1. Entropy, Free Energy, and Equilibrium Thermodynamics

2. Copyright © Cengage Learning. All rights reserved. 18 | 2 Thermodynamics is the study of heat and other forms of energy involved in chemical or physical processes.

3. Concept Review: First Law of Thermodynamics Energy can be converted from one form to another but it cannot be created nor destroyed. Example: Heat given off by the system is equal to the heat absorbed by the surroundings

4. 4 Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing

5. Copyright © Cengage Learning. All rights reserved. 6 | 5 Heat, q The energy that flows into or out of a system because of a difference in temperature between system and its surroundings. Heat flow- higher temperature to a region of lower temperature. q is positive if heat is absorbed by the system - Endothermic q is negative if heat is evolved by a system - Exothermic

6. Copyright © Cengage Learning. All rights reserved. 6 | 6 Enthalpy, H -heat absorbed or evolved in a chemical reaction. -An extensive property of a substance - A State Function Extensive Property A property that depends on the amount of substance. Mass and volume are extensive properties.

7. Copyright © Cengage Learning. All rights reserved. 6 | 7 A state function is a property of a system that depends only on its present state Final – Initial Products - reactants

8. Copyright © Cengage Learning. All rights reserved. 6 | 8 The altitude of a campsite is a state function. It is independent of the path taken to reach it.

9. 9 Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.  H = H (products) – H (reactants)  H = heat given off or absorbed during a reaction at constant pressure H products < H reactants  H < 0 H products > H reactants  H > 0

10. Copyright © Cengage Learning. All rights reserved. 6 | 10 You burn 15.0 g sulfur in air. How much heat evolves from this amount of sulfur? The thermochemical equation is S 8 (s) + 8O 2 (g)  8SO 2 (g);  H = -2.39 × 10 3 kJ

11. Copyright © Cengage Learning. All rights reserved. 6 | 11 S 8 (s) + 8O 2 (g)  8SO 2 (g);  H = -2.39  10 3 kJ Molar mass of S 8 = 256.52 g q = – 1.40  10 2 kJ

12. Standard enthalpies of formation (from thermodynamic table) can be used to calculate the standard enthalpy for a reaction,  H°. Methyl alcohol, CH 3 OH, is toxic because liver enzymes oxidize it to formaldehyde, HCHO, which can coagulate protein. Calculate  H o for the following reaction: 2CH 3 OH(aq) + O 2 (g)  2HCHO(aq) + 2H 2 O(l) Standard enthalpies of formation, CH 3 OH(aq):−245.9 kJ/mol HCHO(aq):−150.2 kJ/mol H 2 O(l):−285.8 kJ/mol

13. Copyright © Cengage Learning. All rights reserved. 6 | 13 We want  H° for the reaction: 2CH 3 OH(aq) + O 2 (aq)  2HCHO(aq) + 2H 2 O(l)

14. 2 nd Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged un an equilibrium process. - Explains why chemical processes tend to favor one direction over another - Will a reaction occur when reactants are brought together under a specific set of conditions (P, T, concentrations..)??? - Is a reaction going to happen spontaneously at a given set of conditions? A spontaneous reaction is a reaction that occurs under specified conditions.

15. 15 Spontaneous Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 o C and ice melts above 0 o C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous

16. 16 spontaneous nonspontaneous

17. 17 Does a decrease in enthalpy mean a reaction proceeds spontaneously? H + (aq) + OH - (aq) H 2 O (l)  H 0 = -56.2 kJ/mol H 2 O (s) H 2 O (l)  H 0 = 6.01 kJ/mol NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq)  H 0 = 25 kJ/mol H2OH2O Spontaneous reactions CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H 0 = -890.4 kJ/mol Enthalpy and Spontaneity Spontaneous reactions can be exothermic and endothermic. There must be another thermodynamic function that can determine spontaneity of a reaction.

18. 18 Entropy (S) is a measure of the randomness or disorder of a system. orderS disorder S  S = S f - S i If the change from initial to final results in an increase in randomness S f > S i  S > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state S solid < S liquid << S gas H 2 O (s) H 2 O (l)  S > 0

19. 19 Processes that lead to an increase in entropy (  S > 0)

20. 20  S > 0 Example: Br 2 (l) Br 2 (g)  S > 0 Example: I 2 (s) I 2 (g)

21. 21 W = number of microstates - The larger the W the greater the entropy. S = k ln W  S = S f - S i Entropy Distribution III is most probable because there are 6 microstates of 6 ways to achieve it. 4 molecules in 2 compartments: Boltzmann equation: As a state function:

