Thermodynamics

1
Thermodynamics

• Honors Unit 5

2
Energy Basic Principles

• Thermodynamics the study of energy changes
• Energy the ability to do work or produce heat
• Note Work is force acting over a distance

3
Energy Basic Principles

• Kinetic Energy energy of motion
• KE
• Potential Energy
• energy due to position
• or composition

4
Law of Conservation of Energy

• A.k.a. first Law of Thermodynamics
• Energy can be converted from one form to another
  but cant be created or destroyed
• This means the total energy of the universe is
  CONSTANT!

5
Heat vs. Temperature

• Temperature measure of the random motion of a
  substance
• Temperature is proportional to kinetic energy (it
  is a measure of the average kinetic energy in a
  substance)
• Heat (q) flow of energy due to a temperature
  difference

6
Important Aspects of Thermal Energy Temperature

• Heat is NOT the same as temperature!
• The more kinetic energy a substance has, the
  greater the temperature of its atoms and
  molecules.
• The total thermal energy in an object is the sum
  of its individual energies of all the molecules.
• For any given substance, its thermal energy
  depends not only on its composition but also on
  the amount of substance

7
System vs. Surroundings

• A system is the part of the universe we are
  studying.
• The surroundings
• are everything else
• outside
• of the system.

8
Direction of Heat Flow

• Heat transfer occurs when two objects are at two
  different temperatures.
• Eventually the two objects reach the same
  temperature
• At this point, we say that the system has reached
  equilibrium.

9
Thermal Equilibrium

• Heat transfer always occurs with heat flowing
  from the HOT object to the COLD object.

10
Thermal Equilibrium

• Transfer of heat continues until both objects are
  at the same temperature!

11
Thermal Equilibrium

• The quantity of heat lost by the hotter object
  and the quantity of heat gained by the cooler
  object are EQUAL.

12
Exothermic vs. Endothermic

• Exothermic process ? heat is transferred from
  system to the surroundings
• Heat is lost from the system (temperature in
  system decreases)
• Endothermic process ? (opposite of exothermic
  process) heat transferred from surroundings to
  the system
• Heat is added to the system (temperature in
  system increases)

13
Exothermic Process
14
Endothermic Process
15
Units of Energy

• Joule (J) is the SI unit of energy heat
• One kilojoule (kJ) 1000 joules (J)
• calorie (cal) heat required to raise the
  temperature of 1.00 g of water by 1 C
• 1 calorie 4.184 J

16
Units of Energy

• Food is measured in Calories (also known as
  kilocalories) instead of calories
• 1 Cal 1 kcal 1000 calories

17
Units of Energy

• 3800 cal __________ Cal _________ J

18
Units of Energy

• The label on a cereal box indicates that 1
  serving provides 250 Cal. What is the energy in
  kJ?

19
Heat Transfer

• Direction and sign of heat flow MEMORIZE!
• ENDOTHERMIC heat is added to the system the
  temperature increases (q)
• EXOTHERMIC heat is lost from the system (added
  to the surroundings) the temperature in the
  system decreases (-q)

20
Heat Capacity

• The quantity of heat required to raise an
  objects temperature by 1 C (or by 1 Kelvin)
• Heat capacity is an extensive property.
• Which will take more heat to raise the
  temperature by 1 C?

21
Specific Heat (Specific Heat Capacity)

• Specific Heat (C) - The quantity of heat required
  to raise the temperature of one gram of a
  substance by 1 C
• Intensive property
• Units
• J/(gC) or J/(gK)
• cal/(gC) or cal/(gK)

22
Examples of Specific Heat

• At the beach, which gets hotter, the sand or the
  water?
• Higher specific heat means the substance takes
  longer to heat up cool down!

