Unit 3: Thermochemistry

1
Unit 3 Thermochemistry

• Chemistry 3202

2
Unit Outline

• Temperature and Kinetic Energy
• Heat/Enthalpy Calculation
• Temperature changes (q mc?T)
• Phase changes (q n?H)
• Heating and Cooling Curves
• Calorimetry (q C?T above formulas)

3
Unit Outline

• Chemical Reactions
• PE Diagrams
• Thermochemical Equations
• Hesss Law
• Bond Energy
• STSE What Fuels You?

4
Temperature and Kinetic Energy

• Thermochemistry is the study of energy changes
  in chemical and physical changes
• eg. dissolving a solid
• burning
• phase changes

5

• Temperature - a measure of the average kinetic
  energy of particles in a substance
• - a change in temperature means particles are
  moving at different speeds
• - measured in either Celsius degrees or degrees
  Kelvin
• Kelvin Celsius 273.15

6
p. 628
7
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8
Kinetic Energy
9

• The Celsius scale is based on the freezing and
  boiling point of water
• The Kelvin scale is based on absolute zero -
  the temperature at which particles in a substance
  have zero kinetic energy.

10
Heat/Enthalpy Calculations

• system - the part of the universe being studied
  and observed
• surroundings - everything else in the universe
• open system - a system that can exchange matter
  and energy with the surroundings
• eg. an open beaker of water
• a candle burning
• closed system - allows energy transfer but is
  closed to the flow of matter.

11

• isolated system a system completely closed to
  the flow of matter and energy
• heat - refers to the transfer of kinetic energy
  from a system of higher temperature to a system
  of lower temperature.
• - the symbol for heat is q
• WorkSheet Thermochemistry 1

12
Part A Thought Lab (p. 631)
13
Part B Thought Lab
14
Heat/Enthalpy Calculations

• specific heat capacity - the amount of energy ,
  in Joules (J), needed to change the temperature
  of one gram (g) of a substance by one degree
  Celsius (C).
• The symbol for specific heat capacity is a
  lowercase c

15

• A substance with a large value of c can absorb or
  release more energy than a substance with a small
  value of c.
• ie. For two substances, the substance with the
  larger c will undergo a smaller temperature
  change with the same amount of heat applied

16
FORMULA

• q heat (J)
• m mass (g)
• c specific heat capacity
• ?T temperature change
• T2 T1
• Tf Ti

• q mc?T

17

• eg. How much heat is needed to raise the
  temperature of 500.0 g of water from 20.0 C to
  45.0 C?
• Solve q m c ?T
• for c, m, ?T, T2 T1
• p. 634 s 1 4
• p. 636 s 5 8
• WorkSheet Thermochemistry 2

18

• heat capacity - the quantity of energy , in
  Joules (J), needed to change the temperature of a
  substance by one degree Celsius (C)
• The symbol for heat capacity is uppercase C
• The unit is J/ C or kJ/ C

19
FORMULA

• C heat capacity
• c specific heat capacity
• m mass
• ?T T2 T1

• C mc
• q C ?T

Your Turn p.637 s 11-14 WorkSheet
Thermochemistry 3
20
Enthalpy Changes

• The difference between the potential energy of
  the reactants and the products during a physical
  or chemical change is the Enthalpy change or ?H.
• AKA Heat of Reaction

21

• Endothermic Reaction

PE
Reaction Progress
22

• Endothermic Reaction

PE
Enthalpy
?H
Reaction Progress
23
Enthalpy
?H is
Endothermic
24
Enthalpy
Exothermic
25
Enthalpy Changes in Reactions

• All chemical reactions require bond breaking in
  reactants followed by bond making to form
  products
• Bond breaking requires energy (endothermic) while
  bond formation releases energy (exothermic)

see p. 639
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Enthalpy Changes in Reactions

• endothermic reaction - the energy required to
  break bonds is greater than the energy released
  when bonds form.
• exothermic reaction - the energy required to
  break bonds is less than the energy released when
  bonds form.

