Chapter 19 Chemical Thermodynamics Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 19 Chemical Thermodynamics John D. Bookstaver St. Charles Community College Cottleville, MO © 2009, Prentice-Hall, Inc.

Sec. 19.2 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. © 2009, Prentice-Hall, Inc.

Entropy Like total energy, E, and enthalpy, H, entropy is a state function. Therefore, S = Sfinal  Sinitial © 2009, Prentice-Hall, Inc.

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 = qrev T © 2009, Prentice-Hall, Inc.

Sample Exercise 19.2 (p. 819) Elemental mercury is a silver liquid at room temperature. Its normal freezing point is -38.9oC, and its molar enthalpy of fusion is DHfusion = 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? Hint: Calculate q, as you have done before, then reverse the sign and divide by T in K. (DSsys = -2.44 J/K) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.2) Do all exothermic phase changes have a negative value for the entropy change of the system? Yes, because the heat transferred from the system has a negative sign. Yes, because the temperature decreases during the phase transition. No, because the entropy change depends on the sign of the heat transferred to or from the system. No, because the heat transferred to the system has a positive sign. More than one of the previous answers is correct. © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.2) Do all exothermic phase changes have a negative value for the entropy change of the system? Yes, because the heat transferred from the system has a negative sign. Yes, because the temperature decreases during the phase transition. No, because the entropy change depends on the sign of the heat transferred to or from the system. No, because the heat transferred to the system has a positive sign. More than one of the previous answers is correct. © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (19.2) The normal boiling point of ethanol, C2H5OH, is 78.3oC, 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 C2H5OH(g) at 1 atm condenses to liquid at the normal boiling point? (-163 J/K) © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

Second Law of Thermodynamics In other words: For reversible processes: Suniv = Ssystem + Ssurroundings = 0 For irreversible processes: Suniv = Ssystem + Ssurroundings > 0 © 2009, Prentice-Hall, Inc.

Second Law of Thermodynamics These last truths mean that as a result of all spontaneous processes the entropy of the universe increases. © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

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.  © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

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. © 2009, Prentice-Hall, Inc.

Entropy on the Molecular Scale The change in entropy for a process, then, is S = k lnWfinal  k lnWinitial lnWfinal lnWinitial S = k ln Entropy increases with the number of microstates in the system. © 2009, Prentice-Hall, Inc.

Entropy on the Molecular Scale The number of microstates and, therefore, the entropy tends to increase with increases in Temperature. Volume. The number of independently moving molecules. © 2009, Prentice-Hall, Inc.

Entropy and Physical States Entropy increases with the freedom of motion of molecules. Therefore, S(g) > S(l) > S(s) © 2009, Prentice-Hall, Inc.

Solutions Generally, when a solid is dissolved in a solvent, entropy increases. © 2009, Prentice-Hall, Inc.

Entropy Changes 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. © 2009, Prentice-Hall, Inc.

Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. © 2009, Prentice-Hall, Inc.

Sample Exercise 19.3 (p. 826) Predict whether DS is positive or negative for each of the following processes, assuming each occurs at constant temperature: a) H2O(l)  H2O(g) b) Ag+(aq) + Cl-(aq)  AgCl(s) c) 4 Fe(s) + 3 O2(g)  2 Fe2O3(s) d) N2(g) + O2(g)  2 NO(g) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.3) Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system: a) CO2(s)  CO2(g) b) CaO(s) + CO2(g)  CaCO3(s) c) HCl(g) + NH3(g)  NH4Cl(s) d) 2 SO2(g) + O2(g)  2 SO3(g) © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (19.3) Since the entropy of the universe increases for spontaneous processes, does it mean that the entropy of the universe decreases for nonspontaneous processes?   © 2009, Prentice-Hall, Inc.

Sample Exercise 19.4 (p. 827) In each pair, choose the system that has greater entropy and explain your choice: a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25oC b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25oC c) 1 mol of HCl(g) or 1 mol or Ar(g) at 298 K © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.4) Which system has the greatest entropy? 1 mol of H2(g) at STP b) 1 mol of H2(g) at 100oC and 0.5 atm c) 1 mol of H2O(s) at 0oC d) 1 mol of H2O(l) at 25oC © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (19.4) Choose the substance with the greater entropy in each case:   1 mol of H2(g) at STP or 1 mol of SO2(g) at STP 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP. © 2009, Prentice-Hall, Inc.

