1. Chapter 16 Spontaneity, entropy and free energy

2. 2 1 st Law of Thermodynamics When methane and oxygen react to form carbon dioxide and water, the products have lower potential energy than the reactants. This change in potential energy results in thermal energy flow (heat) to the surroundings. But the energy of the universe remains “constant”

3. 3 1 st Law of Thermodynamics Keeps track of the energy l How much energy is involved in the change? l What form is the energy in? l Does the energy flow in or out of the system? ***But does not tell us why?**

4. Spontaneous l A reaction that will occur without outside intervention. l We can’t determine how fast. l We need both thermodynamics and kinetics to describe a reaction completely. l Thermodynamics compares initial and final states. l Kinetics describes pathway between (activation energy, [concentration], T, catalysts.

5. The rate of a reaction depends on the pathway from reactants to products; this is the domain of kinetics. Thermodynamics tells us whether a reaction is spontaneous based only on the properties of the reactants and products. The predictions of thermodynamics do not require knowledge of the pathway between reactants and products.

6. Thermodynamics l 1st Law- the energy of the universe is constant. l Keeps track of thermodynamics doesn’t correctly predict spontaneity. l What makes a process spontaneous? – A ball rolls down a hill, but never back up – Steel rusts – A gas fills a container uniformly – Why???? – One common characteristic in all…

7. 7 Entropy The driving force for a spontaneous process is an increase in the entropy of the universe. Entropy, S, can be viewed as a measure of randomness, or disorder.

8. Entropy  Defined in terms of probability.  Substances take the arrangement that is most likely.  The most likely is the most random.  Calculate the number of arrangements for a system. Nature spontaneously proceeds toward states that have the highest probability of existing.

9. 9 The expansion of an ideal gas into an evacuated bulb. Exist in more states, the more spread out the molecules. Why is this Spontaneous?

10. l 4 possible arrangements (microstates) l 25% chance of finding the left empty l 1/2 2 =1/4 l 50 % chance of them being evenly dispersed 2 Physical evidence of entropy

11. l 4 atoms l ½ = 1/16 l 8% chance of finding the left empty l 50 % chance of them being evenly dispersed 4

12. 12 The greater the disorder the larger the entropy.

13. Entropy from physical evidence l S solid <S liquid <<S gas l Gases have a huge number of positions possible. (Positional probability) l Pure solid dissolves in a solvent, the entropy of the substance increases – Except:Carbonates l Entropy increases with increasing molecular complexity (more e- moving) l Increase # moles, increase entropy

14. Concept Check Predict the sign of  S for each of the following, and explain: a) The evaporation of alcohol b) The freezing of water c) Compressing an ideal gas at constant temperature d) Heating an ideal gas at constant pressure e) Dissolving NaCl in water + – – + +

15. 15 The Second Law of Thermodynamics...in any spontaneous process there is always an increase in the entropy of the universe.  S univ > 0 for a spontaneous process.  S univ =  S sys +  S surr

16. Entropy l Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate. ∆S univ = ∆ S sys + ∆ S surr l If ∆ S univ is positive the process is spontaneous. l If ∆ S univ is negative the process is spontaneous in the opposite direction.

17. Entropy Changes in Chemical Reactions N 2 (g) +3H 2 (g)  2NH 3 (g) We go from 4 to 2 gas molecules. What happens to the ∆S, (+,-)? ∆S – The change in positional entropy is dominated by the relative number of molecules of gaseous reactants and products. (Although change is important we can assign absolute entropy values.)

18. 18 Predict the sign of ΔS° a. CaCo 3 (s) → CaO(s) + CO 2 (g) b. 2SO 2 (g) + O 2 (g) → 2SO 3 (g) l What if the # of molecules is the same? Generally, the more complex the molecule, the higher the standard entropy value. + -

19. Third Law of Thermo l The entropy of a pure crystal at 0 K is 0. l Gives us a starting point for finding values of S at other temperatures which must be > 0. l Remember Entropy is a state function.

20. Copyright©2000 by Houghton Mifflin Company. All rights reserved.20 Figure 16.5: (a) A perfect crystal of hydrogen chloride at 0 K. (b) As the temperature rises above 0 K, lattice vibrations allow some dipoles to change their orientations, producing some disorder and an increase in entropy.

21. 21 l Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed. l ∆S is also an extensive property, so it is dependent on the amount.

