conditions.0086

Standard conditions happen to be 25°C or 298 K and 1 atm pressure.0090

It is just we need a particular standard so we choose that standard.0097

It is equal to the sum, I have forgotten how to write this term here.0102

The sum of the δ H is of formation for the products - the sum of the δ H is of formation for the reactants including coefficients.0113

Again, this is something that we already know from general chemistry, it is products – reactants.0130

Whenever you see a standard table of thermo chemical data or thermodynamic data, you are going to have the enthalpy a formation listed.0136

You are going to have the standard free energies and you are going to have the entropy listed.0144

And you can use those tables, you just take the enthalpy of the products, subtract the enthalpy of the reactants making sure to include the coefficients0148

and that gives you the actual enthalpy change for the heat of reaction under constant pressure conditions,0160

because under constant pressure conditions the enthalpy is equal to the heat.0169

Let us just go ahead and calculate this.0176

All we need is a table of thermodynamic data so we have δ H for elements and their standard state is 0.0177

This is chlorine gas CO₂ so that is going to be 0 and O₂ gas that is going to be 0.0193

We only have to worry about that one and that one right there.0198

When we look up the enthalpy for the Ti Cl4, we have -803 kJ/ mol × 1 mol.0203

That was canceled and we add 2 at the enthalpy for this which is 0 and we subtract the sum of the enthalpy of formation for the reactants.0220

And we get, for the titanium dioxide, we look it up and -945 kJ/ mol.0231

Again, the coefficient on that is 1 so it is × 1 mol + 0.0240

It is going to be -803 --945 so we have a δ H for this reaction is going to equal 142 kJ or 142,000 J.0248

Make sure you watch out for the units, it is really important.0265

When you are looking at the table of thermodynamic values, the enthalpy of formation and the standard free energies of formation those are in kJ/ mol.0270

The entropy is going to be in J/ mol/ K.0282

Later on, when we actually get to the equation you are familiar with, δ G = δ H - T δ S, when you are mixing does up and0287

when you are working with entropy, as well as enthalpy and free energy, we want to make sure the units match.0295

We either need to convert to J or we need to convert the entropy to kJ.0301

142 there you go, that is it.0310

This is positive enthalpy which means this is an endothermic reaction.0314

In other words, in order for this reaction to both go forward, it actually has to pull 142 kJ of heat from the surroundings, it has to go into the system.0318

Let us go ahead, that was nice and easy.0333

Let us go ahead and this one says calculate δ U for the reaction in problem 1, assuming the gases behave ideally.0336

This is different, now we are not calculating enthalpy, we are calculating the change in energy.0344

Let us go ahead and write our equation again just so we had on this page.0351

We had a titanium dioxide which was solid + we had 2 Cl2 which was gaseous and it went 2 Ti Cl4 the titanium 4 chloride which was solid + 02 gas.0355

Here is what is interesting.0379

Notice that we have solid gas and a solid gas, when we run these reactions, when calculating heats of reaction,0380

If any of the reactants or products are gaseous as they are in this case, the reaction has to be one of the bomb colorimeter.0399

A bomb colorimeter, if you remember from general chemistry or if you do not I will explain it right now.0436

It is basically is a colorimeter where the volume is held constant.0441

We put everything in sight of this thing called a bomb and it is immersed in water and we run the reaction inside.0445

We generally tend to keep the temperature and we tend to keep the volume.0452

The volume is absolutely fixed because it is a bomb, it is a single volume, there is no expansion.0458

This makes sense because if you are going to produce, for example this reaction produce oxygen gas, you have to contain that gas.0463

You cannot just let it escape into the atm which is why we have to constrict and make sure it stays in one place.0468

There are some things that we can control.0478

In this case, we are controlling volume and we also make sure the temperature stays constant so the only pressure that ends up changing.0480

Let us go ahead and see what is happening here.0489

The transformation is this.0492

Let us go ahead and see, the transformation is we have our reactants and it is going two products.0496

It is going to be at a given pressure, the reactant is going to be at a given pressure P1, a certain temperature, and a certain volume.0511

We are going to end up keeping temperature and volume the same experimentally but we are going to end up with a new pressure.0520

