1. Microstates of Entropy
Chapter 19 cont and 19.4
Microstates of Entropy

2. Entropy on the Molecular Scale
Ludwig Boltzmann described the concept of entropy on the molecular level.
Temperature is a measure of the average kinetic energy of the molecules in a sample.
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3. Entropy on the Molecular Scale
Molecules exhibit several types of motion:
Translational: Movement of the entire molecule from one place to another.
Vibrational: Periodic motion of atoms within a molecule.
Rotational: Rotation of the molecule about an axis or rotation about  bonds. 
and disc from book: D:\Chapter_19\C_Media\molecular-motions\_molecular-motions.html
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4. Entropy on the Molecular Scale
Boltzmann envisioned the motions of a sample of molecules at a particular instant in time.
This would be akin to taking a snapshot of all the molecules.
He referred to this sampling as a microstate of the thermodynamic system. 
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5. Entropy on the Molecular Scale
Each thermodynamic state has a specific number of microstates, W, associated with it.
Entropy is
S = k ln W
where k is the Boltzmann constant, 1.38  1023 J/K. 
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6. Entropy on the Molecular Scale
The change in entropy for a process, then, is
S = k ln Wfinal  k ln Winitial
Wfinal
Winitial
S = k ln
Entropy increases with the number of microstates in the system.
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7. Entropy on the Molecular Scale
The number of microstates and, therefore, the entropy, tends to increase with increases in
Temperature
Volume
The number of independently moving molecules.
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8. Entropy and Physical States
Entropy increases with the freedom of motion of molecules.
Therefore,
S(g) > S(l) > S(s)
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10. In which phase are water molecules least able to have rotational motion?
A. Vapor
B. Liquid
C. Ice
Answer: C

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Solutions
Generally, when a solid is dissolved in a solvent, entropy increases.
However, water becomes more organized. Why?
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Entropy Changes
In general, entropy increases when
Gases are formed from liquids and solids;
Liquids or solutions are formed from solids;
The number of gas molecules increases;
The number of moles increases.
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14. What major factor leads to a decrease in entropy as the reaction shown takes place?
A. Increase in number of molecules during change
B. Decrease in number of molecules during change
C. Change in pressure of system
D. Change in internal energy of system
Answer: B

15. Third Law of Thermodynamics
The entropy of a pure crystalline substance at absolute zero is 0.
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17. Why does the plot show vertical jumps at the melting and boiling points?
A. During melting or boiling at constant temperature entropy dramatically increases because energy is removed from the system during the change.
B. During melting or boiling at constant temperature entropy changes significantly because more microstates exist for the final phase compared to the initial phase.
C. During melting or boiling at constant temperature entropy changes significantly because fewer microstates exist for the final phase compared to the initial phase.
D. During melting or boiling at constant temperature entropy changes significantly because no change in the number of microstates occurs during the phase change.
Answer: B

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Standard Entropies
These are molar entropy values of substances in their standard states.
Standard entropies tend to increase with increasing molar mass.
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Standard Entropies
Larger and more complex molecules have greater entropies.
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21. What might you expect for the value of S° for butane, C4H10?
A. 270 J/mol-K to 300 J/mole-K
B. 280 J/mol-K to 305 J/mole-K
C. 310 J/mol-K to 315 J/mole-K
D. 320 J/mol-K to 340 J/mole-K
Answer: C

22. Entropy Changes S = nS(products) — mS(reactants)
Entropy changes for a reaction can be estimated in a manner analogous to that by which H is estimated:
S = nS(products) — mS(reactants)
where n and m are the coefficients in the balanced chemical equation.
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23. Entropy Changes in Surroundings
Heat that flows into or out of the system changes the entropy of the surroundings.
For an isothermal process:
Ssurr =
qsys T
At constant pressure, qsys is simply H for the system.
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24. Entropy Change in the Universe
The universe is composed of the system and the surroundings.
Therefore,
Suniverse = Ssystem + Ssurroundings
For spontaneous processes
Suniverse > 0
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25. Entropy Change in the Universe
Since
Ssurroundings =
and
qsystem = Hsystem
This becomes:
Suniverse = Ssystem +
Multiplying both sides by T, we get
TSuniverse = Hsystem  TSsystem
qsystem T
Hsystem T
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26. Al2O3(s) + 3 H2(g)  2 Al(s) + 3 H2O(g)
Sample Exercise 19.5 Calculating ΔS° from Tabulated Entropies
Calculate the change in the standard entropy of the system, ΔS° , for the synthesis of ammonia from N2(g)
and H2(g) at 298 K:
N2(g) + 3 H2(g)  2 NH3(g)
Solution
Analyze We are asked to calculate the standard entropy change for the synthesis of NH3(g) from its constituent elements.
Plan We can make this calculation using Equation 19.8 and the standard molar entropy values in Table 19.1 and Appendix C.
Solve
Using Equation 19.8, we have ΔS° = 2S°(NH3)–[S°(N2) + 3S°(H2)]
Substituting the appropriate values from Table 19.1 yields
ΔS° = (2 mol)(192.5 J/mol–K)–[(1 mol)(191.5 J/mol–K) + (3 mol)(130.6 J/mol–K)]
= –198.3 J/K
Check: The value for ΔS° is negative, in agreement with our qualitative prediction based on the decrease in the number of molecules of gas during the reaction.
Practice Exercise
Using the standard molar entropies in Appendix C, calculate the standard entropy change, ΔS°, for the following reaction at 298 K:
Al2O3(s) + 3 H2(g)  2 Al(s) + 3 H2O(g)
Answer: J/K 26

27. Plan today Gibbs Free energy WS Graph problems from WS
Problems from book
Reminder: Quest is a quiz grade, next text is Thursday after break!

28. Whiteboard
Group
Problem 1
2,7 2
3,8 3
4,7 4
5,8 5
6,7 6 7 8