1. AP Chemistry: Chapters 6 and 17 notes
Thermodynamics

2. 6.1 The nature of energy

3. 6.1 The Nature of Energy
In this chapter, we will study energy in terms of chemical potential energy and how that energy can change form to accomplish work. We will focus specifically on thermochemistry, which involves heat and energy transfer.

4. Energy and work Energy- the capacity to do work or to produce heat
Work- force acting over a distance (Work = Fd)
It involves a transfer of energy

5. The 1st LAW OF THERMODYNAMICS: LAW OF CONSERVATION OF ENERGY
1st Law of Thermodynamics- also known as the Law of Conservation of Energy.
States that energy can be converted from one form to another but it can be neither created nor destroyed.
The total amount of energy in the universe is constant.

6. Energy can be classified in two ways:
Potential energy- energy due to position or composition (included chemical potential energy)
KE = ½ mv2
m = mass in kg
v = velocity in m/s
units are J, since J = (kg.m2)/s2
Kinetic energy- energy due to the motion of an object
Kinetic energy is dependent on the mass and velocity of an object

7. Heat
Heat- (q) involves a transfer of energy between two objects due to a temperature difference.
Heat always moves from warmer matter to cooler matter.

8. System vs surroundings
In this chapter, we will describe where heat moves and will use the terms system and surroundings.
In terms of a chemical reaction, the system is our reaction.
The surroundings are everything else, including things like the container the reaction occurs in, the room it sits in, etc.

9. temperature
Temperature- a property that reflects random motions of the particles of a particular substance
Exothermic- reaction which releases heat
Energy flows out of the system
Potential energy is changed to thermal energy
Products have lower potential energy than reactants

10. Endothermic Reaction which absorbs heat energy flows into the system
thermal energy is changed into potential energy
products have higher PE than reactants

11. The value of “e”
Internal energy (E) of a system is the sum of the kinetic and potential energies of all the particles in a system.
Thermodynamic quantities always consist of a number and a sign (+ or -). The sign represents the system’s point of view. (Engineers use the surrounding’s point of view)
Exothermic -E (systems energy is decreasing)
Endothermic +E (systems energy is increasing)

12. Calculating the energy in a system
E = q + w
E is the change in the system’s internal energy
q represents heat
w represents work usually in J or kJ
Example: Calculate E if q = -50 kJ and w = +35kJ.
DE = q + w
= -50 kJ + 35 kJ
= -15 kJ

13. Work on Gases
Units come from: 1L.atm = 0.001m3.101,325 Pa
(1 Pa = N/m2), so N.m or J
For a gas that expands or is compressed, work can be calculated by:
w = - PV
units: L.atm

14. Example: Calculate the work if the volume of a gas is increased from 15 ml to 2.0 L at a constant pressure of 1.5 atm.
w = -PDV
w = -1.5 atm (1.985L)
w = -3.0 L . atm

15. A) - + B) C) D) E) Heat, q Added to the system
Practice Question One version of the first law of thermodynamics is expressed as ∆E = q + w Which gives the sign convention for this relationship that is usually used in chemistry?
Heat, q
Added to the system
added to the surroundings
Work, w done on the system
Work, w, done on the surroundings A) - + B) C) D) E)

16. Let’s Talk Lab…
Solve for the calorimeter constant and write this on your cup. Be sure to hang on to your cup!
Solve for q=mCΔT for the solid you used. The mass used is the solid and the liquid mass. The specific heat used is for water.
Discuss your procedures for Thursday with your lab partner. You determine how you will calculate the amount of heat lost or gained by an entire cold pack.

17. Lab Data Compound Name Heat of Solution Group 1
Heat of Solution Group 2 (if tested)
Avg. Heat of Solution (if tested twice)
NaCl
CaCl2
NaC2H3O2
Na2CO3
LiCl
NH4NO3

18. 6.2 Enthalpy and calorimetry

19. Enthalpy H = E + PV Enthalpy (H) concerns the heat energy in a system.
E is internal energy
P is pressure
V is volume
Enthalpy (H) concerns the heat energy in a system.
H = q at constant pressure only
Reactions that do not involve gases or where moles of gases do not change are considered “at constant pressure”.
At constant pressure, the terms heat of reaction and change in enthalpy are used interchangeably.