This is what happens in terms of the state variables P1 TV to P2 TV.0526

The relation between δ H and δ U, we are looking for δ U.0533

We already calculated δ H from the previous problem.0538

The relation is remember δ H = δ U+ δ PV.0542

Remember the relation H = U + PV, in this case δ H = δ U+ δ PV that is the actual relation.0560

But notice we are holding volume constant so it comes out.0569

What you end up with is δ H = δ U+ δ P × V.0573

Now δ H = δ U, δ P is just the final pressure - the initial pressure inside that bomb × the volume.0585

Again, this says assuming the gases behave ideally.0600

The ideal gas law is PV = nrt therefore P = nrt/ V.0604

Therefore, pressure 2 = n2 RT/ T.0615

Remember, T is constant and V is constant, R is a constant, it is the gas constant.0622

The only thing that changes is the number of moles and P1 = n2 × RT/ V.0625

Therefore, we will put these expressions into here and we end up with δ H = δ U + n2 RT/ V – n1 RT/ V × V.0635

The V's cancel out and we are left with δ H = δ U + n2 – n1.0658

I will pull out the RT δ n × RT.0665

δ n is just the difference in the number of moles of the gaseous components of products and reactants.0673

δ n is just the number of moles and product - the number of moles in the reactant of the gaseous products.0680

The solids they do not count, they do not contribute to anything in terms of actual pressure of things like that.0688

It is the gas that contributes to the pressure.0694

This is our basic equation that we are going to need.0696

What we are going to do is we are going to move this over the other side to get the δ U because δ U is what we want.0698

Let us go ahead and calculate δ n.0707

δ n = the number of moles of gaseous products - the number of moles gaseous reactants.0711

Again that is just the coefficients δ n = 1 -2 = -1.0722

δ H = δ U+ δ N × RT move this over and we are left with δ U = δ H – δ n × RT.0735

We already calculated δ H, I’m going to do this in terms of, that was 142 kJ.0754

We have 142,000 J.0762

I’m just going to go ahead and convert them to J because R is here, I’m going to use the 8.314 J/ K mol - δ n is -1, R is 8.314 J/ mol K, and then RT temperature at 298, 25°.0767

When I do this, I end up with δ U= 144000 J or 144 kJ.0794

What is important was what we derive, this relation right here.0817

It all comes from the definition of enthalpy H = U + PV.0821

δ H = δ U + δ PV.0828

V is held constant so I can pull that out so I just get δ H = δ U+ δ P × V.0832

Pv = nrt so I put that in and I'm left with this final relationship here.0845

Enthalpy energy is the relationship that we used.0852

Let us go ahead and see what else we can do here.0856

This is one of the problems that you probably did not see when you are in general chemistry and it is a very important problem0866

because it is going to allow you to calculate enthalpy is a different temperatures.0874

All your enthalpy is that we calculated in general chemistry where from the table of thermodynamic data and all that is done at 25°C 1 atm pressure.0880

They told you this at the time but of course you did not do anything with it, that enthalpy reaction depends on the temperature which we run the reaction.0891

The reaction was at 25°C vs. Reaction that is 225°C, the enthalpy is going to be different.0900

We need to find a way to calculate the enthalpy at a new temperature.0907

That is what I do with this problem.0911

The following data holds at 1000 K.0915

We have nitrogen gas + 3 mol of hydrogen gas goes to 2 mol of ammonia gas.0919

Sorry about that, it should be right there, the δ H for this reaction at 1000° = -123.8 kJ/ mol.0926

Here is some more data here, the substance nitrogen gas its molar heat capacity is 3.50 × R.0939

Hydrogen has a molar heat capacity of 3.47 R and ammonia has a molar capacity of 4.22 R.0948

We want to calculate the heat of formation of NH3 at 298 K.0955

We want to find the heat formation NH3.0962

In other words, we want to find this.0965

We want to find N2 + 3H2 goes to 2 NH3.0967

We want a find δ H of formation at 298.0976

The δ H of formation is for formation of 1 mol.0981

We actually have a something like this.0987

This is actually what we want, the δ H of formation is the heat of a reaction for the formation of 1 mol of whatever it is that we are looking for.0998

The reactants and their elements in their most basic state, nitrogen gas, hydrogen gas, goes to 1 mol of ammonia gas.1010

This is what we are looking for.1020

Let us get started.1022

We will write this again so δ H values are generally temperature dependent.1025

In other words, the enthalpy change for a reaction depends on the temperature at which a reaction is run.1043

