Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW

1. Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW ENTROPY 8.1 Spontaneous Change 8.2 Entropy and Disorder 8.3 Changes in Entropy 8.4 Entropy Changes Accompanying Changes in Physical State 8.5 A Molecular Interpretation of Entropy 8.6 The Equivalence of Statistical and Thermodynamic Entropies 8.7 Standard Molar Entropies 8.8 Standard Reaction Entropies 2012 General Chemistry I

2. ENTROPY (Sections 8.1-8.8) - The 1st law of thermodynamics says if a reaction takes place, then the total energy of the universe remains unchanged. It cannot be used to predict the directionality of a process. - The natural progression of a system and its surroundings (or “the universe”) is from order to disorder, from organized to random. - A new thermodynamic state function is needed to predict both directionality and extent of disorder.

3. 8.1 Spontaneous Change • Spontaneous change is a change that has a tendency to occur without needing to be driven by an external influence. Heat flow - Spontaneous changes need not be fast: e.g. C(diamond)  C(graphite); Mixing of gases H2(g) + 1/2O2(g)  H2O(l)

4. 8.2 Entropy and Disorder - Energy and matter tend to disperse in a disorderly fashion. • Entropy, S is defined as a measure of disorder. • The second law of thermodynamics: The entropy of an isolated system increases in any spontaneous change. At constant temperature unit: J·K-1 - Entropy is a state function; the change in entropy of a system is independent of the path between its initial and final states.

6. 8.3 Changes in Entropy - Thermal disorder: arising from the thermal motion of the molecules - Positional disorder: related to the locations of the molecules • DS for a process with changing temperature: → (CV if V is constant, CP if P is constant)

7. DS for a reversible, isothermal expansion of an ideal gas

11. EXAMPLE 8.5 In an experiment, 1.00 mol Ar(g) was compressed suddenly (and irreversibly) from 5.00 L to 1.00 L by driving in a piston. and in the process its temperature was increased from 20.0 oC to 25.2 oC. What is the change in entropy of the gas? To solve this problem, we consider two reversible stages between initial and final states. Then DS(irrev) = DS(rev 1) + DS(rev 2).

12. 8.4 Entropy Changes Accompanying Changes in Physical State - Phase transition: solid → liquid, Tf (fusion or melting point) liquid → solid, Tb (boilingpoint) - At the transition temperature (such as Tb), The temperature remains constant as heat is supplied. The transfer of heat is reversible. The heat supplied is equal to the enthalpy change due to the constant pressure (at 1 atm). • Entropy of vaporization, DSvap qrev = DHvap > 0 in all cases

13. - Standard entropy of vaporization, DSvapo : DSvap at 1 bar Some Standard entropies of vaporization at Tb (Table 8.1)

14. Trouton’s rule: DSvapo = ~85 J·K-1mol-1 There is approximately the same increase in positional disorder for most liquids when evaporating. - Exceptions; water, methanol, ethanol, ··· due to extensive hydrogen bonding in liquid phases - Standard entropy of fusion, DSfuso > 0 in all cases

15. Temperature dependence of DSvapo • To determine the entropy of vaporization of water at 25 oC • (not at Tb), we can use an entropy change cycle:

16. 297s Exercise 8.43 Calculate the standard entropy of vaporization of water at 85 oC, given that its standard entropy of vaporization at 100 oC is 109.0 J·K-1·mol-1 and the molar heat capacities at constant pressure of liquid water and water vapor are 75.3 J·K-1·mol-1 and 33.6 J·K-1·mol-1, respectively, in this range. Solution

17. 8.5 A Molecular Interpretation of Entropy • The third law of thermodynamics S → 0 as T → 0 • Boltzmann formula S = kB ln W - statistical entropy: kB = 1.381 × 10-23 J·K-1 = R/NA - W : the number of microstates the number of ways that the atoms or molecules in the sample can be arranged and yet still give rise to the same total energy - When we measure the bulk properties of a system, we are measuring an average taken over the many microstates (ensemble) that the system has occupied during the measurement.