22. 22 Calculate ΔS ° Substance S°(J/K·mol) Al 2 O 3 51 H 2 131 Al 28 H 2 0 189 (56J/K + 567J/K) – (393J/K + 51J/K) = 179J/K The more complex the molecule, the higher Is the standard entropy value. Al 2 O 3 (s) + 3H 2 (g) → 2Al(s) + 3H 2 0(g)

23. Figure 16.6: The H2O molecule can vibrate and rotate in several ways, some of which are shown here.

24. l exothermic processes ∆H- ∆S surr is (+) l endothermic processes ∆H+ ∆ S surr is (-) Consider this process: H 2 O(l)  H 2 O(g) ∆ S sys is positive ∆ S surr is negative ∆ S univ depends on temperature. These are in Opposition Which controls The situation? At 1atm water changes from liquid to gas:  Above 100°C it is spontaneous.  Below 100°C the opposite process is spontaneous Effect of Temperature ∆S univ = ∆ S sys + ∆ S surr

25.  The lower the T, the greater the impact. Δs sys = heat transferred T at which the transfer occured

26. 26 ΔS uni = ΔS sys + ΔS surr ΔS surr = + when E is added (exo) = - when E is used (endo) Express ΔS surr in terms of enthalpy ΔH Heat flow(q)(constant P) = change in enthalpy= ΔH Into and out of the system + endothermic/- exothermic At constant temperature and pressure: (-)Sign changes the point of View from the sys. to the surr.

27. 2H 2 (g) + 2O 2 (g)  H 2 O(g) Ignite & very fast! (spontaneous)  S sys = -88.9 J/K why???  H sys = -483.6 kJ  S surr = ?  S univ = ? Even though the entropy of the system declines, the entropy of the surroundings is so large that the overall change is positive. 1620 J/K 1530J/K

28. Bottom line: A process is spontaneous in spite of a negative entropy change as long as it is extremely exothermic. Sufficient exothermicity offsets system ordering.

29.  S sys -  H/T  S surr  S univ Spontaneous? +++ --- +-? +-? Yes No, Reverse At Low temp. At High temp.

30. 30 Free Energy   G =  H  T  S  (from the standpoint of the system) A process (at constant T, P) is spontaneous in the direction in which free energy decreases:  S (refers to the system)**  G means +  S univ

31. 31  G =  H  T  S To see how this equation relates to spontaneity, Divide by –T At constant T and P

32. 32 This means; l A process at constant T and P is only spontaneous if ΔG is negative. (decreasing) ΔG > 0 entropy of universe decreases ΔG < 0 entropy of universe increases Lets use the free energy equation to predict the spontaneity of melting of ice: H 2 O(s)→H 2 O(l)

33. 33 H 2 O(s)→H 2 O(l) Based on your daily experiences you would say melting ice is spontaneous at T> 0,and not spontaneous at T< 0.  G =  H  T  S HEAT BEING RELEASED ENTROPY INCREASING ENTROPY DECREASING

34. 34 Example At what temperatures is the following process spontaneous at 1 atm? Br 2 (l)→Br 2 (g) ΔH°=31.0 kJ/mol and ΔS°=93.0 J/K ·mol What is the normal boiling point of liquid Br 2 ?

35. 35 1. Vaporization will occur at all T where ΔG is negative 2. ΔS° favors vaporization due to increase in positional entropy. 3. ΔH° favors the opposite process (exothermic) 4. The opposite tendencies will balance at the boiling pt. At this BP temp. liquid and gaseous Br 2 are in equilibrium ΔG °= 0. ΔG °= ΔH °- TΔS ° 0 = ΔH °- T ΔS ° ΔH °= T ΔS ° T= ΔH / ΔS ° 3.1x10 4 J/mol/93.0J/K·mol = 333K Br 2 (l) → Br 2 (g) ΔH°=31.0 kJ/mol and ΔS°=93.0 J/K ·mol

36. 36 l T> 333K, TΔS° has larger magnitiude than ΔH °; ΔG ° is negative. Above 333K, vaporization is spontaneous; opposite occurs spontaneously below this T. l At 333K, liquid and gaseous Br 2 coexist in equilibrium.

37. 37 Summary 1. T > 333K. ΔS° controls. Increase in entropy when Br 2 (l) is vaporized. 2. T < 333K. The process is spontaneous in the direction in which it is exothermic. ΔH° controls. 3. T = 333K. ΔS° and ΔH° are balanced (ΔG° = 0), liquid and gas coexist. Normal BP.

