1. Catalyst

2. Egg Throwing

4. Justify – TPS
Which one of these represents a spontaneous process and which of these represents a process that does not occur spontaneously?

5. Lecture 5.6 – 2nd and 3rd Law of Thermodynamics

6. Today’s Learning Targets
LT 5.12 – I can analyze the Second Law of Thermodynamics and how this relates to the idea of spontaneous process, entropy, and whether a process is reversible or irreversible.
LT 5.13 – I can analyze the Third Law of Thermodynamics and how it relates to entropy of a system.
LT 5.14 – I can interpret the entropy of a chemical reaction and how standard entropies can be use to calculate the overall entropy of the system.

7. Constraints of the 1st Law
1st Law and ΔH allows us to clearly study energy, but it has limitations.
ΔH does not provide a good description of whether a reaction is likely to occur
There are examples of both positive and negative ΔH values leading to reactions that are favorable.

8. Spontaneous Process Spontaneous
A spontaneous process is one that proceeds without any outside assistance.
A non-spontaneous process is one that requires assistance in order to occur
Spontaneity does not mean a reaction occurs “fast”. Only tells us the energy of a reaction
Spontaneous
Provide examples of how water freezing can be both spontaneous and non-spontaneous based on the temperature. Additionally, a gas filling a room is a spontaneous process.
Non – Spontaneous

9. Endothermic and Spontaneous
Why are reactions/processes that have a +ΔH able to occur (e.g. reactions that feel cold)?
Only reactions that have –ΔH should be favored by the universe
The answer is due to:
Entropy!!!

10. Reversible vs. Irreversible Processes
Reversible Process is any process that can be restored to its original state without no overall net change.
Irreversible Process is a process that cannot be simply reversed to return system to its original state
Irreversible Process
Changes in temperature are irreversible processes
Hot
Cold
Not Possible!

11. Isothermal Process

12. Major Idea All real processes are irreversible
Therefore, all spontaneous processes are irreversible

13. ΔS and Spontaneity
Need to be able to characterize a reaction that we are unfamiliar with as spontaneous or non-spontaneous
Entropy is a state function so:
ΔS = Sfinal – Sinitial
For a process under constant pressure, ΔS can be calculated by:
qrev represents the heat required to make the process reversible

14. ΔS and Phase Changes Phase changes are isothermal processes.
Imagine examining the melting of a substance. Therefore, we can say:
qrev = ΔHfusion
This means that for phase changes:

15. Class Example
Elemental mercury is a silver liquid at room temperature. Its normal freezing point is oC, and its molar enthalpy of fusion is ΔHfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg (l) freezes at the normal freezing point?

16. Table Talk
The normal boiling point of ethanol, C2H5OH, is 78.3 oC, and its molar enthalpy of vaporization is kJ/mol. What is the change in entropy in the system when 68.3 g of C2H5OH (g) at 1 atm condenses to liquid at the normal boiling point?

17. ΔS and the 2nd Law of Thermodynamics
Entropy is not conserved (unlike energy in the 1st law)
Example: 1 mole of water freezes at T < 0 oC
H2O (l)  H2O (s)
What is ΔSsys? Assume the surroundings are around 310 K.
We know that Sliquid > Ssolid
Therefore ΔSsys < 0 because solid has less disorder than liquid
This also means that ΔHsys < 0 because heat was released to the surroundings
This ΔHsys leads to an increase in the overall entropy of the surroundings. If you were to calculate the ΔS for both system and surrounding, then you would discover that:
ΔSsys + ΔSsurr = ΔSuniversie > 0 for all spontaneous processes
THE UNIVERSE FAVORS DISORDER!!!!

20. Did Lisa Simpson Violate the 2nd Law?
Homer claims that Lisa has violated the laws of thermodynamics
With your group, read the short overview of perpetual motion.
Applying your knowledge of thermodynamics, draw a conclusion about whether the laws have been violated or not.

22. ΔSsurroundings and ΔHsystem
As we lose energy, entropy increases. So, we can say that:
Furthermore, we know that entropy increases as T increases, so:
Therefore, it is possible to conclude that:

23. Two Important Equations
When does S equal 0?
When does ΔSsurr equal 0?
There is no entropy when there is only one microstate that exists
ΔS = 0 when the temperature reaches 0 K.
This means that at 0 K there is no entropy because there is only one microstate that exists.