20. Enthalpy Change in a System
The change in the enthalpy of a system can be calculated using:
H= ΣmH products - ΣnH reactants
For an exothermic reaction, H is negative
For an endothermic reaction, H is positive

21. SO2 + ½O2  SO3 H= ΣmH products - ΣnH reactants
Example: Sulfur dioxide reacts with oxygen. Write the balanced equation for the reaction of one mole of sulfur dioxide, calculate the enthalpy value for the reaction, the draw an energy diagram for this reaction based on your answer. Be sure to label reactants, products, and the enthalpy of the system.
SO2 + ½O2  SO3
Balanced Equation
Enthalpy Value
Energy Diagram
H= ΣmH products - ΣnH reactants
H= [-395.7kJ/mol] - [-296.8kJ/mol + ½ 0] = kJ/mol

22. Thermodynamic equations
For many reactions, a value is given alongside a balanced equation called a thermochemical equation. This value should be associated with the moles of each substance given in the problem.
1/8 S8(s) + O2(g)  SO2(g) H = kJ
As you can see, 1/8 mole of sulfur would release this amount of energy while one mole of sulfur would release eight times that amount of energy.

23. DH = -28 kJ (or say 28 kJ are released)
Example: For the reaction 2Na + 2H2O  2NaOH + H2 , H = -368 kJ Calculate the heat change that occurs when 3.5 g of Na reacts with excess water.
3.5g Na 1 mol Na kJ =
22.99g Na 2 mol Na
DH = -28 kJ (or say 28 kJ are released)

24. Please have your notes out on your desk. 
Today we will do some quick notes before you begin your lab. You will have plenty of time, don’t worry!
HW 1 and 2 will be checked tomorrow.

25. calorimetry Heat Capacity (C) C = heat absorbed Increase in temp.
C = J/g°C or J/mol°C
Calorimetry- the science of measuring heat flow in a chemical reaction.
It is based on observing the temperature change when a body absorbs or discharges heat.
The instrument used to measure this change is the calorimeter.

26. Here are some helpful tables that are found in the back of your text book.

27. calorimetry q = mCT J = (J/g°C )(g)(°C)
Constant pressure calorimetry- pressure remains constant during the process
Constant pressure calorimetry uses a set up called a coffee cup calorimeter
The primary reaction to calculate heat changes in a system is the “Mcat” equation.
Energy released as heat = (heat capacity) (mass of solution ) (increase in temp)

28. qwater = mCΔT and qwater = qMo So… mwCw ΔTw = mMoCMo ΔTMo
Example: A coffee cup calorimeter contains 150 g H2O at 24.6oc. A 110 g block of molybdenum is heated to 100oc and then placed in the water in the calorimeter. The contents of the calorimeter come to a temperature of 28.0oc. What is the heat capacity per gram of molybdenum?
qwater = mCΔT and qwater = qMo
So… mwCw ΔTw = mMoCMo ΔTMo
(150g)(4.18J/goC)(3.4oC) = (110g) (CMo)(72oC)
CMo = 0.27J/goC

29. 100. 0 ml of 0. 100M silver nitrate is mixed with 100. 0 ml of 0
100.0 ml of 0.100M silver nitrate is mixed with ml of 0.200M sodium chloride. Both solutions start at room temperature (25.0°C) and, once combined into a coffee cup calorimeter, the final temperature reading after mixing is 26.8°c. Find the heat of reaction in kj/mol of silver nitrate.
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
q = mCΔT = (200.0g)(4.18 J/g°C)(1.8°C)
q = 1500 J or 1.5 kJ
(0.1000L)(0.100M AgNO3) = mol AgNO3 in the reaction
1.5kJ/0.0100mol = 150 kJ/mol

30. Extensive vs intensive properties
Let’s Practice!
The number of calories of energy you derive from eating a banana
The number of calories of energy made available to your body when you consume 10.0 g of sugar
The density of your blood
The mass of iron present in blood
The electrical resistance of a piece of 22-gauge copper wire.
The melting point of copper wire. E
In terms of calorimetry, we can describe certain properties of the reaction as an:
Extensive property- this depends on the amount of substance (ex. Heat of reaction)
Intensive property- doesn’t depend on the amount of substance (ex. Temperature) I I I E I I

31. Calorimetry can also be done in a closed, rigid container.
This is called constant volume calorimetry
Ex. Flashbulb in a camera or a bomb calorimeter
No work can be done since the volume doesn’t change
Heat evolved = T x heat capacity of calorimeter (energy required to change the temp 1oC)

32. 6.3 Hess’s Law

33. Hess’s Law
Hess’s Law- States that the change in enthalpy from products to reactants, or H, is the same whether the reaction occurs in one step or in several steps.
H is not dependent on the reaction pathway.
The sum of the H for each step equals the H for the total reaction.
If a reaction is reversed, the sign of H is reversed.
If the coefficients in a reaction are multiplied by an integer, the value of H is multiplied by the same integer.