38.  G=  H-T  S HH SS Spontaneous? +- At all Temperatures ++ At high temperatures, “entropy driven” -- At low temperatures, “enthalpy driven” +- Not at any temperature, Reverse is spontaneous

39. Free Energy in Reactions  Gº = standard free energy change. l Free energy change that will occur if reactants in their standard state turn to products in their standard state.

40. 40 Why Free energy changes? l Tells us relative tendancy of a reaction to occur. l The more - ΔG°, the more to the right the reaction will proceed. l Use standard state because energy varies w/ pressure or concentration l We must compare all reactions under the same pressure or concentration conditions.

41. 41 ΔG° not measured directly (calculated 3 ways)   G ° =  H °  T  S ° (constant T) 2. Free energy is also a state function, therefore, we can use a process similar to finding  H using Hess’s Law. 3. Standard free energy of formation (  G f   G  =  n p  G f  (products)   n r  G f  (reactants)

42. Copyright©2000 by Houghton Mifflin Company. All rights reserved.42  G ° =  H °  T  S ° (constant T) Consider the reaction 2SO 2 (g) + O 2 (g) → 2SO 3 (g) Carried out at 25C and 1 atm. Calculate ΔH°, ΔS°& ΔG° using the following data: SubstanceΔH° f (kJ/mol)S°(J/K.mol) SO 2 (g)-297248 O 2 (g)-396257 SO 3 (g)0205

43. 43 ΔH°? From enthalpies of formation. ΔH°= 2mol(-396kJ/mol) – 2mol(-297kJ/mol) – 0 = -198 kJ ΔS°? Standard entropy values ΔS°= 2mol(257) – 2mol(248) – 1mol(205) = -187J/K & ΔG° using the following data:

44. 44 ΔG°?  G ° =  H °  T  S ° = -198kJ – (298K)(-7J/K)(1kJ/1000J) = -142kJ

45. 45 Use the following data (at 25C) C diamond (s) + O 2 (g) → CO 2 (g)ΔG°= -397kJ C graphite (s) + O 2 (g) → CO 2 (g) ΔG°= -394kJ Calculate ΔG° for the reaction C diamomd (s)→ C graphite (s) Free energy is also a state function, therefore, we can use a process similar to finding Δ H using Hess’s Law.

46. 46 C diamond (s) + O 2 (g) → CO 2 (g) ΔG°=-397kJ CO 2 (g) → C graphite (s) + O 2 (g) ΔG°=394kJ =-3kJ The reaction is spontaneous but very slow at 25C and 1atm. Reverse occurs at high T and P

47. 47  G  =  n p  G f  (products)  n r  G f  (reactants) l Change in free energy that accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products. l Standard free energy of formation of an element in its standard state is zero.

48. 48 Methanol is a high-octane fuel used in high-performance racing engines. Calculate ΔG° for the reaction, 2CH 3 OH(g) + 3O 2 (g) → 2CO 2 (g) + 4H 2 O(g) Free energies of formation are: Substance ΔG°(kJ/mol) CH 3 OH -163 O20O20 CO 2 -394 H 2 O-229 = 2mol(- 394kJ/mol ) + 4mol(-229kJ/mol) – 3(0)+ 2mol(-163kJ/mol) = -1378 kJ

49. Free Energy in Reactions There are tables of  Gº f. l Products-reactants because it is a state function. l The standard free energy of formation for any element in its standard state is 0. l Remember- Spontaneity tells us nothing about rate.

50. 50 Dependence of Free Energy on Pressure A system at constant T and P will proceed spontaneously in the direction to lower G. The equilibrium position represents the lowest G value available to a reaction system. G changes as a reaction proceeds because G is dependent on P and concentration.

51. 51 G = H - TS How does pressure affect the thermodynamic of free energy? For an ideal gas H is not P dependent S depends on P, because of the dependence on V. Therefore G depends on P.

52. 52 Consider 1 mol of ideal gas at a given Temp At V=10.0L there are more positions available than at V=1.0L S larger V > S small V S low P > S high P (P and V are inversely proportional.)

