1. Thermodynamics

2. Spontaneity, Entropy, and Free Energy  Spontaneous Processes, Entropy, and the Second Law of Thermodynamics

3. Basic Concepts and Skills You should be able to: Determine whether the entropy of the system will increase or decrease, given a chemical reaction or a physical process. Determine which one will have the highest (or lowest) entropy, given a choice of chemical substances or physical states

4. Requirements for Spontaneity  Enthalpy Change = ΔH Many (but not all) reactions and processes that are spontaneous in nature are exothermic, ΔH < 0. However, there are important exceptions  NaCl dissolves in water: ΔH = +3 kJ/mol  Gases freely expand into vacuum: ΔH = 0  Conclusion: Enthalpy is not the key factor.  What other factor(s) must be considered in order to predict the spontaneity of a process?

5. Spontaneity  Reminder: Thermodynamics allows us to predict whether a process will occur, but gives not information about the amount of time required for the process (kinetics).  Thermodynamics depends on the relative energies of products and reactants.  Kinetics depends on the activation energy barrier that must be overcome for the process to occur.  A spontaneous process is one that occurs without outside intervention given enough time.

6. Spontaneity The change in Enthalpy (ΔH), is only one of the factors responsible for the Spontaneity or (non spontaneity) of a reaction. Overall: 2O 2 + CH 4  2H 2 O + CO 2 Enthalpy Reactants: 2O 2 + CH 4 H react Products: 2H 2 O + CO 2 H prod

7. Spontaneity An ideal gas spontaneously expands into vacuum. For this process, ΔH = 0.

8. Spontaneity spontaneous nonspontaneous Spontaneity is a one-way street. The gas will not spontaneously recollect itself in the right bulb.

9. Entropy  Here are five ways of arranging four molecules in two separate compartments  Principle: The greater number of arrangements that are possible, the greater the disorder (entropy).

10. Entropy Practice: For each choice of two chemical systems, choose which of the two will have the greater entropy.  1 mol O 2 (g) or 1 mol O 3 (g) Answer:  CH 4 (g) or C 2 H 6 (g) Answer:  NaCl (s) or Na +1 (aq) + Cl -1 (aq) Answer:

11. Entropy Practice: For each of the following reactions, state whether the entropy change is positive or negative. (ΔS > 0 means entropy increases as a result of the reaction.)  2H 2 (g) + O 2 (g)  2H 2 O(l)  H 2 O(l)  H 2 O(g)  Ba(OH) 2 (s) + CO 2  BaCO 3 (s) + H 2 O(l)

12. Entropy Practice: Which of the following processes will be accompanied by the greatest increase in entropy? A. H 2 O(s)  H 2 O(l) B. H 2 O(g)  H 2 O(g) C. 2SO 2 (g) + O 2 (g)  2SO 3 (g) D. C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g) Answer:

13. The Second Law of Thermodynamics  The Second Law of Thermodynamics: Accompanying any spontaneous process there is always an increase in the entropy of the universe. ΔS univ > 0 for a spontaneous process.

14. The Second Law of Thermodynamics  The system is normally the focus of study.  The surroundings include everything else in the universe.  The sum of entropy changes in the system plus those in the surroundings in ΔS univ. Thus, ΔS sys + ΔS surr = ΔS univ  For a spontaneous processes, ΔS sys + ΔS surr > 0.

15. The Second Law of Thermodynamics  Thus far we have considered only the entropy of the system.  We will broaden our discussion to include the entropy of the surroundings.

16. Spontaneity, Entropy, and Free Energy The Gibbs Energy

17. Basic Concepts and Skills  Given the signs of ΔH and ΔS, you should be able to predict whether a chemical reaction will be spontaneous (ie., always, never, or sometimes)  You should be able to calculate the temperature at which a spontaneous reaction will become nonspontaneous (or vise versa).  For a phase transition, you should be able to calculate ΔH f , S, or T, given the other two quantities.

18. The Gibbs Energy Introduction  We learned that for a spontaneous process, ΔS univ > 0; where ΔS univ = ΔS sys + ΔS surr  How can we measure these quantities?  Can we find a simple way to combine the sum of the two terms, ΔS sys + ΔS surr, into a single expression?

19. The Gibbs Energy Consider the term, ΔS surr  The sign of ΔS surr depends on the direction of heat flow. If heat flows from the system into the surroundings, ΔS surr > 0. If heat flows into the system from the surroundings, ΔS surr < 0.  The magnitude of ΔS surr depends on the temperature.