Standard Entropies These are molar entropy values of substances in their standard states. Standard entropies tend to increase with increasing molar mass. © 2009, Prentice-Hall, Inc.

Standard Entropies Larger and more complex molecules have greater entropies. © 2009, Prentice-Hall, Inc.

S = nS(products) — mS(reactants) 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 in the balanced chemical equation. © 2009, Prentice-Hall, Inc.

Entropy Changes in Surroundings Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process: Ssurr = qsys T At constant pressure, qsys is simply H for the system. © 2009, Prentice-Hall, Inc.

Entropy Change in the Universe The universe is composed of the system and the surroundings. Therefore, Suniverse = Ssystem + Ssurroundings For spontaneous processes Suniverse > 0 © 2009, Prentice-Hall, Inc.

Entropy Change in the Universe qsystem T Since Ssurroundings = and qsystem = Hsystem This becomes: Suniverse = Ssystem + Multiplying both sides by T, we get TSuniverse = Hsystem  TSsystem Hsystem T © 2009, Prentice-Hall, Inc.

Sample Exercise 19.5 (p. 829) Calculate the change in the standard entropy of the system, DSo, for the synthesis of ammonia from N2(g) and H2(g) at 298 K. N2(g) + 3 H2(g)  2 NH3(g) (-198.3 J/K)   © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.5) Using the standard molar entropies in Appendix C, calculate the standard entropy change, DSo, for the “water-splitting” reaction at 298 K:   2 H2O(l)  2 H2(g) + O2(g) 326.3 J/K 265.7 J/K 163.2 J/K 88.5 J/K -326.3 J/K © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (19.5) Using the standard entropies in Appendix C, calculate the standard entropy change, DSo for the following reaction at 298 K: Al2O3(s) + 3 H2(g)  2 Al(s) + 3 H2O(g) (180.39 J/K)   © 2009, Prentice-Hall, Inc.

DH, including Hess’s Law – the heat component Where we’ve been: DH, including Hess’s Law – the heat component DS, including #microstates – the entropy component Where we’re going: DH and DS  Gibbs Free Energy – will a reaction occur? Application to equilibrium and electrochemistry © 2009, Prentice-Hall, Inc.

Gibbs Free Energy TDSuniverse is defined as the Gibbs free energy, G. When Suniverse is positive, G is negative. Therefore, when G is negative, a process is spontaneous. © 2009, Prentice-Hall, Inc.

Gibbs Free Energy If DG is negative, the forward reaction is spontaneous. If DG is 0, the system is at equilibrium. If G is positive, the reaction is spontaneous in the reverse direction. © 2009, Prentice-Hall, Inc.

Sample Exercise 19.6 (p. 833) Calculate the standard free energy change for the formation of NO(g) from N2(g) and O2(g) at 298 K: N2(g) + O2(g)  2 NO(g) given that DHo = 180.7 kJ and DSo = 24.7 J/K. Is the reaction spontaneous under these circumstances? (no, DG = 173.3 kJ) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.6) Which of the following statements is true? All spontaneous reactions have a negative enthalpy change, All spontaneous reactions have a positive entropy change, All spontaneous reactions have a positive free-energy change, All spontaneous reactions have a negative free-energy change, All spontaneous reactions have a negative entropy change. © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.6) Which of the following statements is true? All spontaneous reactions have a negative enthalpy change, All spontaneous reactions have a positive entropy change, All spontaneous reactions have a positive free-energy change, All spontaneous reactions have a negative free-energy change, All spontaneous reactions have a negative entropy change. © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (19.6) A particular reaction has DHo = 24.6 kJ and DSo = 132 J/K at 298 K. Calculate DGo. Is the reaction spontaneous under these conditions? (Yes, DGo = -14.7 kJ) © 2009, Prentice-Hall, Inc.

Standard Free Energy Changes Analogous to standard enthalpies of formation are standard free energies of formation, G. f DG = SnDG (products)  SmG (reactants) f where n and m are the stoichiometric coefficients. © 2009, Prentice-Hall, Inc.

