1. Chapter 19 Chemical Thermodynamics

2. 19.1 Spontaneous Processes

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.
∆E = q + w

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.

5. Spontaneous Processes
Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.

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.

7. Sample Exercise 19.1
Predict whether the following processes are spontaneous as described, spontaneous in the reverse direction, or in equilibrium:
A) When a piece of metal is heated to 150ºC is added to water at 40ºC
B) Water at room temperature decomposes into H2(g) and O2(g)
C) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1ºC

8. Practice
Under 1 atm pressure, CO2(s) sublimes at -78ºC. Is the transformation of CO2(s) to CO2(g) a spontaneous process at -100ºC and 1 atm pressure.

9. 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.

10. Irreversible Processes
Irreversible processes cannot be undone by exactly reversing the change to the system.
Spontaneous processes are irreversible

11. 19.1 GIST
If a process is nonspontaneous, does that mean the process cannot occur under any circumstances?
If you evaporate water and then condense it, have you necessarily performed a reversible process?

12. 19.2 Entropy and the Second Law of Thermodynamics

13. 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

14. 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.

15. Entropy
Like total energy, E, and enthalpy, H, entropy is a state function.
Therefore,
S = Sfinal  Sinitial

16. Entropy
For a process occurring at constant temperature (an isothermal process), the change in entropy (ΔS) is equal to the heat that would be transferred if the process were reversible (qrev) divided by the temperature:
Can be used to calculate ΔS for phase changes.
S =
qrev T

17. Sample Exercise 19.2
The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9ºC, and its molar enthalpy of fusion is ΔHfus = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing poing?

18. Practice
The normal boiling point of ethanol, C2H5OH, is 78.3ºC, 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?

19. 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.

20. Second Law of Thermodynamics
In other words:
For reversible processes:
Suniv = Ssystem + Ssurroundings = 0
For irreversible processes:
Suniv = Ssystem + Ssurroundings > 0

21. Second Law of Thermodynamics
These last truths mean that as a result of all spontaneous processes the entropy of the universe increases.

22. 19.2 GIST
How do we reconcile the fact that S is a state function but that ∆S depends on q, which is not a state function?
The rusting of iron 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?

23. 19.3 The Molecular Interpretation of Entropy

24. 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.

25. 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. 

26. 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.

27. 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.

28. 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.

29. 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.

30. Entropy and Physical States
Entropy increases with the freedom of motion of molecules.
Therefore,
S(g) > S(l) > S(s)

31. Solutions
Generally, when a solid is dissolved in a solvent, entropy increases.

32. 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.

33. Sample Exercise 19.3
Predict whether ΔS is positive or negative for each of the following processes, assuming each occurs at constant temperature:
A) H2O(l) → H2O(g)
B) Ag+(aq) + Cl-(aq) → AgCl(s)
C) 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
D) N2(g) + O2(g) → 2 NO(g)

34. Practice
Indicate which of the following processes produces an increase of decrease in the entropy of the system:
A) CO2(s) → CO2(g)
B) CaO(s) + CO2(g) → CaCO3(s)
C) HCl(g) + NH3(g) → NH4Cl(s)
D) 2 SO2(g) + O2(g) → 2 SO3(g)

35. Sample Exercise 19.4
Choose the sample of matter that has greater entropy in each pair, and explain your choice:
A) 1 mol of NaCl(s) of 1 mol of HCl(g) at 25ºC
B) 2 mol of HCl(g) or 1 mol of HCl(g) at 25ºC
C) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K

36. Practice Choose the substance with the greater entropy:
A) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100ºC and 0.5 atm
B) 1 mol of H2O(s) at 0ºC or 1 mol of H2O(l) at 25ºC
C) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP
D) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP

37. Third Law of Thermodynamics
The entropy of a pure crystalline substance at absolute zero is 0.

38. 19.3 GIST
What kinds of motion can a molecule undergo that single atom cannot?
What is the entropy of a system that has only a single microstate?
If you are told the entropy of a certain system is zero, what do you know about the system?

39. 19.4 Entropy Changes in Chemical Reactions

40. Standard Entropies
These are molar entropy values of substances in their standard states.
Standard entropies tend to increase with increasing molar mass.

41. Standard Entropies
Larger and more complex molecules have greater entropies.

42. 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.

43. 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.

44. Entropy Change in the Universe
The universe is composed of the system and the surroundings.
Therefore,
Suniverse = Ssystem + Ssurroundings
For spontaneous processes
Suniverse > 0

45. Sample Exercise 19.5
Calculate ΔSº for the synthesis of ammonia from N2(g) and H2(g) at 298 K: N2(g) + 3 H2(g) → 2 NH3(g)
Using the standard entropies in Appendix C, calculate the standard entropy change, ΔSº, for the following reaction at 298 K: Al2O3(s) + 3 H2(g) → 2 Al2(s) + 3 H2O(g)

46. p.819 GIST
If a process is exothermic, does the entropy of the surroundings (1) always increase, (2) always decrease, or (3) sometimes increase and sometimes decrease, depending on the process.

47. 19.5 Gibbs Free Energy

48. Gibbs Free Energy
The best method of determining spontaneity is to calculate Gibb’s free energy, G
G = H – TS
∆G = ∆H – T∆S
∆Go = ∆Ho – T∆So

49. 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.