23
Examples of Specific Heat

• Specific heat (C) the heat required to raise the
  temperature of 1 gram of a substance by 1 C
• Cwater 4.184 J/(g?C)
• Csand 0.664 J/(g?C)

24
Calculating Changes in Thermal E

• q m x C x ?T
• q mC?T
• q heat (cal or J)
• m mass (g)
• C specific heat capacity, J/(g?C)
• ?T change in temperature, Tfinal Tinitial
• (C or K)
• All units must match up!!!

25
Example
q mC?t

• How much heat in J is given off by a 75.0 g
  sample of pure aluminum when it cools from 84.0C
  to 46.7C? The specific heat of aluminum is
  0.899 J/(gC).

26
Example
q mC?t

• What is the specific heat of benzene if 3450 J of
  heat are added to a 150.0 g sample of benzene and
  its temperature increases from 22.5 C to 35.8 C?

27
Example
q mC?t

• A 50.0 g sample of water gives off 1.025 kJ as it
  is cooled. If the initial temperature of the
  water was 85.0 C, what was the final temperature
  of the water? The specific heat of water is 4.18
  J/(gC).

28
Calorimetry

• Calorimetry measurement of quantities of heat
• A calorimeter is the device in which heat is
  measured.

29
Calorimetry

• Assumptions
• Heat lost -heat gained by the system
• In a simple calorimeter, no heat is lost to the
  surroundings

30
Coffee Cup Calorimetry

• Use styrofoam instead of a beaker to keep heat
  in
• Steps
• Add hot solid metal to cool water
• Water will heat up (T rises) as metal cools
• Eventually, water metal are at same T.
• qmetal qwater 0
• qmetal -qwater
• (Heat lost by metal -heat gained by the water)

31
Calorimetry

• qmetal -qwater
• Heat lost by metal -heat gained by water
• Since q mC?T,
• mmCm?Tm -mH2OCH2O?TH2O

32
Sample Problem

• A 358.11 g piece of lead was heated in water to
  94.1 C. It was removed from the water and
  placed into 100. mL of water in a Styrofoam cup.
  The initial temperature of the water was 18.7 C
  and the final temperature of the lead and water
  was 26.1 C. What is the specific heat of lead
  according to this data?

33

• A 358.11 g piece of lead was heated in water to
  94.1 C. It was removed from the water and
  placed into 100. mL of water in a Styrofoam cup.
  The initial temperature of the water was 18.7 C
  and the final temperature of the lead and water
  was 26.1 C. What is the specific heat of lead
  according to this data?

34
Bomb Calorimeter

• Constant volume bomb calorimeter
• Burn sample in O2
• Some heat from reaction warms water
• qwater mCH2O?T
• Some heat from reaction warms bomb
• qbomb Cbomb?T
• qrxn qH2O qbomb 0

35
Energy Changes of State

• All changes of state involve energy changes (more
  in Unit 9)
• Note that fusion melting

36
State Functions

• A property where the change from initial to final
  state does not depend on the path taken
• Ex.) The change in elevation from the top to
  bottom of a ski slope is independent of the path
  taken to go down from the slope

37
Enthalpy Changes for Chemical Rxns.

• Heat of reaction
• The heat absorbed or given off when a chemical
  reaction occurs at constant T (temp.) and P
  (pressure)

38
Enthalpy

• Enthalpy (H)
• The heat content of a reaction (chemical energy)
• ?H change in enthalpy
• The amount of energy absorbed by or lost from a
  system as heat during a chemical process at
  constant P
• ?H Hfinal - Hinitial

39
Properties of Enthalpy

• Enthalpy is an extensive property
• It does depend on quantity
• Enthalpy is a state function
• Depends only on the final initial values
• Every reaction has a unique enthalpy value since
  ?H Hproducts - Hreactants

40
Representation of Enthalpy as a Graph
41
Two Ways to Designate Thermochemical Equations

• Endothermic
• H2 (g) I2 (s) ? 2 HI (g) ?H 53.0 kJ
• H2 (g) I2 (s) 53.0 kJ ? 2 HI (g)