18. EXAMPLE 8.7 Calculate the entropy of a tiny solid made up of four diatomic molecules of a compound such as carbon monoxide, CO, at T = 0 when (a) the four molecules have formed a perfectly ordered crystal in which all molecules are aligned with their C atoms on the left and (b) the four molecules lie in random orientations, but parallel. (a) 4 CO molecules perfectly ordered: (b) 4 CO in random, but parallel: (c) 1 mol CO in random, but parallel:

19. - Residual entropy at T = 0, arising from positional disorder 4.6 J·K-1 for the entropy of 1 mol CO < 5.76 J·K-1 Nearly random arrangementdue to a small electric dipole moment - Solid HCl; S ~ 0 at T = 0 due to the bigger dipole moment leading strict head-to-tail arrangement

20. EXAMPLE 8.8 The entropy of 1.00 mol FClO3(s) at T = 0 is 10.1 J·K-1. Interpret it. 4 orientations possible nearly random arrangement

21. 8.6 The Equivalence of Statistical and Thermodynamic Entropies = Thermodynamic entropy DS = qrev/T behavior of bulk matter Statistical entropy S = k ln W behavior of molecules - Consider a one-dimensional box, for the statistical entropy, • At T = 0, only the lowest energy level occupied • → W = 1 and S = 0 • At T > 0, W > 1 and S > 0 • When the box length is increased at constant T, • the molecules are distributed across more levels. • → W and S increase.

22. - W = constant × V - For N molecules, - The change when a sample expands isothermally from V1 to V2 is, V2 = nR ln V1 - By raising the temperature, The molecules have access to larger number of energy levels → W and S increase.

23. - The equations used to calculate changes in the statistical entropy and the thermal entropy lead to the same result. 1. Both are state functions. Number of microstates depends only on its current state. 2. Both are extensive (dependent on “extent”) properties. 2 × no. of molecules = entropy changes from k ln W to 2k ln W 3. Both increase in a spontaneous change. In any irreversible change, the overall disorder increases → no. of microstates increases. 4. Both increase with temperature. When T increases, more microstates become accessible.

24. 302s Exercise 8.27 If SO2F2 adopts a disordered arrangement in its crystal form, what would its residual molar entropy be? Solution

25. 302s Exercise 8.29 8.29 Which substance in each of the following pairs has the higher molar entropy at 298 K: (a) HBr(g) or HF(g); (b) NH3(g) or Ne(g); (c) I2(s) or I2(l); (d) 1.0 mol Ar(g) at 1.00 atm or 1.0 mol Ar(g) at 2.00 atm? Solution

26. 302s Exercise 8.31 List the following substances in order of increasing molar entropy at 298 K: H2O(l), H2O(g), H2O(s), C(s, diamond). Explain your reasoning. Solution

27. 302s Exercise 8.33 Which substance in each of the following pairs would you expect to have the higher standard molar entropy at 298 K? Explain your reasoning: (a) Iodine vapor or bromine vapor; (b) the two liquids cyclopentane and 1-pentene; (c) ethene (ethylene) or an equivalent mass of polyethylene, a substance formed by the polymerization of ethylene. Solution

28. 302s Exercise 8.35 • Without performing any calculations, predict whether there is an Increase or a decrease in entropy for each of the following processes: • Cl2(g) + H2O(l) → HCl(aq) + HClO(aq); • Cu3(PO4)2(s) → 3 Cu2+(aq) + 2 PO43-(aq); • SO2(g) + Br2(g) + 2 H2O(l) → H2SO4(aq) + 2 HBr(aq). Solution

29. 8.7 Standard Molar Entropies Molar entropy, S(T), can be determined from measurement of Cp at different temperatures. For heating at constant P, CP and CP/T → 0 as T→ 0

30. Standard molar entropy, Smo is the molar entropy of the pure substance at 1 bar.

31. Standard molar entropy, Smo : light heavy - Diamond (2.4 J·K-1) vs. lead (64.8 J·K-1): rigid bonds vs. vibrational energy levels - H2 (130.7 J·K-1) vs. N2 (191.6 J·K-1): the greater the mass, the closer energy levels - CaCO3 (92.9 J·K-1) vs. CaO (39.8 J·K-1): large, complex vs. smaller, simpler - In general, Smo: gases >> liquids > solids Related to freedom of movement and disordered state

32. 8.8 Standard Reaction Entropies • Standard reaction entropy, DSo, is the difference between the standard molar entropies of the products and those of the reactants, taking into account their stoichiometric coefficients.

