1. CH 19: Thermodynamics

2. What happens to particles as they undergo phase change?
Consider:
Kinetic energy
Enthalpy
Entropy
Reversible?
Spontaneous?

3. 19.1 Spontaneous Processes
– process that occurs without any outside intervention, the internal energy alone determines if a reaction will occur

4. Reversible process
When a system and surroundings is changed in such a way that they can be restored by reversing the change exactly
Whenever a chemical system is in equilibrium, reactants and products can interconvert reversibly
H2O(s) H2O (l)

5. Irreversible Process
- A change to a system and surroundings that can be restored by taking a new path to get back and changing the surroundings

6. 19.2 Entropy
disorder of a system
The more disorder of a system, the larger the entropy
DS = Sfinal-Sinitial

7. Entropy
When DS > 0, the system is disordered
When DS < 0, the system is more ordered or less random

8. Second Law of Thermodynamics
states the entropy of the universe increases in any spontaneous process
(entropy is not conserved)

9. 19.3 The Molecular Interpretation of Entropy
Entropy changes are related to the way the particles in a system can be arranged
Particles change through translational, vibrational, and rotational motion
Entropy and movement decrease with decreasing T

10. Third Law of Thermodynamics

11. Third Law of Thermodynamics
entropy of a pure crystalline substance at absolute zero (0 Kelvin) is zero

12. Entropy Increases when:
1. Liquids or solutions are formed from solids
2.Gases are formed from either solids or liquids
3. The number of molecules of gas increases during a chemical rxn

13. Sample Exercise 19.3 Predicting the Sign of ΔS
Predict whether ΔS is positive or negative for each of the following processes, assuming each occurs at constant temperature:

14. 1 mol of NaCl(s) or 1 mol of HCl(g) at 25 °C,
Sample Exercise 19.4 Predicting Which Sample of Matter Has the Higher Entropy
Choose the sample of matter that has greater entropy in each pair, 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.

15. 19.4 Entropy Changes in Rxns
DS=qrev/T
(note: T must be in Kelvin)

16. 19.4 Entropy Changes in Rxns
Standard Molar Entropies (So)
- the molar entropy values of substances in their standard states, pure substance at 1 atm; found on appendix C

17. Standard Molar Entropies (So)
Characteristics:
1. Molar entropies of elements are not 0
2. So (g) > So(l) >So (s)
3. So increases w/ increasing MM
4. Increase w/ increasing atoms

18. Entropy Change
DSo = SnSo(products) - SmSo(reactants) , where m and n are coefficients of the balanced chemical eq

19. Sample Exercise 19.5 Calculating ΔS from Tabulated Entropies
Calculate ΔSº for the synthesis of ammonia from N2(g) and H2(g) at 298 K:
N2(g) + 3 H2(g) → 2 NH3(g)

20. 19.5 Gibbs Free Energy
DG = DH-TDS at constant temperature

21. Relationship between G and spontaneity
1. If DG is negative, the rxn is spontaneous in the forward direction (thermodynamically favored)
2. If DG is zero, the rxn is at equilibrium
3. If DG is positive, the rxn in the forward direction is nonspontaneous; work must be supplied from the surroundings to make it occur. The reverse rxn is spontaneous.

22. 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?

23. Standard Free-Energy Changes
DGo = SnGfo(products) - SmGfo(reactants)

24. Standard Free-Energy Changes
When DG>0 (nonspontaneous), it is the measure of the minimum amount of work that must be done to cause the process to occur.

25. 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)

26. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ΔH° = –2220 kJ
Sample Exercise 19.8 Estimating and Calculating ΔG°
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ΔH° = –2220 kJ
(a) Without using data from Appendix C, predict whether ΔG° for this reaction is more negative or less negative than ΔH°.
(b) Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your prediction from part (a) correct?

27. 19.6 Free Energy and Temperature
The value of T plays a major role in determining DG

28. 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°(-92.38kJ) and ΔS°(-198.3J/K) 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° for the reaction at 25 °C and 500 °C.

29. 19.7 Free Energy and the Equilibrium Constant
DG at nonstandard conditions:
DG= DGo + RTlnQ, where R is the ideal gas constant 8.314J/mol-K, T is abs T, and Q is the reaction quotient
At Equilibrium: DG=0 and Q = Keq
DGo = - RT lnKeq
Keq= e-DG/RT

30. Sample Exercise 19.12 Calculating an Equilibrium Constant for ΔG°
Use standard free energies of formation to calculate the equilibrium constant, Keq, at 25 °C for the reaction involved in the Haber process:
(recall 19.9 DGo=-33.3kJ)