1. Chapter 17TMHsiung©2015Slide 1 of 66 Chapter 17 Free Energy and Thermodynamics

2. Chapter 17TMHsiung©2015Slide 2 of 66 1.Nature’s Heat Tax: You Can’t Win and You Can’t Break Even 2.Spontaneous and Nonspontaneous Processes 3.Entropy and the Second Law of Thermodynamics 4.Heat Transfer and Changes in the Entropy of the Surroundings 5.Gibbs Free Energy 6.Entropy Changes in Chemical Reactions: Calculating ΔS o rxn 7.Free Energy Changes in Chemical Reactions: Calculating ΔG o rxn Contents

3. Chapter 17TMHsiung©2015Slide 3 of 66 8.Free Energy Changes for Nonstandard States: The Relationship between ΔG o rxn and ΔG rxn 9.Free Energy and Equilibrium: Relating ΔG o rxn to the Equilibrium Constant (K) Continued

4. Chapter 17TMHsiung©2015Slide 4 of 66 1.Nature’s Heat Tax: You Can’t Win and You Can’t Break Even 1)First Law of Thermodynamics The total energy of the universe is constant that energy cannot be created or destroyed, but you can transfer it from one place to another: ΔE universe = 0 = ΔE system + Δ E surroundings There are two ways energy is “lost” from a system: ̶ Converted to heat, q ̶ Used to do work, w The energy change in the system: Δ E = q + w or Δ E = Δ H + P Δ V Internal energy change ( Δ E) is a state function, depends only on the state of the system, not on how the system got to that state. *****

5. Chapter 17TMHsiung©2015Slide 5 of 66 2)Heat Tax Conversion of energy to heat, which is “lost” by heating up the surroundings. *****

6. Chapter 17TMHsiung©2015Slide 6 of 66 2.Spontaneous and Nonspontaneous Processes *If a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction. 1)Spontaneous and Nonspontaneous Spontaneous process: A process that occurs without outside intervention. Nonspontaneous: A process that occurs must be assisted by outside intervention.

7. Chapter 17TMHsiung©2015Slide 7 of 66 2)Chemical Thermodynamics Information Stability of particular substances The Spontaneity of a chemical reaction Equilibrium constant (K eq ) of a chemical reaction Predict the proportions of products and reactants at equilibrium. The optimum temperature, pressure, and choice of solvent etc. for a particular reaction.

8. Chapter 17TMHsiung©2015Slide 8 of 66 3)Kinetics versus Thermodynamics Thermodynamics tells us what processes are possible. Kinetics tells us whether the process is practical. (high activation energy can effectively block a reaction although that is thermodynamically favored)

9. Chapter 17TMHsiung©2015Slide 9 of 66 4)Exothermic reaction versus Endothermic reaction Most spontaneous processes are exothermic, however, some spontaneous processes are endothermic. For example: - the melting of ice above 0 o C - the evaporation of liquid water to gaseous water - the dissolution of sodium chloride in water

10. Chapter 17TMHsiung©2015Slide 10 of 66 5)Natural tendencies for spontaneous processes To achieve a lower energy state, i.e., decreasing enthalpy (H) -The enthalpy change, ΔH, is the difference in the sum of the internal energy and PV work energy of the reactants to the products. Toward a more disordered state, i.e., increasing entropy (S) -The entropy change, ΔS, is the difference in randomness of the reactants compared to the products.

11. Chapter 17TMHsiung©2015Slide 11 of 66 3.Entropy and the Second Law of Thermodynamics 1)Entropy (S) A thermodynamic function that is proportional to the number of energetically equivalent ways to arrange the components of a system to achieve a particular state; a measure of the energy randomization or energy dispersal in a system. S = k ln W k = Boltzmann constant = 1.38 × 10 −23 J/K W: the number of energetically equivalent ways a system can exist, unitless S, unit generally in J/mol The greater spreading of the energy of the microscopic particles, the greater the S of the system. *****

12. Chapter 17TMHsiung©2015Slide 12 of 66 Continued Each system has the same total energy (4 J), but System A has only one possible microstate while System B has two possible microstate.