28
Enthalpy Changes in Reactions

• ?H can represent the enthalpy change for a
  number of processes
• Chemical reactions
• ?Hrxn enthalpy of reaction
• ?Hcomb enthalpy of combustion
• (see p. 643)

29

• Formation of compounds from elements
• ?Hof standard enthalpy of formation
• The standard molar enthalpy of formation is the
  energy released or absorbed when one mole of a
  compound is formed directly from the elements in
  their standard states.
• ( see p. 642)

30

• Phase Changes (p.647)
• ?Hvap enthalpy of vaporization
• ?Hfus enthalpy of melting
• ?Hcond enthalpy of condensation
• ?Hfre enthalpy of freezing
• Solution Formation
• ?Hsoln enthalpy of solution

31

• There are three ways to represent any enthalpy
  change
• 1. thermochemical equation - the energy term
  written into the equation.
• 2. enthalpy term is written as a separate
  expression beside the equation.
• 3. enthalpy diagram.

32

• eg. the formation of water from the elements
  produces 285.8 kJ of energy.
• 1. H2(g) ½ O2(g) ? H2O(l) 285.8 kJ
• 2. H2(g) ½ O2(g) ? H2O(l) ?Hf -285.8
  kJ/mol

thermochemical equation
33
enthalpy diagram

• 3.

Enthalpy (H)
examples pp. 641-643 questions p. 643 s 15-18
WorkSheet Thermochemistry 4
34
Calculating Enthalpy Changes

• FORMULA
• q n?H

• q heat (kJ)
• n of moles
• ?H molar enthalpy
• (kJ/mol)

35

• eg. How much heat is released when 50.0 g of
  CH4 forms from C and H ?
• p. 642

q n?H (3.115 mol)(-74.6 kJ/mol)
-232 kJ
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• eg. How much heat is released when 50.00 g of
  CH4 undergoes complete combustion?

q n?H (3.115 mol)(-965.1 kJ/mol)
-3006 kJ
38
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39

• eg. How much energy is needed to change 20.0 g
  of H2O(l) at 100 C to steam at 100 C ?
• Mwater 18.02 g/mol ?Hvap 40.7 kJ/mol

q n?H (1.110 mol)(40.7 kJ/mol)
45.2 kJ
40
?Hfre and ?Hcond have the opposite sign of the
above values.
41

• eg. The molar enthalpy of solution for ammonium
  nitrate is 25.7 kJ/mol. How much energy is
  absorbed when 40.0 g of ammonium nitrate
  dissolves?

q n?H (0.4996 mol)(25.7 kJ/mol)
12.8 kJ
42

• What mass of ethane, C2H6, must be burned to
  produce 405 kJ of heat?
• ?H -1250.9 kJ
• q - 405 kJ

q n?H
n 0.3238 mol m n x M (0.3238
mol)(30.08 g/mol) 9.74 g
43

• Complete p. 645 s 19 23
• pp. 648 649 s 24 29
• 19. (a) -8.468 kJ (b) -7.165 kJ
• 20. -1.37 x103 kJ
• 21. (a) -2.896 x 103 kJ
• 21. (b) -6.81 x104 kJ
• 21. (c) -1.186 x 106 kJ
• 22. -0.230 kJ
• 23. 3.14 x103 g

44

• 24. 2.74 kJ
• 25.(a) 33.4 kJ (b) 33.4 kJ
• 26.(a) absorbed (b) 0.096 kJ
• 27.(a) NaCl(s) 3.9 kJ/mol ? NaCl(aq)
• (b) 1.69 kJ
• (c) cool heat absorbed from water
• 28. 819.2 g
• 29. 3.10 x 104 kJ

45

• p. 638 4 8
• pp. 649, 650 s 3 8
• p. 657, 658 s 9 - 18

WorkSheet Thermochemistry 5
46
Heating and Cooling Curves

• Demo Cooling of p-dichlorobenzene

47
Cooling curve for p-dichlorobenzene
80
Temp. (C )
liquid
50
freezing
solid
20
Time
48
Heating curve for p-dichlorobenzene
80
Temp. (C )
50
20
Time
49

• What did we learn from this demo??
• During a phase change temperature remains
  constant and PE changes
• Changes in temperature during heating or cooling
  means the KE of particles is changing

50
p. 651
51
p. 652
52
p. 656
53
Heating Curve for H20(s) to H2O(g)

• A 40.0 g sample of ice at -40 C is heated until
  it changes to steam and is heated to 140 C.
• Sketch the heating curve for this change.
• Calculate the total energy required for this
  transition.