20. The Gibbs Energy At constant temperature and pressure, ΔS surr = -ΔH sys /T ** where ΔH is measured in the system.  This relationship provides a way to measure ΔS surr by concentrating solely on changes that take place in the system. **Note: This relationship can be derived from the thermodynamic definition of entropy, ΔS = Δq rev /T

21. The Gibbs Energy  Substitute –ΔH/T for ΔS surr in the expression, ΔS univ = ΔS sys + ΔS surr : ΔS univ = ΔS sys – ΔH sys /T sys ΔS univ can now be determined by making measurements only on the system.

22. The Gibbs Energy ΔS univ = ΔS sys – ΔH sys /T sys From here on, we will drop the “sys” subscripts. Keep in mind, however, that these variables refer to the system. Multiply both sides by –T: –TΔS univ = ΔH - T ΔS T > 0 always. (why?) ΔS univ > 0 implies that - T ΔS univ < 0 for a spontaneous process.

23. The Gibbs Energy ΔS univ > 0  -Δ S univ < 0 for a spontaneous process. Define: ΔG  ΔH – T ΔS (constant T, P) ΔG is called the Gibbs Energy. (Some texts will refer to it as the Gibbs Free Energy, or simply the Free Energy.) A process or reaction will be spontaneous at constant temperature and pressure if and only if ΔG < 0. Reminder: ΔG is a property of the system!

24. The Gibbs Energy ΔG = ΔH – TΔS A process (at constant T, P) is always spontaneous in the direction in which the Gibbs free energy decreases. ΔG 0.

25. The Gibbs Energy  Under what conditions of ΔH and ΔS is a process spontaneous? (That is, when is ΔH – TΔS < 0?) ΔH 0, then ΔG < 0 __________ ΔH > 0 and ΔS 0 __________ ΔH < 0 and ΔS < 0, then ΔG ??? __________ ΔH > 0 and ΔS > 0, then ΔG ??? __________

26. The Gibbs Energy ΔH ΔS Result – +Spontaneous at all temps + +Spontaneous at high temps – –Spontaneous at low temps + – Not spontaneous at any temps

27. The Gibbs Energy  Problem Type: Calculate the boiling point, given the enthalpy and entropy of vaporization.  Note: at the boiling point, vaporization is an equilibrium process.  At equilibrium, ΔG = 0 (constant T, P).  Because ΔG = ΔH – TΔS this implies that ΔH – TΔS = 0 for every equilibrium process. so ΔH = TΔS (Constant T, P, equilibrium)  T = ΔH/ΔS for an equilibrium process.

28. The Gibbs Energy  Sample Problem.  Consider the vaporization of mercury. At its boiling point, the enthalpy of vaporization is 58.51 kJ/mol, and the entropy of vaporization is 92.92 J/K-mol. Determine the normal boiling point of Hg(l) in C.

29. The Gibbs Energy Solution.  Given: ΔH vap = ________________; ΔS vap = ________________.  For equilibrium process, T bp = ΔH vap / ΔS vap  Substitute, taking care to convert units so that they match. T = _________  Convert temperature to C. (C = K – 273)  T = _________

30. The Gibbs Energy  Problem Type: Calculate the temperature above which a reaction becomes spontaneous (or nonspontaneous)  Key concept  If nonspontaneous at low temperatures ΔG > 0, for low temperatures.  If spontaneous at high temperatures ΔG < 0, for high temperatures.

31. The Gibbs Energy  At the temperature where the reaction becomes spontaneous, ΔG = 0 (constant T, P).  This implies that T = ΔH/ΔS for the “inversion” temperature. (The derivation of this result is identical to the equation for the boiling point calculation.)

32. Gibbs Energy  Sample Problem. Consider the reaction, A(g) + B(g)  C(l) ΔH for the reaction is -107 kJ/mol, and ΔS is -201 J/K-mol. The reaction is spontaneous at room temperature. At what temperature (in C) will it become nonspontaneous?

33. The Gibbs Energy  Solution.  Given ΔH = -107 kJ/mol; ΔS = -201 J/K- mol.  T = ΔH/ΔS  Substitute, and apply unit conversion. T = ____  Convert temperature to C.

34. Spontaneity, Entropy, and Free Energy  Entropy changes in Chemical Reactions.

35. Basic Concepts and Skills You should be able to  Calculate ΔS for a reaction, given the standard entropies of both reactants and products.  Given a chemical reaction or a physical process, determine whether the entropy of the system will increase or decrease.

36. Entropy Changes in Reactions  Predicting the sign of ΔS For a given reaction, predict the sign of ΔS. If ΔS > 0, entropy in the system increases. If ΔS < 0, entropy in the system decreases. Strategy: qualitatively compare the expected entropies of the products to those of the reactants.

37. Entropy Changes in Reactions CaCO 3 (s)  CaO(s) + CO 2 (g) 2SO 2 (g) + O 2 (g)  2SO 3 (g) Al 2 O 3 (s) + 3H 2 (g)  2Al(s) + 3H 2 O(g)

38. Entropy Changes in Reactions How can we calculate a numerical value for the ΔS of a reaction?  Before we can do this, we must first learn how to assign a standard entropy, S, to the reactants and products of the reaction.  The Third Law of Thermodynamics gives us the foundation for establishing the values of the standard entropy of a substance, S.

39. Entropy Changes in Reactions  The Third Law of Thermodynamics... the entropy of a perfect crystal at 0 K is zero.  Because S is explicitly known (S=0) at T = 0 K, S values at other temperatures can be precisely calculated.  Recall that absolute internal energies (E) and enthalpies (H) cannot be measured. However, absolute entropies (S) can be determined.

40. Entropy Changes in Reactions Third Law Entropy as a Function of Temperature 0 0

41. Selected Standard Molar Entropies (25  C) Substance S  (J/K-mol) Substance S  (J/K-mol) Fe(s)27.2H 2 (g)130.6 NaCl (s)72.3H 2 O(g)188.8 FeCl 3 (s)142.3N 2 (g)191.5 H 2 O(l)69.9O 2 (g)205.0 CH 3 OH(l)126.8CH 3 OH(g)237.6 C 6 H 6 (l)172.8C 6 H 6 (g)269.2

42. Entropy Changes in Reactions How do we use the tabulated values of S to calculate the standard entropy change in a chemical reaction, ΔS? The process is essentially the same as you used in the previous chapter to calculate ΔH from a table of ΔH f . The key difference: the standard entropy of an element in its standard state is not zero.

43. Entropy Changes in Reactions Review: Calculate the standard enthalpy of reaction, ΔH for the following reaction: Al 2 O 3 (s) + 3H 2 (g)  2Al(s) + 3H 2 O(g) Recall: ΔH = (n P ΔH f  P ) – (n r ΔH f  r ) n P represents the coefficient in the balanced equation for the product; ΔH f  r represents the standard enthalpy of formation for the reactant.

44. Entropy Changes in Reactions Al 2 O 3 (s) + 3H 2 (g)  2Al(s) + 3H 2 O(g) The tabulated standard enthalpies of formation (in kJ/mol) are: ΔH f  for Al 2 O 3 (s) = -1676 kJ/mol ΔH f  for H 2 O(g) = -242 kJ/mol ΔH = (n P ΔH f  P ) – (n r ΔH f  r ) ΔH = [2(0)+3(-242)] – [1(-1676)+3(0)] = _____________ kJ/mol

45. Entropy Changes in Reactions  To calculate the standard entropy change in a reaction, we use an equivalent process.  ΔS = (n P S P ) – (n r S r )  The difference is that we use tabulated values of standard entropies, because we are calculating a change in entropy.

46. Entropy Changes in Reactions Problem: Calculate the standard entropy change in the following reaction. Al 2 O 3 (s) + 3H 2 (g)  2Al(s) + 3H 2 O(g) The tabulated standard entropies (in J/K-mol) are: S for Al 2 O 3 (s) = 51 J/K-mol S for 3H 2 (g) = 131 J/K-mol S for 2Al(s) = 28 J/K-mol S for 3H 2 O(g) = 189 J/K-mol Important: Remember that standard entropies of pure elements in their standard states are not equal to zero!

47. Entropy Changes in Reactions Solution: ΔS = (n P S P ) – (n r S r ) Al 2 O 3 (s) + 3H 2 (g)  2Al(s) + 3H 2 O(g) 51 131 28 189 ΔS = [2(28)+3(189)] – [1(51)+3(131)] ΔS = ___________ J/K

48. Spontaneity, Entropy, and Free Energy  Gibbs Free Energy Changes in Chemical Reactions

49. Basic Concepts and Skills You should be able to:  Calculate ΔG from ΔH f  and S informations  Calculate either ΔH f  or S, given ΔG and the other quantity.  Use values of ΔG f  values to calculate ΔG for a reaction

50. Gibbs Energy Changes in Reactions Introduction  The goal at hand is to calculate the standard Gibbs energy change in a chemical reaction.  Motivation: For constant (p, T) systems, ΔG < 0 guarantees that a reaction is spontaneous.