13. Chapter 17TMHsiung©2015Slide 13 of 66 The Spontaneous Mixing of Ideal Gases Intermolecular forces are negligible, no significant enthalpy change, total energy of the system remains unchanged However, the number of possibilities for the distribution of that energy increases Continued

14. Chapter 17TMHsiung©2015Slide 14 of 66 Spontaneous Formation of An Ideal Solution Entropy of the mixture is higher than the entropies of the two substances separated. Continued

15. Chapter 17TMHsiung©2015Slide 15 of 66 Vaporization of Water Evaporation is spontaneous because of the increase in entropy Continued

16. Chapter 17TMHsiung©2015Slide 16 of 66 2)Second Law of Thermodynamics ***** Spontaneous process: ΔS univ = ΔS sys + ΔS surr > 0 *ΔS = S final − S initial System Surrounding Universe ***** All spontaneous process are accompanied by an increase in entropy (S) of the universe.

17. Chapter 17TMHsiung©2015Slide 17 of 66 3)Absolute Entropy of a Substance The particular S of a substance is calculated from the amount of heat required to raise the temperature from 0 K. Standard molar entropy (S o ): The entropy of one mole (or 1 M) of a substance in its standard state. (Appendix IIB) In a particular phase, S slow increase with increasing T, the temperature dependence of S: S (T2) = S (T1) + C p ln(T 2 –T 1 ) C p : Heat capacity of the substance S sharply change at phase change/transition: S (phase 2) = S (phase 1) + ΔH/T example: ΔS fusion = ΔH fusion /T m )

18. Chapter 17TMHsiung©2015Slide 18 of 66 4)Entropy Change Associated with a Change in State

19. Chapter 17TMHsiung©2015Slide 19 of 66 5)Predicting the Trends of Entropy S Solid < S Liquid < S Gas S (small atom or molecule) < S (large atom or molecule) S (simple molecule) < S (complex molecule) S (low temperature) < S (high temperature) S (solute) + S (solvent) < S (solution) S (less amount of gases) < S (more amount of gases) *****

20. Chapter 17TMHsiung©2015Slide 20 of 66 Example 17.1 Predicting the Sign of Entropy Change a. Since a gas has a greater entropy than a liquid, the entropy decreases and ΔS is negative. b. Since a solid has a lower entropy than a gas, the entropy increases and ΔS is positive. c. Since the number of moles of gas increases, the entropy increases and ΔS is positive. Solution Predict the sign of ΔS for each process: a. H 2 O (g) → H 2 O (l) b. Solid carbon dioxide sublimes. c. 2 N 2 O (g) → 2 N 2(g) + O 2(g) *****

21. Chapter 17TMHsiung©2015Slide 21 of 66 4.Heat Transfer and Changes in the Entropy of the Surroundings 1)Entropy of the Surroundings (ΔS surr ) a)Observation: Water freezes at temperatures below 0 o C, the entropy of the water decreases, the process is spontaneous. Water vapor in air condenses into fog on a cold night, the entropy of the water decreases, the process is spontaneous.

22. Chapter 17TMHsiung©2015Slide 22 of 66 b)Explanation: When the entropy change in a system is unfavorable (negative), the entropy change in the surroundings must be favorable (positive), and large in order to allow the process to be spontaneous. *An exothermic process increases the entropy of the surroundings. *An endothermic process decreases the entropy of the surroundings.

23. Chapter 17TMHsiung©2015Slide 23 of 66 2)The Temperature Dependence of ΔS surr a)Observation: Water freezes at temperatures below 0 o C spontaneously, however, freezing of water becomes nonspontaneous above 0 o C. b)Explanation: The greater the temperature, the smaller the increase in entropy for a given amount of energy dispersed into the surroundings.

24. Chapter 17TMHsiung©2015Slide 24 of 66 Continued

25. Chapter 17TMHsiung©2015Slide 25 of 66 3)Quantifying Entropy Changes in the Surroundings At Constant P and T *****

26. Chapter 17TMHsiung©2015Slide 26 of 66 Example 17.2 Calculating Entropy Changes in the Surroundings a. Solution Consider the combustion of propane gas: C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O(g) ΔH rxn = –2044 kJ a. Calculate the entropy change in the surroundings associated with this reaction occurring at 25 o C. b. Determine the sign of the entropy change for the system. c. Determine the sign of the entropy change for the universe. Will the reaction be spontaneous? *****

27. Chapter 17TMHsiung©2015Slide 27 of 66 b. An increase in the number of moles of gas implies a positive ΔS sys. c. ΔS univ is positive and the reaction is spontaneous. Continued

28. Chapter 17TMHsiung©2015Slide 28 of 66 5.Gibbs Free Energy 1)The Equation *****

29. Chapter 17TMHsiung©2015Slide 29 of 66 ***** ΔG < 0 :spontaneous process ΔG = 0 :at equilibrium ΔG > 0 :nonspontaneous process 2)A Summary *****

30. Chapter 17TMHsiung©2015Slide 30 of 66 Continued

31. Chapter 17TMHsiung©2015Slide 31 of 66 3)The Effect of ΔH, ΔS, and T on Spontaneity ΔG = ΔH − TΔS *****

32. Chapter 17TMHsiung©2015Slide 32 of 66 Example 17.3 Computing Gibbs Free Energy Changes and Predicting Spontaneity from ΔH and ΔS a. The reaction is not spontaneous. Solution Consider the reaction for the decomposition of carbon tetrachloride gas: CCl 4(g) → C (s, graphite) + 2 Cl 2(g) ΔH = + 95.7 kJ; ΔS = +142.2 J/K a. Calculate ΔG at 25 o C and determine whether the reaction is spontaneous. b. If the reaction is not spontaneous at 25 o C, determine at what temperature (if any) the reaction becomes spontaneous. *****

33. Chapter 17TMHsiung©2015Slide 33 of 66 b. Since ΔS is positive, ΔG becomes more negative with increasing temperature. Continued The reaction is spontaneous above this temperature. *****

34. Chapter 17TMHsiung©2015Slide 34 of 66 6.Entropy Changes in Chemical Reactions: Calculating ΔS rxn 1)Standard State For Gas: pure gas at exactly 1 atm pressure. For Solid or Liquid: 1 mole pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest. Usually 25 o C Solution: Substance in a solution with concentration 1 M. *****

35. Chapter 17TMHsiung©2015Slide 35 of 66 2)Third Law of Thermodynamics For a perfect crystal at absolute zero (0 K), the absolute entropy = 0 J/mol ∙ K. S = k ln W = k ln 1 = 0 *****

36. Chapter 17TMHsiung©2015Slide 36 of 66 3)Factors that affect the standard entropy State of the substance Molar mass of the substance Particular allotrope Molecular complexity Dissolution

37. Chapter 17TMHsiung©2015Slide 37 of 66 State of the substance Molar mass of the substance

38. Chapter 17TMHsiung©2015Slide 38 of 66 Particular allotrope *Allotrope: Those elements can exist in two or more forms. Diamond: three-dimensional crystal structure. Graphite: the atoms bond together in sheets, but the sheets have freedom to slide past each other.

39. Chapter 17TMHsiung©2015Slide 39 of 66 Molecular complexity Ar: only translational motion. NO, translational motion, rotational motion, and vibrational motions of the molecules *****

40. Chapter 17TMHsiung©2015Slide 40 of 66 Dissolution

41. Chapter 17TMHsiung©2015Slide 41 of 66 4)The Standard Entropy Change, ΔS The standard entropy change is the difference in absolute entropy between the reactants and products under standard conditions. ΔSº reaction = (∑n p Sº products ) − (∑n r Sº reactants ) The standard enthalpy of formation, ΔH f °, of an element is 0 kJ/mol. The absolute entropy (S o ) at 25 o C is always positive. *****

42. Chapter 17TMHsiung©2015Slide 42 of 66 Standard Absolute Entropies *More complete list can be found in Appendix IIB **Intensive property: A property such as density that is independent of the amount of a given substance. Extensive property: A property that depends on the amount of a given substance, such as mass. *****

43. Chapter 17TMHsiung©2015Slide 43 of 66 Example 17.4 Calculating Standard Entropy Changes (ΔS rxn ) Appendix IIB. Solution Calculate ΔS rxn for the balanced chemical equation: 4 NH 3(g) + 5 O 2(g) → 4 NO (g) + 6 H 2 O (g) ° ° ° * ΔS rxn is positive, the number of moles of gas increases. OK. *****

44. Chapter 17TMHsiung©2015Slide 44 of 66 7.Free Energy Changes in Chemical Reactions: Calculating ΔG o rxn 1)Standard free energies of formation (  G o f ) The change in free energy when 1 mol (or 1 M) of a compound in its standard state forms from its constituent elements in their standard states. The free energy of formation listed in Appendix IIB. Same as standard formation enthalpy (  H f o ),  G o f of the pure elements in their standard states is zero.