42
Two Ways to Designate Thermochemical Equations

• Exothermic
• ½ CH4 (g) O2 (g) ? ½ CO2 (g) H2O (l)
  ?H -445.2 kJ
• ½ CH4(g) O2(g) ? ½ CO2(g) H2O(l) 445 .2 kJ

43
Two Ways to Designate Thermochemical Equations

• Note the meaning of the sign in ?H in the
  equations above!!
• Endothermic ?H
• Exothermic ?H -

44
Two Ways to Designate Thermochemical Equations

• Note the important of designating the physical
  state or phase of matter. Why??
• Because this will change the heat of reaction!
  (?H)

45
Thermochemical Equations

• What do the coefficients stand for? How can they
  differ from the ones we have used before?
• Coefficients the number of moles (as before)
• BUT
• We can use fractional coefficients now!

46
Thermochemical Equations

• What is the standard state? How do we designate
  conditions of temperature and pressure that are
  not at standard state?
• Standard state 1 atm pressure 25 C
• ?H ?H at standard state
• Must show conditions over arrow if not at
  standard state!

47
Thermochemical Equations

• How can we find the enthalpy of reaction when we
  reverse it?
• Reverse the reaction, reverse the sign of ?H!
• Example
• CO (g) ½ O2 (g) ? CO2 (g) ?H -283 kJ
• CO2 (g) ? CO (g) ½ O2 (g) ?H 283 kJ

48
Example

• Given Rxn. 1, find the ?H for Rxns. 2 3
• Reaction 1
• 2 SO2 (g) O2 (g) ? 2 SO3 (g) ?H 197.8 kJ
• Reaction 2
• SO2 (g) ½ O2 (g) ? SO3 (g) ?H
• Reaction 3
• 4 SO3 (g) ? 4 SO2 (g) 2 O2(g) ?H

49
?H as a Stoichiometric Quantity

• Given the reaction below, how much heat is
  produced when 15.0 g of NO2 are produced?
• 2 NO (g) O2 (g) ? 2 NO2 (g) ?H -114.1 kJ

50
?H as a Stoichiometric Quantity

• Given ?H -283 kJ
• CO (g) ½ O2 (g) ? CO2 (g)
• (a) Calculate the enthalpy of the above reaction
  when 3.00 g of product are formed

51
?H as a Stoichiometric Quantity

• Given ?H -283 kJ
• CO (g) ½ O2 (g) ? CO2 (g)
• (b) If only 10.0 grams of oxygen and an unlimited
  supply of CO are available to run this reaction,
  how much heat will be given off?

52
?H as a Stoichiometric Quantity

• Given ?H -283 kJ
• CO (g) ½ O2 (g) ? CO2 (g)
• (c) How many grams of carbon monoxide are
  necessary (assuming oxygen is unlimited) to
  produce 500 kJ of energy in this reaction?

53
?H as a Stoichiometric Quantity

• Given ?H -283 kJ
• CO (g) ½ O2 (g) ? CO2 (g)
• (d) Calculate the heat of decomposition of two
  moles of carbon dioxide.

54
Hesss Law

• The heat of a reaction (?H) is constant, whether
  the reaction is carried out directly in one step
  or indirectly through a number of steps.
• The heat of a reaction (?H) can be determined as
  the sum of heats of reaction of several steps.

55
Hesss Law Example

• Consider the formation of water
• H2(g) ½ O2(g) ? H2O(g) 241.8 kJ
• (Exothermic Rxn ? ?H -241.8 kJ)

56
Hesss Law
57
Hesss Law
58
Hesss Law

• S ?H along one path
• S ?H along another
• Since ?H is a state function!!