50. Gibbs Free Energy
Another useful equation – since ∆G = 0 when a system is at equilibrium, the equation can be rearranged to solve for the temperature of the rxn at equilibrium.
T = ∆ Ho
∆ So

51. Is the reaction spontaneous under these conditions?
Sample Exercise 19.6 Calculating Free-Energy Change from ΔH°, T, ΔS°
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 ΔH° = kJ and ΔS° = 24.7 J/K. Is the reaction spontaneous under these circumstances?
A particular reaction has ΔH° = 24.6 kJ and ΔS° = 132 J/K at 298 K. Calculate ΔG°.
Is the reaction spontaneous under these conditions?
Practice Exercise

52. Standard Free Energy Changes
Analogous to standard enthalpies of formation are standard free energies of formation, G.
DG = SnDG (products)  SmG (reactants) f
where n and m are the stoichiometric coefficients.

53. (b) What is ΔG° for the reverse of the above reaction?
Sample Exercise 19.7 Calculating Standard Free-Energy Change from
Free Energies of Formation
(a) Use data from Appendix C to calculate the standard free-energy change for the following reaction at
298 K:
P4(g) + 6 Cl2(g) → 4 PCl3(g)
(b) What is ΔG° for the reverse of the above reaction?

54. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g).
By using data from Appendix C, calculate ΔG° at 298 K for the combustion of methane:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g).
Practice Exercise

55. Sample Exercise 19.8
In section 5.7 we used Hess’s law to calculate ΔHº for the combustion of propane gas at 298K: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ΔHº = kJ
A) Without using the data from Appendix C, predict whether ΔGº for this reaction is more negative or less negative than ΔHº.
B) Use the date from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your prediction correct?

56. Consider the combustion of propane to form CO2(g) and H2O(g) at 298 K:
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g). Would you expect ΔG° to be more negative or less negative than ΔH°?
Practice Exercise

57. 19.5 GIST
Give the criterion for spontaneity first in terms of entropy and then in terms of free energy.
What does the superscript ° indicate when associated with a thermodynamic quantity, as in ΔH°, ΔS°, or ΔG°?

58. 19.6 Free Energy and Temperature

59. 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.
p.825 GIST: The normal boiling point of benzene is 80°C. At 100°C and 1 atm, which term is greater for the vaporization of benzene, ΔH or TΔS?

60. Free Energy and Temperature

61. Sample Exercise 19.9 Determining the Effect of Temperature on Spontaneity
The Haber process for the production of ammonia involves the equilibrium
Assume that ΔH° and ΔS° for this reaction do not change with temperature.
(a) Predict the direction in which ΔG° for this reaction changes with increasing temperature. (b) Calculate the values ΔG° of for the reaction at 25 °C and 500 °C.

62. (b) Using the values obtained in part (a), estimate ΔG° at 400 K.
(a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate ΔH° and ΔS° at 298 K for the following reaction: 2 SO2(g) + O2(g) → 2 SO3(g).
(b) Using the values obtained in part (a), estimate ΔG° at 400 K.
Practice Exercise

63. 19.7 Free Energy and the Equilibrium Constant

64. Free Energy and Equilibrium
Under any conditions, standard or nonstandard, the free energy change can be found this way:
G = G + RT lnQ
R is J/mol K
(Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.)

65. Sample Exercise 19.10 Relating ΔG to a Phase change at Equilibrium
As we saw in Section 11.5, the normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at a pressure of 1 atm. (a) Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl4(l). (b) What is the value of ΔG° for the equilibrium in part (a)? (c) Use thermodynamic data in Appendix C and Equation to estimate the normal boiling point of CCl4.

66. Use data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br2(l). (The experimental value is given in Table 11.3.)
Practice Exercise

67. Sample Exercise 19.11 Calculating the Free-Energy Change under
Nonstandard Conditions
We will continue to explore the Haber process for the synthesis of ammonia:
Calculate ΔG at 298 K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3.

68. Calculate ΔG at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3.
Practice Exercise

69. 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

70. Sample Exercise 19.12 Calculating an Equilibrium Constant for ΔG°
Use standard free energies of formation to calculate the equilibrium constant, K, at 25 °C for the reaction involved in the Haber process:
The standard free-energy change for this reaction was calculated in Sample Exercise 19.9:
ΔG° = –33.3 kJ/mol = –33,300 J/mol.

71. Use data from Appendix C to calculate the standard free-energy change, ΔG°, and the equilibrium constant, K, at 298 K for the reaction
Practice Exercise

72. Integrative Exercise
Consider the equilibria of NaCl(s) and AgCl(s) dissolving in water to form ions: NaCl(s) ↔ Na+(aq) + Cl-(aq) AgCl(s) ↔ Ag+(aq) + Cl-(aq)
A) Calculate the value of ΔGº at 298 K for each
B) Are the two values different due to the enthalpy term or the entropy term of the standard free energy change?
C) Use the values of ΔGº to calculate the Ksp values
D) Sodium chloride is soluble and silver chloride is insoluble – is this consistent with your results in part C?
E) How will ΔGº for the solution process of these salts change with increasing T? What effect should this change have on the solubility of the salts?