51. When calculating standard enthalpy changes, we used the expression, ΔH = n P ΔH f  P (prod) – n r ΔH f  r (react) N is the coefficient of the chemical species in the balanced chemical equation. ΔH f  is the standard enthalpy of formation of a chemical species from its elements in their standard states. (Recall that ΔH f  = 0 for all pure elements in their standard state.) Gibbs Energy Changes in Reactions

52. ΔH f  is the enthalpy change that occurs when elements in their standard states are combined to form one mole of product. H 2 (g) + ½ O 2 (g)   H 2 O(l) The enthalpy change of the above reaction is the molar enthalpy of formation of water.

53. Standard States (Review) The standard state of a substance is defined by the following principles: A gas compound is at 1.00 atm partial pressure. A solid or liquid is pure. A solute has a concentration of 1.00 M. A pure element is in the physical form (solid, liquid, or gas) that is thermodynamically most stable at 1 atm and the temperature of interest (normally 25C. Note: Standard states of compounds can be defined for temperatures other than 298 K.

54. When calculating standa4rd entropy changes, we used the expression, ΔS = n P S P (prod) – n r S r (react) S is the standard entropy of the chemical species. (Recall that S for pure elements in their standard states is not equal to zero.) Gibbs Energy Changes in Reactions

55. The process for calculating ΔG is similar. ΔG = standard free energy change that occurs if reactants in their standard state are converted to products in their standard state. ΔG = n P ΔG f  P (prod) – n r ΔG f  r (react) Gibbs Energy Changes in Reactions

56. Again, n is the coefficient of the chemical species in the balanced chemical equation. ΔG f  is the standard Gibbs energy of formation of a chemical species from its elements in their standard states. Gibbs Energy Changes in Reactions

57. H 2 (g) + ½ O 2 (g)  H 2 O(l) The Gibbs energy change of the above reaction is defined as the molar Gibbs energy of formation of water. Standard states are the same as those previously discussed. ΔG f  = 0 for a pure element in its standard state. e.g., for H 2 (g)  H 2 (g), ΔG f  = 0 Gibbs Energy Changes in Reactions

58. Sample Problem. Determine the value of standard Gibbs energy change, ΔG, for the following reaction. 2NO 2 (g)  N 2 O 4 (g) Gibbs Energy Changes in Reactions

59. Solution: 2NO 2 (g)  N 2 O 4 (g) Find values of ΔG f  from tables: ΔG f  for NO 2 (g) = 52 kJ/mol ΔG f  for N 2 O 4 (g) = 98 kJ/mol ΔG = n P ΔG f  P (prod) – n r ΔG f  r (react) ΔG = __________ Is the reaction spontaneous? Gibbs Energy Changes in Reactions

60. Sample Problem. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l) The standard molar enthalpies and Gibbs energies of formation for these species (in kJ/mol) are CH 4 (g) 2O 2 (g) CO 2 (g) 2H 2 O(l) ΔH f  -75 0 -393.5 -286 ΔG f  -54 0 -394 -237 Use these data to calculate the standard entropy change of this reaction, ΔS, in units of J/K at 298 K. Gibbs Energy Changes in Reactions

61. Solution: Recall that ΔG = ΔH – TΔS TΔS = ΔH – ΔG ΔS = Strategy: Use the given data to solve for both ΔG and ΔH, then make the appropriate substitutions into the above equation. Gibbs Energy Changes in Reactions

62. Solution: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l) CH 4 (g) 2O 2 (g) CO 2 (g) 2H 2 O(l) ΔH f  -75 0 -393.5 -286 ΔG f  -54 0 -394 -237 ΔH = = ΔG = = ΔS = = Gibbs Energy Changes in Reactions

63. Sample Problem: C 6 H 12 O 6 (s) + 6O 2 (g)  6CO 2 (g) + 6H 2 O(l) The standard molar enthalpies of formation (kJ/mol) and standard enthalpies for these species (J/K-mol) are: C 6 H 12 O 6 (s) 6O 2 (g)  6CO 2 (g) 6H 2 O(l) ΔH f  -1275 0 -393.5 -286 ΔS 212 205 214 70 Use these data to calculate the standard Gibbs energy change of this reaction, ΔG, in units of kJ at 298 K. Gibbs Energy Changes in Reactions

64. Solution: C 6 H 12 O 6 (s) + 6O 2 (g)  6CO 2 (g) + 6H 2 O(l) Recall that ΔG = ΔH – TΔS Strategy: Use the given data to solve for both ΔH and ΔS, then make the appropriate substitutions into the above equation. Gibbs Energy Changes in Reactions

65. Solution: C 6 H 12 O 6 (s) + 6O 2 (g)  6CO 2 (g) + 6H 2 O(l) ΔH = = ΔS ΔS = = ΔG = = Note: the second term (T ΔS) must be converted from J to kJ! (divide by 1000). Gibbs Energy Changes in Reactions

66. Spontaneity, Entropy, and Free Energy Gibbs Free Energy and Equilibrium

67. Basic Concepts and Skills You should be able to : Use the value of ΔG to predict the spontaneous direction of a chemical reaction. Calculate K from ΔG, and vice versa.

