1. Chapter 19 Chemical Thermodynamics
Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Chapter 19 Chemical Thermodynamics
John D. Bookstaver
St. Charles Community College
Cottleville, MO
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2. Aim: What are spontaneous and irreversible reactions?
Do Now: What did the first law of thermodynamics tell us about energy?
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3. First Law of Thermodynamics
You will recall from Chapter 5 that energy cannot be created nor destroyed.
Therefore, the total energy of the universe is a constant.
Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.
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4. Spontaneous Processes
Spontaneous processes are those that can proceed without any outside intervention.
The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return to vessel B.
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5. Spontaneous Processes
Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.
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6. Spontaneous Processes
Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures.
Above 0 C it is spontaneous for ice to melt.
Below 0 C the reverse process is spontaneous.
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7. Reversible Processes
In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process.
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8. Irreversible Processes
Irreversible processes cannot be undone by exactly reversing the change to the system.
Spontaneous processes are irreversible.
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9. Aim: What is Entropy, and how does it impact the second law of thermodynamics?
Do Now: When Water boils to become a gas, does entropy increase or decrease?
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10. Entropy
Entropy (S) is a term coined by Rudolph Clausius in the 19th century.
Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, q T
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11. Entropy
Entropy can be thought of as a measure of the randomness of a system.
It is related to the various modes of motion in molecules.
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12. Entropy
Like total energy, E, and enthalpy, H, entropy is a state function.
Therefore,
S = Sfinal  Sinitial
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13. Entropy
For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature:
S =
qrev T
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14. Sample Exercise 19.2 (Pg. 819) Calculating ∆S for a Phase Change
Elemental Mercury is a silver liquid at room temperature. Its normal freeing point is ∘C, and its molar enthalpy of fusion is ∆Hfusion= 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at its normal freezing point.
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15. Second Law of Thermodynamics
The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes, and the entropy of the universe does not change for reversible processes.
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16. Second Law of Thermodynamics
In other words:
For reversible processes:
Suniv = Ssystem + Ssurroundings = 0
For irreversible processes:
Suniv = Ssystem + Ssurroundings > 0
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17. Second Law of Thermodynamics
These last truths mean that as a result of all spontaneous processes the entropy of the universe increases.
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18. Give It Some Thought (Pg. 821)
The rusting of iron is spontaneous and is accompanied by a decrease in the entropy of the system (the iron and oxygen). What can we conclude about the entropy change of the surroundings?
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19. Aim: How is entropy seen at the molecular level, and what is the third law of thermodynamics?
Do Now: In general, why would you say that the entropy of a sample of gas is greater than that of a liquid?
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20. 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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21. 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 on about an axis or rotation about  bonds. 
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22. Give it some Thought. Pg. 824
Can Argon atoms undergo vibrational motion?
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23. 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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24. Entropy on the Molecular Scale
Each thermodynamic state has a specific number of microstates, W, associated with it.
Entropy is
S = k lnW
where k is the Boltzmann constant, 1.38  1023 J/K.
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25. Entropy on the Molecular Scale
The change in entropy for a process, then, is
S = k lnWfinal  k lnWinitial
lnWfinal
lnWinitial
S = k ln
Entropy increases with the number of microstates in the system.
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26. 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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27. Entropy and Physical States
Entropy increases with the freedom of motion of molecules.
Therefore,
S(g) > S(l) > S(s)
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28. Solutions
Generally, when a solid is dissolved in a solvent, entropy increases.
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29. 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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30. Sample Exercise 19.3 (Pg. 826)
Predict whether ∆S is positive or negative for each process, assuming each occurs at constant temperature:
H2O (l)  H2O (g)
Ag+ (aq) + Cl- (aq)  AgCl (s)
4Fe (s) + 3O2 (g)  2Fe2O3 (s)
N2 (g) + O2(g)  2NO (g)
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31. Sample Exercise 19.4 (Pg. 827)
In each pair, choose the system that has greater entropy and explain your choice:
1 mol of NaCl (s) or 1 mol of HCl(g) at 25∘ C
2 mol of HCl (g) or 1 mol of HCl (g) at 25∘ C
1 mol of HCl (g) or 1 mol of Ar(g) at 298 K
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32. Third Law of Thermodynamics
The entropy of a pure crystalline substance at absolute zero is 0.
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33. 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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34. Standard Entropies
Larger and more complex molecules have greater entropies.
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35. S = nS(products) — mS(reactants)
Entropy Changes
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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36. Sample Exercise 19.5 (Pg. 829)
Calculate the change in the standard entropy of the system, ∆S∘, for the synthesis of ammonia from N2 (g) and H2 (g) at 298 K.
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37. Practice
Calculate the standard entropy change, ∆S∘, for the “water splitting” reaction at 298K.
2H2O (l)  2H2(g) + O2 (g)
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38. Aim: What is Gibbs Free Energy, and how is it related to Enthalpy and Entropy?
Do Now: Spontaneous processes that result in a decrease in entropy, proceed only if the overall process is _________.
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39. 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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40. 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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41. Entropy Change in the Universe
qsystem T
Since Ssurroundings =
and qsystem = Hsystem
This becomes:
Suniverse = Ssystem +
Multiplying both sides by T, we get
TSuniverse = Hsystem  TSsystem
Hsystem T
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42. Gibbs Free Energy
TDSuniverse is defined as the Gibbs free energy, G.
When Suniverse is positive, G is negative.
Therefore, when G is negative, a process is spontaneous.
G = H T∆S
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43. Gibbs Free Energy
If DG is negative, the forward reaction is spontaneous.
If DG is 0, the system is at equilibrium.
If G is positive, the reaction is spontaneous in the reverse direction.
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44. Sample Exercise 19.6 (Pg. 833)
Calculate the standard free-energy change for the formation of NO (g) from N2 (g) and O2 (g) at 298 K: N2 (g) + O2 (g)  2 NO (g) Given that ∆H∘ =180.7 kJ and ∆S∘ = 24.7 J/K. Is the reaction spontaneous under these conditions?
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45. Standard Free Energy Changes
Analogous to standard enthalpies of formation are standard free energies of formation, G. f
DG = SnDG (products)  SmG (reactants) f
where n and m are the stoichiometric coefficients.
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46. Sample Exercise 19.7 (Pg. 834)
a) Calculate the standard free-energy change for the reaction P4(g) + 6Cl2 (g) 4 PCl3 (g) b) What is ∆G∘ for the reverse reaction? c) What do these values tell us?
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47. Sample Exercise 19.8 ( Pg.835)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) ∆H∘= kJ Without using any numerical data, will ∆G∘ be more or less negative than ∆H∘ at T= 298K.
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48. Free Energy Changes At temperatures other than 25°C, DG° = DH  TS
How does G change with temperature?
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49. Free Energy and Temperature
There are two parts to the free energy equation:
H— the enthalpy term
TS — the entropy term
The temperature dependence of free energy, then comes from the entropy term.
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50. Free Energy and Temperature
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51. Free Energy and Equilibrium
Under any conditions, standard or nonstandard, the free energy change can be found this way:
G = G + RT lnQ
(Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.)
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52. Free Energy and Equilibrium
At equilibrium, Q = K, and G = 0.
The equation becomes
0 = G + RT lnK
Rearranging, this becomes
G = RT lnK
or,
K = e
-G RT
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