33. EXAMPLE 8.9 Calculate DSo for N2(g) + 3H2(g) → 2NH3(g) at 25 oC.

34. 307s Exercise 8.35 Use data in Table 8.3 or Appendix 2A to calculate the standard reaction entropy for each of the following reactions at 25 C. Interpret the sign and magnitude of the reaction entropy. (a) The formation of 1.00 mol H2O(l) from the elements in their most stable state at 298 K. (b) The oxidation of 1.00 mol CO(g) to carbon dioxide. (c) The decomposition of 1.00 mol calcite, CaCO3(s), to carbon dioxide gas and solid calcium oxide. (d) The decomposition of potassium chlorate: 4 KClO3 (s)  3 KClO4 (s) + KCl (s)

35. 307s Solutions

36. 307s

37. Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW GLOBAL CHANGES IN ENTROPY 8.9 The Surroundings 8.10 The Overall Change in Entropy 8.11 Equilibrium GIBBS-FREE ENERGY 8.12 Focusing on the System 8.13 Gibbs Free Energy of Reaction 8.14 The Gibbs Free Energy and Nonexpansion Work 8.15 The Effect of Temperature 8.16 Impact on Biology: Gibbs Free Energy Changes in Biological Systems 2012 General Chemistry I

38. GLOBAL CHANGES IN ENTROPY(Sections 8.9-8.11) 8.9 The Surroundings - The second law refers to an isolated system (system + surroundings = universe). - Only if the total entropy change is positive will the process be spontaneous.

39. Sometimes DSsurr can be difficult to compute, but in general it can be obtained from the enthalpy change for the process (that is, of the system).

41. 8.10 The Overall Change in Entropy - To use the entropy to judge the direction of spontaneous change, we must consider the change in the entropy of the system plus the entropy change in the surroundings: • Spontaneous exothermic • (DH<0) reactions:

42. EXAMPLE 8.11 Is the reaction spontaneous at 298 K? 2 Mg(s) + O2(g) → 2 MgO(s) DSo = -217 J·K-1 DHo = -1202 kJ The reaction is spontaneous

43. Spontaneous endothermic • (DH>0) reactions: There can still be an overall increase in entropy if the disorder of the system increases enough. Summary

44. - A process produces maximum work if it takes place reversibly. • Clausius inequality DS = > DS - For an isolated system (universe), q = 0 The entropy of an isolated system cannot decrease. - For two given states of the system,ΔS is a state function (path-independent) but ΔStot is not. (See EXAMPLE 8.12)

45. EXAMPLE 8.12 Calculate DS, DSsurr, and DStot for (a) the isothermal, reversible expansion and (b) the isothermal, free expansion of 1.00 mol of ideal gas molecules from 8.00 L to 20.00 L at 292 K. Explain any differences between the two paths. (a) Isothermal reversible expansion at 292 K

46. (b) Isothermal free expansion 292 K

47. 312s Exercise 8.49 • Initially a sample of ideal gas at 323 K occupies 1.67 L at 4.95 atm. • The gas is allowed to expand to 7.33 L by two pathways: • isothermal, reversible expansion (b) isothermal, irreversible free expansion. • Calculate ΔStot, ΔS, ΔSsurr for these pathways.

49. 8.11 Equilibrium • Dynamic equilibrium is one where there is no net tendency to change but microscopic forward and reverse • processes occur at matching rates. Thermal equilibrium: no net flow of energy as heat Mechanical equilibrium: no tendency to expand or contract Chemical equilibrium: no net change in composition at thermodynamic equilibrium

50. GIBBS FREE ENERGY(Sections 8.12-8.16) 8.12 Focusing on the System - at constant T and P • Gibbs free energy, G G = H - TS - Change in Gibbs free energy - at constant T and P The direction of spontaneous change is the direction of decreasing Gibbs free energy.