1. Jeffrey Mack California State University, Sacramento Chapter 19 Principles of Chemical Reactivity: Entropy and Free Energy

2. Chemical Kinetics provides us with information about the rate of reaction and how the reaction proceeds. Chemical Thermodynamics provides us with information about equilibrium and whether or not a reaction is Spontaneous. Entropy & Free Energy

3. Spontaneous Chemical Thermodynamics provides us with information about equilibrium and whether or not a reaction is Spontaneous. Spontaneous changes occur only in the direction that leads to equilibrium. Systems never change spontaneously in a direction that takes them farther from equilibrium. Example: Heat Transfer Heat (thermal energy) always flows spontaneously from a hot object to a cold object. The process occurs until thermal equilibrium is achieved, that is when both objects are at the same temperature. Entropy & Free Energy

4. Additional Examples: Solvation of a Soluble salt: NH 4 NO 3 (s) dissolves spontaneously even the process is endothermic (  H > 0) Expansion of a gas and Diffusion. Gasses mix to form homogeneous mixtures spontaneously. Certain Phase Changes: Ice melts spontaneously above 0 o C. Water evaporates even though the enthalpy of vaporization is endothermic. Certain Chemical Reactions: Na(s) reacts vigorously when dropped in water. Iron rusts when exposed to the atmosphere. Spontaneous Process

5. But many spontaneous reactions or processes are endothermic or even have ∆H  0. NH 4 NO 3 (s) + heat  NH 4 NO 3 (aq) ∆H = 0 Spontaneous Process

6. A Review of Concepts of Thermodynamics First law of thermodynamics: The law of conservation of energy; energy cannot be created or destroyed. State Function: Quantity in which its determination is path independent.  U = q + w: The change in internal energy of a system is a function of heat and work done on or by the system.  H: Heat transferred at constant pressure. Exothermic Process:  H < 0 Endothermic Process:  H > 0Thermodynamics

7. Enthalpy alone does not predict spontaneity: Some processes are energetically favored (  r H < 0) but not spontaneous. Equilibrium alone cannot determine spontaneity: Some processes are favored based on Equilibrium (K >> 1) yet they are non-spontaneous. There must be another factor that plays a role in determination of spontaneity! Thermodynamics

8. Diamond is thermodynamically favored to convert to graphite, but not kinetically favored. Paper burns once the reaction is initiates. The process is product-favored & kinetically favored. Thermodynamics & Kinetics

9. Reactants Products Kinetics Thermodynamics Thermodynamics & Kinetics

10. Factors that Affect Spontaneity (Thermodynamic favorability): 1.Enthalpy: Comparison of bond energy (  H) 2.Entropy: Randomness vs. Order of a system (  S) In general, enthalpy is more important than entropy. Thermodynamics & Kinetics

11. dispersedIn a spontaneous processes the energy of the final state is more dispersed. disorderThe system moves to a higher state of disorder. ENTROPY, S.The thermodynamic quantity associated with disorder and energy dispersal is called ENTROPY, S. The 2 nd law of thermodynamics states that a spontaneous process results in an increase in the entropy of the universe.  S > 0 Reaction of K and water Dispersal of Energy: Entropy

12. Observation of a spontaneous process shows that it is associated with a dispersal of energy. Energy Dispersal Dispersal of Energy: Entropy

13. The change in entropy for a spontaneous process is given by: Where q rev is the heat gained or lost by the system during the process and T is the absolute temperature. A reversible process can be returned to its original state. (chemical equilibrium), an irreversible process cannot. Example: The breaking of a coffee mug into many pieces is an irreversible process. Dispersal of Energy: Entropy

14. To begin, particle 1 has 2 units of energy and 2-4 have none. Dispersal of Energy: Entropy

15. Particle 1 can transfer one unit of energy to particle 2, then the other to 3 or 4. Dispersal of Energy: Entropy

16. Particle 2 can initially have two units of energy. Dispersal of Energy: Entropy

17. Particle 2 transfer one unit to particles 4 or 3. (2 to 1 has already been counted.) Dispersal of Energy: Entropy

18. Particle 4 can initially have two units of energy. Dispersal of Energy: Entropy

19. Particle 4 transfer one unit to particle 3. (4 to 1 & 4 to 2 have already been counted.) Dispersal of Energy: Entropy

20. Particle 3 can start with two units of energy. Energy transfers between particle 3 were previously counted. Dispersal of Energy: Entropy

