1. 1 November 2011 Objective: You will be able to:
describe enthalpy and differentiate between endothermic and exothermic reactions and calculate the enthalpy of a reaction.
Homework: p. 263 #21, 22, 23, 24, 25: tomorrow

3. Thermochemistry

4. Thermodynamics: study of interconversion of heat and other kinds of energy
First Law of Thermodynamics: energy can be converted from one form to another, but can not be created or destroyed
We can not accurately measure total energy of a system
Instead, we measure changes in energy ΔE

5. Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g) H2O (l) + energy
H2O (g) H2O (l) + energy
Endothermic process is any process in which heat has to be supplied to the system from the surroundings.
energy + 2HgO (s) Hg (l) + O2 (g)
energy + H2O (s) H2O (l)

6. Enthalpy of Chemical Reactions
Specifically, we’re interested in heat flow that occurs under constant pressure

7. DH = H (products) – H (reactants)
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
Enthalpy change (∆H) is the heat absorbed during a physical or chemical process.
DH = H (products) – H (reactants)
Hproducts < Hreactants
Hproducts > Hreactants
DH < 0
DH > 0
6.4

8. Enthalpies of Reaction
Expressed in kJ or kJ/mol
endothermic: ∆H is always positive
endothermic changes absorb heat
exothermic: ∆H is always negative
exothermic changes liberate heat
Calculating ∆H is the same for a wide variety of processes (see table)

9. Thermochemical Equations
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.
H2O (s) H2O (l)
DHfus = 6.01 kJ

10. Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0
890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)
DHcomb = kJ

11. Thermochemical Equations
The stoichiometric coefficients always refer to the number of moles of a substance
H2O (s) H2O (l)
DH = 6.01 kJ/mol ΔH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l) H2O (s)
DH = kJ
If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.
2H2O (s) H2O (l)
DH = 2 mol x 6.01 kJ/mol = 12.0 kJ
The physical states of all reactants and products must be specified in thermochemical equations.

12. Note on “implied signs”
In a sentence, the sign of ∆H is often implied.
“6.9 kJ of heat was liberated from the system” = −6.9 kJ
“6.9 kJ of heat was absorbed by the system” = +6.9 kJ

13. Phase transitions solid to liquid; liquid to gas absorb heat
endothermic
positive ∆H
H2O(l) → H2O(g) ∆Hvap= +44 kJ
gas to liquid; liquid to solid
liberate heat
exothermic
negative ∆H
H2O(g) → H2O(l) ∆Hvap= −44 kJ

14. Example 1 Given the thermochemical equation
2SO2(g) + O2(g)  2SO3(g) DH = kJ/mol
calculate the heat evolved when 87.9 g of SO2 is converted to SO3.

15. Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air:
P4(s) + 5O2(g)  P4O10(s) DH= -3013kJ/mol

16. Example 3
How many kJ of heat are absorbed by the surroundings when 25.0 g of methane (CH4) burns in air?

17. Problem
Determine the amount of heat (in kJ) given off when 1.26x104 g of NO2 are produced according to the equation 2NO(g) + O2(g) → 2NO2(g) ∆H = −114.6 kJ/mol

18. What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs? Explain.

19. Why are phase changes from solid to liquid and from liquid to gas always endothermic?
Is the process of sublimation endothermic or exothermic? Explain.

20. 2 November 2011 Objective: You will be able to:
describe calorimetry and calculate heat change and specific heat.
Homework Quiz: Week of Oct. 31
Determine the amount of heat (in kJ) given off when 1.26x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) → 2NO2(g) ∆H = −114.6 kJ/mol

21. Agenda Homework Quiz Go over homework
Calorimetry calculations examples
Practice Problems
Pre-lab questions and lab set up
Homework: Finish pre-lab questions, lab notebook set up
p. #

22. Calorimetry Calorimetry: measurement of heat flow
Calorimeter: an apparatus that measures heat flow
Heat capacity, C, heat required to raise the temperature of an object by 1 K (units are J/K)
Molar heat capacity, Cmolar: amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are J/mol K)

23. Specific heat capacity: the heat capacity of one gram of a substance (units: J/g K)
water: 1 cal/g K = J/ gK

24. Calorimetry q=CΔt q = heat absorbed in Joules
C = specific heat capacity of a substance: the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree. (=mass*specific heat)
q=msΔt
Ex. 1: A 466 g sample of water is heated from 8.50oC to 74.60oC. Calculate the amount of heat absorbed (in kJ) by the water.