59
Hesss Law

• Given
• C(s) O2(g) ? CO2(g) ?H -393.5 kJ
• 2 CO(g) O2(g) ? 2 CO2(g) ?H -577.0 kJ
• Determine the heat of reaction for
• C(s) ½ O2(g) ? CO(g)

60
Hesss Law

• Given
• C(s) O2(g) ? CO2(g) ?H -393.5 kJ
• C2H4(g)3 O2(g) ? 2 CO2(g)2 H2O(l) ?H -1410.9
  kJ
• H2(g) ½ O2(g) ? H2O(l) ?H -285.8 kJ
• Determine the heat of reaction for
• 2 C(s) 2 H2(g) ? C2H4(g)

61
Standard Enthalpies of Formation

• NIST (National Institute for Standards and
  Technology) gives values for
• ?Hf standard molar heat of formation
• Definition
• The heat content or enthalpy change when one
  mole of a compound is formed at 1.0 atm pressure
  and 25 C from its elements under the same
  conditions.

62
Examples of Formation Equations

• H2(g) ½ O2(g) ? H2O(g)
• ?Hf(H2O, g) -241.8 kJ/mol
• C(s) ½ O2(g) ? CO(g)
• ?Hf(CO, g) -111 kJ/mol
• Elements/reactants ? 1 mol of compound
• Notice units are per mole

63
Standard Enthalpy of Formation Values

• Can look up values of in reference book or
  textbook
• By definition, ?Hf 0 for elements in their
  standard states
• Example Cl2 (g)
• H2 (g)
• Ca (s)

64
Summation Equation

• In general, when all enthalpies of formation are
  known
• ?Hrxn S?Hf(products) - S?Hf(reactants)
• Must multiply all Hf values by coefficient from
  balanced equation!!!

65
Summation Equation Example

• Use the summation equation to determine the
  enthalpy of the following reaction
• 4 NH3(g) 5 O2(g) ? 4 NO(g) 6 H2O(g)
• ?Hreaction S?Hf(products) - S?Hf(reactants)

66
Unit 5 Part II Thermodynamics Spontaneity,
Entropy and Free Energy
67
Spontaneous Change

• What is a spontaneous process?
• A process that occurs by itself without an
  outside force helping it.
• A spark to start a process is OK though

68
Spontaneous Change

• Which of the following are spontaneous processes?
• Snowman melting in the sun
• Assembling a jigsaw puzzle
• Rusting of an iron object in humid air
• Recharging of a camera battery

69
Spontaneous Reactions and Energy

• Many spontaneous reactions are exothermic, but
  not all!
• Example
• H2O (s) ? H2O (l)
• is spontaneous and an ENDOTHERMIC reaction!
• (? H 6.0 kJ)

70
What other Factor Influences Spontaneity?

• The Randomness Factor!
• Nature tends to move spontaneously to a more
  random state.

71
Entropy Disorder and Spontaneity

• What is entropy?
• A measure of the randomness (disorder) of a
  system
• it is a state function)!

Reaction of K with water
72
The Second Law of Thermodynamics

• The Second Law of Thermodynamics states
• In a spontaneous process, there is a net increase
  of entropy (taking into account system and
  surroundings).

73
Spontaneous Processes result in more random
states (more disorder).

• EXAMPLE
• H2O (s) ? H2O (l)
• Water molecules are more disordered as a liquid
  than as a solid.

74
Sample Problem

• Predict which of the following processes have a
  positive change in entropy
• (an increase in the randomness or disorder)
• a. Taking dry ice from a freezer and allowing
  it to warm from -80oC to room temperature
• b. dissolving blue food coloring in water
• c. freezing water into ice cubes

75
Entropy

• Entropy is used to quantify randomness or
  disorder.
• Like enthalpy, entropy is also a state function.

76
The Third Law of Thermodynamics

• The Third Law of Thermodynamics states
• A completely ordered pure crystalline solid has
  an entropy of zero at 0 K.