54
q mc?T
140
100
q n?H
Temp. (C )
q mc?T
q n?H
0
q mc?T
-40
Time
55

• Data
• cice 2.01 J/g.C
• cwater 4.184 J/g.C
• csteam 2.01 J/g.C
• ?Hfus 6.02 kJ/mol
• ?Hvap 40.7 kJ/mol

56

• warming ice
• q mc?T
• (40.0)(2.01)(0 - -40)
• 3216 J
• warming water
• q mc?T
• (40.0)(4.184)(100 0)
• 16736 J

• warming steam
• q mc?T
• (40.0)(2.01)(140 -100)
• 3216 J

57
n 40.0 g 18.02
g/mol 2.22 mol
moles of water

• melting ice
• q n?H
• (2.22 mol)(6.02 kJ/mol)
• 13.364 kJ
• boiling water
• q n?H
• (2.22 mol)(40.7 kJ/mol)
• 90.354 kJ

58
Total Energy

• 90.354 kJ
• 13.364 kJ
• 3216 J
• 3216 J
• 16736 J

• 127 kJ

59
Practice

• p. 655 s 30 34
• pp. 656 s 1 - 9
• p. 657 s 2, 9
• p. 658 s 10, 16 20

WorkSheet Thermochemistry 6

• 30.(b) 3.73 x103 kJ
• 31.(b) 279 kJ
• 32.(b) -1.84 x10-3 kJ
• 33.(b) -19.7 kJ -48.77 kJ
• 34. -606 kJ

60
Law of Conservation of Energy (p. 627)

• The total energy of the universe is constant
• ?Euniverse 0
• Universe system surroundings
• ?Euniverse ?Esystem ?Esurroundings
• ?Euniverse ?Esystem ?Esurroundings 0
• OR ?Esystem -?Esurroundings
• OR qsystem -qsurroundings

First Law of Thermodynamics
61
Calorimetry (p. 661)

• calorimetry - the measurement of heat changes
  during chemical or physical processes
• calorimeter - a device used to measure changes
  in energy
• 2 types of calorimeters
• 1. constant pressure or simple calorimeter
  (coffee-cup calorimeter)
• 2. constant volume or bomb calorimeter.

62
SimpleCalorimeter
p.661
63

• a simple calorimeter consists of an insulated
  container, a thermometer, and a known amount of
  water
• simple calorimeters are used to measure heat
  changes associated with heating, cooling, phase
  changes, solution formation, and chemical
  reactions that occur in aqueous solution

64

• to calculate heat lost or gained by a chemical or
  physical change we apply the first law of
  thermodynamics
• qsystem -qcalorimeter
• Assumptions
• the system is isolated
• c (specific heat capacity) for water is not
  affected by solutes
• heat exchange with calorimeter can be ignored

65

• eg.
• A simple calorimeter contains 150.0 g of water.
  A 5.20 g piece of aluminum alloy at 525 C is
  dropped into the calorimeter causing the
  temperature of the calorimeter water to increase
  from 19.30C to 22.68C.
• Calculate the specific heat capacity of the
  alloy.

66

• eg.
• The temperature in a simple calorimeter with a
  heat capacity of 1.05 kJ/C changes from 25.0 C
  to 23.94 C when a very cold 12.8 g piece of
  copper was added to it.
• Calculate the initial temperature of the
  copper. (c for Cu 0.385 J/g.C)

67
Homework

• p. 664, 665 s 1b), 2b), 3 4
• p. 667, s 5 - 7

68

• (60.4)(0.444)(T2 98.0) -(125.2)(4.184)(T2
  22.3)
• 26.818(T2 98.0) -523.84(T2 22.3)
• 26.818T2 - 2628.2 -523.84T2 11681
• 550.66T2 14309.2
• T2 26.0 C

