1. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 1 Thermodynamics: Spontaneity, Entropy, and Free Energy Chapter Seventeen

2. General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 2 Thermodynamics examines the relationship between heat and work. Spontaneity is the notion of whether or not a process can take place unassisted. Entropy is a mathematical concept describing the distribution of energy within a system. Free energy is a thermodynamic function that relates enthalpy and entropy to spontaneity, and can also be related to equilibrium constants. Introduction

3. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 3 With a knowledge of thermodynamics and by making a few calculations before embarking on a new venture, scientists and engineers can save themselves a great deal of time, money, and frustration. –“To the manufacturing chemist thermodynamics gives information concerning the stability of his substances, the yield which he may hope to attain, the methods of avoiding undesirable substances, the optimum range of temperature and pressure, the proper choice of solvent.…” - from the introduction to Thermodynamics and the Free Energy of Chemical Substances by G. N. Lewis and M. Randall Thermodynamics tells us what processes are possible. –(Kinetics tells us whether the process is practical.) Why Study Thermodynamics?

4. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 4 A spontaneous process is one that can occur in a system left to itself; no action from outside the system is necessary to bring it about. A nonspontaneous process is one that cannot take place in a system left to itself. If a process is spontaneous, the reverse process is nonspontaneous, and vice versa. Example: gasoline combines spontaneously with oxygen. However, “spontaneous” signifies nothing about how fast a process occurs. A mixture of gasoline and oxygen may remain unreacted for years, or may ignite instantly with a spark. Spontaneous Change

5. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 5 Thermodynamics determines the equilibrium state of a system. Thermodynamics is used to predict the proportions of products and reactants at equilibrium. Kinetics determines the pathway by which equilibrium is reached. A high activation energy can effectively block a reaction that is thermodynamically favored. Example: combustion reactions are thermodynamically favored, but (fortunately for life on Earth!) most such reactions also have a high activation energy. Spontaneous Change (cont’d)

6. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 6 Example 17.1 Indicate whether each of the following processes is spontaneous or nonspontaneous. Comment on cases where a clear determination cannot be made. (a)The action of toilet bowl cleaner, HCl(aq), on “lime” deposits, CaCO 3 (s). (b)The boiling of water at normal atmospheric pressure and 65 °C. (c)The reaction of N 2 (g) and O 2 (g) to form NO(g) at room temperature. (d)The melting of an ice cube.

7. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 7 Example 17.1 Indicate whether each of the following processes is spontaneous or nonspontaneous. Comment on cases where a clear determination cannot be made. (a) The action of toilet bowl cleaner, HCl(aq), on “lime” deposits, CaCO 3 (s). (b) The boiling of water at normal atmospheric pressure and 65 °C. (c) The reaction of N 2 (g) and O 2 (g) to form NO(g) at room temperature. (d) The melting of an ice cube. Solution (a) When we add the acid, the fizzing that occurs—the escape of a gas—indicates that the reaction occurs without any further action on our part. The net ionic equation is CaCO 3 (s) + 2 H 3 O + (aq)  Ca 2+ (aq) + 3 H 2 O(l) + CO 2 (g) The reaction is spontaneous. (b) We know that the normal boiling point of a liquid is the temperature at which the vapor pressure is equal to 1 atm. For water, this temperature is 100 °C. Thus, the boiling of water at 65 °C and 1 atm pressure is nonspontaneous and not possible. (c) Nitrogen and oxygen gases occur mixed in air, and there is no evidence of their reaction to form toxic NO at room temperature. This makes their reaction to form NO(g) appear to be nonspontaneous. However, this could be an example of a spontaneous reaction that occurs extremely slowly, like that of H 2 and O 2 to form H 2 O. We need additional criteria before we can answer this question. (d) We know that at 1 atm pressure ice melts spontaneously at temperatures above its normal melting point of 0 °C. Below this temperature, it does not. To answer this question, we would have to know the temperature.

8. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 8 Example 17.1 continued Indicate whether each of the following processes is spontaneous or nonspontaneous. Comment on cases where a clear determination cannot be made. (a)The decay of a piece of wood buried in soil. (b)The formation of sodium, Na(s), and chlorine, Cl 2 (g), in a vigorously stirred aqueous solution of sodium chloride, NaCl(aq). (c)The ionization of hydrogen chloride when HCl(g) dissolves in liquid water. Exercise 17.1A Indicate whether each of the following processes is spontaneous or nonspontaneous. Comment on cases where a clear determination cannot be made. (a)The dissolution of 1.00 mL of liquid ethanol in 100 mL of liquid water. (b)The formation of lime, CaO(s), and carbon dioxide gas at 1 atm pressure from limestone, CaCO 3 (s), at 600 °C. (c)The condensation of CO 2 (g) at 0.50 atm to produce CO 2 (s). (d)The displacement of H 2 (g) at 1 atm from 1 M HCl(aq) by Cu(s). Exercise 17.1B

9. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 9 Early chemists proposed that spontaneous chemical reactions should occur in the direction of decreasing energy. It is true that many exothermic processes are spontaneous and that many endothermic reactions are nonspontaneous. However, enthalpy change is not a sufficient criterion for predicting spontaneous change … Spontaneous Change (cont’d)

10. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 10 Spontaneous Change (cont’d) Water falling (higher to lower potential energy) is a spontaneous process. H 2 and O 2 combine spontaneously to form water (exothermic) BUT … … liquid water vaporizes spontaneously at room temperature; an endothermic process. Conclusion: enthalpy alone is not a sufficient criterion for prediction of spontaneity.

11. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 11 The Concept of Entropy When the valve is opened … … the gases mix spontaneously. There is no significant enthalpy change. Intermolecular forces are negligible. So … why do the gases mix?

12. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 12 The other factor that drives reactions is a thermodynamic quantity called entropy. Entropy is a mathematical concept that is difficult to portray visually. The total energy of the system remains unchanged in the mixing of the gases … The Concept of Entropy (cont’d) … but the number of possibilities for the distribution of that energy increases.

13. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 13 Benzene and toluene have similar intermolecular forces, so there is no enthalpy change when they are mixed. They mix completely because entropy of the mixture is higher than the entropies of the two substances separated. Formation of an Ideal Solution

14. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 14 Increase in Entropy in the Vaporization of Water Evaporation is spontaneous because of the increase in entropy.

15. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 15 The spreading of the energy among states, and increase of entropy, often correspond to a greater physical disorder at the microscopic level (however, entropy is not “disorder”). There are two driving forces behind spontaneous processes: the tendency to achieve a lower energy state (enthalpy change) and the tendency for energy to be distributed among states (entropy). In many cases, however, the two factors work in opposition. One may increase and the other decrease or vice versa. In these cases, we must determine which factor predominates. The Concept of Entropy

16. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 16 The difference in entropy (S) between two states is the entropy change (  S). The greater the number of configurations of the microscopic particles (atoms, ions, molecules) among the energy levels in a particular state of a system, the greater is the entropy of the system. Entropy generally increases when: –Solids melt to form liquids. –Solids or liquids vaporize to form gases. –Solids or liquids dissolve in a solvent to form nonelectrolyte solutions. –A chemical reaction produces an increase in the number of molecules of gases. –A substance is heated. Assessing Entropy Change

17. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 17 Example 17.2 Predict whether each of the following leads to an increase or decrease in the entropy of a system. If in doubt, explain why. (a) The synthesis of ammonia: N 2 (g) + 3 H 2 (g)  2 NH 3 (g) (b) Preparation of a sucrose solution: C 12 H 22 O 11 (s) C 12 H 22 O 11 (aq) (c) Evaporation to dryness of a solution of urea, CO(NH 2 ) 2, in water: CO(NH 2 ) 2 (aq)  CO(NH 2 ) 2 (s) H 2 O(l)

18. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 18 Solution (a)Four moles of gaseous reactants produce two moles of gaseous product. Because the reaction leads to a decrease in the number of gas molecules, we predict a decrease in entropy. (b)The sucrose molecules are confined in highly ordered crystals in the solid state, whereas they become widely distributed in the aqueous solution. The number of available energy levels for dispersal of the system’s energy increases, and we therefore predict an increase in entropy. (c)Crystallization of the urea from an aqueous solution is the reverse of the process described in part (b), thus indicating a decrease in entropy. At the same time, however, in the evaporation of water a liquid is converted to a gas, suggesting an increase in entropy (recall Figure 17.4). Without further information, we cannot say whether the entropy of the system increases or decreases. Example 17.2 Predict whether each of the following leads to an increase or decrease in the entropy of a system. If in doubt, explain why. (a) The synthesis of ammonia: (b) Preparation of a sucrose solution: (c) Evaporation to dryness of a solution of urea, CO(NH 2 ) 2, in water:

19. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 19 Sometimes it is necessary to obtain quantitative values of entropy changes.  S = q rxn /T where q rxn is reversible heat, a state function. Entropy Change A reversible process can be reversed by a very small change, as in the expansion of this gas. A reversible process is never more than a tiny step from equilibrium. The expansion can be reversed by allowing the sand to return, one grain at a time.

20. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 20 Entropy as a Function of Temperature Entropy always increases with temperature … … and it increases dramatically during a phase change.

21. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 21 According to the Third Law of Thermodynamics, the entropy of a pure, perfect crystal can be taken to be zero at 0 K. The standard molar entropy, S°, is the entropy of one mole of a substance in its standard state. Since entropy increases with temperature, standard molar entropies are positive—even for elements.  S =  v p S°(products) –  v r S°(reactants) Standard Molar Entropies Does the form of this equation look familiar? (remember calculating enthalpy change from ∆H f °?)

22. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 22 Example 17.3 Use data from Appendix C to calculate the standard entropy change at 25 °C for the Deacon process, a high-temperature, catalyzed reaction used to convert hydrogen chloride (a by-product from organic chlorination reactions) into chlorine: 4 HCl(g) + O 2 (g)  2 Cl 2 (g) + 2 H 2 O(g)

23. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 23 Example 17.3 Use data from Appendix C to calculate the standard entropy change at 25 °C for the Deacon process, a high-temperature, catalyzed reaction used to convert hydrogen chloride (a by-product from organic chlorination reactions) into chlorine: 4 HCl(g) + O 2 (g)  2 Cl 2 (g) + 2 H 2 O(g) Strategy This problem requires a straightforward application of Equation (17.2), keeping in mind that we should expect a decrease in entropy ( ∆ S° < 0) because five moles of gaseous reactants yield four moles of gaseous products. Solution We relate ∆ S° for the reaction to the standard molar entropies, S°, of reactants and products as follows:

24. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 24 Example 17.3 continued Assessment Just as we expected, ∆ S° is indeed negative; entropy decreases. Use data from Appendix C to calculate the standard molar entropy change at 25 °C for the reaction CO(g) + H 2 O(g)  CO 2 (g) + H 2 (g) Note that this is reaction (c) in Exercise 17.2A, for which your answer should have been that you could not predict whether entropy increases or decreases. Exercise 17.3A The following reaction may occur in the high-temperature, catalyzed oxidation of NH 3 (g). Use data from Appendix C to calculate the standard molar entropy change for this reaction at 25 °C. NH 3 (g) + O 2 (g)  N 2 (g) + H 2 O(g) (not balanced) Exercise 17.3B

25. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 25 Entropy can be used as a sole criterion for spontaneous change … … but the entropy change of both the system and its surroundings must be considered. The Second Law of Thermodynamics establishes that all spontaneous processes increase the entropy of the universe (system and surroundings). If entropy increases in both the system and the surroundings, the process is spontaneous. –Is it possible for a spontaneous process to exhibit a decrease in entropy? Yes, if the surroundings ____________________. The Second Law of Thermodynamics

26. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 26 Therefore: Since heat is exchanged with the surroundings: For a spontaneous process: Multiply by –T: The Second Law of Thermodynamics ∆S total = ∆ S universe = ∆ S system + ∆ S surroundings ∆ S universe > 0 q surr = –q p = ∆ H sys q surr –∆H sys ∆ S surr = –––– = –––––– T T and: ∆H sys ∆ S univ = ∆ S sys – –––––– T –T ∆ S univ = ∆ H sys – T ∆ S sys

27. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 27 What is the significance of: –T ∆ S univ = ∆ H sys – T ∆ S sys ? The entropy change of the universe—our criterion for spontaneity—has now been defined entirely in terms of the system. The quantity –T ∆ S univ is called the free energy change (  G). For a process at constant temperature and pressure:  G sys =  H sys – T  S sys Free Energy and Free Energy Change

28. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 28 If  G < 0 (negative), a process is spontaneous. If  G > 0 (positive), a process is nonspontaneous. If  G = 0, neither the forward nor the reverse process is favored; there is no net change, and the process is at equilibrium. Free Energy and Free Energy Change

29. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 29 Case 3 illustrated ∆ H is (+) and is more- or-less constant with T. Since ∆ S is (+), the slope T ∆ S is also (+). At low T, the size of T ∆ S is small, and ∆ H (+) predominates. At high T, the size of T ∆ S is large, and –T ∆ S predominates.

30. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 30 Example 17.4 Predict which of the four cases in Table 17.1 you expect to apply to the following reactions. (a) C 6 H 12 O 6 (s) + 6 O 2 (g)  6 CO 2 (g) + 6 H 2 O(g)  H = -2540kJ (b) Cl 2 (g)  2 Cl(g) Example 17.5 A Conceptual Example Molecules exist from 0 K to a few thousand kelvins. At elevated temperatures, they dissociate into atoms. Use the relationship between enthalpy and entropy to explain why this is to be expected.

31. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 31 Example 17.4 Predict which of the four cases in Table 17.1 you expect to apply to the following reactions. (a) C 6 H 12 O 6 (s) + 6 O 2 (g)  6 CO 2 (g) + 6 H 2 O(g) ∆ H = –2540 kJ (b) Cl 2 (g)  2 Cl(g) Solution (a)The reaction is exothermic ( ∆ H 0. Because ∆ H < 0 and ∆ S > 0, we expect this reaction to be spontaneous at all temperatures (case 1 of Table 17.1). (b)One mole of gaseous reactant produces two moles of gaseous product, and therefore we expect ∆ S > 0. No value is given for ∆ H. However, we can determine the sign of ∆ H by examining what occurs during the reaction. Bonds are broken, but no new bonds are formed. Because molecules must absorb energy for bonds to break, the reaction must be endothermic. Thus, ∆ H must be positive. Because ∆ H and ∆ S are both positive, this reaction is an example of case 3 in Table 17.1. We expect the reaction to be nonspontaneous at low temperatures and spontaneous at high temperatures.

32. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 32 Example 17.4 continued Predict which of the four cases in Table 17.1 is likely to apply to each of the following reactions. (a) N 2 (g) + 2 F 2 (g)  N 2 F 4 (g) ∆ H = –7.1 kJ (b) COCl 2 (g)  CO(g) + Cl 2 (g) ∆ H = +110.4 kJ Exercise 17.4A Predict which of the four cases in Table 17.1 is likely to apply to the following reaction. Use the value –93.85 kJ/mol for as necessary. Br 2 (g) + BrF 3 (g)  3 BrF(g) Exercise 17.4B for BrF(g) and additional data from Appendix C

33. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 33 The standard free energy change,  G°, of a reaction is the free energy change when reactants and products are in their standard states. The standard free energy of formation,  G f °, is the free energy change for the formation of 1 mol of a substance in its standard state from the elements in their standard states.  G° =  v p  G f °(products) –  v r  G f °(reactants) Standard Free Energy Change, ∆G° The form of this equation should appear very familiar by now!

34. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 34 Example 17.6 Calculate  G at 298 K for the reaction 4 HCl(g) + O 2 (g)  2 Cl 2 (g) + 2 H 2 O(g) ∆H° = -114.4 kJ (a) using the Gibbs equation (17.8) and (b) from standard free energies of formation.

35. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 35 Example 17.6 Calculate ∆ G° at 298 K for the reaction 4 HCl(g) + O 2 (g)  2 Cl 2 (g) + 2 H 2 O(g) ∆ H° = –114.4 kJ (a) using the Gibbs equation (17.8) and (b) from standard free energies of formation. Strategy (a)To use the standard-state form of the Gibbs equation, we need values of both ∆ H° and ∆ S°. The value of ∆ H° is given, and we can obtain ∆ S° from standard molar entropies, which we did in Example 17.3. (b)We must look up standard free energies of formation in Appendix C and use them in Equation (17.9). Solution (a)The value of ∆ S° we determined in Example 17.3 was –128.8 J K –1. To express ∆ H° and ∆ S° in the same energy unit, we can convert –128.8 J K –1 to –0.1288 kJ K –1. Then ∆ G° = ∆ H° – T ∆ S° = –114.4 kJ – [298 K x (–0.1288 kJ K –1 )] = –114.4 kJ + 38.4 kJ = –76.0 kJ (b)Equation (17.9) takes the form

36. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 36 Example 17.6 continued Assessment The principal check on part (a) is that the T ∆ S° product should be about 40 kJ (that is, 300 x 0.13), which, when added to the value of ∆ H°, should yield a value of ∆ G° ≈ –75 kJ. The principal check on part (b) is that it should give the same answer as part (a), and it does. Use the standard-state form of the Gibbs equation to determine the standard free energy change at 25 °C for these reactions: (a) 2 NO(g) + O 2 (g)  2 NO 2 (g) ∆ H° = –114.1 kJ ∆ S° = –146.2 J K –1 (b) 2 CO(g) + 2 NH 3 (g)  C 2 H 6 (g) + 2 NO(g) ∆ H° = +409.0 kJ ∆ S° = –129.1 J K –1 Exercise 17.6A Use standard free energies of formation from Appendix C to determine the standard free energy change at 25 °C for these reactions: (a) CS 2 (l) + 2 S 2 Cl 2 (g)  CCl 4 (l) + 6 S(s) (b) NH 3 (g) + O 2 (g)  N 2 (g) + H 2 O(g) (not balanced) Exercise 17.6B

37. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 37 At equilibrium,  G = 0 (reaction is neither spontaneous nor nonspontaneous). Therefore, at the equilibrium temperature, the free energy change expression becomes:  H = T  Sand  S =  H/T Trouton’s rule states that the entropy change is about the same when one mole of a substance is converted from liquid to vapor (at the normal boiling point).  S° vapn for many substances is about 87 J mol –1 K –1. This rule works best with nonpolar substances. It generally fails for liquids with a more ordered structure, such as those with extensive hydrogen bonding. Free Energy Change and Equilibrium

38. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 38 Illustrating Trouton’s Rule The three substances have different entropies and different boiling points, but  S of vaporization is about the same for all three.

39. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 39 Example 17.7 At its normal boiling point, the enthalpy of vaporization of pentadecane, CH 3 (CH 2 ) 13 CH 3, is 49.45 kJ/mol. What should its approximate normal boiling point temperature be?

40. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 40 Example 17.7 At its normal boiling point, the enthalpy of vaporization of pentadecane, CH 3 (CH 2 ) 13 CH 3, is 49.45 kJ/mol. What should its approximate normal boiling point temperature be? Strategy Trouton’s rule provides an approximate value (87 J mol –1 K –1 ) for the molar entropy of vaporization of a nonpolar liquid at its normal boiling point. Pentadecane, an alkane, is a nonpolar liquid. When we apply Equation (17.10), we will solve for the boiling point, T bp. Solution Assessment To test the validity of our estimate, we could consult a chemistry handbook. We would find the experimentally determined boiling point of pentadecane to be 543.8 K. Our estimate is within about 5% of the true value.

41. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 41 Raoult’s Law Revisited Entropy of vaporization of the solvent is about the same in each case, which means … A pure solvent has a lower entropy than a solution containing the solvent. … higher entropy for the vapor from the solution than from the pure solvent. Entropy of a vapor increases if the vapor expands into a larger volume—lower vapor pressure.

42. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 42  G = 0 is a criterion for equilibrium at any temperature.  G° = 0 is a criterion for equilibrium at a single temperature, that temperature at which the equilibrium state has all reactants and products in their standard states.  G and  G o are related through the reaction quotient, Q:  G =  G° + RT ln Q When  G = 0, then Q = K eq, and the equation becomes:  G° =  RT ln K eq Relationship of  G° to the Equilibrium Constant, K eq

43. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 43 The concentrations and partial pressures we have used in K eq are approximations. Activities (a) are the correct variables for K eq. But activities are very difficult to determine. That is why we use approximations. For pure solid and liquid phases: a = 1. For gases: Assume ideal gas behavior, and replace the activity by the numerical value of the gas partial pressure (in atm). For solutes in aqueous solution: Replace solute activity by the numerical value of the solute molarity (M). The Equilibrium Constant, K eq

44. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 44 Example 17.8 Write the equilibrium constant expression, K eq, for the oxidation of chloride ion by manganese dioxide in an acidic solution: MnO 2 (s) + 4 H + (aq) + 2 Cl – (aq)  Mn 2+ (aq) + Cl 2 (g) + 2 H 2 O(l)

45. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 45 Strategy First, we can write the K eq expression in terms of activities (a), and then we can substitute the numerical values of the appropriate quantities for these activities in accordance with the conventions just introduced. Example 17.8 Write the equilibrium constant expression, K eq, for the oxidation of chloride ion by manganese dioxide in an acidic solution: Solution We start with the equation In substituting for these activities according to our conventions, 1. No term will appear for MnO 2 (s) because it is a pure solid phase with a = 1. 2. For Mn 2+, H +, and Cl –, ionic species in dilute aqueous solution, activities are replaced by the numerical values of the molarities. 3. For Cl 2 (g), the activity is replaced by the numerical value of the partial pressure. 4. Because H 2 O is the preponderant species in the dilute aqueous solution, the activity of H 2 O is essentially the same as in pure liquid water, which means that a = 1. Making these substitutions, we get

46. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 46 Rearranging Equation (17.12): Calculating Equilibrium Constants  G° ln K eq =  ––––– RT  G° =  RT ln K eq The units of  G° and R must be consistent; both must use kJ or both must use J.

47. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 47 Example 17.9 Determine the value of K eq at 25 °C for the reaction 2 NO 2 (g) N 2 O 4 (g)

48. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 48 Strategy Our first task is to obtain a value of ∆ G°, which we can do from tabulated standard free energies of formation in Appendix C. Then we can calculate K eq using Equation (17.12). Example 17.9 Determine the value of K eq at 25 °C for the reaction Solution The standard free energies of formation of N 2 O 4 (g) and NO 2 (g) from Appendix C are substituted into Equation (17.9). Next we list the data required for Equation (17.12), first changing the unit of ∆ G° from kJ to J/mol. Then we can rearrange Equation (17.12) to solve for K eq.

49. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 49 Example 17.9 continued Assessment In this problem, the value of ∆ G° is negative, and so ln K eq must be greater than 1. If we had made a rather common error and forgotten about the minus sign on the right in Equation (17.12), the value of K eq would have been less than 1 (that is, e –1.93 = 0.145)—a tip-off that an error had been made. Use data from Appendix C to determine K eq at 25 °C for the reaction Exercise 17.9A Use data from Appendix C to determine K eq at 25 °C for the reaction If NO(g) at 1 atm is allowed to react with an excess of Br 2 (l), what will be the partial pressures of NO(g) and NOBr(g) at equilibrium? Exercise 17.9B

50. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 50 The Sign and Magnitude of  G° Large, negative  G° ; equilibrium lies far to right. Large, positive  G° ; equilibrium lies far to left. Intermediate  G° ; equilibrium lies in intermediate position.

51. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 51 To obtain equilibrium constants at different temperatures, it will be assumed that  H° does not change much with temperature. To obtain K eq at the desired temperature, the van’t Hoff equation is used: The Dependence of  G° and K eq on Temperature  H° ln K eq = ––––– + constant or RT K 2  H° 1 1 ln ––– = –––– ––– – ––– K 1 R T 1 T 2 [ ] The form used depends on whether we have a single value of K eq available, or multiple values.

52. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 52 Example 17.10 Consider this reaction at 298 K: CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) ∆H° 298 = – 41.2 kJ Determine K eq for the reaction at 725 K.

53. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 53 Strategy This is the reaction for which K eq is plotted as a function of 1/T in Figure 17.13. We cannot simply use the graph to obtain ln K eq when 1/T = 1/725 K because this point falls outside the range of data plotted. However, we can select a data point listed in the caption of Figure 17.13 as our K 1 and T 1 for use in Equation (17.14). The other required data are given in the statement of the problem. Example 17.10 Consider this reaction at 298 K: Determine K eq for the reaction at 725 K. Solution Suppose these are the data we choose for the van’t Hoff equation:

54. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 54 Solution continued Example 17.10 continued The solution of Equation (17.14) forK 2 then follows: Assessment Notice that in the data table accompanying Figure 17.13, the value of K eq becomes progressively smaller as the temperature increases. Because T 2 is greater than the highest temperature in the table, K 2 must be somewhat less than 16. The calculated value of 6.6 seems reasonable. Another point to notice in this problem is that in order to avoid an inadvertent loss of significant figures, it is best to store the value of (1/T 1 – 1/T 2 ) in your calculator rather than to write intermediate results, such as the 1.56 x 10 –3 and 1.38 x 10 –3 shown in the solution.

55. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 55 Cumulative Example Waste silver from photographic solutions or laboratory operations can be recovered using an appropriate redox reaction. A 100.0-mL sample of silver waste is 0.200 M in Ag + (aq) and 0.0200 M in Fe 3+ (aq). Enough iron(II) sulfate is added to make the solution 0.200 M in Fe 2+ (aq). When equilibrium is established at 25 °C, how many moles of solid silver will be present? Ag + (aq) + Fe 2+ (aq)  Ag(s) + Fe 3+ (aq)

56. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 56 Strategy The Ag(s) must come at the expense of some of the Ag + (aq) originally present. We can determine [Ag + ] at equilibrium, and the amount of Ag(s) that forms will be related to the difference [Ag + ] init – [Ag + ] equil. The core of the equilibrium calculation will be the ICE tabular format introduced in Chapter 14, but first we need an equilibrium constant for the reaction. Because this information is not given, we must turn to Equation (17.12) relating ln K eq to ∆ G° for the reaction. We can establish ∆ G° from data in Appendix C. In writing out the solution to this problem, we will reverse the order of the steps described here. Solution To determine ∆ G°, we write the appropriate data from Appendix C under the chemical equation and then apply Equation (17.9). Cumulative Example Waste silver from photographic solutions or laboratory operations can be recovered using an appropriate redox reaction. A 100.0-mL sample of silver waste is 0.200 M in Ag + (aq) and 0.0200 M in Fe 3+ (aq). Enough iron(II) sulfate is added to make the solution 0.200 M in Fe 2+( aq). When equilibrium is established at 25 °C, how many moles of solid silver will be present?

57. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 57 Cumulative Example continued Solution continued Then we can use this value of ∆ G° in Equation (17.12) to solve for ln K eq. We solve for K eq by taking the exponential of both sides of the previous expression. We could establish the direction of net change by calculating Q c, but that is not necessary. Because there is no silver metal initially, the reaction must proceed to the right to form silver metal. Now, we enter the solution concentration data into the ICE format.

58. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 58 Cumulative Example continued Assessment In choosing the correct root of the quadratic equation, we were guided by the fact that x < 0.200. That is, [Ag + ] at equilibrium is (0.200 – x) M, and this value must be a positive quantity. The rejected root had x = 0.6. We can verify that the equilibrium concentrations are correct by showing that Q c based on equilibrium concentrations is equal to the value of K eq used in the calculation. Here, we obtain Q c = 0.08/(0.14) 2 = 4. This matches K eq = 3.3 quite well, given the rather limited precision of the calculation. The limitation in precision was introduced in calculating ∆ G°, where we had to work with the difference between two numbers of nearly equal magnitude.

59. Chapter Seventeen General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 59 Cumulative Example continued Solution continued Next we substitute the equilibrium concentrations from the ICE table into the K eq expression and express the equation in the customary quadratic format. We now solve the quadratic equation for x, keeping in mind that x must be less than 0.200. Then we determine the equilibrium concentrations of the ions. Finally, we can determine the number of moles of Ag(s) displaced from solution and appearing as Ag(s) by taking the difference between initial and equilibrium concentrations and multiplying by the volume.