Let us begin with this one, let us begin with δ H = H of the products - the H of the reactants.1074

That is what δ is, just products – reactants that is the definition of the δ.1087

I will go ahead and differentiate this with respect to T.1090

We have differential with respect to temperature not time.1094

Differential with respect to temperature so we have DDT of this δ H = DH DT – DH sub R DT.1101

Products and reactants, DH DT we know what that is.1120

Anytime you have an enthalpy divided by a temperature that is the definition of heat capacity.1132

DH DT that is the definition of constant pressure heat capacity.1138

These are standard, let us just go ahead and put the standard, in other words, the heat capacity at 25°C and 1 atm pressure.1146

Since that is the case, this D of the δ H with respect to temperature, this thing right here that = δ of the heat capacity,1155

in other words, the heat capacity of all of the products - the heat capacity of the reactants.1180

That is what we are saying here, that is what this is.1186

Because DH DT = heat capacity is δ H is actually δ CP.1190

We are just taking the heat constant pressure heat capacity of the products - the constant pressure heat capacity of the reactants.1197

This is this and this is this.1208

Since we have that, let us go ahead and move this over there and we have the following.1217

We have the D of δ H standard = δ CP × DT.1222

Let us go ahead integrate this.1235

If we integrate this from temperature 1 to temperature 2, we integrate this from temperature 1 to temperature 2.1238

Here, what this becomes is just δ H at temperature 2 - δ H at temperature 1.1246

Write the integral of the D just go away and you are left with δ H and you do the T2 - T1 = the integral from T1 to T2, the difference in the heat capacity DT.1264

Therefore, I just take this and move it over here and I get the change in enthalpy of the given reaction1282

at a particular temperature T2 = the change in enthalpy at a given temperature.1291

The way I keep this standard is actually the standard does not refer to temperature, the standards refers to pressure.1302

When you see this thing right here, it actually means 1 atm pressure.1309

It just happens that most of the work that we do in general chemistry happens at 25°C.1313

We also include that 25°C and its degree sign but you often see the degree sign for different temperatures1319

because that really just refers to the 1 atm pressure.1326

It is when we change the pressure that this degree sign tends to disappear.1328

I will move this over + the integral from T1 to T2 of δ CP DT.1333

This is the equation that we wanted.1343

This is the equation that we are going to use.1344

Here is what is going on, it is telling me that if I want to calculate the change in enthalpy of the reaction at a given temperature like 1000°,1348

I start by calculating the δ H at the temperature that I do know which is 25°.1359

Then, I integrate the difference in the heat capacities of the products and the reactants from one temperature to the other.1365

That is what this says.1375

I have a way of calculating the enthalpy change for any reaction if I already know the change at 25°C1376

and if I know the constant pressure heat capacities for the reactants and the products.1385

Constant pressure heat capacities are tabulated for everything.1390

These are just because I have to look them up in a table, the way you would in other bit of information in chemistry and physics.1395

Let us go ahead and do this problem.1401

This is the import the equation.1404

Again δ CP, let me write that down, where this δ CP = the sum of the heat capacities of the products - the sum of the heat capacities of the reactants.1406

That is all this is.1430

Something very important to remember that is why I put several asterisk by it.1433

When we calculate enthalpy for elements are going to be 0 like the O₂ and the Cl₂ it was 0.1441

Heat capacity is never 0.1451

Therefore, when you are dealing with elements you cannot ignore the heat capacity.1453

The heat capacity for every product in every reacted must appear in this equation.1458

The heat capacity for elements is not 0 like it is for the δ H of formation.1468

The δ H of formation of elements is 0.1484

All heat capacities must be accounted for and again multiplied by the respective coefficients.1491

Here is what we have, we have that δ H at a given temperature = the δ H at some initial temperature + the integral from T0 to T of the change in the δ CP DT.1530

We can set up this integral up in two ways.1549

I will go ahead and show you the two different ways to set up and choose one to actually work with.1558