69

• 6. System (Mg)
• m 0.50 g or 0.02057 mol
• Find ?H
• Calorimeter
• 100 ml or m 100 g
• c 4.184
• T2 40.7
• T1 20.4

70

• 7. System
• ?H -53.4 kJ/mol
• n CV (0.0550L)(1.30 mol/L) 0.0715 mol
• Calorimeter
• 110 ml or m 110 g
• c 4.184
• T1 21.4
• Find T2

71

• 6.
• qMg -qcal
• n?H -mc?T

72

• Bomb Calorimeter

73
Bomb Calorimeter

• used to accurately measure enthalpy changes in
  combustion reactions
• the inner metal chamber or bomb contains the
  sample and pure oxygen
• an electric coil ignites the sample
• temperature changes in the water surrounding the
  inner bomb are used to calculate ?H

74

• to accurately measure ?H you need to know the
  heat capacity (kJ/C) of the calorimeter.
• must account for all parts of the calorimeter
  that absorb heat
• Ctotal Cwater Cthermom. Cstirrer
  Ccontainer
• NOTE C is provided for all bomb
• calorimetry calculations

75

• eg. A technician burned 11.0 g of octane in a
  steel bomb calorimeter. The heat capacity of the
  calorimeter was calibrated at 28.0 kJ/C. During
  the experiment, the temperature of the
  calorimeter rose from 20.0 C to 39.6 C.
• What is the enthalpy of combustion for octane?

76

• Eg.
• 1.26 g of benzoic acid, C6H5COOH(s) (?Hcomb
  -3225 kJ/mol), is burned in a bomb calorimeter.
  The temperature of the calorimeter and its
  contents increases from 23.62 C to 27.14 C.
  Calculate the heat capacity of the calorimeter.

77
Homework

• p. 675 s 8 10

WorkSheet Thermochemistry 6
78
Hesss Law of Heat Summation

• the enthalpy change (?H) of a physical or
  chemical process depends only on the beginning
  conditions (reactants) and the end conditions
  (products)
• ?H is independent of the pathway and/or the
  number of steps in the process
• ?H is the sum of the enthalpy changes of all the
  steps in the process

79
eg. production of carbon dioxide

• Pathway 1 2-step mechanism
• C(s) ½ O2(g) ? CO(g) ?H -110.5 kJ
• CO(g) ½ O2(g) ? CO2(g) ?H -283.0 kJ

• C(s) O2(g) ? CO2(g) ?H -393.5 kJ

80
eg. production of carbon dioxide

• Pathway 2 formation from the elements
• C(s) O2(g) ? CO2(g) ?H -393.5 kJ

81
Using Hesss Law

• We can manipulate equations with known ?H to
  determine the enthalpy change for other
  reactions.
• NOTE
• Reversing an equation changes the sign of ?H.
• If we multiply the coefficients we must also
  multiply the ?H value.

82

• eg.
• Determine the ?H value for
• H2O(g) C(s) ? CO(g) H2(g)
• using the equations below.
• C(s) ½ O2(g) ? CO(g) ?H -110.5 kJ
• H2(g) ½ O2(g) ? H2O(g) ?H -241.8 kJ

83

• eg.
• Determine the ?H value for
• 4 C(s) 5 H2(g) ? C4H10(g)
• using the equations below.
• ?H (kJ)
• C4H10(g) 6½ O2(g) ? 4 CO2(g) 5 H2O(g)
  -110.5
• H2(g) ½ O2(g) ? H2O(g) -241.8
• C(s) O2(g) ? CO2(g) -393.5

Switch
Multiply by 5
Multiply by 4
84

• 4 CO2(g) 5 H2O(g) ? C4H10(g) 6½ O2(g)
  110.5
• 5(H2(g) ½ O2(g) ? H2O(g) -241.8)
• 4(C(s) O2(g) ? CO2(g) -393.5)

4 CO2(g) 5 H2O(g) ? C4H10(g) 6½ O2(g)
110.5 5 H2(g) 2½ O2(g) ? 5
H2O(g) -1209.0 4C(s) 4 O2(g) ? 4
CO2(g) -1574.0
Ans -2672.5 kJ
85
Practice