22. 2 nd Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged un an equilibrium process. Can be expressed in the following equations: For a spontaneous process: ∆S univ = ∆S sys + ∆S surr > 0 For an equilibrium process: ∆S univ = ∆S sys + ∆S surr = 0 For a spontaneous reaction ∆S rxn > 0 or +

23. 23 Entropy Changes in the System (  S sys ) aA + bB cC + dD S0S0 rxn dS 0 (D) cS 0 (C) = [+] - bS 0 (B) aS 0 (A) [+] S0S0 rxn nS 0 (products) =  mS 0 (reactants)  - The standard entropy of reaction (  S 0 ) is the entropy change for a reaction carried out at 1 atm and 25 0 C. rxn What is the standard entropy change for the following reaction at 25 0 C? 2CO (g) + O 2 (g) 2CO 2 (g) S 0 (CO) = 197.9 J/K mol S 0 (O 2 ) = 205.0 J/K mol S 0 (CO 2 ) = 213.6 J/K mol S0S0 rxn = 2 x S 0 (CO 2 ) – [2 x S 0 (CO) + S 0 (O 2 )] S0S0 rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K mol

24. From the standard entropy values in Appendix 3, calculate the standard entropy changes for the following reactions at 25 0 C: (a)CaCO 3 (s)  CaO(s) + CO 2 (g) (b)N 2 (g) + 3H 2 (g)  2NH 3 (g) (c)H 2 (g) + Cl 2 (g)  2HCl(g) If a reaction produces more gas molecules than it consumes,  S 0 > 0. If the total number of gas molecules diminishes,  S 0 < 0. If there is no net change in the total number of gas molecules, then  S 0 may be positive or negative BUT  S 0 will be a small number. What is the sign of the entropy change for the following reaction? 2Zn (s) + O 2 (g) 2ZnO (s)

25. 25 Entropy Changes in the Surroundings (  S surr ) Exothermic Process  S surr > 0 Endothermic Process  S surr < 0 ∆S surr is proportional to -∆H sys

26. The change in entropy is also dependent o the Temperature: If the temperature of the surroundings is high, and the reaction is exothermic, there will be little impact on the molecular motion or disorderliness (entropy) If the temperature of the surroundings is low and the reaction is exothermic, the impact will be greater N 2 (g) + 3H 2 (g)  2NH 3 (g) ΔH 0 rxn = -92.6 kJ/mol (this is from the thermodynamic table ) From the same table, ∆S sys = -199 J/K.mol ΔS surr = -(-92.6 x100) J/mol = 311 J/K.mol 298 K Δs univ = ΔS surr + ∆S sys = 112 J/K.mol Therefor the reaction is spontaneous since the entropy is positive.

27. 3 rd Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero of temperature. S= klnW since it is a perfect crystal, W= 1 S= k ln 1 = 0 The 3 rd law allows us to determine the absolute entropies of substances from 0 K to any higher temperature.

28. Gibbs Free Energy From the second law of thermodynamics we learned that: ∆S univ = ∆S sys + ∆S surr > 0 What if we just consider the system only? How do we determine the spontaneity of a rxn? ∆S univ = ∆S sys + ∆S surr > 0 T∆S univ = -ΔH sys + T∆S sys > 0 Expression only in terms of the system Can be rewritten as: -T∆S univ = ΔH sys - T∆S sys < 0 At constant P and T, if the changes in enthalpy of a system are such that ΔH sys - T∆S sys is < 0, the reaction is spontaneous A new thermodynamic function called Gibbs free energy is used to determine absolute spontaneity of a reaction: G= H-TS or ∆G= ∆H - T ∆S Free energy is energy available to do work. Therefore if reaction is accompanied by usable energy, (- ∆G), reaction is spontaneous.

29. ΔG < 0 The reaction is spontaneous in the forward direction ΔG > 0 the reaction is nonspontaneous. The reaction is spontaneous in the opposite direction. ΔG = 0 The system is at equilibrium. There is no net change. aA + bB cC + dD G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - Standard free energy of formation (  G 0 ) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f  G 0 of any element in its stable form is zero. f

30. Copyright © Cengage Learning. All rights reserved. 18 | 30 Using standard enthalpies of formation,  H f ° and the value of  S° from the previous problem, calculate  G° for the oxidation of ethyl alcohol to acetic acid. CH 3 CH 2 OH(l) + O 2 (g)  CH 3 COOH(l) + H 2 O(l)

31. Copyright © Cengage Learning. All rights reserved. 18 | 31 CH 3 CH 2 OH(l) + O 2 (g)  CH 3 COOH(l) + H 2 O(l)  H f ° kJ/mol –277.6 0 –487.0 –285.8 n mol1111 n  H f ° kJ –277.60–487.0 –285.8 –277.6 kJ –772.8 kJ  H° = –495.2 kJ  S° = –136 J/K T = 298 K

32. Copyright © Cengage Learning. All rights reserved. 18 | 32  H° = –495.2 kJ  S° = –136 J/K = –0.136 kJ/K T = 298 K  G° =  H° – T  S°  G° = –495.2 kJ – (298 K)(–0.136 kJ/K)  G° = –495.2 kJ + 40.5 kJ  G° = –454.7 kJ The reaction is spontaneous.