We can set up this way.1562

We are looking for the δ H in this particular problem, we are looking for the δ H of formation at 298°K,1564

because the information they give us is 4000 K so we can set up this way.1570

We can go δ H at 298 = δ H at 1000 + the integral from 1000 to 298.1577

It is going to be from this temperature to that temperature.1608

Generally, we tend to increase temperature.1611

In this case, they gave us the δ H at a higher temperature.1613

They want to know what it is for the normal or lower temperature.1616

It is what you are given to where you are going, that is the integral.1619

This is going to end up being negative of the δ CP DT or I can set up this way.1624

I can set up as δ H 1000 = δ H of 298 + the integral of 298 to 1000 of δ CP DT.1634

In this case, some people like to go, they like the lower limit integration to be a lower number, it is a personal choice.1651

It does not really matter.1657

In this case because this is the number you are looking for, you are given this one, you are going to calculate this one.1658

When you do this arithmetic, you are just going to move this over to actually get that value.1663

Let us go ahead and do it this way.1670

The δ H at 298 is going to equal δ H at 1000 - the integral 298 to 1000 of δ CP DT.1677

Let us go ahead and calculate δ CP.1695

δ CP is equal to the capacities of the products - the capacities of the reactants multiplied by the coefficients.1698

We will do this in red.1708

Our equation was, I’m just going to write the equation right here.1709

I had N2 + 3H2 = 2NH3.1715

For NH3= 4.22 × R.1725

Remember it was 4.22 R.1732

So × 8.314 and then there are 2 mol × 2, that is our products.1734

We will do the N2.1743

The N2 was 3.50 R, 8.314, and there is 1 mol of it + for H2 it was 3.47 R 8.314.1746

And of course, there are 3 mol of it, that is that.1763

Our δ CP δ of the heat capacities of products and reactants = -45.48 J K.1767

We can actually run the reaction.1784

Our δ H of 298 = the δ H.1787

That is what we are doing, we decided to use this version of it.1800

Not this version but this version, δ H of 298 it was the δ H to 1000.1804

That was given to us, the δ H 1,000 was -123.8 kJ.1808

So -123,800 J - the integral from 298 to 1000 of the δ CP which we just calculated as - 45.48 DT.1814

You can do this by hand or put a mathematical software, you end up with the following.1840

When you actually do this calculation, what you are going to end up with is - 91,873 J but notice this is for the formation of 2 mol of the NH3.1851

We were looking for the δ H of formation which by definition is the formation of 1 mol.1870

We just divide that by 2.1874

Our δ H of formation, sorry about that not the reaction.1877

The δ H of formation is going to be at 298 going to be - 45.9 kJ/ mol, that is our final answer, that is what we wanted.1884

And it is all based on this right there.1905

This the δ H at 1 temperature is the δ H at any given temperature is that 298 - 25°C that we normally get from thermodynamic cables +1912

the integral up the difference of the heat capacities of products and reactants.1925

It is very important equation.1930

Let us move on to the next one.1936

This look like it is going to be long.1937

Hopefully, it is not too bad.1939

Part A, using a table standard thermodynamic data, calculate the heat of a precision of water at 25°C.1942

This one is very easy, not a problem.1948

Calculate the work done in vaporizing 2 mol of water at 25°C under constant pressure of 1 atm.1951

It should not be too bad because we know what the definition of work is.1958

We know the work is external pressure × change in volume.1961

I will go ahead and write that down here so work = external pressure × change in volume, that is just the definition of work.1968

Find a δ U for the vaporization of water.1978

We are probably end up using that relationship δ H = δ U+ δ PV and we are going to probably work with that and manipulate that little bit.1984

The CP for water vapor is this, the CP in other words the constant pressure heat capacity for liquid water is this.1997

Find the enthalpy of vaporization of water at 100°C.2003

We are going to do what we just did.2007

We are going to find enthalpy in a given temperature and we are going to use it to find enthalpy at a higher temperature.2009

Let us jump right on in and see what we can do.2016

Part A, use a table standard thermodynamic data, calculate the heat of a precision of water at 25°C.2020

Part A, our reaction is this, we are going H to a liquid to H2O gas and we want to find δ H of this reaction.2030

This is the δ H of vaporization, in this reaction we are taking the liquid and we are converting it into a gas.2043

Δ H of vaporization happens to equal the δ H of the reaction which is this reaction which is equal to.2050

Let us look this one up, it is this – this, what you end up with is -241.814 - -285.830 so you end up with δ H of vaporization = 44.016 kJ/ mol.2062

