Chapter 16 Thermodynamics: Entropy, Free Energy, and Equilibrium Chemistry: McMurry and Fay, 6th Edition Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 6:19:03 PM Chapter 16 Thermodynamics: Entropy, Free Energy, and Equilibrium Copyright © 2011 Pearson Prentice Hall, Inc.

16.1 Spontaneous Processes Chemistry: McMurry and Fay, 6th Edition Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 16.1 Spontaneous Processes 11/29/2018 6:19:03 PM Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence. Will the reverse reaction spontaneous? Copyright © 2011 Pearson Prentice Hall, Inc.

16.1 Spontaneous process Spontaneity reaction always moves a system toward equilibrium Both forward and reverse reaction depends on Temperature Pressure Composition of reaction mixture Q < K; reaction proceeds in the forward direction Q>K; reaction proceeds in the reverse direction Spontaneity of a reaction does not identify the speed of reaction

Gibbs free energy and spontaneity Gibbs free energy and Reaction spontaneity i. The spontaneity of a chemical reaction is the possibility of the reaction occurring or not without the aid of outside energy. ii. If a reaction occurs without a net input of energy, it is called “spontaneous.” This type of reaction actually releases (“gives off”) useful energy. iii. If a reaction requires energy in order to occur, it is called “nonspontaneous.” This type of reaction “takes in” energy when it occurs. iv. Depiction of spontaneity with a reaction energy diagram v. Gibbs free energy is the energy that is “free” to do work. “Free” doesn’t mean “costs nothing,” it means “available,” as in “Are you free tonight at around 8:00?

Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 Free Energy Using the second law and DG = DH - TDSsys DG < 0 The reaction is spontaneous. DG > 0 The reaction is nonspontaneous. DG = 0 The reaction mixture is at equilibrium. Thermodynamic Standard State: Most stable form of a substance at 1 atm pressure and at a specified temperature, usually 25 °C; 1 M concentration for all substances in solution DG° = DH° - TDS°sys Copyright © 2008 Pearson Prentice Hall, Inc.

The effect of H, S and T on Spontaneity Low Temperature High Temperature - + Spontaneous (G<0) Nonspontaneous (G > 0) Spontaneous (G< 0) nonSpontaneous (G>0) Nonspontaneous (G>0)

Entropy and Gibbs free energy Like Horxn, at standard condition DSo = So(products) - So(reactants) DGo = DGof (products)  DGof (reactants)

Chemistry: McMurry and Fay, 6th Edition Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 6:19:03 PM Copyright © 2011 Pearson Prentice Hall, Inc.

16.5 Standard Molar Entropies and Standard Entropies of Reaction Chemistry: McMurry and Fay, 6th Edition Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 6:19:03 PM 16.5 Standard Molar Entropies and Standard Entropies of Reaction Calculate ΔSo for the following reaction CO(g) + 2H2(g) = CH3OH(l) Predicted Sign for Δ So? __________ p652 Copyright © 2011 Pearson Prentice Hall, Inc.

Standard Free Energies of Formation Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium Standard Free Energies of Formation 11/29/2018 Using table values, calculate the standard free-energy change (ΔGo) at 25 °C for the reduction of iron(III) oxide with carbon monoxide: 2Fe(s) + 3CO2(g) Fe2O3(s) + 3CO(g) Fe2O3(s) 3CO(g) 2Fe(s) 3CO2(g) Gof (kJ/mol) -743.5 -137.2 -394.4 Copyright © 2008 Pearson Prentice Hall, Inc.

Example What is the standard free-energy change, ΔoG, for the following reaction at 25.0oC? 2KClO3(g)  2KCl(s) + 3O2(g) ΔHo = -78.0kJ ΔSo =493.9J/K Is the reaction spontaneous at standard-state condition? Does the reaction become nonspontaneous at higher temperature?

16.9 Standard Free-Energy Changes for Reactions Chemistry: McMurry and Fay, 6th Edition Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 6:19:03 PM 16.9 Standard Free-Energy Changes for Reactions Calculate the standard free-energy change at 25 oC for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes: 2NH3(g) N2(g) + 3H2(g) DHo = 92.2 kJ DSo = 198.7 J/K p658 Negative value for DG° means it’s spontaneous. Spontaneous doesn’t mean fast, however. Rate of reaction is kinetics. Will the reaction become nonspontanous at higher temperature? If so, then determine at what temperature will it become nonspontanous? Copyright © 2011 Pearson Prentice Hall, Inc.

Entropy and the Second Law of Thermodynamics Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium Entropy and the Second Law of Thermodynamics 11/29/2018 DStotal = DSsystem + DSsurroundings or DStotal = DSsys + DSsurr DStotal > 0 The reaction is spontaneous. DStotal < 0 The reaction is nonspontaneous. DStotal = 0 The reaction mixture is at equilibrium. Copyright © 2008 Pearson Prentice Hall, Inc.

Entropy and the Second Law of Thermodynamics Chemistry: McMurry and Fay, 6th Edition Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 6:19:03 PM Entropy and the Second Law of Thermodynamics DSsurr a  DH DSsurr a T 1 DSsurr = T  DH Copyright © 2011 Pearson Prentice Hall, Inc.

Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 Free Energy: a thermodynamic quantity defined by G = H - TS DG = DH - TDSsys Using: DStotal = DSsys + DSsurr Dssurr = T - DH DG = -TDStotal DStotal = DSsys - DH T Free energy was introduced in Ch 8 (Thermochemistry: Chemical Energy). Constant pressure and temperature Copyright © 2008 Pearson Prentice Hall, Inc.

Example Consider the combustion of propane gas: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) a. Calculate the entropy change in the surrounding associated with this reaction occurring at 25.0oC b. Determine the entropy change for the system c. Determine the entropy change for the universe (entropy total). Will the reaction be spontaneous? So (J/K mol) ΔHof (kJ/mol) C3H8(g) 270.2 -103.8 O2(g) 205.0 CO2(g) 213.6 -393.5 H2O(g) 188.7 -241.8

16.3 Entropy and Probability Chemistry: McMurry and Fay, 6th Edition Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 6:19:03 PM 16.3 Entropy and Probability S = k ln W k = Boltzmann’s constant = 1.38 x 1023 J/K W = The number of ways that the state can be achieved. Copyright © 2011 Pearson Prentice Hall, Inc.

16.4 Entropy and Temperature Chemistry: McMurry and Fay, 6th Edition Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 6:19:03 PM 16.4 Entropy and Temperature ΔS increases when increasing the average kinetic energy of molecules Total energy is distributed among the individual molecules in a number of ways Botzman- the more way (W) that the energy can be distributed the greater the randomness of the state and higher the entropy Copyright © 2011 Pearson Prentice Hall, Inc.

16.4 Entropy and Temperature Chemistry: McMurry and Fay, 6th Edition Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 6:19:03 PM 16.4 Entropy and Temperature Third Law of Thermodynamics: The entropy of a perfectly ordered crystalline substance at 0 K is zero. Lower temperature Less molecular motion Narrower distribution of individual molecular energies Less randomness Lower entropy Higher temperature Greater molecular motion Broader distribution of individual molecular energies More randomness Higher entropy Look back at S = k ln W. If W = 1 (only one way to order a perfectly ordered crystalline substance), then ln W = 0. Copyright © 2011 Pearson Prentice Hall, Inc.

16.6 Entropy and the Second Law of Thermodynamics Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 16.6 Entropy and the Second Law of Thermodynamics First Law of Thermodynamics: In any process, spontaneous or nonspontaneous, the total energy of a system and its surroundings is constant. Helps keeping track of energy flow between system and the surrounding Does not indicate the spontaneity of the process Second Law of Thermodynamics: In any spontaneous process, the total entropy of a system and its surroundings always increases. Provide a clear cut criterion of spontaneity Direction of spontaneous change is always determined by the sign of the total entropy change The first law was defined in Ch 8 (Thermochemistry: Chemical Energy). Entropy was also introduced in that chapter. Copyright © 2008 Pearson Prentice Hall, Inc.

16.11 Free Energy and Chemical Equilibrium Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 16.11 Free Energy and Chemical Equilibrium 11/29/2018 DG = DG° + RT ln Q When the reaction mixture is mostly reactants: Q << 1 RT ln Q << 0 DG < 0 The total free energy decreases as the reaction proceeds spontaneously in the forward direction. When the reaction mixture is mostly products: Q >> 1 RT ln Q >> 0 DG > 0 The total free energy decreases as the reaction proceeds spontaneously in the reverse direction. Copyright © 2008 Pearson Prentice Hall, Inc.

Chemistry: McMurry and Fay, 6th Edition Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 6:19:03 PM Copyright © 2011 Pearson Prentice Hall, Inc.

Free Energy and Chemical Equilibrium Chemistry: McMurry and Fay, 6th Edition Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 6:19:03 PM Free Energy and Chemical Equilibrium DG = DGo + RT ln Q DG = Free-energy change under nonstandard conditions. At equilibrium, DG = 0 and Q = K. DGo = RT ln K Copyright © 2011 Pearson Prentice Hall, Inc.

Free Energy Changes and the Reaction Mixture Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 Free Energy Changes and the Reaction Mixture Calculate DG for the formation of ethylene (C2H4) from carbon and hydrogen at 25 °C when the partial pressures are 100 atm H2 and 0.10 atm C2H4. C2H4(g) 2C(s) + 2H2(g) Is the reaction spontaneous in the forward or the reverse direction? C(s) H2(g) C2H4(g) ∆ Gof (kJ/mol) 61.8 Copyright © 2008 Pearson Prentice Hall, Inc.

Free Energy and Chemical Equilibrium Chemistry: McMurry and Fay, 6th Edition Chapter 16: Thermodynamics: Entropy, Free Energy, and Equilibrium 11/29/2018 6:19:03 PM Free Energy and Chemical Equilibrium Calculate Kp at 25 oC for the following reaction: CaO(s) + CO2(g) CaCO3(s) CaCO3(s) CaO(s) CO2(g) Gof (kJ/mol) -1128.8 -603.5 -394.4 Problem 16.15 p668 Copyright © 2011 Pearson Prentice Hall, Inc.

Example The value of ΔGof at 25oC for gaseous mercury is 31.85 kJ/mol. What is the vapor pressure of mercury at 25oC? Hg(l) Hg(g)