21. microstate Each unique combination that results in a dispersion of energy is called a microstate. There are 10 microstates in this system. The greater the number of microstates, the greater the entropy of the system. Dispersal of Energy: Entropy

23. As the size of the container increases, the number of microstates accessible to the system increases. Therefore the entropy of the system increases. Dispersal of Energy: Entropy

24. The entropy of liquid water is greater than the entropy of solid water (ice) at 0˚ C. Energy is more dispersed in liquid water than in solid water due to the lack of an ordered network as in the solid state. Dispersal of Energy: Entropy

25. S o (J/Kmol) H 2 O(liq)69.95 H 2 O(gas)188.8 Energy dispersal S (solids) < S (liquids) < S (gases) Entropy & States of Matter

26. S˚(Br 2 liq) < S˚(Br 2 gas)S˚(H 2 O sol) < S˚(H 2 O liq) Entropy & States of Matter

27. Entropy and Microstates: As the number of microstates increases, so does the entropy of the system. S = k  lnW k = Boltzman’s constant (1.381  10  23 J/K) W = the number of microstates Dispersal of Energy: Entropy

28. Entropy and Microstates: The change in entropy associated with a process is a function of the number of final and initial microstates of the system. If W final > W inital,  S > 0 If W final < W inital,  S < 0 Entropy, Entropy Change, & Energy Dispersal: A Summary

29. When a solute dissolves in a solvent the process is spontaneous owing to the increase in entropy. Matter (and energy) are more dispersed. The number of microstates is increased. Dispersal of Energy: Entropy

30. The entropy of a substance increases with temperature. Molecular motions of heptane, C 7 H 16 Molecular motions of heptane at different temps. Entropy Measurements & Values

31. An Increase in molecular complexity generally leads to increase in S. Entropy Measurements & Values

32. Entropies of ionic solids depend on coulombic attractions. S° (J/Kmol) MgO26.9 NaF51.5 Mg 2+ & O 2- Na + & F - Entropy Measurements & Values

33. Defined by Ludwig Boltzmann, the third law states that a perfect crystal at 0 K has zero entropy; that is, S =0. The entropy of an element or compound under any other temperature and pressure is the entropy gained by converting the substance from 0 K to those conditions. To determine the value of S, it is necessary to measure the energy transferred as heat under reversible conditions for the conversion from 0 K to the defined conditions and then to use Equation 19.1 Because it is necessary to add energy as heat to raise the temperature, all substances have positive entropy values at temperatures above 0 K. Standard Molar Entropies

34. The standard molar entropy, S°, of a substance is the entropy gained by converting 1 mol of it from a perfect crystal at 0 K to standard state conditions (1 bar, 1m for a solution) at the specified temperature. Standard Molar Entropies

35. For a phase change, where q = heat transferred in the phase change Entropy Changes for Phase Changes

36. S increases slightly with T S increases a large amount with phase changes Entropy & Temperature

37. Standard molar entropy values can be used to calculate the change in entropy that occurs in various processes under standard conditions. The standard entropy change for a reaction (  r S°) can be found in the same manner as  r H° were: Where n & m are the stoichiometric balancing coefficients. This calculation is valid only under reversible conditions. Determining Entropy Changes in Physical & Chemical Processes

38. Problem: Consider the reaction of hydrogen and oxygen to form liquid water. What is the standard molar entropy for the reaction? Determining Entropy Changes in Physical & Chemical Processes

39. Problem: Consider the reaction of hydrogen and oxygen to form liquid water. What is the standard molar entropy for the reaction? Determining Entropy Changes in Physical & Chemical Processes

40. Problem: Consider the reaction of hydrogen and oxygen to form liquid water. What is the standard molar entropy for the reaction? Determining Entropy Changes in Physical & Chemical Processes

41. Problem: Consider the reaction of hydrogen and oxygen to form liquid water. What is the standard molar entropy for the reaction? The enthalpy change is negative (net decrease in dispersion) due to the change in number of moles: 3 reduced to 2 Determining Entropy Changes in Physical & Chemical Processes

42. The second law of thermodynamics states:..  S° universe = ∆S° system + ∆S° surroundings Any change in entropy for the system plus the entropy change for the surroundings must equal the overall change in entropy for the universe. A process is considered to be spontaneous under standard conditions if  S°(universe) is greater than zero. 2 nd Law of Thermodynamics

43. entropy driven The solution process for NH 4 NO 3 (s) in water is an entropy driven process. ∆S° universe = ∆S° system + ∆S° surroundings 2 nd Law of Thermodynamics