That is our answer.2097

A 25°C 44.016 kJ of heat needs to be put into the liquid water in order to convert all of it to gaseous water, that is all the says.2099

The δ H of vaporization is, how much heat is required to affect this transformation.2112

If I'm at 25°C, in order to convert all that water I need to put in this much heat.2118

Part B, what is the work required to affect this transformation?2125

We said that the work = the external pressure × the change in volume.2133

We know what the external pressures is 1 atm, this is happening under atmospheric conditions.2139

You are just boiling water, it is what you are doing.2143

Our δ V = the volume of the gas - the volume of liquid.2147

In other words liquid has a certain volume, you are converting that all to water vapor gas.2155

The change in volume is the difference between the volume that the gas occupies and the volume of liquid occupies.2160

We have a little bit of calculation to do to find out what the volume of the liquid is.2166

We only know what the volume of the gas is, we can just go ahead and use what we know from general chemistry which is 22.4 L/ mol.2172

Let us see what we can do as far as the volume of liquid is concerned.2179

The volume of liquid, volume of the liquid water, here is how you calculate it.2183

25°C, the density of water is 0.9970 g/ ml.2190

There are 1000 ml in 1 L and there are in 1 mol of water 18 g.2200

G cancels g, ml cancels ml, I'm left with mol and L.2214

The number is going to be 55.39 mol/ L.2219

I need the molar volume so I need L/ ml.2228

I’m going to this, when I reciprocated it I end up getting 1/ 55.39 mol/ L gives me 0.01805 L/ mol.2232

1 mol liquid water at 25°C occupies 0.01805 L.2255

This is just a simple conversion.2262

I have 2 mol of water so I just multiply by 2 and I get 0.03611 L.2267

2 mol of water at 25°C liquid water occupies 0.03611 L.2282

That is the volume of liquid water.2288

Let us go ahead and find the volume of the gas.2293

The volume of the gas is very simple.2296

We have 22.414 L/ mol.2299

I have 2 mol of water so I have 44.828.2306

Please confirm my arithmetic, notorious for arithmetic mistakes.2315

The volume of our gas is that, δ V our change in volume is the volume of the gas - the volume of liquid.2318

It is equal this - that so 44.828 -0.03611.2328

I'm left with 44.792 L.2341

Our δ V is that, now we can go to our problem.2347

How our work = our external pressure × change in volume.2352

Our external pressure is 1 atm, in this particular case our change in volume was 44.792 L.2360

Therefore, our work = 44.792 L atm that is a unit of energy but we need to convert that J.2370

Therefore, 44.792 L atm is × 8.314 J =.08206 L atm.2383

Those are just the two different versions of R, the gas constant in J/ mol K L atm/ mol K.2400

That is your conversion factor.2407

When you do this, you end up with the work = 4.5 kJ/ 2 mol.2410

Per mol = 2.25 kJ/ mol.2422

There we go.2430

What this says that if I have 1 mol water at 25°C, the amount of work that I do in converting that liquid water to gas,2432

When I'm converting it to gas I’m pushing away the atm, the gas, the liquid, the water vapor is expanding.2444

It is doing work, it is pushing against the atm.2451

The amount of work it is doing per mol of water is 2.25 kJ.2453

That is a lot of work.2457

Part C, it wants you to find the change in energy for this transformation.2462

We know that δ H = δ U.2468

I will just start with the definition of enthalpy H = U + PV.2473

Therefore, δ H = δ U+ δ PV.2482

Pressure is constant, it is the atmospheric pressure so that P comes out.2493

Therefore, what you have is now δ H = δ U+ P δ V.2498

δ H = δ U + δ V that is just the work.2510

We just did that, that is just work.2516

Therefore, our δ U = δ H – W.2519

We already calculated δ H that was 44.016 kJ/ mol.2526

We calculate the work which was 2.25 kJ/ mol.2534

Therefore, the change in energy is 41.766 kJ/ mol.2540

In going from liquid to gas at 25°C the energy of the system increased by 41.76 kJ, that is a huge amount of energy, that is all this is saying.2555

This is all based on just simple basic mathematical relationship that we already know,2568

the handful of equations that we already know from our work with work and heat and energy in the first law of thermodynamics.2572