24. 3rd Law of Thermodynamics
The 3rd Law of Thermodynamics claims that the entropy of a pure crystalline substance at absolute zero is 0.
There is only one possible microstate because all motion has ceased, which means there is 0 entropy in the system.
This is one of the proofs that absolute zero is a quantity that must exist.

26. Entropy Change for Chemical Reactions
We use calorimetry to measure ΔH
We do not have a nice way to measure ΔS in a reaction
We do know that every substance must have a point where S = 0 due to the 3rd Law
As temperature increases from 0 K, the value of S also increases. We therefore can determine standard entropies (So)
Standard Entropies is the entropy gained by taking 1 mol of perfect crystalline substance at 0 K and bringing it to standard conditions
Measured in J/(mol x K)

27. Important Points on So Unlike ΔHf, So is not zero for elements
So is greatest in gas, than liquids, and least in solids
So increases as substances have larger molar masses
So increases as the number of atoms in a chemical formula increases
Increased number of possible microstates

28. Calculating ΔSo
We can use So values to calculate ΔSo through the equation:

29. Class Example For the reaction: N2 (g) + 3 H2 (g)  2 NH3 (g)
Calculate ΔSo for the reaction. You know that So (N2) = J/(mol x K), that So (H2) = J/(mol x K), and that So (NH3) = J/(mol x K).

30. Al2O3 (s) + 3 H2 (g)  2 Al (s) + 3 H2O (g)
Table Talk
Calculate ΔSo for the following reaction:
Al2O3 (s) + 3 H2 (g)  2 Al (s) + 3 H2O (g)
You know that So (Al2O3) = J/ (mol x K), So (H2) = J/(mol x K), So (Al) = J/(mol x K), and So (H2O ) = J/(mol x K).

31. Entropy Changes and Spontaneity
We now have a tool to calculate ΔSsystem, but how do we combine this with ΔSsurroundings and ΔSuniverse?
We know that:
We then can add surroundings and system to determine if the reaction is spontaneous.

32. Class Example For the reaction: N2 (g) + 3 H2 (g)  2 NH3 (g)
Determine whether or not the reaction is spontaneous under standard conditions. You know that ΔHf for NH3 is kJ.
Additionally you know that
So (N2) = J/(mol x K)
So (H2) = J/(mol x K)
So (NH3) = J/(mol x K).

33. Al2O3 (s) + 3 H2 (g)  2 Al (s) + 3 H2O (g)
Table Talk
For the following reaction:
Al2O3 (s) + 3 H2 (g)  2 Al (s) + 3 H2O (g)
Determine whether or not the reaction is spontaneous under standard conditions. You know that
ΔHf (Al2O3) = kJ/mol and ΔHf (H2O) = kJ/mol
You know that:
So (Al2O3) = J/ (mol x K)
So (H2) = J/(mol x K)
So (Al) = J/(mol x K)
So (H2O ) = J/(mol x K).

34. White Board Questions

35. Question 1
Which molecule should have a higher So? Why?
C2H6 or C2H2

36. Question 2 Calculate ΔSo for the following chemical reaction:
C2H4 (g) + H2 (g)  C2H6 (g)
So (C2H4) = 219 J/ (mol x K)
So (H2) = J/ (mol x K)
So (C2H6) = J/ (mol x K)

37. Question 3 Calculate ΔSo for the following chemical reaction:
N2O4 (g)  2NO2
So (N2O4) = J/ (mol x K)
So (NO2) = J/ (mol x K)

38. Be(OH)2 (s)  BeO (s) + H2O (g)
Question 4
Determine if the following reaction is spontaneous at 298 K:
Be(OH)2 (s)  BeO (s) + H2O (g)
So (Be(OH)2) = J/ (mol x K)
So (BeO) = J/ (mol x K)
So (H2O) = J/ (mol x K)
ΔHf (Be(OH)2) = kJ/mol
ΔHf (BeO) = kJ/mol
ΔHf (H2O) = kJ/mol

39. 3 CH3OH (g) + O2 (g)  2 CO2 (g) + 4 H2O (g)
Question 5
Determine if the following reaction is spontaneous at 298 K:
3 CH3OH (g) + O2 (g)  2 CO2 (g) + 4 H2O (g)
So (CH3OH) = J/ (mol x K)
So (O2) = J/ (mol x K)
So (H2O) = J/ (mol x K)
So (CO2) = J/ (mol x K)
ΔHf (CH3OH) = kJ/mol
ΔHf (CO2) = kJ/mol
ΔHf (H2O) = kJ/mol

41. Closing Time
Read 19.1, 19.2, and 19.4
Do book problems: