2. Chapter 16 Spontaneity, Entropy and Free Energy

3. First Law of Thermodynamics You will recall from Chapter 6 that energy cannot be created nor destroyed. Therefore, the total energy of the universe is a constant. Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.

4. Spontaneous (Thermodynamically Favored) Processes Spontaneous (Thermodynamically favored) processes are those that can proceed without any outside force. Two driving forces for ALL chemical reactions are enthalpy and entropy. The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not act spontaneously

5. Spontaneous Processes Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.

6. Spontaneous Processes Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. Above 0  C it is spontaneous for ice to melt. Below 0  C the reverse process is spontaneous.

7. Reversible Processes In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process.

8. Irreversible Processes Irreversible processes cannot be undone by exactly reversing the change to the system. Spontaneous processes are irreversible, under identical conditions.

9. SAMPLE EXERCISE Identifying Spontaneous Processes Predict whether the following processes are spontaneous as described, spontaneous in the reverse direction, or in equilibrium: (a) When a piece of previously heated metal at 150°C is added to water at 40°C, the water gets hotter. (b) Water at room temperature decomposes into H 2 (g) and O 2 (g). (c) Benzene vapor, C 6 H 6 (g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1°C. Solve: (a) This process is spontaneous. (b) Experience tells us that this process is not spontaneous—we certainly have never seen hydrogen and oxygen gas bubbling up out of water! Rather, the reverse process—the reaction of H 2 and O 2 to form H 2 O—is spontaneous once initiated by a spark or flame. (c) By definition, the normal boiling point is the temperature at which a vapor at 1 atm is in equilibrium with its liquid. Thus, this is an equilibrium situation. Neither the condensation of benzene vapor nor the reverse process is spontaneous. If the temperature were below 80.1°C, condensation would be spontaneous.

10. SAMPLE EXERCISE Identifying Spontaneous Processes PRACTICE EXERCISE Under 1 atm pressure CO 2 (s) sublimes at –78°C. Is the transformation of CO 2 (s) to CO 2 (g) a spontaneous process at –100ºC and 1 atm pressure? at –63ºC and 1 atm pressure? Answer: No, the reverse process is spontaneous at this temperature. Yes at 63ºC and 1 atm pressure?

11. Entropy Entropy (symbol: S) is a term coined by Rudolph Clausius in the 19th century. Clausius described Entropy as the ratio of heat delivered and the temperature at which it is delivered, qTqT

12. Entropy Entropy can be thought of as a measure of the randomness of a system. It is related to the various modes of motion in molecules.

13. Entropy Like total energy, E, and enthalpy, H, entropy is a state function. Therefore,  S = S final  S initial

14. Entropy For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature:  S = q rev T Units: J/K ***Temperature in Kelvin

15. SAMPLE EXERCISE Calculating  S for a Phase Change The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is –38.9ºC, and its molar enthalpy of fusion is  fusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point?

16. SAMPLE EXERCISE Calculating  S for a Phase Change Solution Analyze: We first recognize that freezing is an exothermic process; heat is transferred from the system to the surroundings when a liquid freezes (q < 0). The enthalpy of fusion is  for the melting process. Because freezing is the reverse of melting, the enthalpy change that accompanies the freezing of 1 mol of Hg is –  fusion = 2.29 kJ/mol. Plan: We can use –  fusion and the atomic weight of Hg to calculate q for freezing 50.0 g of Hg: We can use this value of q as q rev in Equation. We must first, however, convert the temperature to K: Check: The entropy change is negative because heat flows from the system making q rev negative. Comment: The procedure we have used here can be used to calculate  S for other isothermal phase changes, such as the vaporization of a liquid at its boiling point. Solve: We can now calculate the value of  S sys :

17. SAMPLE EXERCISE continued Answer: –163 J/K PRACTICE EXERCISE The normal boiling point of ethanol, C 2 H 5 OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy in the system when 68.3 g of C 2 H 5 OH(g) at 1 atm condenses to liquid at the normal boiling point?

18. Second Law of Thermodynamics The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes, and the entropy of the universe does not change for reversible processes.