53. 53 G = G° + RT ln(P) l G°= P at 1atmR= gas constant l G = P at P atmT = K To demonstrate how the  G is pressure dependent: N 2 (g) + 3H 2 (g) → 2NH 3 (g)  G  =  n p  G f  (products)   n r  G f  (reactants)

54. 54 N 2 (g) + 3H 2 (g) → 2NH 3 (g)  G =  n p  G (products)   n r  G (reactants)  G =  2G NH3   1G N2 + 3G H2  G NH3 = G° NH3 + RT ln(P NH3 )  G N2 = G° N2 + RT ln(P N2 )  G H2 = G° H2 + RT ln(P H2 )

55. 55  G = 2[G° NH3 + RT ln(P NH3 )]  [G° N2 + RT ln(P N2 )] -3[G° H2 + RT ln(P H2 )]  G =(2G° NH3 - G° N2 -3G° H2 ) + RT[2 ln(P NH3 )- ln(P N2 )- 3ln(P H2 )]  G=  G° + RT[2 ln(P NH3 )- ln(P N2 )- 3ln(P H2 )]

56. 56 N 2 (g) + 3H 2 (g) → 2NH 3 (g)  G=  G° + RT[2ln(P NH3 )- ln(P N2 )- 3ln(P H2 )] 2lnP NH3 = lnP NH3 2 -3ln(P H2 )= ln(1/P H2 ) -ln(P N2 )= ln(1/P N2 ) Reaction quotient Q=K no shift Q>K left Q<K right 3

57. 57 Free Energy and Pressure  G =  G  + RT ln(Q)

58. 58 One method for synthesizing methanol (CH 3 OH) involves reacting carbon monoxide and hydrogen gases: CO(g) +2H 2 (g) → CH 3 OH(l) Calculate ΔG at 25°C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol.

59. 59  G =  G  + RT ln(Q)  G  f(CH3OH) = -166kJ  G  f(H2) = 0  G  f(CO) = -137kJ = -166 – (-137) – 0= -29kJ = -2.9x10 4 J ΔG°= -2.9x10 4 J/mol rxn. R= 8.3145J/K molT=298K Q=1/(P (co) )(P H2 2 )= 1/(5.0)(3.0 2 )= 2.2x10 -2 ΔG= -2.9x10 4 J/mol rxn + (8.3145J/K mol)(298K)ln(2.2x10 -2 ) = -38kJ/mol rxn **G is more negative than G° CO(g) +2H 2 (g) → CH 3 OH(l)

60. 60 We have seen that the formation of CH 3 OH is spontaneous. But will it really form in a flask under these conditions? -no! -it is true that 1mol of CH 3 OH has lower G than CO or H 2. However, when they are mixed under these conditions there is an even lower free energy available to this system than 1.0 mol of pure CH 3 OH. **The system can achieve lowest possible free energy by going to equilibrium, not going to completion.

61. 61 Figure 16.7: Schematic representations of balls rolling down two types of hills. Analogous to Phase change Chemical Change Lowest possible Free energy The system can achieve lowest possible free energy by going to equilibrium, not going to completion

62. 62 Free Energy and Equilibrium Equilibrium- when the forward and reverse reaction rates are equal. Thermodynamics view (equilibrium point) occurs at the lowest value of free energy available to the reaction system.

63. 63 A(g)↔ B(g) 1.0 mol of A placed in a reaction vessel at a pressure of 2.0 atm. Free energy of A = G A = G° A + RTln(P A ) Free energy of B = G B = G° B + RTln(P B ) Total free energy of system =G =G A + G B

64. Figure 16.8: (a) The initial free energies of A and B. (b) As A(g) changes to B(g), the free energy of A decreases and that of B increases. (c) Eventually, pressures of A and B are achieved such that G A = G B, the equilibrium position.

65. 65 Suppose, minimum free energy is reached when 75% of A has been changed to B. At this point, P A =.25 its original P. (.25)(2.0atm)=0.50atm P B = (0.75)(2.0atm)= 1.5atm Since this is the equilibrium position, we use P to find K.

66. 66 Figure 16.9: (a) The change in free energy to reach equilibrium, beginning with 1.0 mol A(g) at P A = 2.0 atm. (b) The change in free energy to reach equilibrium, beginning with 1.0 mol B(g) at P B = 2.0 atm. (c) The free energy profile for A(g) → B(g) in a system containing 1.0 mol (A plus B) at P TOTAL = 2.0 atm. Each point on the curve corresponds to the total free energy of the system for a given combination of A and B.