45. Chapter 17TMHsiung©2015Slide 45 of 66 2)Calculating  G o rxn with Tabulated Values of Free Energies of Formation (  G o f ) ΔG o rxn = ∑n p ΔG f o (products) − ∑n r ΔG f o (reactants) *****

46. Chapter 17TMHsiung©2015Slide 46 of 66 Example 17.7 Appendix IIB *  G o f of a pure element is zero. Solution Ozone in the lower atmosphere is a pollutant that can form by the following reaction involving the oxidation of unburned hydrocarbons: CH 4(g) + 8 O 2(g) → CO 2(g) + 2 H 2 O (g) + 4 O 3(g) Use the standard free energies of formation to determine ΔG rxn for this reaction at 25 o C. ° ° ° *****

47. Chapter 17TMHsiung©2015Slide 47 of 66 3)Calculating Standard Free Energy Changes with ΔG o rxn = ΔH o rxn − TΔS o rxn Example 17.5 At 25 o C Appendix IIB Solution One of the possible initial steps in the formation of acid rain is the oxidation of the pollutant SO 2 to SO 3 by the reaction: Calculate ΔG rxn at 25 o C and determine whether the reaction is spontaneous. *****

48. Chapter 17TMHsiung©2015Slide 48 of 66 The reaction is spontaneous at this temperature because ΔG rxn is negative. Continued

49. Chapter 17TMHsiung©2015Slide 49 of 66 Example 17.6 Other than 25 o C ΔG o rxn at 125 o C is less negative (or more positive) than the value of ΔG o rxn at 25 o C (which is –70.9 kJ), the reaction is less spontaneous. For the reaction in Example 17.5, estimate the value of ΔG o rxn at 125 o C. Is the reaction more or less spontaneous at this elevated temperature; that is, is the value of ΔG o rxn more negative (more spontaneous) or more positive (less spontaneous)? ° Solution * The different temperature resulted in relative small variations of ΔS o rxn and ΔH o rxn (25 o C). *****

50. Chapter 17TMHsiung©2015Slide 50 of 66 4)Calculating ΔG o rxn for a Stepwise Reaction Example 17.8 Calculating ΔG rxn for a Stepwise Reaction Find ΔG rxn for the reaction. 3 C(s) + 4 H 2 (g) → C 3 H 8 (g) Use the following reactions with known ΔG’s: ° Solution °

51. Chapter 17TMHsiung©2015Slide 51 of 66 5)Coupled Reactions Coupled reactions: Use a thermodynamically favorable reaction to drive an unfavorable one. Example1:

52. Chapter 17TMHsiung©2015Slide 52 of 66 Example 2 (a biochemical cycle) C 6 H 12 O 6(s) + 6O 2(g) → 6CO 2(g) + 6H 2 O (l) ΔG o = –2280 kJ ADP + H 3 PO 4 → ATP + H 2 O (l) ΔG o = +31 kJ C 6 H 12 O 6(s) + 6O 2(g) + ADP + H 3 PO 4 → 6CO 2(g) + ATP + 7H 2 O (l) ΔG o = –2249 kJ Alnaine + Glycine → AlnanylglycineΔG o = +29 kJ ATP + H 2 O → ADP + H 3 PO 4 ΔG o = –31 kJ ATP + H 2 O + Alnaine + Glycine → Ananylglycine + ADP + H 3 PO 4 ΔG o = –2 kJ

53. Chapter 17TMHsiung©2015Slide 53 of 66 8.Free Energy Changes for Nonstandard States: The Relationship between ΔG o rxn and ΔG rxn Q.Spilled water spontaneously evaporates even though ΔG o rxn for the vaporization of water is positive. Why? A.Because ordinary conditions are not standard conditions and ΔG o rxn applies only to standard conditions.  G o rxn describes the free energy change at standard state.  G rxn describes the free energy change at any specified state.