19. Second Law of Thermodynamics In other words: For reversible processes:  S univ =  S system +  S surroundings = 0 For irreversible processes:  S univ =  S system +  S surroundings > 0

20. Second Law of Thermodynamics These last truths mean that as a result of all spontaneous processes the entropy of the universe constantly increases.

21. 1.The enthalpy of the system must increase by a greater amount than the entropy of the system decreases. 2.The entropy of the universe for the process must increase by the same amount as the entropy of the system decreases. 3.The entropy of the universe for the process must increase by a greater amount than the entropy of the system decreases. 4.The entropy of the universe for the process must decrease by a greater amount than the entropy of the system decreases.

22. 1.The enthalpy of the system must increase by a greater amount than the entropy of the system decreases. 2.The entropy of the universe for the process must increase by the same amount as the entropy of the system decreases. 3.The entropy of the universe for the process must increase by a greater amount than the entropy of the system decreases. 4.The entropy of the universe for the process must decrease by a greater amount than the entropy of the system decreases.

23. Entropy on the Molecular Scale Ludwig Boltzmann described the concept of entropy on the molecular level. Temperature is a measure of the average kinetic energy of the molecules in a sample.

24. Entropy on the Molecular Scale Molecules exhibit several types of motion:  Translational: Movement of the entire molecule from one place to another.  Vibrational: Periodic motion of atoms within a molecule.  Rotational: Rotation of the molecule on about an axis or rotation about  bonds.

25. Entropy on the Molecular Scale Boltzmann envisioned the motions of a sample of molecules at a particular instant in time.  This would be akin to taking a snapshot of all the molecules. He referred to this sampling as a microstate of the thermodynamic system.

26. 1.Molecules and single atoms can experience all the same types of motion. 2.A molecule can vibrate (atoms moving relative to one another) and rotate (tumble); a single atom can do neither 3.A molecule can vibrate (atoms moving relative to one another) and rotate (tumble); a single atom can only rotate. 4.A molecule can translationally move and rotate (tumble); a single atom can do neither.

27. 1.Molecules and single atoms can experience all the same types of motion. 2.A molecule can vibrate (atoms moving relative to one another) and rotate (tumble); a single atom can do neither 3.A molecule can vibrate (atoms moving relative to one another) and rotate (tumble); a single atom can only rotate. 4.A molecule can translationally move and rotate (tumble); a single atom can do neither.

28. Entropy on the Molecular Scale Each thermodynamic state has a specific number of microstates, W, associated with it. Entropy is S = k lnW where k is the Boltzmann constant, 1.38  10  23 J/K.

29. Entropy on the Molecular Scale The change in entropy for a process, then, is  S = k lnW final  k lnW initial W final W initial  S = k ln Entropy increases with the number of microstates in the system.

30. 1.S = 0 2.S < H 3.S = W 4.S > H

31. 1.S = 0 2.S < H 3.S = W 4.S > H

32. Entropy on the Molecular Scale The number of microstates and, therefore, the entropy tends to increase with increases in  Temperature. more energy = more movement  Volume. more space = more possible positions  The number of independently moving molecules. more particles = more possible positions

33. Entropy and Physical States Entropy increases with the freedom of motion of molecules. Therefore, S (g) > S (l) > S (s)

34. Solutions Generally, when a solid is dissolved in a solvent, entropy increases.

35. Entropy Changes (Reactions) In general, entropy increases when  Gases are formed from liquids and solids.  Liquids or solutions are formed from solids.  The number of gas molecules increases.  The number of moles increases.

36. Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0

37. 1.It is a substance at the triple point. 2.It is pure liquid at 0°K (absolute zero). 3.It is an elemental standard state. 4.It must be a perfect crystal at 0°K (absolute zero).

38. 1.It is a substance at the triple point. 2.It is pure liquid at 0K (absolute zero). 3.It is an elemental standard state. 4.It must be a perfect crystal at 0K (absolute zero).

39. Standard Entropies These are molar entropy values of substances in their standard states. Standard entropies tend to increase with increasing molar mass. WHY?

40. Standard Entropies Larger and more complex molecules have greater entropies.

41. Entropy Changes Entropy changes for a reaction can be estimated in a manner analogous to that by which  H is estimated:  S° =  n  S° (products) -  m  S° (reactants) where n and m are the coefficients (mol) in the balanced chemical equation.