34. Hess’s Law applied
N2 + O2  2NO
2NO + O2  2NO2
N2 + 2O2  2NO2

35. Example: given the following reactions and their respective enthalpy changes, calculate H for the reaction: 2C + H2  C2H2.
C2H2 + 5/2 O2  2CO2 + H2O
H = kJ/mol C2H2
C + O2  CO2
H = kJ/mol C
H2 + ½ O2  H2O
H = kJ/mol H2
2C + 2O2  2 CO H = 2(-393.5) kJ
H2 + ½ O2  H2O H = kJ
2CO2 + H2O  C2H2 + 5/2 O2 H = kJ
2C + H2 C2H  H = 226.7kJ

36. Example: the heat of combustion of C to CO2 is -393
Example: the heat of combustion of C to CO2 is kJ/mol of CO2, whereas that for combustion of CO to CO2 is kJ/mol of CO2. Calculate the heat of combustion of C to CO.
First, write the equations given, as well as your goal equation.
Goal: C + ½O2  CO
Given: C + O2  CO DH = kJ
CO + ½ O2  CO DH = kJ
Now, rearrange to find the goal equation!
C + O2  CO DH= kJ
CO2  CO + 1/2 O2 DH= kJ
C + 1/2O2  CO DH= kJ

37. 6.4 Standard enthalpies of formation

38. ΔHf° Standard enthalpy of formation ( 𝛥𝐻 𝑓 0 )
The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states at 25oC.
The degree sign on a thermodynamics function indicates that the process it represents has been carried out at standard state conditions.

39. Standard state conditions
for gases, pressure is 1 atm
for a substance in solution, the concentration is 1 M
for a pure substance in a condensed state (liquid or solid), the standard state is the pure liquid or solid.
for an element, the standard state is the form in which the element exists under conditions of 1 atm and 25oC.
Values of 𝛥𝐻 𝑓 0 are found in Appendix 4 and at the end of these notes
𝛥𝐻 𝑓 0 reaction =  𝛥𝐻 𝑓 0 products -  𝛥𝐻 𝑓 0 reactants

40. Example: The standard enthalpy change for the reaction CaCO3(s)  CaO + CO2(g) is kJ. Calculate the 𝛥𝐻 𝑓 0 for CaCO3(s).
Solve for the heat of a single compound by using 𝛥𝐻 𝑓 0 reaction =  𝛥𝐻 𝑓 0 products -  𝛥𝐻 𝑓 0 reactants
178.1 kJ = [-635 kJ ] - [x]
X = kJ

41. 6.5 Present Sources of Energy
In this section, we will discuss some sources of energy, including fossil fuels, and their effects on the environment.

42. petroleum
Petroleum- A thick, dark liquid composed of hydrocarbons chains of 5-25 carbons
Refining petroleum involves the process of pyrolytic cracking, or distilling the fractions of petroleum from the main sample based on their molecular mass and boiling point.

43. gasoline
Gasoline C5-C12
Gasoline, when first used in car engines, caused a dramatic knocking sound and was thus treated with tetraethyl lead, (C2H5)4Pb, an antiknock agent. This introduced lead into the atmosphere as the fuel was spent and increased the amount of ingested lead in the human and animal populations until 1960 when “leaded” gas was finally phased out.
kerosene & jet fuel C10-C18
heating and lubricating oil and diesel fuel C15-C25
asphalt >C25

44. Natural gas
Natural gas- This substance is usually found alongside petroleum and is composed mostly of methane. It also contains ethane, propane, and butane.

45. Coal
Coal- formed from the remains of plants buried under pressure for many years. Cellulose, CH2Ox, gradually loses its H and O.
Coal develops through 4 stages:
lignite (least valuable)
subbituminous
bituminous (high sulfur)
anthracite (most valuable)
Coal provides 20% of our energy in the U.S.