25. Ex. 2
1.435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter. The temperature of the water rose from 20.28oC to 25.95oC. If the heat capacity of the bomb plus water was kJ/oC, calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion).

26. Ex. 3
A lead pellet having a mass of g at oC was placed in a constant pressure calorimeter of negligible heat capacity containing mL of water. The water temperature rose from 22.50oC to 23.17oC. What is the specific heat of the lead pellet?

27. Ex. 4
What is the molar heat of combustion of liquid ethanol if the combustion of grams of ethanol causes a calorimeter to increase in temperature by 3.19 K?
heat capacity of the calorimeter is kJ/K

28. Pre-Lab Molar heat of crystallization a.k.a. Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook!)
Make a procedure summary and set up a data table.

29. Homework

30. 3 November 2011 Objective: You will be able to:
calculate the molar heat of crystallization for a chemical handwarmer
Homework Quiz: A 10.0 gram piece of copper (specific heat = J/g oC) has been heated to 100oC. It is then added to a sample of water at 22.0oC in a calorimeter. If the final temperature of the water is 28.0oC, calculate the mass of the water in the calorimeter.
specific heat of water = J/ g oC

31. Lab With your partner, carefully follow the directions.
Hand warmer pouch and metal disk: grams
Collect all data, results and answers to questions in your lab notebook.

32. 7 November 2011 Objective: You will be able to
calculate the standard enthalpy of formation and reaction for compounds.
Homework Quiz: (Week of Nov. 7)
An iron bar of mass 869 g cools from 94oC to 5oC. Calculate the heat released (in kJ) by the metal.
specific heat of iron=0.444 J/g·oC

33. Agenda Homework Quiz Return Tests
Standard Enthalpy of Formation/Reaction notes/problems
Homework: p. 264 #46, 49, 51, 53, 56, 57, 59, 62, 64: Tuesday
Test on Thermochem Thurs.
Lab notebook: due tomorrow (checklist!)
Test Corrections: Mon.

34. 8 November 2011 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)
Objective: You will be able to
review thermochemistry
Homework Quiz: (Week of Nov. 7)
Calculate the heat of combustion for the following reaction from the standard enthalpies of formation in Appendix 3.
2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)

35. Agenda Homework quiz Collect lab notebooks
Homework answers and more enthalpy problems
Problem set work time
Homework: Problem Set: Thurs.
Test on Thermochem Thurs.
Test Corrections: Mon.

36. 9 November 2022 Objective: You will be able to:
practice thermochemistry for a test.
Homework Quiz:
Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
S(rhombic) + O2 (g) SO2 (g) DH0 = kJ
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ

37. Agenda Homework Quiz Problem Set work time
Homework: Problem set due tomorrow
Thermochemistry test tomorrow

38. Test Corrections Required. 2nd quarter quiz grade
Show work, including for multiple choice questions.
On a separate sheet of paper or using a different color pen.
Due Mon.

39. Standard Enthalpy of Formation and Reaction

40. Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
½H2(g) + ½I2(s) → HI(g) ∆H = KJ

41. Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC. f
The standard heat of formation of any element in its most stable form is zero.
DH0 (O2) = 0 f
DH0 (O3) = 142 kJ/mol f
DH0 (C, graphite) = 0 f
DH0 (C, diamond) = 1.90 kJ/mol f

43. Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table. (Hint: Be sure to make only one mole of product!)

44. Based on the definition of a formation reaction, explain why the standard heat of formation of an element in its most stable thermodynamic state is zero. Write a chemical equation to illustrate your answer.