33. 33 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - What is the standard free-energy change for the following reaction at 25 0 C? G0G0 rxn 6  G 0 (H 2 O) f 12  G 0 (CO 2 ) f = [+] - 2  G 0 (C 6 H 6 ) f [] G0G0 rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ/mol Is the reaction spontaneous at 25 0 C?  G 0 = -6405 kJ/mol < 0 spontaneous

34. 34  G =  H - T  S

35. 35 CaCO 3 (s) CaO (s) + CO 2 (g)  H 0 = 177.8 kJ/mol  S 0 = 160.5 J/K·mol  G 0 =  H 0 – T  S 0 At 25 o C,  G 0 = 130.0 kJ/mol  G 0 = 0 at 835 o C Temperature and Spontaneity of Chemical Reactions Equilibrium Pressure of CO 2 A chemist will always ask: at what temperature will the decomposition of CaCO 3 becomes favorable? T =  H 0 /  S 0 T = 835 0 C, at which  G 0 = 0, therefore T must be higher than 835 0 C

36. Free Energy and Chemical Equilibrium We rewrite the equation into : ∆G = ∆G 0 + RT lnQ ΔG vs. ΔG 0 ΔG is nonstandard condition ΔG 0 is standard condition Standard condition is at 1 M, 25 0 C, and 1 atm. As soon as reaction starts 1M is no longer existing. The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol. Respectively. Calculate the entropy changes for the solid  liquid and liquid to vapor transitions for benzene. At 1 atm, benzene melts at 5.5 0 and boils at 80.1 0 C.

37. Free Energy and Chemical Equilibrium At equilibrium, ΔG= 0 and Q = K, where K is the equilibrium constant. This: ΔG 0 = - RT ln K * The larger the K, the more negative the ΔG 0.

38. Copyright © Cengage Learning. All rights reserved. 18 | 38 Calculate the value of the thermodynamic equilibrium constant at 25°C for the reaction N 2 O 4 (g)  2NO 2 (g) The standard free energy of formation at 25°C is 51.30 kJ/mol for NO 2 (g) and 97.82 kJ/mol for N 2 O 4 (g).  G° = 2 mol(51.30 kJ/mol) – 1 mol(97.82 kJ/mol)  G° = 102.60 kJ – 97.82 kJ  G° = 4.78 kJ

39. Copyright © Cengage Learning. All rights reserved. 18 | 39  G = –RT ln K

40. 1. Using the data in Appendix 3, calculate the equilibrium constant (K p ) for the following reaction at 25 0 C: 2H 2 O (l)  2H 2 (g) + O 2 (g) 2. Calculate ΔG 0 for the following process at 25 0 C: BaF 2 (s)  Ba 2+ (aq) + 2F - (aq) 3. The equilibrium constant (K p ) for the reaction N 2 O 4 (g)  2NO 2 (g) is 0.113 at 298 K, which corresponds to a standard free-energy change of 5.40 kJ/mol. IN a certain experiment, the initial pressures are P NO2 = 0.112 atm and P N2O4 = 0.453 atm. Calculate ∆G for the reaction at these pressures and predict the direction of the net reaction toward equilibrium.

41. Copyright © Cengage Learning. All rights reserved. 18 | 41 Sodium carbonate, Na 2 CO 3, can be prepared by heating sodium hydrogen carbonate, NaHCO 3 : 2NaHCO 3 (s)  Na 2 CO 3 (s) + H 2 O(g) + CO 2 (g) Estimate the temperature at which the reaction proceeds spontaneously at 1 atm. * See thermodynamic table for data.

42. Copyright © Cengage Learning. All rights reserved. 18 | 42 2NaHCO 3 (s)  Na 2 CO 3 (s) + H 2 O(g) + CO(g)  H f °, kJ/mol–947.7–1130.8–241.8–393.5 n, mol2111 n  H f °, kJ–1895.4–1130.8–241.8–393.5 –1895.4 kJ–1766.1 kJ  H° = 129.3 kJ S f °, J/mol  K102139188.7213.7 n, mol2111 nS f °, J/K204139188.7213.7 204 J/K541.4 J/K  S° = 337.4 J/K

43. Copyright © Cengage Learning. All rights reserved. 18 | 43

44. Practice exercises 17. 12; 17.14; 17.25; 17.32; 17.29