77
Standard Molar Entropies ?So (1 mole, standard
conditions)

• Tells you entropy at 25oC and 1 atm (standard
  state conditions)
• Units J/mol K
• Note Elements DO NOT have ?So 0!
• (like they did with ?Ho)

78
Standard Molar Entropies
79
For a substance, Entropy generally increases as

• Phase change occurs from s? l ? g
• moles of gas increase from reactants to
  products
• T increases (KE increases)

80
For a reaction, entropy generally increases as

• 1. Reactants (solids or liquids)? Products
  (gases)
• 2. Total moles of products gt Total
  moles of reactants
• 3. Total moles of gaseous products gt Total
  moles of gaseous reactants
• 4. T is increasing.

81
Sign of ?So for a reaction means

• ?S ? Entropy increases
• S prod gt S react
• -?S ? Entropy decreases
• S react gt S prod

82
Example

• Predict the sign of ?S in each of the following
  reaction, and explain your prediction.
• NH3 (g) HCl (g) ? NH4Cl (s)
• 2 KClO3 (s) ? 2 KCl (s) 3 O2 (g)
• CO (g) H2O (g) ? CO2 (g) H2 (g)

83
Calculating ?S for a Reaction
?So ? So (products) - ? So (reactants)

• Calculation is similar to ?Ho
• (from Part I)
• Note units are JOULES not kJ as before!

84

• Example Calculate ?So for the following reaction
  using the tables in your reference book and the
  summation equation.
• 2 H2 (g) O2 (g) ? 2H2O (l)

85
Gibbs Free Energy and Free Energy Change

• The Gibbs (also known as Gibbs-Helmholtz)
  Equation shows relationship between Energy,
  Entropy and Spontaneity
• ?G ?H - T ?S
• Change in Free Energy Change in Enthalpy
  (Temp. x Change in Entropy)

86
What is free energy?

• Free energy
• AVAILABLE energy

87
The Relationship between ?Greaction and
Spontaneity

• 1. If ?G positive, reaction is NONSPONTANEOUS.
• 2. If ?G zero, reaction is at equilibrium
  (balanced).
• 3. If ?G negative, reaction is SPONTANEOUS.

88
Gibbs Free Energy, G

• Spontaneous Processes
• Must Have
• a Negative Free Energy!

J. Willard Gibbs1839-1903
89
How are these factors and spontaneity related?
Case ?H ?S ?G Result
1 - - spontaneous at all T
2 - - - - - spontaneous toward low T HOWEVER nonspontaneous toward high T
3 - nonspontaneous toward low T HOWEVER spontaneous toward high T
4 - nonspontaneous at all T
EXOTHERMIC reactions with Increasing Entropy are
Always spontaneous!
90
Example
Predict if the reaction will be spontaneous or
not. Use ?H as given and your estimate of the
sign of ?S. a. C6H12O6 (s) 6 O2 (g) ? 6 CO2
(g) 6 H2O (g)
?H -2540 kJ b. Cl2 (g) ? 2
Cl (g) ?H is positive

91
Two methods of calculating ?Go

• ?Go ?Ho - T?So
• a) Determine ?Horxn and ?Sorxn and use Gibbs
  equation.
• b) Use tabulated values of free energies
• of formation, ?Gfo.
• (we will not do this calculation, since it is
  similar to the ?Ho one we did in Part I)

92
Standard Free Energy Change, ?Go ?Go ?Ho -
T?So

• Note
• The units for ?Ho are generally in kJ
• The units for ?So generally are in J
• You must convert FIRST before beginning the
  problem!
• T is in K (oC 273)

93
Example Gibbs Equation

• Calculate ?Go for the reaction below, and predict
  whether the reaction is spontaneous at 25oC.
• C (s) 2H2 (g) ? CH4 (g)
• ?So -80.8 J/mol K ?Ho -74.8 kJ/mol
  T 298 K

94
Standard Free Energy of Formation

• ?Gorxn. SGof(products) - SGof(reactants)
• SAME SUMMATION EQUATION as
• ?Ho !!!!!
• (Use reference book for values)