1. Chapter 19 Chemical Thermodynamics

2. First Law of Thermodynamics
You will recall from Chapter 5 that energy cannot be created or 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.

3. Spontaneous Processes
Spontaneous processes are those that can proceed without any outside intervention.
The gas in vessel A will spontaneously effuse into vessel B, but once the gas is in both vessels, it will not spontaneously return to vessel A.

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

5. 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.

6. Solution: Sample Exercise 19.1 Identifying Spontaneous Processes
Predict whether each process is spontaneous as described, spontaneous in the reverse direction, or in equilibrium: (a) Water at 40 °C gets hotter when a piece of metal heated to 150 °C is added. (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, 80.1 °C.
Solution:
This process is spontaneous. Whenever two objects at different
temperatures are brought into contact, heat is transferred from the hotter
object to the colder one. Thus, heat is transferred from the hot metal to the cooler water. The final temperature, after the metal and water achieve the same temperature (thermal equilibrium), will be somewhere between the initial temperatures of the metal and the water.
(b) Experience tells us that this process is not spontaneous we certainly have
never seen hydrogen and oxygen gases spontaneously bubbling up out of
water! Rather, the reverse process—the reaction of H2 and O2 to form
H2O—is spontaneous.
(c) The normal boiling point is the temperature at which a vapor at 1 atm is in
equilibrium with its liquid. Thus, this is an equilibrium situation. If the
temperature were below 80.1 °C, condensation would be spontaneous.

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.

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

9. Entropy
Entropy (S) is a term coined by Rudolph Clausius in the nineteenth century.
Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, . q T
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.

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

11. Solution: Sample Exercise 19.2 Calculating ΔS for a Phase Change
Elemental mercury is a silver liquid at room temperature. Its normal freezing point is
–38.9 °C, 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?
Solution:
– 38.9 °C = – K = K 11

12. 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.
In other words:
For reversible processes:
Suniv = Ssystem + Ssurroundings = 0
For irreversible processes:
Suniv = Ssystem + Ssurroundings > 0
These last truths mean that as a result of all spontaneous processes, the entropy of the universe increases.

13. 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.

14. 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. 

15. 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.
A microstate is a particular microscopic arrangement of the atoms or molecules of the system that corresponds to the given state of the system. 

16. 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. 

17. 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.

18. 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.

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

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

21. 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.

22. Solution: Sample Exercise 19.3 Predicting the Sign of ΔS
Predict whether is positive or negative for each process, assuming each occurs at constant ΔS 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)
Solution:
Evaporation is positive.
(b) This process is negative.
This process is negative.
The sign of ΔS is impossible to predict based on our discussions thus far,
but we can predict that ΔS will be close to zero. 22

23. Solution: Sample Exercise 19.4 Predicting Relative Entropies
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
(c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.
Solution:
HCl(g) has the higher entropy because the particles in gases are more disordered and have more freedom of motion than the particles in solids.
When these two systems are at the same pressure, the sample containing 2 mol of HCl has twice the number of molecules as the sample containing 1 mol. Thus, the 2 mol sample has twice the number of microstates and twice the entropy.
The HCl system has the higher entropy because the number of ways in which an HCl molecule can store energy is greater than the number of ways in which an Ar atom can store energy. (Molecules can rotate and vibrate; atoms cannot.) 23

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

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

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

27. 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.

28. ΔS°= 2S°(NH3) – [S°(N2) + 3S°(H2)]
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:
ΔS°= 2S°(NH3) – [S°(N2) + 3S°(H2)]
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 28

29. 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.

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

31. 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

32. 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.
Start here 4/1/10

33. 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.

34. Sample Exercise 19.6 Calculating Free–Energy Change from ΔH°, T, and 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 conditions?
Solution:
Because is positive, the reaction is not spontaneous under standard conditions at 298 K. 34

35. 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.

36. Free Energy Changes At temperatures other than 25 C, DG = DH  TS
How does G change with 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.