42. Entropy Changes in Surroundings Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process:  S surr =  q sys T At constant pressure, q sys is simply  H  for the system.

43. Determining ΔS surr In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore. For example, iron is used to reduce antimony in sulfide ores: Sb 2 S 3(s) +3Fe (s)  2Sb (s) + 3FeS (s) ΔH = -125 kJ Carbon is used as the reducing agent for oxide ores: Sb 4 O (s) +6C (s)  4Sb (s) +6CO (g) ΔH = 778 kJ

44. Practice: Calculate ΔS surr for the following reactions at 25 o C and 1 atm. Indicate where entropy of surroundings is increasing or decreasing. 1. Combustion of propane, C 3 H 8, where enthalpy of formation is -2221 kJ 2. Decomposition of Nitrogen dioxide into nitrogen monoxide and oxygen where enthalpy of formation is 112 kJ

45. Entropy Change in the Universe The universe is composed of the system and the surroundings. Therefore,  S universe =  S system +  S surroundings For spontaneous processes  S universe > 0

46. Entropy Change in the Universe This becomes:  S universe =  S system + Multiplying both sides by  T,  T  S universe =  H system  T  S system  H system T

47. 1.always increase 2.always decrease 3.sometimes increase and sometimes decrease, depending on the process

48. 1.always increase 2.always decrease 3.sometimes increase and sometimes decrease, depending on the process

49. Determine ΔS univ for the combustion of 2.5 moles of glucose.

50. Gibbs Free Energy Defined as a thermodynamic equation equal to the enthalpy (of a system or process) minus the product of the entropy at a given temperature. Used to predict thermodynamic favorability of a reaction. From previously derived equation,  T  S universe =  system  T  S system.  T  S universe is defined as the Gibbs free energy,  G. When  S universe is positive,  G is negative. Therefore, when  G is negative, a process is spontaneous or thermodynamically favorable.

51. Gibbs Free Energy 1.If  G is negative, the forward reaction is spontaneous. 2.If  G is 0, the system is at equilibrium. 3.If  G is positive, the reaction is spontaneous in the reverse direction.

52. 1.Entropy of universe increases and free energy of the system decreases. 2.Entropy of system decreases and free energy of the universe increases. 3.Entropy of system increases and free energy of the universe decreases. 4.Entropy of universe decreases and free energy of the system increases.

53. 1.Entropy of universe increases and free energy of the system decreases. 2.Entropy of system decreases and free energy of the universe increases. 3.Entropy of system increases and free energy of the universe decreases. 4.Entropy of universe decreases and free energy of the system increases.

54. Standard Free Energy Changes Analogous to standard enthalpies of formation are standard free energies of formation,  G . f  G  =  n  G  (products)   n  G  (reactants) f f where n is the stoichiometric coefficients.

55. 1.It indicates the process is spontaneous under standard conditions. 2.It indicates the process has taken place under standard conditions. 3.It indicates the process has taken place at 273K and 1 barr. 4.It indicates the process has taken place at 1 atm and 0K.

56. 1.It indicates the process is spontaneous under standard conditions. 2.It indicates the process has taken place under standard conditions. 3.It indicates the process has taken place at 273K and 1 barr. 4.It indicates the process has taken place at 1 atm and 0K.

57. Free Energy Changes At temperature of 25 o C, use appendix 4 At temperatures other than 25°C,  G° =  H   T  S  How does  G  change with temperature?

58. Free Energy and Temperature How does  G  change with temperature? There are two parts to the free energy equation:   H  : the enthalpy term  T  S  : the entropy term The temperature dependence of free energy, then comes from the entropy term.

59. Free Energy and Temperature

60. 1.  H = T  S 2.  H < T  S 3.  H > T  S 4.We cannot determine without additional information.

61. 1.  H = T  S 2.  H < T  S 3.  H > T  S 4.We cannot determine without additional information.

62. Free Energy and Equilibrium Under any conditions, standard or nonstandard, the free energy change can be found this way:  G =  G  + RT lnQ (Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.)

63. Free Energy and Equilibrium At equilibrium, Q = K, and  G = 0. The equation becomes 0 =  G  + RT lnK Rearranging, this becomes  G  =  RT lnK or, K = e  G  /RT