45. Hess’s Law
Hess’s Law: if a reaction is carried out in a series of steps, DH of the overall reaction is equal to the sum of the DH’s for each individual step.
(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

46. Example
Calculate DH for the following reaction 2S(s) + 3O2(g) → 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) → SO2(g) DH = −296.9 kJ
2SO2(g) + O2(g) → 2SO3(g) DH = −196.6 kJ

47. Standard Heat of Reaction
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atm.
rxn
aA + bB cC + dD
DH0
rxn
dDH0 (D) f
cDH0 (C) = [ + ] -
bDH0 (B)
aDH0 (A)
DH0
rxn = S
DH0 (products) f - S
DH0 (reactants) f

48. Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol. (Hint: Be really careful with your math, and the + and − signs!)

49. Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)

50. Benzene (C6H6) burns in air to produce carbon dioxide and liquid water
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
DH0 (products) f = S
DH0 (reactants) -
DH0
rxn
6DH0 (H2O) f
12DH0 (CO2) = [ + ] -
2DH0 (C6H6)
DH0
rxn =
[ 12 × × ] – [ 2 × ] = kJ
-6535 kJ
2 mol
= kJ/mol C6H6

51. Practice Problem 1
Calculate the heat of combustion for the following reaction:
C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l)
The standard enthalpy of formation of ethylene (C2H4) is 52.3 kJ/mol.
Then, calculate per mole of ethylene combusted.

52. Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron.
This reaction is highly exothermic and the liquid iron formed is used to weld metals. Calculate the heat released in kJ/g of Al.
∆Hof for Fe(l) is kJ/mol.

53. What if the reaction is not a single step? Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
rxn
S(rhombic) + O2 (g) SO2 (g) DH0 = kJ
rxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ
rxn

54. Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
rxn
S(rhombic) + O2 (g) SO2 (g) DH0 = kJ
rxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ
rxn
1. Write the enthalpy of formation reaction for CS2
2. Add the given rxns so that the result is the desired rxn.

55. Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
rxn
S(rhombic) + O2 (g) SO2 (g) DH0 = kJ
rxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ
rxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
rxn
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
2S(rhombic) + 2O2 (g) SO2 (g) DH0 = x2 kJ
rxn +
CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = kJ
rxn
C(graphite) + 2S(rhombic) CS2 (l)
DH0 = (2x-296.1) = 86.3 kJ
rxn

56. Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements:
2C(graphite) + H2(g)  C2H2(g)
Here are the equations for each step:
C(graphite) + O2(g)  CO2(g) DH0= kJ/mol
H2(g) + 1/2O2(g)  H2O(l) DH0= kJ/mol
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l)
DH0= kJ/mol

57. Homework
p. 264 #46, 49, 51, 53, 56, 57, 59, 62, 64: Tuesday

58. 4 Nov. 2010
Take out homework: p. 264 #42, 43, 46, 51, 53, 58, 63
Objective: SWBAT review thermochemistry for a test on Monday!
Do now: Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements:
2C(graphite) + H2(g)  C2H2(g)
Here are the equations for each step:
C(graphite) + O2(g)  CO2(g) DH0= kJ/mol
H2(g) + 1/2O2(g)  H2O(l) DH0= kJ/mol
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l)
DH0= kJ/mol

59. Agenda Do now Homework Chapter 6 Problem Set Work Time
Homework: Finish problem set: Mon.
Test: Mon.
Bonus: Tues.

60. Chapter 6 Problem Set Due Mon.
p. 263 #16, 20, 33, 37, 38, 52, 54b, 57, 64, 70, 73, 80
Please show all your work!
BONUS: p. 269 #119 Tues. (Up to +5% on your quarter grade!)

61. Read p. 258-261: Heats of solution and dilution, on your own

62. 8 Nov. 2010 Hand in Ch. 6 Problem Set
Objective: SWBAT show what you know about thermochemistry on a test!
Do Now: Review the sign conventions for q and w.
A reaction absorbs heat from its surroundings and has work done on it by the surroundings. Is it endothermic or exothermic? What are the signs for q and w?

63. Agenda Collect ch. 6 problem set Questions? Thermodynamics Test
Homework: Bonus problem due tomorrow! p. 269 #119
Read p
p. 829#1-5

64. 9 Nov. 2010 Take out homework: Bonus problem
Objective: SWBAT predict and calculate entropy changes to systems.
Do now: Give one example of a spontaneous process. Is it spontaneous in the opposite direction?

65. Agenda Do now Notes on spontaneous processes and entropy
Practice Problems
Homework: p. 829 # (SR)

66. Thermodynamics Part 2
1st Law of Thermodynamics: energy can be converted from one form to another, but can not be created or destroyed.
How to measure these changes?
∆H: Change in enthalpy: amount of heat given off or absorbed by a system

67. 2nd Law of Thermodynamics: explains why chemical processes tend to favor one direction

68. Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature, pressure, concentration, etc.)

69. Spontaneous Physical and Chemical Processes
A waterfall runs downhill
A lump of sugar dissolves in a cup of coffee
At 1 atm, water freezes below 0 0C and ice melts above 0 0C
Heat flows from a hotter object to a colder object
A gas expands in an evacuated bulb
Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
Processes that occur spontaneously in one direction can not, under the same conditions, also take place spontaneously in the opposite direction!

70. spontaneous
nonspontaneous

71. Does a decrease in enthalpy mean a reaction proceeds spontaneously?
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = kJ/mol
H+ (aq) + OH- (aq) H2O (l) ∆H0 = kJ/mol
H2O (s) H2O (l) ∆H0 = 6.01 kJ/mol
NH4NO3 (s) NH4+(aq) + NO3- (aq) ∆H0 = 25 kJ/mol
H2O

72. Does a decrease in enthalpy mean a reaction proceeds spontaneously?
No!
Being exothermic (“exothermicity”) favors spontaneity, but does not guarantee it!
An endothermic reaction can be spontaneous.
Not all exothermic reactions are spontaneous.
So how do we predict if a reaction is spontaneous under given conditions if ∆H doesn't help us?!

73. Entropy (S) is a measure of the randomness or disorder of a system.
If the change from initial to final results in an increase in randomness,
∆S > 0
For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state
Ssolid < Sliquid << Sgas
H2O (s) H2O (l)
∆S > 0

74. Processes that lead to an increase in entropy (∆S > 0)

75. Example: Br2(l) Br2(g)
Example: I2(s) I2(g)
∆S > 0
∆S > 0

76. Entropy
State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.
Examples:
energy, enthalpy, pressure, volume, temperature
, entropy
Review
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

78. How does the entropy of a system change for each of the following processes?
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice

79. How does the entropy of a system change for each of the following processes?
(a) Condensing water vapor
Randomness decreases
Entropy decreases (∆S < 0)
(b) Forming sucrose crystals from a supersaturated solution
Randomness decreases
Entropy decreases (∆S < 0)
(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S > 0)
(d) Subliming dry ice
Randomness increases
Entropy increases (∆S > 0)

80. Practice
Predict whether the entropy change is greater or less than zero for each of the following processes:
freezing ethanol
evaporating a beaker of liquid bromine at room temperature
dissolving glucose in water
cooling nitrogen gas from 80oC to 20oC

81. First Law of Thermodynamics
Energy can be converted from one form to another but energy cannot be created or destroyed.
Second Law of Thermodynamics
The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.
Spontaneous process:
∆Suniv = ∆Ssys + ∆Ssurr > 0
Equilibrium process:
∆Suniv = ∆Ssys + ∆Ssurr = 0

82. Entropy Changes in the System (∆Ssys)
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C.
rxn
aA + bB cC + dD
∆S0
rxn
dS0(D)
cS0(C) = [ + ] -
bS0(B)
aS0(A)
∆S0
rxn
nS0(products) = Σ
mS0(reactants) -
What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) CO2 (g)
S0(CO) = J/K•mol
S0(CO2) = J/K•mol
S0(O2) = J/K•mol

83. Entropy Changes in the System (∆Ssys)
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C.
rxn
aA + bB cC + dD
∆S0
rxn
dS0(D)
cS0(C) = [ + ] -
bS0(B)
aS0(A)
∆S0
rxn
nS0(products) = Σ
mS0(reactants) -
What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) CO2 (g)
S0(CO) = J/K•mol
S0(CO2) = J/K•mol
S0(O2) = J/K•mol
DS0
rxn
= 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
DS0
rxn
= – [ ] = J/K•mol