Standard Conditions Solid pure solid Liquid pure liquid Gas 1 atm pressure Solution 1 M concentration Elements standard free energy of formation of an element in its free state = zero © 2009, Prentice-Hall, Inc.

Sample Exercise 19.7 (p. 834) Use data from Appendix C to calculate the standard free-energy change for the following reaction at 298 K: P4(g) + 6 Cl2(g)  4 PCl3(g) (-1102.8 kJ) b) What is DGo for the reverse of the above reaction? (+1102.8 kJ) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.7) The following chemical equations describe the same chemical reaction. How do the free energies of these two chemical equations compare? (1) 2 H2O(l)  2 H2(g) + O2(g) (2) H2O(l)  H2(g) + ½ O2(g)   DG1o = DG2o DG1o = 2 DG2o 2 DG1o = DG2o none of the above © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.7) The following chemical equations describe the same chemical reaction. How do the free energies of these two chemical equations compare? (1) 2 H2O(l)  2 H2(g) + O2(g) (2) H2O(l)  H2(g) + ½ O2(g)   DG1o = DG2o DG1o = 2 DG2o 2 DG1o = DG2o none of the above © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (19.7) Using the data from Appendix C, calculate DGo at 298 K for the combustion of methane: CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) (-800.7 kJ) © 2009, Prentice-Hall, Inc.

Sample Exercise 19.8 (p. 835) In Section 5.7 we used Hess’s law to calculate DHo for the combustion of propane gas at 298 K: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DHo = -2220 kJ Without using data from Appendix C, predict whether DGo for this reaction is more negative or less negative than DHo. Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your prediction from part (a) correct? (-2108 kJ) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.8) If a reaction is exothermic and its entropy change is positive, which statement is true? The reaction is spontaneous at all temperatures, The reaction if nonspontaneous at all temperatures, The reaction is spontaneous only at higher temperatures, The reaction is spontaneous only at lower temperatures. © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.8) If a reaction is exothermic and its entropy change is positive, which statement is true? The reaction is spontaneous at all temperatures, The reaction if nonspontaneous at all temperatures, The reaction is spontaneous only at higher temperatures, The reaction is spontaneous only at lower temperatures. © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (19.8) Consider the combustion of propane to form CO2(g) and H2O(g) at 298 K:   C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) DHo = -2220 kJ Would you expect DGo to be more negative or less negative than DHo? © 2009, Prentice-Hall, Inc.

Free Energy Changes At temperatures other than 25°C, DG° = DH  TS How does G change with temperature? © 2009, Prentice-Hall, Inc.

Free Energy and 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. © 2009, Prentice-Hall, Inc.

Free Energy and Temperature © 2009, Prentice-Hall, Inc.

Sample Exercise 19.9 (p. 837) The Haber process for the production of ammonia involves the following equilibrium: N2(g) + 3 H2(g) D 2 NH3(g) Assume that DHo and DSo for this reaction do not change with temperature. a) Predict the direction in which DGo for this reaction changes with increasing temperature. b) Calculate the values of DGo for the reaction at 25oC and 500oC. (-33.3 kJ, 61 kJ) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.9) What is the temperature above which the Haber ammonia process becomes nonspontaneous? 25oC 47oC 61oC 193oC 500oC   © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (19.9) a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate DHo and DSo at 298 K for the following reaction: 2 SO2(g) + O2(g)  2 SO3(g) (DHo = -196.6 kJ, DSo = -189.6 J/K) b) Using the values obtained in part (a), estimate DGo at 400 K. (DGo = -120.8 kJ) © 2009, Prentice-Hall, Inc.

Free Energy and Equilibrium In this section, we are going to 1. rearrange DGo = DHo – TDSo to find T for a phase change (equilibrium) 2. use DGo (free energy under standard conditions) to calculate DG under nonstandard conditions 3. directly relate DGo and Keq (equilibrium constant), in order to solve several types of problems. © 2009, Prentice-Hall, Inc.