37. 4 PCl3(g)  P4(g) + 6 Cl2(g) ΔG° = +1102.8 kJ
Sample Exercise 19.7 Calculating Standard Free–Energy Change from Free Energies of Formation
Use data from Appendix C to calculate the standard free–energy change for
the reaction P4(g) + 6 Cl2 (g)  4 PCl3 (g) run at 298 K.
(b) What is ΔG° for the reverse of this reaction?
Solution:
(a)
(b)
When we reverse the reaction, we reverse the roles of the reactants and products. Thus, reversing the reaction changes the sign of ΔG in Equation 19.14, just as reversing the reaction changes the sign of ΔH.
Hence, using the result from part (a), we have
4 PCl3(g)  P4(g) + 6 Cl2(g) ΔG° = kJ 37

38. Solution: Sample Exercise 19.8 Estimating and Calculating ΔG°
In Section 5.7 we used Hess’s law to calculate ΔH° for the combustion of propane gas at 298 K: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) ΔH° = –2220 kJ
Without using data from Appendix C, predict whether ΔG° for this reaction is more negative or less negative than ΔH°.
Use data from Appendix C to calculate ΔG° for the reaction at 298 K. Is your prediction from part (a) correct?
Solution:
(a) By using the general rules discussed in Section 19.3, we expect a decrease in the number of gas molecules to lead to a decrease in the entropy of the system—the products have fewer possible microstates than the reactants. We therefore expect ΔS°and TΔS° to be negative. Because we are subtracting TΔS°, which is a negative number, we predict that ΔG° is less negative than ΔH°.
(b) Using Equation and values from Appendix C, we have: . 38

39. Free Energy and Temperature
Under any conditions, standard or nonstandard, the free energy change can be found this way:
G = G + RT ln Q
(Under standard conditions, all concentrations are 1 M, so Q = 1 and ln Q = 0; the last term drops out.)

40. 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 the reaction changes with increasing temperature. (b) Calculate ΔG° at 25 °C and 500 °C.
Solution:
For this reaction to be negative because the number of molecules of gas is smaller in
the products. Because ΔS° is negative, –TΔS° is positive and increases with increasing temperature. As a result, ΔG°becomes less negative (or more positive) with increasing temperature. Thus, the driving force for the production of NH3 becomes smaller with increasing temperature.
At T = 25 °C = 298 K, we have
At T = 500 °C = 773 K, we have 40

41. Free Energy and Equilibrium
At equilibrium, Q = K, and G = 0.
The equation becomes
0 = G + RT ln K
Rearranging, this becomes
G = RT ln K or
K = e
G/RT

42. Sample Exercise 19.10 Relating ΔG to a Phase Change at Equilibrium
Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl4(l).
What is the value of ΔG° for the equilibrium in part (a)?
Use data from Appendix C and Equation to estimate the normal boiling point of CCl4.
Solution:
The normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at a pressure of 1 atm:
(b) At equilibrium, ΔG = 0. In any normal boiling–point equilibrium, both liquid and vapor are in their standard state of pure liquid and vapor at 1 atm (Table 19.2). Consequently, Q = 1, ln Q = 0, and
ΔG = ΔG° for this process. We conclude that ΔG° = 0 for the equilibrium representing the normal boiling point of any liquid. (We would also find that ΔG° = 0 for the equilibria relevant to normal melting points and normal sublimation points.) 42

43. Tb = (32.6 kJ / 95.0 J/k) * (1000 J / 1 kJ) = 343 k = 70 oC
Continued
(c) Combining Equation with the result from part (b), we see that the equality at the normal boiling point, Tb, of CCl4(l) (or any other pure liquid) is:
ΔG° = ΔH° – Tb ΔS° = 0
Solving the equation for Tb, we obtain Tb = ΔH°/ ΔS°
Tb = ΔH°/ ΔS°
Tb = (32.6 kJ / 95.0 J/k) * (1000 J / 1 kJ) = 343 k = 70 oC

44. Sample Exercise 19.11 Calculating the Free–Energy Change under Nonstandard Conditions
Calculate DG at 298 K for a mixture of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3 being used in the Haber process:
Solution: 44

45. Sample Exercise 19.12 Calculating an Equilibrium Constant from ΔG°
The standard free–energy change for the Haber process at was obtained in Sample Exercise 19.9 for the Haber reaction:
Use this value of ΔG° to calculate the equilibrium constant for the process at 25 °C.
Solution:
K = e -DG°/RT = e –(-33,300 J/mol)/(8.314 J/mol-K)(298 K) = e13.4 = 7 x105 45