68. Gibbs Energy and Equilibrium So far we have dealt with ΔG; the Gibbs energy change under standard conditions. Reminder: Standard conditions:  A gas is at 1.00 atm partial pressure.  A solid or liquid compound is pure.  A pure element is in the physical form (solid, liquid, gas) that is thermodynamically most stable at 1 atm and the temperature of interest (25 C).

69. Gibbs Energy and Equilibrium Conditions change throughout the course of a chemical reaction; therefore, the Gibbs energy is continuously changing as the reaction proceeds. What can we say about the Gibbs energy under non-standard conditions? How can we use this information to describe the equilibrium condition?

70. Gibbs Energy and Equilibrium If a reaction path were linearly downhill as in the dashed path,the reaction would go to completion. If the path has a minimum Gibbs energy position at some intermediate composition, equilibrium will occur at the Gibbs energy minimum. How is a reaction equilibrium related to the Gibbs Energy?

71. Gibbs Energy and Equilibrium Equilibrium occurs where G reaches a minimum.

72. Gibbs Energy and Equilibrium Gibbs energy vs reaction coordinate for the ammonia (NH 3 ) system. N 2 (g) + 3H 2 (g) ⇄ 2 NH 3 (g)

73. Gibbs Energy and Equilibrium Equilibrium occurs at the point of the reaction coordinate when the Gibbs energy is a minimum. Regardless of which direction the system moves from equilibrium, ΔG > 0; therefore, departure from the equilibrium composition is nonspontaneous.

74. Gibbs Energy and Equilibrium (a) ΔG is normally not zero at the beginning of a reaction. (a) The actual (nonstandard) Gibbs energies of reactant and product move toward each other as the reaction proceeds. (b) ΔG = 0 at equilibrium; at equilibrium, the (nonstandard) Gibbs energies of reactants and products are equal. Movement away from equilibrium is always nonspontaneous.

75. Gibbs Energy and Equilibrium The change in Gibbs energy for a reaction can be calculated under nonstandard conditions. ΔG = ΔG + RTln(Q) Q = reaction quotient R = universal gas constant: 8.314 J/mol-K T = temperature (Kelvin) ΔG = standard Gibbs energy change

76. Gibbs Energy and Equilibrium ΔG = ΔG + RTln(Q)  We can use this equation to find the relationship between ΔG and the equilibrium constant, K.  Recall: At equilibrium, ΔG = 0 At equilibrium, Q = K. We will substitute these values into the above equation for the special case when the system is at equilibrium

77. Gibbs Energy and Equilibrium ΔG = ΔG + RTln(Q) At equilibrium 0 = ΔG + RTln(K) Therefore ΔG = -RTln(K) Rearranging: ln K = - ΔG/RT K = exp(- ΔG/RT)

78. Gibbs Energy and Equilibrium Sample Problem. Calculate the equilibrium constant at 298 K for the dissolution reaction, NaCl  Na +1 (aq) + Cl -1 (aq) Strategy K = exp(- ΔG/RT) Calculate ΔG from tables Substitute and evaluate

79. Gibbs Energy and Equilibrium Solution. NaCl(s)  Na +1 (aq) + Cl -1 (aq) ΔG f  for NaCl(s) = -384 kJ/mol ΔG f  for Na +1 (aq) = -262 kJ/mol ΔG f  for Cl -1 (aq) = -131 kJ/mol  K = exp(- ΔG/RT)  Calculate ΔG from tables. ΔG = Substitute and evaluate (Beware of units!) K = =

80. Gibbs Energy and Equilibrium Sample Problem HNO 2 (aq) ⇄ H +1 (aq) + NO -1 (aq) K a for HNO 2 is 4.0 x 10 -4  Determine the standard Gibbs energy change (in kJ/mol) for the aqueous dissociation of nitrous acid at 298 K. Strategy ΔG = -RTln(K) Substitute and evaluate.

81. Gibbs Energy and Equilibrium Solution. ΔG = -RTln(K) Substitute and evaluate ΔG