67. 67 Summary The reaction proceeds to the minimum G (equilibrium), which corresponds to the point where: G prod. = G react. ΔG = G prod – G react. = 0

68. 68 Free Energy and Equilibrium  G =  G   RT ln(Q) at equilibrium  G =0 and Q = K So  G =0 =  G  +RT ln(K)  G  =  RT ln(K) K = equilibrium constant

69. 69  G  =  RT ln(K)   G  =  The free energies of the reactants and products are equal when all components are in the standard states (1 atm for gases). The system is at equilibrium when the P react and prod. =1atm. K=1

70. 70 Case 2  G  < 0  G  = (  G  prod -  G  react ) is negative  G  prod <  G  react the system will adjust to the right to reach equilibrium. K>1 since the P prod. at equilibrium is >1atm and P react. <1atm

71. 71 Case 3  G   G  = (  G  prod -  G  react ) is positive  G  react <  G  prod the system will adjust to the left to reach equilibrium. K<1 since the P react. >1atm and P prod <1atm at equilibrium

72. 72

73. 73 Consider the ammonia synthesis reaction N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) Where ΔG°= -33.3kJ/mol of N 2 consumed at 25C. For each of the following mixtures of reactants and products at 25C, predict the direction in which the system will shift to reach equilibrium. a. P NH3 = 1.00atm, P N2 =1.47atm, P H2 =1.00x10 - 2 atm b. P NH3 = 1.00atm, P N2 =1.00atm, P H2 =1.00atm

74. 74 N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)  G =  G   RT ln(Q) P NH3 = 1.00atm, P N2 =1.47atm, P H2 =1.00x10 - 2 atm Q=(1.00) 2 /(1.47)(1.00x10 -2 ) 3 = 6.80x10 5 T=298K R = 8.3145 J/K mol ΔG° = -33.3kJ/mol ΔG= -3.33x10 4 J/mol + (8.3145 J/K mol ) (298K)(ln 6.80x10 5 ) = 0 At equilibrium no shift

75. 75 N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)  G =  G   RT ln(Q) P NH3 = 1.00atm, P N2 =1.00atm, P H2 =1.00atm Q=(1.00) 2 /(1.00)(1.00) 3 = 1.00atm T=298K R = 8.3145 J/K mol ΔG° = -33.3kJ/mol ΔG=  G   RT ln(1) =  G  ΔG=  G  kJ/mol Since it’s (-), shift right, K>1

76. 76 The overall reaction for the corrosion (rusting) of iron by oxygen is l 4Fe(s) + 3O 2 (g)↔ 2Fe 2 O 3 (s) l Using the following data, calculate the equilibrium constant for this reaction at 25C. SubstanceΔH f ° (kJ/mol)S° (J/K. Mol) Fe 2 O 3 -82690 Fe027 O2O2 0205

77. Copyright©2000 by Houghton Mifflin Company. All rights reserved.77 Temperature Dependence of K l y = mx + b y=ln(K), m=-ΔH/R=slope, xx=1/T, and b= ΔS/R=intercept ; will be linear if plotted (  H  and S   independent of temperature over a small temperature range) -

78. 78 Free Energy and Work using a chemical process to do work Maximum possible useful work obtainable from a process at constant T and P is equal to the change in free energy. w max = ΔG l ΔG for a spontaneous process: energy that is free to do useful work l ΔG for a non-spontaneous process: minimum amount of work that must be expended to make the process occur.

79. 79 l Knowing ∆G tells us about efficiency. l Amount of work actually obtained from a spontaneous process is always less than the max possible amount. l Energy is always wasted.

80. 80 Chemical change in a battery can do work by sending current to a starter motor. ( Hypothetical Reversible process) Current flowing through a wire: Energy is wasted as heat Force current in opposite direction To recharge battery

81. 81 Reversible v. Irreversible Processes l Reversible: The universe is exactly the same as it was before the cyclic process. l Irreversible: The universe is different after the cyclic process. l All real processes are irreversible -- (some work is changed to heat). Surroundings have less of an ability to do the work. Entropy is increasing!!!!

82. At 1500°C for the reaction CO(g) + 2H 2 (g) → CH 3 OH(g) the equilibrium constant is K p = 1.4 x 10 -7. Is  H° at this temperature: A. positive B. negative C. zero D. can not be determined

83. The standard free energy (  G rxn 0 for the reaction N 2(g) + 3H 2(g) → 2NH 3(g) is -32.9 kJ. Calculate the equilibrium constant for this reaction at 25 o C. A. 13.3 B. 5.8 x 10 5 C. 2.5 D. 4.0 x 10 -6 E. 9.1 x 10 8