46. CO2 and the Environment Effects of CO2 on the Climate
Greenhouse effect - H2O and CO2 molecules in the atmosphere reflect IR radiation and send it back to earth thus raising the earth’s temperature.
The CO2 concentration has increased by about 16% in the past 100 years because of increase in the use of fossil fuels.

47. New Energy Sources
Coal gasification- treating coal with oxygen and steam at high temperatures to break down many of the C-C bonds and form C-O and C-H bonds.
The products are syngas (CO + H2) and methane gas. Syngas may be converted to methanol.
CO(g) + 2H2(g)  CH3OH(l)

48. New energy sources Hydrogen as a Fuel H2 (g) + ½O2(g)  H2O(l)
Ho = -286 kJ
~2.5 times the energy of natural gas
3 problems: production (too expensive), transport (too volatile), and storage (large volume, decomposes to H atoms on metal surfaces, makes metal brittle-forms metal hydrides)

49. Other Alternatives:
oil shale
ethanol
gasohol
seed oil (sunflower)

50. CHAPTER 17: Spontaneity, Entropy and Free Energy

51. 17.1 & 17.2 Spontaneous Processes and Entropy and the Second Law of Thermodynamics

52. 1st Law of Thermodynamics
The first Law of Thermodynamics states that energy is neither created nor destroyed; it is constant in the universe. We can measure energy changes in chemical reactions to help determine exactly what is happening and how it is occurring.
Keep in mind that thermodynamics deals with the reactants and products while kinetics deals with how reactants become products.

53. Spontaneous… What??
Spontaneous process- occurs without outside intervention (ex. Rusting)
This may be fast or slow (CO2 sublimes at room temperature vs. iron rusting when in the presence of oxygen)

54. entropy Low Entropy! High Entropy!
Entropy (S)- a measure of randomness or disorder
This is associated with probability. (there are more ways for something to be disorganized than organized)
Entropy increases going from a solid to a liquid to a gas and when solutions are formed.
Entropy increases in a reaction when more atoms or molecules are formed.
Entropy increases with increasing temperature.

55. Entropy increases…
SLG
More molecules produced
Temperature increases

56. 2nd Law of thermodynamics
2nd Law of Thermodynamics- In any spontaneous process there is always an increase in the entropy of the universe. The energy of the universe is constant but the entropy of the universe is increasing.

57. 17.3 & 17.4 Entropy and free energy

58. Free energy
G = H -TS
Free energy (G)- the amount of energy available to do work.
Free energy change is a measure of the spontaneity of a reaction. It is the maximum work available from the system.
A spontaneous reaction carried out as constant temperature and pressure has a negative G. For example, when ice melts H is positive (endothermic), S is positive and G = 0 at 0˚C.

59. G = H -TS + - - + + ? - - ? + + + S H G Spontaneous? Yes, always
Yes, always
Yes, at high T
Yes, at low T
No, never + - - + + ? - - ? + + +

60. 17.5 entropy changes in chemical reactions

61. 3rd law of thermodynamics
Third Law of Thermodynamics- The entropy of a perfect crystal at 0 K is zero.
Any substance that is not at 0 Kelvin must have a value for entropy! This sets entropy apart for enthalpy. Enthalpy values are just changes, while entropy values are absolute and cannot drop below zero.

62. Ex. Given the following standard molar entropies, calculate So for the reaction: 2Al(s) + 3MgO(s)  3Mg(s) + Al2O3(s)
Mg(s) = 33.0 J/K Al2O3(s) = 51.0 J/K
Al(s) = 28.0 J/K MgO(s) = 27.0 J/K
Soreaction = Soproducts - Soreactants
So = [3(33.0J/K) J/K] – [2(28.0J/K) + 3(27.0J/K)]
So = 13.0 J/Kmol.rxn

63. 17.6 free energy and chemical reactions

64. The Standard Free Energy Change
Standard free energy change (G°) -The change in free energy that occurs if the reactants in their standard states are converted to products in their standard states.
G° =Gfoproducts - Gforeactants at standard conditions
Gfo for a free element in its standard state is zero.

65. Ex. Given the equation N2O4(g)  2NO2(g) and the following data, calculate Go.
Gfo for N2O4(g) = kJ/mol,
Gfo for NO2(g) = kJ/mol
Go = [2(51.30kJ/mol)] – [ kJ/mol]
= 4.78 kJ/mol rxn

66. The Gibbs-Helmholtz equation
The Gibbs-Helmholtz equation works like the Gibbs Free Energy equation, just at standard state conditions.
Go =Ho -TSo
(When working this, change the units for S to kJ).