84. Practice
From the standard entropy values in Appendix 3, calculate the standard entropy changes for the following reactions at 25oC:
CaCO3(s) CaO(s) + CO2(g)
N2(g) + 3H2(g) NH3(g)
H2(g) + Cl2(g) HCl(g)

85. Entropy Changes in the System (∆Ssys)
When gases are produced (or consumed)
If a reaction produces more gas molecules than it consumes, ∆S0 > 0.
If the total number of gas molecules diminishes, ∆S0 < 0.
If there is no net change in the total number of gas molecules, then ∆S0 may be positive or negative BUT ∆S0 will be a small number.
What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) ZnO (s)
The total number of gas molecules goes down, ∆S is negative.

86. Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negative:
2H2(g) + O2(g) H2O(l)
NH4Cl(s) NH3(g) + HCl(g)
H2(g) + Br2(g) HBr(g)
I2(s) I(g)
2Zn(s) + O2(g) ZnO(s)

87. Homework
p. 829 # (SR)

88. 10 Nov. 2010
Objective: SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction.
Do now: Give one example of a change that increases entropy.

89. Agenda Do now Homework: SR
Gibbs Free Energy and Predicting Spontaneity
HW: p. 829 #15, 16, 17, 18, 19, 20 (TTL) (Tues.)
Read lab and answer pre-lab questions in lab notebook (Mon.)
AP Test Thermo questions (Tues.)

90. Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process
∆Ssurr > 0
Endothermic Process
∆Ssurr < 0

91. Third Law of Thermodynamics
The entropy of a perfect crystalline substance is zero at the absolute zero of temperature.
S = k ln W
W = 1
S = 0

92. Gibbs Free Energy Spontaneous process: ∆Suniv = ∆Ssys + ∆Ssurr > 0
Equilibrium process:
∆Suniv = ∆Ssys + ∆Ssurr = 0
For a constant-temperature process:
Gibbs free energy (G)
∆G = ∆Hsys -T ∆Ssys
∆G < The reaction is spontaneous in the forward direction.
∆G > The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
∆G = The reaction is at equilibrium.

93. ∆G0 of any element in its stable form is zero.
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.
rxn
aA + bB cC + dD
∆G0
rxn
d ∆G0 (D) f
c ∆G0 (C) = [ + ] -
b ∆G0 (B)
a ∆G0 (A)
∆G0
rxn
n ∆G0 (products) f = S
m ∆G0 (reactants) -
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f
∆G0 of any element in its stable form is zero. f

95. What is the standard free-energy change for the following reaction at 25 0C?
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)

96. What is the standard free-energy change for the following reaction at 25 0C?
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DG0
rxn
nDG0 (products) f = S
mDG0 (reactants) -
DG0
rxn
6DG0 (H2O) f
12DG0 (CO2) = [ + ] -
2DG0 (C6H6)
DG0
rxn =
[ 12x– x–237.2 ] – [ 2x124.5 ] = kJ/mol
Is the reaction spontaneous at 25 0C?
DG0 = kJ/mol
< 0
spontaneous

97. Practice
Calculate the standard free-energy changes for the following reactions at 25oC
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
2MgO(s) Mg(s) + O2(g)
H2(g) + Br2(l) HBr(g)

98. ∆G = ∆H - T ∆S

99. Recap: Signs of Thermodynamic Values
Negative
Positive
Enthalpy (ΔH)
Exothermic
Endothermic
Entropy (ΔS)
Less disorder
More disorder
Gibbs Free Energy (ΔG)
Spontaneous
Not spontaneous

100. Homework
p. 829 #15, 16, 17, 18, 19, 20 (TTL) (Tues.) Read lab and answer pre-lab questions in lab notebook (Mon.) AP Test Thermo questions (Tues.)

101. 16 Nov. 2010
Objective: SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems.
Do now: If enthalpy is positive and entropy is positive, at what temperatures will a reaction proceed spontaneously?

102. Agenda Do now Homework solutions (TTL)
AP Test question solutions - collected
Thermochemistry review problems
Homework: Lab notebook: Thurs.
Come after school to mass your filter paper + M2CO3!