Sample Exercise 19.10 (p. 839) Phase Changes As we saw in Section 11.5, the normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at a pressure of 1 atm. Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl4(l). b) What is the value of DGo for the equilibrium in (a)? c) Use thermodynamic data in Appendix C and DGo = DHo – TDSo to estimate the normal boiling point of CCl4. (70oC) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.10) If the normal boiling point of a liquid is 67oC, and the standard molar entropy change for the boiling process is +100 J/K, estimate the standard molar enthalpy change for the boiling process. +6700 J -6700 J +34,000 J -34,000 J   © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.10) If the normal boiling point of a liquid is 67oC, and the standard molar entropy change for the boiling process is +100 J/K, estimate the standard molar enthalpy change for the boiling process. +6700 J -6700 J +34,000 J -34,000 J   © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (19.10) Use data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br2(l). (The experimental value is given in Figure 11.5). (330 K)   © 2009, Prentice-Hall, Inc.

Free Energy and Equilibrium Reminders: DGo and Keq apply to standard conditions (pure solid or liquid, gases at 1 atm and 1M solutions). DG and Q (equilibrium quotient) apply to any conditions. © 2009, Prentice-Hall, Inc.

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.) R = 8.314 J/mol.K, T is in K © 2009, Prentice-Hall, Inc.

Sample Exercise 19.11 (p. 840) We will continue to explore the Haber process for the synthesis of ammonia: N2(g) + 3 H2(g) D 2 NH3(g) Calculate DG at 298 K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3. (-44.9 kJ/mol) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.11) Which of the following statements is true?   The larger the Q, the larger the DGo. If Q = 0, the system is at equilibrium. If a reaction is spontaneous under standard conditions, it is spontaneous under all conditions. The free-energy change for a reaction is independent of temperature. If Q > 1, DG > DGo. © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.11) Which of the following statements is true?   The larger the Q, the larger the DGo. If Q = 0, the system is at equilibrium. If a reaction is spontaneous under standard conditions, it is spontaneous under all conditions. The free-energy change for a reaction is independent of temperature. If Q > 1, DG > DGo. © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (19.11) Calculate DG at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3. (-26.0 kJ/mol) © 2009, Prentice-Hall, Inc.

© 2009, Prentice-Hall, Inc.

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 © 2009, Prentice-Hall, Inc.

Free Energy and Equilibrium If DGo < 0, then K >1 when ln K is +  more negative DGo  larger K  products are favored If DGo = 0, then K = 1 If DGo > 0, then K <1 when ln K is -  more positive DGo  smaller K  reactants are favored © 2009, Prentice-Hall, Inc.

Next problems: Application of DGo = -RTlnQ © 2009, Prentice-Hall, Inc.

Sample Exercise 19.12 (p. 841) N2(g) + 3 H2(g) D 2 NH3(g) The standard free energy change for the Haber process at 25oC was obtained in Sample Exercise 19.9 for the Haber reaction (-33.3 kJ): N2(g) + 3 H2(g) D 2 NH3(g) Use this value of DGo to calculate the equilibrium constant for the process at 25oC. (7 x 105) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.12) The Ksp for a very insoluble salt is 4.2 x 10-47 at 298 K. What is DGo for the dissolution of the salt in water? -265 kJ/mol -115 kJ/mol -2.61 kJ/mol +115 kJ/mol +265 kJ/mol © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (19.12) The Ksp for a very insoluble salt is 4.2 x 10-47 at 298 K. What is DGo for the dissolution of the salt in water? -265 kJ/mol -115 kJ/mol -2.61 kJ/mol +115 kJ/mol +265 kJ/mol © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (19.12) Use data from Appendix C to calculate the standard free-energy change, DGo, and the equilibrium constant, K, at 298 K for the following reaction: H2(g) + Br2(l) D 2 HBr(g) (-106.4 kJ/mol; 4 x 1018) © 2009, Prentice-Hall, Inc.

Sample Integrative Exercise 19 (p. 843) Consider the simple salts NaCl(s) and AgCl(s). We will examine the equilibria in which these salts dissolve in water to form aqueous solutions of ions: NaCl(s) D Na+(aq) + Cl-(aq) AgCl(s) D Ag+(aq) + Cl-(aq) a) Calculate the value of DGo at 298 K for each of the preceding reactions. b) The two values from part (a) are very different. Is this difference primarily due to the enthalpy term or the entropy term of the standard free-energy change? c) Use the values of DGo to calculate Ksp values for the two salts at 298 K. d) Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part (c)? e) How will DGo for the solution process of these salts change with increasing T? What effect should this change have on the solubility of the salts? © 2009, Prentice-Hall, Inc.