54. Chapter 17TMHsiung©2015Slide 54 of 66 The Free Energy Change of a Reaction under Nonstandard Conditions  G rxn =  G o rxn + RTlnQ Q:reaction quotient R:8.314 Jmol –1 K –1 T:K *****

55. Chapter 17TMHsiung©2015Slide 55 of 66 Example (a)Standard Conditions, P H 2 O = 1 atm, i.e., Q = 1  G rxn =  G o rxn + RTlnQ = +8.59 kJ/mol + RT ln(1) = +8.59 kJ/mol (spontaneous in the reverse direction ) (b)At pressure of 0.0313 atm, i.e., Q = K p = 0.0313  G rxn =  G o rxn + RTlnQ = +8.59 kJ/mol + RT ln (0.0313) = +8.59 kJ/mol - 8.59 kJ/mol = 0 (c)At pressure of 5.00x10 -3 atm, i.e., Q = K p = 5.00x10 -3 atm  G rxn =  G o rxn + RTlnQ = +8.59 kJ/mol + RT ln (5.00x10 -3 ) = +8.59 kJ/mol – 13.1 kJ/mol = – 4.5 kJ/mol  G rxn =  G o rxn + RTlnQ

56. Chapter 17TMHsiung©2015Slide 56 of 66 Continued

57. Chapter 17TMHsiung©2015Slide 57 of 66 Example 17.9 Calculating ΔG rxn under Nonstandard Conditions Consider the reaction at 298 K: Calculate ΔG rxn under these conditions: P NO = 0.100 atm; P O 2 = 0.100 atm; P NO 2 = 2.00 atm Is the reaction more or less spontaneous under these conditions than under standard conditions? Solution The reaction is spontaneous under these conditions, but less spontaneous than it would be under standard conditions. ° ° ° *****

58. Chapter 17TMHsiung©2015Slide 58 of 66 9.Free Energy and Equilibrium: Relating ΔG o rxn to the Equilibrium Constant ( K )Free En ergy  G rxn =  G o rxn + RTlnQ At equilibrium,  G rxn = 0 and Q = K 0 =  G o rxn + RTlnQ =  G o rxn + RTlnK  G o rxn = － RTlnK = － 2.303RTlogK eq lnK = －  G o /RT 1)Equation and Explanation *****

59. Chapter 17TMHsiung©2015Slide 59 of 66 a)When K < 1  G o rxn is positive. Under standard conditions (when Q = 1 ), the reaction is spontaneous in the reverse direction.  G o rxn = － RTlnK

60. Chapter 17TMHsiung©2015Slide 60 of 66 b)When K > 1  G o rxn is negative. Under standard conditions (when Q = 1 ), the reaction is spontaneous in the forward direction.  G o rxn = － RTlnK

61. Chapter 17TMHsiung©2015Slide 61 of 66 c)When K = 1  G o rxn is zero. Under standard conditions (when Q = 1 ), the reaction is equilibrium.  G o rxn = － RTlnK

62. Chapter 17TMHsiung©2015Slide 62 of 66 Example 17.10 The Equilibrium Constant and ΔG rxn Appendix IIB Use tabulated free energies of formation to calculate the equilibrium constant for the reaction at 298 K: Solution ° *****

63. Chapter 17TMHsiung©2015Slide 63 of 66 2)The Temperature Dependence of the Equilibrium Constant For a particular reaction,  H and  S do not change much for different temperature. At standard state:  G o rxn = – RTlnK(1)  G o rxn =  H o rxn – T  S o rxn (2) Combing (1) and (2) – RTlnK =  H o rxn – T  S o rxn =  G o rxn y = mx + b

64. Chapter 17TMHsiung©2015Slide 64 of 66 Example: For the reaction: CO (g) + H 2 O (g) → CO 2(g) + H 2(g) lnK eq vs. 1/T plot Slope  4810 = –  H o /R (R=8.3145 J mol –1 K –1 )  H o  – 40 kJ (tabulated data is – 41.2 kJ).

65. Chapter 17TMHsiung©2015Slide 65 of 66 Calculating K at different temperature

66. Chapter 17TMHsiung©2015Slide 66 of 66 End of Chapter 17