103. CaCO3(s)  CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking, producing calcium metal, paper making, water treatment and pollution control.
It is made by decomposing limestone (CaCO3) in a kiln at high temperature.
But, the reaction is reversible! CaO(s) readily combines with CO2 to form CaCO3.
So, in the kiln, CO2 is constantly removed to shift equilibrium to favor the formation of CaO.

104. At what temperature does this reaction favor the formation of CaO?
Calculate ∆Ho and ∆So for the reaction at 25oC.
Then plug in to ∆Go=∆Ho – T∆So to find ∆Go
What sign is ∆Go? What magnitude?
Set ∆Go=0 to find the temperature at which equilibrium occurs.
At what temperature can we expect this reaction to proceed spontaneously?
Choose a temperature, plug in and check.

105. Keep in mind that this does not mean that CaO is only formed above 835oC.
Some CaO and CO2 will be formed, but the pressure of CO2 will be less than 1 atm (its standard state).

106. Practice Problems
Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous:
∆H = -126 kJ/mol
∆S = 84 J/K·mol
∆H = kJ/mol
∆S = -105 J/K·mol

107. Solutions to AP test problems

108. 17 Nov. 2010 Objective: SWBAT review enthalpy and entropy for a quiz.
Do now:
The reaction between nitrogen and hydrogen to form ammonia is represented below.
N2(g) + 3 H2(g) → 2 NH3(g) ∆H˚ = –92.2 kJ
a. Predict the sign of the standard entropy change, ∆S˚, for the reaction. Justify your answer.
b. The value of ∆G˚ for the reaction represented in part (a) is negative at low temperatures but positive at high temperatures. Explain.

109. Do now solutions
(a) (–); the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ – T∆S˚; enthalpy favors spontaneity (∆H < 0), negative entropy change does not favor spontaneity. Entropy factor becomes more significant as temperature increases. At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G > 0).

110. Thermochemistry review
p. 831 #42, 52, 54, 56, 57, 60, 78
Quiz Thursday
Entropy (∆S)
Free energy (∆G)
Study homework assignments from chapter 18, AP test questions, these review problems.

111. stop

112. During the course of a chemical reaction, not all the reactants and products will be at their standard states (1.0 atm and 25oC).
How do you calculate free energy (and thus, the spontaneity of the reaction)?

113. Gibbs Free Energy and Chemical Equilibrium
∆G = ∆ G0 + RT lnQ
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
Q is the reaction quotient
At Equilibrium
∆G = 0
Q = K
0 = ∆G0 + RT lnK
∆G0 = - RT lnK

114. What is Q?
Reaction quotient

115. What is K?
equilibrium constant

116. ∆G0 = - RT lnK
18.6

117. The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.
C = ms
Heat (q) absorbed or released:
q = msDt
q = CDt
Dt = tfinal - tinitial
6.5

118. Phase Changes
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure.
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm.
11.8

119. The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure.
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature.
11.8

120. Where’s Waldo? Can you find… The Triple Point? Critical pressure?
Critical temperature?
Where fusion occurs?
Where vaporization occurs?
Melting point (at 1 atm)?
Boiling point (at 6 atm)?
Where’s Waldo?
Carbon Dioxide

121. H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Melting
Freezing
11.8

122. H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid.
Sublimation
Deposition
DHsub = DHfus + DHvap
( Hess’s Law)
11.8

123. Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance.
11.8

124. 11.8

125. Sample Problem
How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C?
Step 1: Heat the ice Q=mcΔT
Q = 36 g x 2.06 J/g deg C x 8 deg C = J = 0.59 kJ
Step 2: Convert the solid to liquid ΔH fusion
Q = 2.0 mol x 6.01 kJ/mol = 12 kJ
Step 3: Heat the liquid Q=mcΔT
Q = 36g x J/g deg C x 100 deg C = J = 15 kJ

126. Sample Problem
How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C?
Step 4: Convert the liquid to gas ΔH vaporization
Q = 2.0 mol x kJ/mol = kJ
Step 5: Heat the gas Q=mcΔT
Q = 36 g x 2.02 J/g deg C x 20 deg C = J = 1.5 kJ
Now, add all the steps together
0.59 kJ + 12 kJ + 15 kJ + 88 kJ kJ = 118 kJ