44. Calculating  S° Surroundings : 2 nd Law of Thermodynamics

45. Calculating  S° Surroundings : 2 nd Law of Thermodynamics

46. Calculating  S° Surroundings : 2 nd Law of Thermodynamics

47. Calculating  S° Surroundings : at constant pressure, q system =  r H° system 2 nd Law of Thermodynamics

48. Calculating  S° Surroundings : at constant pressure, q system =  r H° system 2 nd Law of Thermodynamics

49. Problem: Calculate the entropy change for the surroundings for the reaction: 2H 2 (g) + O 2 (g)  2H 2 O(l) 2 nd Law of Thermodynamics

50. Problem: Calculate the entropy change for the surroundings for the reaction: 2H 2 (g) + O 2 (g)  2H 2 O(l) The enthalpy change for the reaction is calculated using standard molar enthalpies of formation: 2 nd Law of Thermodynamics

51. Problem: Calculate the entropy change for the surroundings for the reaction: 2H 2 (g) + O 2 (g)  2H 2 O(l) The enthalpy change for the reaction is calculated using standard molar enthalpies of formation: 2 nd Law of Thermodynamics

52. Problem: Calculate the entropy change for the surroundings for the reaction: 2H 2 (g) + O 2 (g)  2H 2 O(l) As calculated previously:  S° universe =  327 J/K + 1917 J/K =1590. J/K 2 nd Law of Thermodynamics

53. Problem: Calculate the entropy change for the surroundings for the reaction: 2H 2 (g) + O 2 (g)  2H 2 O(l)  S° universe =  327 J/K + 1917 J/K = 1590. J/K The entropy of the universe increases ( > 0) therefore the process is spontaneous at standard state conditions. The process is spontaneous due to the entropy change in the surroundings, not the system. 2 nd Law of Thermodynamics

54. Spontaneous or Not?

55. The method used so far to determine whether a process is spontaneous required evaluation of two quantities,  S°(system) and  S°(surroundings). J. Willard Gibbs asserted that that the maximum non-PV work available to the system must be a function of Enthalpy and Entropy: G = H – TS G = Gibbs Free energy of the system, H = system enthalpy and S = the entropy of the system. J. Willard Gibbs 1839-1903 Gibbs Free Energy, G

56. Since it is impossible to measure individual values of enthalpy, we often express free energy in terms of the changes of thermodynamic quantities.  (G = H – TS) At constant temperature:  G° = Gibbs Free energy change of the system,  H° = enthalpy change and  S° = the entropy change at SS conditions. Gibbs Free Energy, G

57. If the reaction is exothermic (  H < 0) And the change in entropy is positive (  H > 0) at a given temperature, then  S < 0. We then can assert that if  G < 0 that the reaction is spontaneous as well as product favored! Spontaneity is a function of energy and dispersion! Gibbs Free Energy, G

58. HH SS GG Reaction  +  Product Favored +  + Reactant Favored  ? Temperature dependant ++? Gibbs Free Energy, G

59. Since  r G° is related to  S° universe, it follows that: If  r G° < 0: The process is spontaneous in the direction written under standard conditions. If  r G° = 0: The process is at equilibrium under standard conditions and K=1 If  r G° > 0: The process is non-spontaneous in the direction written under standard conditions. Conclusion: A reaction proceeds spontaneously toward the minimum in free energy, which corresponds to equilibrium.  r G o & Equilibrium

60. Product Favored Reactions, ∆G° negative, K > 1 Q < K: Heading to equilibrium  G < 0 Q = K: At equilibrium  G = 0 Q > K: Heading away from equilibrium  G > 0 ∆G, ∆G°, Q, & K

61. Product-favored 2 NO 2  N 2 O 4 ∆ r G° = – 4.8 kJ State with both reactants and products present is more stable than complete conversion. K > 1, more products than reactants. ∆G, ∆G°, Q, & K

62. Reactant Favored Reactions, ∆G° positive, K < 1 Q < K: Heading to equilibrium  G < 0 Q = K: At equilibrium  G = 0 Q > K: Heading away from equilibrium  G > 0 ∆G, ∆G°, Q, & K

63. Reactant-favored N 2 O 4  2 NO 2 ∆ r G° = +4.8 kJ State with both reactants and products present is more stable than complete conversion. K < 1, more reactants than products ∆G, ∆G°, Q, & K

64.  r G° represents the free energy change for a process at standard state conditions. (Equilibrium) What if this is not the case? Under nonstandard conditions: Where R is the gas law constant and Q = ∆G, ∆G°, Q, & K

