1. Thermodynamics (That’s Hot!!! Or…. Not!)

2. The Nature of Energy Kinetic Energy Potential Energy
Physics and Chemistry Agree
Sometimes

3. First Law of Thermodynamics
Energy in the universe is constant
You’ve heard this stated another way
Law of conservation of energy
Energy can be neither created nor destroyed
It is transferred.

4. What is Energy?
When you think of energy, what comes to mind?

5. “Forms” of Energy We associate energy with “enthalpy”
The “force” that holds things together
Whenever there is a “change”, we can describe it as a change in Energy (enthalpy)
How much of an energy change is there?
How do we measure the change?
Atoms →molecules
molecules→ cells
cells → plants / animals
Plants / animals→ ecosystems
Solid
Liquid
Gas
Plasma
Get students to mention the ways to measure. Include temperature and mass. Set up mentally for calorimetry

6. Foreshadowing Energy
Entropy and Free energy will be discussed in the next chapter
Enthalpy is what we are studying now
Describe what bonds atoms and molecules to each other
The production of heat (or absorbing)

7. Measuring Energy Temperature, Heat and Work
Temperature - random motion of particles
average kinetic energy
Heat - transfer of energy
due to a temperature difference
Heat is not contained by an object
If your hand is the same temperature as the radiator, would it feel hot?
It is the temperature difference that feels hot
The fire has a lot of heat.
Work is force acting over a distance.

8. When Energy Transfers Occur, Does the Pathway Matter
When Energy Transfers Occur, Does the Pathway Matter? Or Is It a State Function?
Energy of a “system” is independent of where the energy came from or how it got there
Energy of a system depends on what is in the system and their temperatures.
Energy is a STATE FUNCTION

9. Internal Energy of a System
Sum of the potential and kinetic energies
A change in energy is done by work (motion) or heat
E = q + w
q = heat
w = work

10. Work
Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas).
Work = Force x distance = Pressure x area x height
= Pressure x Volume
Pressure and Volumes are State Functions

11. Work
We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston.
w = -PV

12. Enthalpy
If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy (H) of the system.
Enthalpy is the internal energy plus the product of pressure and volume:
H = E + PV
Since E, P, and V are State functions, then
Enthalpy is a state function as well!

13. State Functions However, q and w are not state functions.
Whether this battery is shorted out or is discharged by running the fan, its E is the same.
But q and w are different in the two cases.

14. Heat and Work More proof that heat and work are not state functions
Ex. 2: Imagine two magic boxes
Each has 1g sugar or 1g of magnesium.
There is a different amount of work (gases expanding against the atmosphere), light and heat produced for each gram of material burned.

15. Enthalpy
When the system changes at constant pressure, the change in enthalpy, H, is
H = (E + PV)
This can be written
H = E + PV

16. Enthalpy
Since E = q + w and w = -PV, we can substitute these into the enthalpy expression:
H = E + PV
H = (q+w) − w
H = q
So, at constant pressure, the change in enthalpy is the heat gained or lost.

17. Endothermicity and Exothermicity
A process is endothermic when H is positive.

18. Endothermicity and Exothermicity
A process is endothermic when H is positive.
A process is exothermic when H is negative.

19. Signs and Thermodynamics
Energy Change
E < 0, or negative
system loses energy
E > 0, or positive
system gains energy

20. Sign Conventions in Thermodynamics
Heat
q > 0
Increase in temp
q < 0
Decrease in temp

21. Sign Conventions in Thermodynamics
Work is more complicated
W = - PV
P = Pressure on the system
V = Change in volume
If a piston increases in size, it must move against the universe. So energy has left the system and gone into the universe. This is exothermic so E would be negative. Since the change in volume is positive, the sign is -PV. (Pg 247 in Book)

22. Sign Conventions
So any work done by the system on the universe will be exothermic meaning w = -
If work is done on the system by the universe, then it is endothermic meaning w = + .
Pressure is measured in N/m2 which = 1Pa
If you have 1kPa of Pressure and Volume is in L, then PV = J
1L•kPa = 1J
101.3kPa = 1 atm
1L•atm = J
(in case you were wondering)

23. Sample Problem
A balloon is heated with 1.3 x 108 J and expands from 4.00 x 106 L to 4.5 x 106 L. Assuming constant outside pressure of 1.0 atm, calculate the change in energy
E = q + w
q = +1.3 x 108 J
W = -P V = -1.0 atm x (4.50 x 106 L x 106 L)
x 106 L atm x (101.3 J / L ATM ) = -5.1 x 107 J
E = +1.3 x 108 J x 107J = 8.0 x 107 J

24. H= qp E = Eprod - Ereact CH4 + 2O2  CO2 + 2 H2O + energy
qp = H = Kj/mol
notice negative value (energy left system)
Stoichiometry type problems
Energy is a reactant or product

25. Chemical Energy energy CH4 + O2 CO2 + 2 H2O 
CH4 + O2  CO H2O + energy
We must define “THE SYSTEM”
The container with the methane and oxygen is a system
The rest of the Universe is not the system.
Heat flows out of the system – exothermic

26. Chemical Energy CH4 + 2O2  CO2 + 2 H2O + energy
Chemical (potential) energy
is converted to thermal (random kinetic energy)
CH4 + 2O2 has more potential energy
CO H2O has less potential energy
The difference is the energy released

27. Chemical Energy N2O2 + energy 2 NO
N2O2 has less potential energy than
2 NO
The difference is the energy absorbed from universe
This is endothermic

28. Enthalpy Problems CH4 + 2O2  CO2 + 2 H2O + energy
qp = H = Kj/mol
A 5.8g sample of methane is burned in oxygen at a constant pressure. How much heat is given off? What is the enthalpy change?
5.8g x 1 mol x kJ = -322 kJ g mol
E = Eprod - Ereact = q = -322kJ
q = 322 kJ

29. Ways to determine Enthalpies of Reactions
Calorimetry – determine through experimentation and calculating heat flow.
Hess’ Law – Calculate enthalpies of a series of reactions that when added together give the desired reaction
Heats of Formations – Use determined energies of compounds to determine the overall enthalpy change.
Bond Energies – Bonds broken – bonds formed (later)

30. Calorimetry
Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

31. Heat Capacity and Specific Heat
The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity.
We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K.
Molar Heat Capacity is the amount of energy required to raise the temperature of 1 mole of a substance 1K.
Remember a change of 1K and 1oC are the same!