67. Ex. For the given reaction and the following information, calculate Go at 25°C. 2PbO(s) + 2SO2(g)  2PbS(s) + 3O2(g)
Ho (kJ/mol)
So (J/mol.K)
PbO (s)
-218.0
70.0
SO2 (s)
-297.0
248.0
PbS (s)
-100.0
91.0
O2 (g) ?
205.0
Ho = [2(-100.0) + 3(0)]- [2(-218.0) + 2(-297.0)]
= kJ/mol rxn
So = [2(91.0) + 3(205.0)] – [2(70.0)+2(248.0)]
= J/K.mol rxn = kJ/mol
Go = Ho - T So
Go= kJ/mol -298K( kJ/mol)
= 782 kJ/mol rxn

68. 17.7 The Dependence of Free Energy on Pressure

69. Entropy and pressure/ volume
In an ideal gas, enthalpy does not depend on pressure. Entropy, however, does depend on pressure because it depends upon volume.
Gases in large volumes have greater entropy than in a small volume.
Gases at a low pressure have greater entropy than at a high pressure.
Because entropy depends on pressure, G of an ideal gas depends on its pressure.

70. 17.7 The Dependence of Free Energy on Pressure
If we incorporate PV = nRT with the equation for free energy, we end up with an equation to calculate free energy in relation to temperature and pressure variables.
G = G° + RT ln (Q)
Q = reaction quotient (partial pressure of products/reactants raised to the power of their coefficients)-only pressures of gases are included.
T = temperature in Kelvin
R = gas constant J/K.mol
G° = free energy change at 1 atm (be sure to change to Joules!)

71. Ex. Calculate G at 298K for the following reaction if the reaction mixture consists of 1.0 atm N2, 3.0 atm H2, and 1.0 atm NH3.
N2(g) + 3H2(g)  2NH3(g) Go = kJ/mol
G = Go + RT lnQ
G = -33,320J/mol J/Kmol(298K) ln (1.0𝑎𝑡𝑚) 2 (1.0 𝑎𝑡𝑚) 1 (3.0𝑎𝑡𝑚) 3
G = -41,500J/mol rxn or
-41.5 kJ/mol rxn

72. 17.8 Free Energy and Equilibrium

73. Free energy and equilibrium
The equilibrium point in terms of kinetics occurs when the forward and reverse reactions were occurring at an equal rate.
In terms of free energy, the equilibrium point occurs at the lowest value of free energy available to the reaction system.

74. Free energy and equilibrium
These two definitions are the same!
G = Gproducts - Greactants = 0
If a process has just shifted from nonspontaneous to spontaneous, then at the point where it changes, the value for G is zero. If G is zero, then Ho = TSo.

75. Ex. Given for the reaction Hg(l)  Hg(g) that Ho = 61
Ex. Given for the reaction Hg(l)  Hg(g) that Ho = 61.3 kJ/mol and So = J/mol.K, calculate the temperature of the normal boiling point of Hg.
Go =Ho -T So
G = 0 at phase change
0 = 61.3 –T(0.1000)
T = 613K or 340oC

76. The “rat link” equation
We can utilize the previous two equations (G = Gproducts - Greactants = 0) and (G = Go + RT ln (Q)) to form an equation that describes the relationship between free energy and the value of the equilibrium constant.
Go = -RT ln(K) (the “rat link” equation)
When Go = 0, free energy of reactants and products are equal when all components are in their standard states. During a phase change, G = 0.
When Go < 0, Go products < Go reactants The reaction is not at equilibrium, K > 1 since pressure of products is > 1 and the pressure of reactants is < 1.
When Go > 0, Go reactants < Go products The reaction is not at equilibrium, K < 1 since pressure of products is < 1 and the pressure of reactants is > 1.

77. Go = -RT ln K Go = -8.314(298)ln (1.0 x 10-3) = 1.7 x 104J or 17 kJ
Ex. Calculate the approximate standard free energy for the ionization of hydrofluoric acid, HF (Ka = 1.0 x 10-3), at 25oC.
Go = -RT ln K
Go = (298)ln (1.0 x 10-3)
= 1.7 x 104J or 17 kJ

78. Free energy and equilibrium
We can use G =Go + RT ln(Q) to calculate the direction that a reaction will shift to reach equilibrium.
Free energy is energy available to do useful work. Wmax = G