We are just applying it to chemical situations.2580

Let us see what we have for D.2585

Let us see if I have another page actually available.2587

I do, great.2590

Part D, they want to know what the δ H is at 100°C instead of 25°C.2593

The δ H that we actually calculated was this thing right here was the first that we get part A.2605

This is a 25°C, that is the δ H in 100 ℃.2612

First of all, let us go ahead and work in K so 25°C = is the same as 298 K and 100°C =373 K.2619

Therefore, from the equation that we have from the previous problem our δ H at 373 K, 1 atm pressure its equal to the δ H at 298 K +2637

the integral from 298 to 373 of the difference in the heat capacities of products and reactants, gas or water vapor, liquid water, DT.2660

That is it, nice and simple.2676

Let us go ahead and do this.2678

Let us go ahead and write up the whole thing δ H 373.2683

Let us go ahead and do that is going to equal δ H at 373 is going to equal δ H of,2693

Let me write this again to keep it on one page + the integral from T0 to δ CP DT.2705

Therefore, δ H 373 = δ H at 298 which we calculated which was part A 44.016 + the integral from 298 to 373.2718

And now the difference in the heat capacities, they gave us those in the beginning part of a problem in part D.2735

Gas, water, that was 33.577 and that is the molar heat capacity for water vapor - 75.291 molar heat capacity for the liquid water.2741

We end up with 44.016 + -41.714 × 75.2763

Just doing this by hand, this is this number and this is a constant so it pulls out.2780

This integral of the DT is just δ T or δ T, between 290 - 373 is 75.2785

What you end up with is δ H at 373 = 40.9 kJ/ mol.2791

This tells me now at 25°C if I'm converting liquid water to water vapor, I have to put it that much heat 44.016 kJ/ mol of water.2806

At 100°C, I only have to put in 40.9 kJ/ mol.2818

Notice, this is less and that makes sense.2824

I'm at a higher temperature so I do not have to work as hard.2826

In order boil water, I have to get it to 100.2829

A part of this energy here, about 4 kJ is used just to get my water from 25°C to 100°C and I can start boiling it off.2834

If I’m already at 100℃, all that heat that I put in automatically goes towards just boiling it off.2843

I do not have to raise it because it is boiling point temperature.2848

This makes sense, empirically it makes sense.2851

We know from experience.2854

Let me write that down.2856

Notice, it takes less energy to vaporize water at 100°C than it does at 25°C.2859

And this makes sense.2887

If you ended up with a higher number than 44 then something happened, something went wrong.2893

A lot of times we get so wrapped up in the mathematics and I'm guilty of this just as much as anybody else,2898

it is probably more so because to the mathematics that we do not pull back and stop and take a look if this actually make sense physically.2902

It is very important that we want you to be able to develop mathematical skills that are important to solve these problems2910

but we want you to understand what is happening conceptually.2917

We want you to realize, wait a minute I have a 25°C of vaporizing water, a certain amount of heat is going to required for that, that is the δ H.2920

If I’m at 100°C that is the boiling point of water already.2930

I should stand back and say I should require less energy because I’m already at the boiling point.2936

Therefore, the number that I get should be lower, sure enough it is lower 40.9 vs. 44.016.2942

It makes sense physically.2952

Let us round of this discussion of thermal chemistry with one more problem here and let us see what we can do.2957

Liquid water has a molar volume of 18.0 cc/ mol, if the temperature is held constant.2966

Temperature is constant and the pressure is increased by 15 atm.2975

We take it from 1 atm to 15 atm, calculate the enthalpy change then calculate the enthalpy change by 15° increase in temperature when the pressure is held constant.2980

Here what we are doing is we have a certain volume of water, we are asking if I hold the temperature constant in the first case and increase the pressure,2993

what is the enthalpy change for that particular process?3004

In the second process, if I hold the pressure constant the change in temperature what is the enthalpy change?3011

That is what is happening here and the molar heat capacity 75.3 J/ mol K.3018

Enthalpy change, remember one of the basic equations is the following.3029

Let us go back to blue on this one , DH = the constant pressure heat capacity DT + DH DP.3032

DP this is the general equation, remember that the change in enthalpy of the system, I can change the temperature,3049

I can change the pressure, that is how I can change the enthalpy.3056

The change is equal to the constant pressure heat capacity × the differential change in temperature + this pressure dependence in enthalpy.3063