32. Heat Capacity and Specific Heat
Specific heat (Cp), then, is used mathematically by the following equation:

33. Calorimetry Why does a piece of metal feel cold to your hand?
Why does the sand heat up faster than the water at the beach?
Why does the center of the United States have more temperature extremes than the coasts?
Why do ceramic tiles prevent a spacecraft from burning up upon reentry into the atmosphere?

34. Calorimetry Constant Pressure Calorimetry (open)
q = H
q = mCp T.
Constant Volume Calorimetry (Bomb)
q = E (need to know heat capacity of calorimeter)
qrxn = Ccal ΔT

35. Hess’ Law or State Functions
NO2(g)  1NO(g) + 1/2O2(g) H = 56 kJ/mol
Double the reaction = double the enthalphy
2NO2(g)  2NO(g) + O2(g)  H = 112 kJ/mol
The reverse of the reaction, “negative”
2NO(g) + O2(g)  2NO2(g) H =- 112 kJ/mol

36. Hess’ Law N2(g) + 2O2(g)  2NO2(g) H = 68 kJ/mol
Can be determined by:
N2(g) + O2(g)  2NO(g)  H = 180 kJ/mol
2NO(g) + O2(g)  2NO2(g)  H =- 112 kJ/mol
N2(g) + 2O2(g)  2NO2(g)  H = 68 kJ/mol

37. Hess’s Law Works because Enthalpy is a state function
The pathway is not important
How we get there is not important
You can use a series of known equations to “add up” to equal the equation you desire.
Arrange the steps to equal the overall equation
Add up the enthalpies

38. 2B(s) + 3H2(g)  B2H6 Pg 258 kJ 2B(s) + 3/2 O2(g)  B2O3(S) -1273
B2O3(s) + 3H2O(g)  B2H6(g) + 3O
3H2(g) + 3/2O2(g)  3H2O(l)
3H2O(l)  3H2O(g)
2B(s) + H2(g)  B2H
Pg 258

39. 2B(s) + 3H2(g)  B2H6 Opposite sign OK Backwards Not enough Not enoughkJ
2B(s) + 3/2 O2(g)  B2O3(S)
B2H6(g) + 3O2  B2O3(s) + 3H2O(g)
H2(g) + 1/2O2(g)  H2O(l)
H2O(l)  H2O(g)
Backwards
Not enough
Not enough
X 3
X 3

40. 2B(s) + 3H2(g)  B2H6 +36 2B(s) + 3/2 O2(g)  B2O3(S) -1273kJ
2B(s) + 3/2 O2(g)  B2O3(S)
B2O3(s) + 3H2O(g)  B2H6(g) + 3O
3H2(g) + 3/2O2(g)  3H2O(l)
3H2O(l)  3H2O(g)
+36
2B(s) + 3H2(g)  B2H6

41. Hess’s Law Calculate ΔH for the reaction 2 C(s) + H2(g)  C2H2(g)
given the following chemical equations and their respective enthalpy changes:
C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(l) ΔH = kJ
C(s) + O2(g)  CO2(g) ΔH = kJ
H2(g) + ½ O2(g)  H2O(l) ΔH = kJ

42. Enthalpies of Formation
An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which one mole of a compound is made from its constituent elements in their elemental forms.
Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure).
See Appendix of textbook for tables of Hf°.

43. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)

44. Calculation of H We can use Hess’s law in this way:
H = nHf°products – mHf° reactants
where n and m are the stoichiometric coefficients.
Take the sum of the products’ heats of formation and subtract the sum of the reactants’ heats of formation.
The Standard Heat of Formation of any element in it’s standard state is 0!

45. Practice
For which of the following reactions at 25 ºC would the enthalpy change represent a standard enthalpy of formation?
2 Na(s) + ½ O2(g)  Na2O (s)
2 K(l) + Cl2(g)  2 KCl(s)
C6H12O6(s)  6 C(diamond) + 6 H2(g) + 3O2(g)

46. Practice
Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to form CO2(g) and H2O(l).
ΔHfº: CO2(g) = kJ/mol
H2O (l) = kJ/mol
C6H6(l) = 49.0 kJ/mol
O2 (g) = 0 kJ/mol

47. Practice
Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to form CO2(g) and H2O(l).
ΔHfº: CO2(g) = kJ/mol
H2O (l) = kJ/mol
C6H6(l) = 49.0 kJ/mol
O2 (g) = 0 kJ/mol
-3267 kJ/mol

48. Practice The standard enthalpy change for the reaction
CaCO3(s)  CaO(s) + CO2(g)
is kJ. From the values for the standard enthalpies of formation of CaO(s) ( kJ) and CO2(g) ( kJ), calculate the standard enthalpy of formation of CaCO3(s).

49. Practice The standard enthalpy change for the reaction
CaCO3(s)  CaO(s) + CO2(g)
is kJ. From the values for the standard enthalpies of formation of CaO(s) ( kJ) and CO2(g) ( kJ), calculate the standard enthalpy of formation of CaCO3(s).
kJ/mol