• pg. 681 s 11-14

WorkSheet Thermochemistry 7
86
Review

• ?Hof (p. 642, 684, 848)
• The standard molar enthalpy of formation is the
  energy released or absorbed when one mole of a
  substance is formed directly from the elements in
  their standard states.
• ?Hof 0 kJ/mol
• for elements in the standard state
• The more negative the ?Hof, the more stable the
  compound

87
Using Hesss Law and ?Hf

• Determine the ?H value for
• C4H10(g) 6½ O2(g) ? 4 CO2(g) 5 H2O(g)
• using the formation equations below.
• ?Hf (kJ/mol)
• 4 C(s) 5 H2(g) ? C4H10(g) -2672.5
• H2(g) ½ O2(g) ? H2O(g) -241.8
• C(s) O2(g) ? CO2(g) -393.5

88
Using Hesss Law and ?Hf

• ?Hrxn ??Hf (products) - ??Hf (reactants)
• eg. Calculate the enthalpy of reaction for the
  complete combustion of glucose.
• C6H12O6(s) 6 O2(g) ? 6 CO2(g) 6 H2O(g)

89

• Use the molar enthalpys of formation to
  calculate ?H for the reaction below
• Fe2O3(s) 3 CO(g) ? 3 CO2(g) 2 Fe(s)
• p. 688 s 21 22

90

• Eg.
• The combustion of phenol is represented by the
  equation below
• C6H5OH(s) 7 O2(g) ? 6 CO2(g) 3 H2O(g)
• If ?Hcomb -3059 kJ/mol, calculate the heat of
  formation for phenol.

91
Bond Energy Calculations (p. 688)

• The energy required to break a bond is known as
  the bond energy.
• Each type of bond has a specific bond energy
  (BE).
• (table p. 847)
• Bond Energies may be used to estimate the
  enthalpy of a reaction.

92
Bond Energy Calculations (p. 688)

• ?Hrxn ?BE(reactants) - ?BE (products)
• eg. Estimate the enthalpy of reaction for the
  combustion of ethane using BE.
• 2 C2H6(g) 7 O2(g) ? 4 CO2(g) 6 H2O(g)
• Hint Drawing the structural formulas for all
  reactants and products will be useful here.

93
? 4 OCO 6 H-O-H
7 O O
2
2(347) 2(6)(338) 7(498)
- 4(2)(745) 6(2)(460)
8236 - 11480
-3244 kJ
p. 690 s 23,24, 26 p. 691 s 3, 4, 5, 7
94
Energy Comparisons

• Phase changes involve the least amount of energy
  with vaporization usually requiring more energy
  than melting.
• Chemical changes involve more energy than phase
  changes but much less than nuclear changes.
• Nuclear reactions produce the largest ?H
• eg. nuclear power, reactions in the sun

95
STSE

• What fuels you? (Handout)
• s 1 4, 7, 9, 10, 13, 14, 16
• (List only TWO advantages for 2)

96

• aluminum alloy water
• m 5.20 g m 150.0 g
• T1 525 ºC T1 19.30 ºC
• T2 ºC T2 22.68 ºC
• FIND c for Al c 4.184 J/g.ºC
• qsys - qcal
• mc?T - mc ?T
• (5.20)(c)(22.68 - 525 ) -(150.0)(4.184)(22.68
  19.30)
• -2612 c -2121
• c 0.812 J/g.C

97

• copper
• m 12.8 g
• T2 ºC
• c 0.385 J/g.C
• FIND T1 for Cu
• qsys - qcal
• mc?T - C?T
• (12.8)(0.385)(23.94 T1) -(1050)(23.94
  25.0)
• 4.928 (23.94 T1) 1113
  
• 23.94 T1 1113/4.928
• 23.94 T1 225.9
• T1 -202 ºC

calorimeter C 1.05 kJ/C T1 25.00 ºC T2
23.94 ºC
98
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99

• Group 2 Total Mass 2.05 g
• Group 5 Total Mass 1.86 g