65. At equilibrium, we know that  r G = 0 and Q = K Therefore: So knowing one quantity yields the other. ∆G, ∆G°, Q, & K

66. Problem:  r G° for the formation of ammonia at SS conditions is  16.37 kJ/mol. What is the value of Kp at this temperature and pressure? (1 mol, 20 °C and 1 atm) ∆G, ∆G°, Q, & K

67. Problem:  r G° for the formation of ammonia at SS conditions is  16.37 kJ/mol. What is the value of Kp at this temperature and pressure? (1 mol, 20 °C and 1 atm) ∆G, ∆G°, Q, & K

68. Problem:  r G° for the formation of ammonia at SS conditions is  16.37 kJ/mol. What is the value of Kp at this temperature and pressure? (1 mol, 20 °C and 1 atm) ∆G, ∆G°, Q, & K

69. Problem:  r G° for the formation of ammonia at SS conditions is  16.37 kJ/mol. What is the value of Kp at this temperature and pressure? (1 mol, 20 °C and 1 atm) K >> 1, product favored as predicted by  G ∆G, ∆G°, Q, & K

70. The relation of ∆ r G, ∆ r G°, Q, K, reaction spontaneity, and product- or reactant favorability.Summary

71. Calculating & Using Free Energy Standard Free Energy of Formation

74.  r G° is negative at 298 K, so the reaction is predicted to be spontaneous under standard conditions at this temperature. It is also predicted to be product-favored at equilibrium. Calculating & Using Free Energy Standard Free Energy of Formation

75. Under reversible conditions, both enthalpy and entropy are state functions. It follows that the Gibbs free energy must also be. Therefore we can write that : Gibbs Free Energy, G

76. Note that ∆ f G° for an element = 0 Free Energies of Formation

77. Using the  f G° found in appendix L of your text, calculate  r G° for the following reaction: 4NH 3 (g) + 7O 2 (g) → 4NO 2 (g) + 6H 2 O(g) Free Energies of Formation

78. Using the  f G° found in appendix L of your text, calculate  r G° for the following reaction: 4NH 3 (g) + 7O 2 (g) → 4NO 2 (g) + 6H 2 O(g) Free Energies of Formation

79. Using the  f G° found in appendix L of your text, calculate  r G° for the following reaction: 4NH 3 (g) + 7O 2 (g) → 4NO 2 (g) + 6H 2 O(g) Free Energies of Formation

80. Using the  f G° found in appendix L of your text, calculate  r G° for the following reaction: 4NH 3 (g) + 7O 2 (g) → 4NO 2 (g) + 6H 2 O(g)  r G° = –1101.14 kJ/mol (product-favored Free Energies of Formation

81. 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) Calculating ∆ r G°

82. By definition: G = H − TS Indicating that free energy is a function of temperature.  r G° will therefore change with temperature. A consequence of this temperature dependence is that, in certain instances, reactions can be product-favored at equilibrium at one temperature and reactant- favored at another. Free Energy & Temperature

83. When a reaction has  r H° < 0 &  r S° > 0 all at all temperatures  r G° is negative. (Product favored) Free Energy & Temperature

84. When a reaction has  r H° > 0 &  r S° > 0 high at high temperatures  r G° is negative. (Product favored) When a reaction has  r H° < 0 &  r S° < 0 low at low temperatures  r G° is negative. (Product favored) Free Energy & Temperature

86. At what temperature will a reaction that is non-spontaneous turn over to a reaction that is? (  r G changes from + to  ) Free Energy & Temperature

87. At what temperature will a reaction that is non-spontaneous turn over to a reaction that is? (  r G changes from + to  ) Example: Example: The reaction, Has the following thermodynamic values. ∆ r H° = +470.5 kJ ∆ r S° = +560.3 J/K ∆ r G° = +301.3 kJ Reaction is reactant-favored at 298 K Free Energy & Temperature

88. At what temperature will a reaction that is non-spontaneous turn over to a reaction that is? (  r G changes from + to  ) Example: Example: The reaction, When  r G  0, the reaction begins to become spontaneous.

89. At what temperature will a reaction that is non-spontaneous turn over to a reaction that is? (  r G changes from + to  ) Example: Example: The reaction, When  r G  0, the reaction begins to become spontaneous. Free Energy & Temperature

90. At what temperature will a reaction that is non-spontaneous turn over to a reaction that is? (  r G changes from + to  ) Example: Example: The reaction, ∆ r H ° = +467.9 kJ ∆ r S ° = +560.3 J/K Free Energy & Temperature