That is the rate at which the enthalpy changes per unit change in pressure × DP.3076

If I hold the temperature constant this just goes to 0 only this matters.3082

If I hold pressure constant this goes to 0 only this matters.3086

We are going to do both.3089

This equation is not valid for gas, it is valid for liquids and for solids.3093

All of the equations that we developed so far they are valid for every single state of aggregation solid, liquid, gas, for every single phase.3097

It does not really matter, we just need to have the particular values that are important to be able to solve the problem.3106

The only thing that we need to recall is that for a liquid this DH DP at constant temperature it is actually equal to the volume of a particular sample.3113

Let us go ahead and do the first part of temperatures held constant and the pressure is increased at 15 atm.3137

In this particular case, the temperature is being held constant so that is going to go to 0.3143

What I had is DH = DH DP T × DP.3162

The DH DP = V so I have is DH = V DP.3174

If I integrate this I end up with the following.3183

I end up with δ H = V × the change in pressure so that is what I'm going to look at.3187

I have the volume and I'm going to find the change in pressure.3199

I already know what the change in pressure is, it is just 1 atm to 15 atm.3203

I just need to know what the volume is.3208

Let us go ahead and make some conversions here.3212

I have 18.0 cm³ × 10⁻⁶ m³/ cm³ I need to convert this cm³ to m³ because when I multiply by a pressure3216

which is going to be in a Pascal, the Pascal is defined in terms of the cubic meter not the cm³ .3240

We have to watch for the units.3247

When I go ahead and do that, I would actually change the 15 atm to Pascal, 15 atm × 1.01 × 10⁵ Pascal/ 1 atm.3250

I’m just making some conversions here, = 15. 15 × 10⁵ Pascal and this is going to be 18 × 10⁻⁶ m³.3273

Therefore, my δ H = V × δ P.3291

My V is this, my δ P is this, I will just multiply them.3294

So δ H = 18.0 × 10⁻⁶ m³/ mol and this is going to be 15.15 × 10⁵ Pascal, when I do this, my δ H going to be 27.27 J.3300

I use my basic equation in this particular case temperature was held constant so this term is unimportant.3337

This for a liquid happens to equal V so I get this and I’m left with δ H = V δ P.3344

And what the volume is, I just have to make sure to be in appropriate unit.3352

I know what the change in pressure is, I just need to make sure it is in the appropriate unit.3355

I will go ahead and calculate the J.3359

Notice, this is not very much 27.27 J.3361

For constant pressure, let us do the same thing.3366

We go ahead and move this back up here.3370

For constant pressure, let us rewrite the equation.3380

The basic equation it never hurts to actually write the equations that you are supposed to know and memorize over and over again.3383

DH = CP DT + DH DP constant temperature DP.3389

And this particular case, it is the pressure that is held constant so this goes to 0 that means there is no change in pressure.3399

DP is 0 that means this is 0.3405

What I have is DH=CP DT.3409

When I integrate this, I end up with δ H = the constant pressure heat capacity × δ T the change in temperature.3414

Therefore, the δ H I should go ahead and put P to let me know that this is happening under constant pressure = CP they gave it to us already.3427

This is 75.3 J/ mol K and the difference in temperature was 50°.3440

50°C the degree increment, the δ of Celsius is the same as the δ of K.3455

Therefore, this is 50 K.3462

What we end up with is 1130 J/ mol.3466

That is a very big difference between 27 J/ mol and 1130 J/ mol.3474

Basically, what this says is that if I have a liquid and if I put some pressure on it, it does not really change the enthalpy of the system.3481

In other words, we are not really investing a lot.3489

If I just change the temperature by 50° all of a sudden invested a lot of energy into this.3493

The enthalpy has gone up by 1130 J.3502

Clearly, for liquid and for a solid, temperature has a much greater affect on the enthalpy of the system than pressure.3507

Obviously, if I put like 10,000 atm pressure on it, that is going to be different but normally with the levels of pressure3515

that we work with the laboratory which is maybe 1 to 20, 30, 40, something like that.3524

It really does not matter.3529

The effect of pressure on enthalpy is for the most part negligible, that is really all that this is saying.3531

That took care of the thermal chemistry via examples.3541

Thank you so much for joining us here at www.educator.com.3543

We will see you next time, bye.3545

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