1 / 35

Chapter 16

Chapter 16. Thermodynamics: Entropy, Free Energy, and Equilibrium. 16.1 Spontaneous Processes. Spontaneous Process : A process that, once started, proceeds on its own without a continuous external influence. Will the reverse reaction spontaneous?. 16.1 Spontaneous process.

bree
Download Presentation

Chapter 16

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 16 Thermodynamics: Entropy, Free Energy, and Equilibrium

  2. 16.1 Spontaneous Processes Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence. Will the reverse reaction spontaneous?

  3. 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

  4. Spontaneous Processes

  5. 16.2 Enthalpy, Entropy, and Spontaneous Processes DS = SfinalSinitial

  6. 16.2Enthalpy, Entropy, and Spontaneous Processes

  7. 16.2 Enthalpy, Entropy, and Spontaneous Processes

  8. 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.

  9. 16.4 Entropy and Temperature Third Law of Thermodynamics: The entropy of a perfectly ordered crystalline substance at 0 K is zero. • Higher temperature • Greater molecular motion • Broader distribution of • individual molecular energies • More randomness • Higher entropy • Lower temperature • Less molecular motion • Narrower distribution of • individual molecular energies • Less randomness • Lower entropy

  10. 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

  11. 16.5 Standard Molar Entropies and Standard Entropies of Reaction Standard Molar Entropy (So): The entropy of 1 mole of a pure substance at 1 atm pressure and a specified temperature.

  12. aA + bB cC + dD Standard Molar Entropies and Standard Entropies of Reaction DSo = So(products) - So(reactants) DSo = [cSo(C) + dSo (D)]  [aSo (A) + bSo (B)] Products Reactants

  13. N2O4(g) 2NO2(g) 16.5 Standard Molar Entropies and Standard Entropies of Reaction Using standard entropies, calculate the standard entropy change for the decomposition of N2O4.

  14. Example • Calculate the standard entropy of reaction at 25oC for the decomposition of calcium carbonate: CaCO3(s)  CaO(s) + CO2(g)

  15. 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

  16. Entropy and the Second Law of Thermodynamics 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.

  17. 1 DSsurra T DH DSsurr = T Entropy and the Second Law of Thermodynamics DSsurraDH

  18. Example • Consider the oxidation of iron metal 4 Fe(s) + 3 O2(g)  2 Fe2O3(s) Determine whether the reaction is spontaneous at 25oC

  19. 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. Will the reaction be spontaneous?

  20. - DH Dssurr = T 16.7 Free Energy Free Energy: a thermodynamic quantity defined by G = H - TS DG = DH - TDSsys DStotal = DSsys + DSsurr Using: DG = -TDStotal DStotal = DSsys - DH T Constant pressure and temperature

  21. Free Energy Using the second law and DG = DH - TDSsys = -TDStotal DG < 0 The reaction is spontaneous. DG > 0 The reaction is nonspontaneous. DG = 0 The reaction mixture is at equilibrium.

  22. 16.8 Standard Free-Energy Changes for Reactions 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°

  23. The effect of H, S and T on Spontaneity

  24. Example • What is the standard free-energy change, ΔoG, for the following reaction at 25oC? 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?

  25. N2(g) + 3H2(g) 2NH3(g) 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: DHo = 92.2 kJ DSo = 198.7 J/K Will the reaction become nonspontanous at higher temperature? If so, then determine at what temperature will it become nonspontanous?

  26. aA + bB cC + dD Standard Free Energies of Formation DGo = DGof (products) DGof (reactants) DGo = [cDGof (C) + dDGof (D)]  [aDGof (A) + bDGof (B)] Products Reactants

  27. Standard Free Energies of Formation

  28. Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) Standard Free Energies of Formation Using table values, calculate the standard free-energy change at 25 °C for the reduction of iron(III) oxide with carbon monoxide:

  29. N2(g) + 3H2(g) 2NH3(g) 2 P NH3 P P H2 N2 Qp = 3 16.10 Free Energy Changes and the Reaction Mixture DG = DG° + RT ln Q DG = Free-energy change under nonstandard conditions. For the Haber synthesis of ammonia:

  30. 2C(s) + 2H2(g) C2H4(g) 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. Is the reaction spontaneous in the forward or the reverse direction?

  31. 16.11 Free Energy and Chemical Equilibrium 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.

  32. Free Energy and Chemical Equilibrium DG = DGo + RT ln Q At equilibrium, DG = 0 and Q = K. DGo = RT ln K

  33. CaCO3(s) CaO(s) + CO2(g) Free Energy and Chemical Equilibrium Calculate Kp at 25 oC for the following reaction:

  34. 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)

More Related