1. Entropy and Free Energy How to predict if a reaction can occur, given enough time? THERMODYNAMICS How to predict if a reaction can occur at a reasonable rate? KINETICS Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

2. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved ThermodynamicsThermodynamics Is the state of a chemical system such that a rearrangement of its atoms and molecules would decrease the energy of the system?Is the state of a chemical system such that a rearrangement of its atoms and molecules would decrease the energy of the system? If yes, system is favored to react — a product-favored system.If yes, system is favored to react — a product-favored system. Most product-favored reactions are exothermic.Most product-favored reactions are exothermic. Often referred to as spontaneous reactions.Often referred to as spontaneous reactions. Spontaneous does not imply anything about time for reaction to occur.Spontaneous does not imply anything about time for reaction to occur.

3. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Diamond is thermodynamically favored to convert to graphite, but not kinetically favored. Thermodynamics and Kinetics

4. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Thermodynamics and Kinetics Diamond is thermodynamically favored to convert to graphite, but not kinetically favored. Paper burns — a product-favored reaction. Also kinetically favored once reaction is begun.

5. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Product-Favored Reactions In general, product- favored reactions are exothermic. Fe 2 O 3 (s) + 2 Al(s) ---> 2 Fe(s) + Al 2 O 3 (s) H = - 848 kJ  H = - 848 kJ

6. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Product-Favored Reactions But many spontaneous reactions or processes are endothermic or even have H = 0. But many spontaneous reactions or processes are endothermic or even have  H = 0. NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq)

7. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Entropy, S One property common to product-favored processes is that the final state is more DISORDERED or RANDOM than the original. Spontaneity is related to an increase in randomness. The thermodynamic property related to randomness is ENTROPY, S. Reaction of K with water

8. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The entropy of liquid water is greater than the entropy of solid water (ice) at 0  C.

9. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved How probable is it that reactant molecules will react? PROBABILITY suggests that a product-favored reaction will result in the dispersal of energy or of matter or both. Directionality of Reactions

10. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Probability suggests that a product- favored reaction will result in the dispersal of energy or of matter or both. Matter Dispersal Directionality of Reactions

11. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Probability suggests that a product- favored reaction will result in the dispersal of energy or of matter or both. Matter Dispersal Directionality of Reactions

12. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Probability suggests that a product- favored reaction will result in the dispersal of energy or of matter or both. Energy Dispersal Directionality of Reactions

13. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Probability suggests that a product- favored reaction will result in the dispersal of energy or of matter or both. Energy Dispersal Directionality of Reactions

14. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Directionality of Reactions — Energy Dispersal Exothermic reactions involve a release of stored chemical potential energy to the surroundings. The stored potential energy starts out in a few molecules but is finally dispersed over a great many molecules. The final state—with energy dispersed—is more probable and makes a reaction product-favored.

15. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved S (gases) > S (liquids) > S (solids) S o (J/Kmol) H 2 O(liq)69.91 H 2 O(gas)188.8 S o (J/Kmol) H 2 O(liq)69.91 H 2 O(gas)188.8 Entropy, S

16. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Entropy of a substance increases with temperature. Molecular motions of heptane, C 7 H 16 Molecular motions of heptane at different temps. Entropy, S

17. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Increase in molecular complexity generally leads to increase in S. S o (J/Kmol) CH 4 248.2 C 2 H 6 336.1 C 3 H 8 419.4 S o (J/Kmol) CH 4 248.2 C 2 H 6 336.1 C 3 H 8 419.4 Entropy, S

18. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Entropies of ionic solids depend on coulombic attractions. S o (J/Kmol) MgO26.9 NaF51.5 S o (J/Kmol) MgO26.9 NaF51.5 Entropy, S

19. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Entropy usually increases when a pure liquid or solid dissolves in a solvent. Entropy, S

20. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Entropy Changes for Phase Changes For a phase change, S = q/T For a phase change,  S = q/T where q = heat transferred in phase change For H 2 O (liq) ---> H 2 O(g) H = q = +40,700 J/mol  H = q = +40,700 J/mol

21. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Entropy Changes for Phase Changes For a phase change, S = q/T For a phase change,  S = q/T where q = heat transferred in phase change For H 2 O (liq) ---> H 2 O(g) H = q = +40,700 J/mol  H = q = +40,700 J/mol

22. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Consider 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) S o = 2 S o (H 2 O) - [2 S o (H 2 ) + S o (O 2 )]  S o = 2 S o (H 2 O) - [2 S o (H 2 ) + S o (O 2 )] S o = 2 mol (69.9 J/Kmol) - [2 mol (130.7 J/Kmol) + 1 mol (205.3 J/Kmol)]  S o = 2 mol (69.9 J/Kmol) - [2 mol (130.7 J/Kmol) + 1 mol (205.3 J/Kmol)] S o = -326.9 J/K  S o = -326.9 J/K Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid. Calculating S for a Reaction Calculating  S for a Reaction S o =  S o (products) -  S o (reactants)  S o =  S o (products) -  S o (reactants)

23. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 2nd Law of Thermodynamics A reaction is spontaneous (product- favored) if ²S for the universe is positive. S universe = S system + S surroundings  S universe =  S system +  S surroundings S universe > 0 for product-favored process  S universe > 0 for product-favored process First calc. entropy created by matter dispersal (S system ) First calc. entropy created by matter dispersal (  S system ) Next, calc. entropy created by energy dispersal (S surround ) Next, calc. entropy created by energy dispersal (  S surround )

24. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Dissolving NH 4 NO 3 in water—an entropy driven process. 2nd Law of Thermodynamics

25. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) S o system = -326.9 J/K  S o system = -326.9 J/K 2nd Law of Thermodynamics

26. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) S o system = -326.9 J/K  S o system = -326.9 J/K 2nd Law of Thermodynamics

27. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) S o system = -326.9 J/K  S o system = -326.9 J/K Can calc. that H o rxn = H o system = -571.7 kJ Can calc. that  H o rxn =  H o system = -571.7 kJ 2nd Law of Thermodynamics

28. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) S o system = -326.9 J/K  S o system = -326.9 J/K Can calc. that H o rxn = H o system = -571.7 kJ Can calc. that  H o rxn =  H o system = -571.7 kJ 2nd Law of Thermodynamics

29. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) S o system = -326.9 J/K  S o system = -326.9 J/K Can calc. that H o rxn = H o system = -571.7 kJ Can calc. that  H o rxn =  H o system = -571.7 kJ S o surroundings = +1917 J/K  S o surroundings = +1917 J/K 2nd Law of Thermodynamics

30. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) S o system = -326.9 J/K  S o system = -326.9 J/K S o surroundings = +1917 J/K  S o surroundings = +1917 J/K S o universe = +1590. J/K  S o universe = +1590. J/K The entropy of the universe is increasing, so the reaction is product- favored. 2nd Law of Thermodynamics

31. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) S o system = -326.9 J/K  S o system = -326.9 J/K S o surroundings = +1917 J/K  S o surroundings = +1917 J/K S o universe = +1590. J/K  S o universe = +1590. J/K The entropy of the universe is increasing, so the reaction is product- favored. 2nd Law of Thermodynamics

32. Gibbs Free Energy, G ²S univ = ²S surr + ²S sys Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

33. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Gibbs Free Energy, G ²S univ = ²S surr + ²S sys

34. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Gibbs Free Energy, G ²S univ = ²S surr + ²S sys Multiply through by -T

35. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Gibbs Free Energy, G S univ = S surr + S sys  S univ =  S surr +  S sys Multiply through by -T -T S univ = H sys - T S sys -T  S univ =  H sys - T  S sys  S univ =  H sys T +  S

36. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Gibbs Free Energy, G S univ = S surr + S sys  S univ =  S surr +  S sys Multiply through by -T -T S univ = H sys - T S sys -T  S univ =  H sys - T  S sys -T S univ = change in Gibbs free energy for the system = G system -T  S univ = change in Gibbs free energy for the system =  G system

37. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Gibbs Free Energy, G S univ = S surr + S sys  S univ =  S surr +  S sys Multiply through by -T -TS univ = H sys - TS sys -T  S univ =  H sys - T  S sys -TS univ = change in Gibbs free energy for the system = G system -T  S univ = change in Gibbs free energy for the system =  G system Under standard conditions — G o = H o - TS o  G o =  H o - T  S o  S univ =  H sys T +  S

38. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Gibbs Free Energy, G G o = H o - T S o  G o =  H o - T  S o Gibbs free energy change = total energy change for system - energy lost in disordering the system total energy change for system - energy lost in disordering the system If reaction is exothermic (H o negative) and entropy increases (S o is +), then G o must be negative and reaction product-favored. If reaction is exothermic (  H o negative) and entropy increases (  S o is +), then  G o must be negative and reaction product-favored. If reaction is endothermic (H o is +), and entropy decreases (S o is -), then G o must be + and reaction is reactant-favored. If reaction is endothermic (  H o is +), and entropy decreases (  S o is -), then  G o must be + and reaction is reactant-favored.

39. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Gibbs Free Energy, G G o = H o - TS o  G o =  H o - T  S o H o S o G o Reaction  H o  S o  G o Reaction exo(-)increase(+)-Prod-favored endo(+)decrease(-)+React-favored exo(-)decrease(-)?T dependent endo(+)increase(+)?T dependent

40. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Gibbs Free Energy, G G o = H o - TS o  G o =  H o - T  S o Two methods of calculating G o Two methods of calculating  G o

41. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Gibbs Free Energy, G G o = H o - TS o  G o =  H o - T  S o Two methods of calculating G o Two methods of calculating  G o a)Determine H o rxn and S o rxn and use GIbbs equation. a)Determine  H o rxn and  S o rxn and use GIbbs equation.

42. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Gibbs Free Energy, G G o = H o - TS o  G o =  H o - T  S o Two methods of calculating G o Two methods of calculating  G o a)Determine H o rxn and S o rxn and use GIbbs equation. a)Determine  H o rxn and  S o rxn and use GIbbs equation. b)Use tabulated values of free energies of formation, G f o. b)Use tabulated values of free energies of formation,  G f o.

43. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Gibbs Free Energy, G G o = H o - TS o  G o =  H o - T  S o Two methods of calculating G o Two methods of calculating  G o a)Determine H o rxn and S o rxn and use GIbbs equation. a)Determine  H o rxn and  S o rxn and use GIbbs equation. b)Use tabulated values of free energies of formation, G f o. b)Use tabulated values of free energies of formation,  G f o. ²G o rxn =  ²G f o (products) -  ²G f o (reactants)

44. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculating G o rxn Calculating  G o rxn Combustion of acetylene C 2 H 2 (g) + 5/2 O 2 (g) --> 2 CO 2 (g) + H 2 O(g) Use enthalpies of formation to calculate H o rxn = -1238 kJ  H o rxn = -1238 kJ Use standard molar entropies to calculate S o rxn = -97.4 J/K or -0.0974 kJ/K  S o rxn = -97.4 J/K or -0.0974 kJ/K G o rxn = -1238 kJ - (298 K)(-0.0974 J/K)  G o rxn = -1238 kJ - (298 K)(-0.0974 J/K) = -1209 kJ = -1209 kJ Reaction is product-favored in spite of negative S o rxn. Reaction is product-favored in spite of negative  S o rxn. Reaction is “enthalpy driven”

45. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculating G o rxn Calculating  G o rxn Is the dissolution of ammonium nitrate product- favored? If so, is it enthalpy- or entropy-driven? NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq)

46. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculating G o rxn Calculating  G o rxn From tables of thermodynamic data we find H o rxn = +25.7 kJ  H o rxn = +25.7 kJ S o rxn = +108.7 J/K or +0.1087 kJ/K  S o rxn = +108.7 J/K or +0.1087 kJ/K G o rxn = +25.7 kJ - (298 K)(+0.1087 J/K)  G o rxn = +25.7 kJ - (298 K)(+0.1087 J/K) = -6.7 kJ = -6.7 kJ Reaction is product-favored in spite of negative H o rxn. Reaction is product-favored in spite of negative  H o rxn. Reaction is “entropy driven” NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq)

47. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculating G o rxn Calculating  G o rxn Combustion of carbon C(graphite) + O 2 (g) --> CO 2 (g) G o rxn = G f o (CO 2 ) - [G f o (graph) + G f o (O 2 )]  G o rxn =  G f o (CO 2 ) - [  G f o (graph) +  G f o (O 2 )] G o rxn = -394.4 kJ - [ 0 + 0]  G o rxn = -394.4 kJ - [ 0 + 0] Note that free energy of formation of an element in its standard state is 0. G o rxn = -394.4 kJ  G o rxn = -394.4 kJ Reaction is product-favored as expected. G o rxn =  ²G f o (products) -  G f o (reactants)  G o rxn =  ²G f o (products) -   G f o (reactants)

48. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Free Energy and Temperature 2 Fe 2 O 3 (s) + 3 C(s) ---> 4 Fe(s) + 3 CO 2 (g) H o rxn = +467.9 kJ S o rxn = +560.3 J/K  H o rxn = +467.9 kJ  S o rxn = +560.3 J/K G o rxn = +300.8 kJ  G o rxn = +300.8 kJ Reaction is reactant-favored at 298 K At what T does G o rxn just change from being (+) to being (-)? At what T does  G o rxn just change from being (+) to being (-)? When G o rxn = 0 = H o rxn - TS o rxn When  G o rxn = 0 =  H o rxn - T  S o rxn

49. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Free Energy and Temperature 2 Fe 2 O 3 (s) + 3 C(s) ---> 4 Fe(s) + 3 CO 2 (g) H o rxn = +467.9 kJ S o rxn = +560.3 J/K  H o rxn = +467.9 kJ  S o rxn = +560.3 J/K G o rxn = +300.8 kJ  G o rxn = +300.8 kJ Reaction is reactant-favored at 298 K At what T does G o rxn just change from being (+) to being (-)? At what T does  G o rxn just change from being (+) to being (-)? When G o rxn = 0 = H o rxn - TS o rxn When  G o rxn = 0 =  H o rxn - T  S o rxn

50. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved K eq is related to reaction favorability. When G o rxn < 0, reaction moves energetically “downhill” When  G o rxn < 0, reaction moves energetically “downhill” ²G o rxn is the change in free energy as reactants convert completely to products. But systems often reach a state of equilibrium in which reactants have not converted completely to products. In this case G rxn is < G o rxn, so state with both reactants and products present is more stable than complete conversion. In this case  G rxn is <  G o rxn, so state with both reactants and products present is more stable than complete conversion. Thermodynamics and K eq

51. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Product-favored reaction. 2 NO 2 ---> N 2 O 4 G o rxn = -4.8 kJ  G o rxn = -4.8 kJ Here G rxn is less than G o rxn, so the state with both reactants and products present is more stable than complete conversion. Here  G rxn is less than  G o rxn, so the state with both reactants and products present is more stable than complete conversion. Thermodynamics and K eq

52. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Reactant-favored reaction. N 2 O 4 --->2 NO 2 G o rxn = +4.8 kJ N 2 O 4 --->2 NO 2  G o rxn = +4.8 kJ Here G o rxn is greater than G rxn, so the state with both reactants and products present is more stable than complete conversion. Here  G o rxn is greater than  G rxn, so the state with both reactants and products present is more stable than complete conversion. Thermodynamics and K eq

53. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved K eq is related to reaction favorability and so to G o rxn. K eq is related to reaction favorability and so to  G o rxn. The larger the value of G o rxn the larger the value of K. The larger the value of  G o rxn the larger the value of K. G o rxn = - RT lnK  G o rxn = - RT lnK where R = 8.31 J/Kmol Thermodynamics and K eq

54. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculate K for the reaction N 2 O 4 --->2 NO 2 G o rxn = +4.8 kJ N 2 O 4 --->2 NO 2  G o rxn = +4.8 kJ G o rxn = - RT lnK  G o rxn = - RT lnK Thermodynamics and K eq

55. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculate K for the reaction N 2 O 4 --->2 NO 2 G o rxn = +4.8 kJ N 2 O 4 --->2 NO 2  G o rxn = +4.8 kJ G o rxn = +4800 J = - (8.31 J/K)(298 K) ln K  G o rxn = +4800 J = - (8.31 J/K)(298 K) ln K G o rxn = - RT lnK  G o rxn = - RT lnK Thermodynamics and K eq

56. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculate K for the reaction N 2 O 4 --->2 NO 2 G o rxn = +4.8 kJ N 2 O 4 --->2 NO 2  G o rxn = +4.8 kJ G o rxn = +4800 J = - (8.31 J/K)(298 K) ln K  G o rxn = +4800 J = - (8.31 J/K)(298 K) ln K G o rxn = - RT lnK  G o rxn = - RT lnK Thermodynamics and K eq

57. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculate K for the reaction N 2 O 4 --->2 NO 2 G o rxn = +4.8 kJ N 2 O 4 --->2 NO 2  G o rxn = +4.8 kJ G o rxn = +4800 J = - (8.31 J/K)(298 K) ln K  G o rxn = +4800 J = - (8.31 J/K)(298 K) ln K K = 0.14 G o rxn = - RT lnK  G o rxn = - RT lnK Thermodynamics and K eq

58. Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Calculate K for the reaction N 2 O 4 --->2 NO 2 G o rxn = +4.8 kJ N 2 O 4 --->2 NO 2  G o rxn = +4.8 kJ G o rxn = +4800 J = - (8.31 J/K)(298 K) ln K  G o rxn = +4800 J = - (8.31 J/K)(298 K) ln K K = 0.14 When G o rxn > 0, then K 0, then K < 1 G o rxn = - RT lnK  G o rxn = - RT lnK Thermodynamics and K eq