Introduction

1. Introduction • 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.

2. Why Study Thermodynamics? • 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.)

3. Spontaneous Change • A spontaneousprocess is one that can occur in a system left to itself; no action from outside the system is necessary to bring it about. • A nonspontaneousprocess 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.

4. Spontaneous Change (cont’d) • 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.

5. Sample problems… Indicate whether each of the following processes is spontaneous or nonspontaneous. Comment on cases where a clear determination cannot be made. • The action of toilet bowl cleaner, HCl(aq), on “lime” deposits, CaCO3(s). • The boiling of water at normal atmospheric pressure and 65 °C. • The reaction of N2(g) and O2(g) to form NO(g) at room temperature. • The melting of an ice cube.

6. Spontaneous Change (cont’d) • 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 … • There is no significant enthalpy change. • Intermolecular forces are negligible. • So … why do the gases mix?

7. The Concept of Entropy (cont’d) • 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 … • … but the number of possibilities for the distribution of that energy increases.

8. Assessing Entropy Change • The difference in entropy (S) between two states is the entropy change (DS). • 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.

9. Example 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: N2(g) + 3 H2(g)  2 NH3(g) (b) Preparation of a sucrose solution: C12H22O11(s) C12H22O11(aq) (c) Evaporation to dryness of a solution of urea, CO(NH2)2, in water: CO(NH2)2(aq)  CO(NH2)2(s) H2O(l)

10. Entropy and Temperature Third Law of Thermodynamics: The entropy of a perfectly ordered crystalline substance at 0 K is zero.

11. Entropy and Temperature

12. Standard Molar Entropies • 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. DS = SvpS°(products) – SvrS°(reactants) Does the form of this equation look familiar? (remember calculating enthalpy change from ΔHf°?)

13. Standard Molar Entropies and Standard Entropies of Reaction Standard Molar Entropy (S°): The entropy of 1 mole of a pure substance at 1 atm pressure and a specified temperature.

14. aA + bB cC + dD Standard Molar Entropies and Standard Entropies of Reaction DS° = S°(products) - S°(reactants) You’ve seen this before!! DS° = [cS°(C) + dS°(D)] - [aS°(A) + bS°(B)] Products Reactants

15. N2O4(g) 2NO2(g) 304.2 240.0 J J K mol K mol Standard Molar Entropies and Standard Entropies of Reaction Using standard entropies, calculate the standard entropy change for the decomposition of N2O4. DS° = 2 S°(NO2(g)) - S°(N2O4(g)) DS° = (2 mol) - (1 mol) DS° = 175.8 J/K

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. - DH Dssurr = T Free Energy Free Energy: G = H - TS DG = DH - TDS DStotal = DSsys + DSsurr Using: DS = DSsys DG = DH - TDS = -TDStotal

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

20. Free Energy Think about this from the equation ΔG = ΔH - TΔS

21. 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°

22. N2(g) + 3H2(g) 2NH3(g) -198.7 J 1 kJ K 1000 J Standard Free-Energy Changes for Reactions Calculate the standard free-energy change at 25 °C for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes: DH° = -92.2 kJ DS° = -198.7 J/K DG° = DH° - TDS° -92.2 kJ - 298 K = x x = -33.0 kJ

23. aA + bB cC + dD Standard Free Energies of Formation DG° = DG°f (products) - DG°f (reactants) DG° = [cDG°f (C) + dDG°f (D)] - [aDG°f (A) + bDG°f (B)] Products Reactants Third verse…same as the first.

24. Standard Free Energies of Formation

25. 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: DG° = [2 DG°f (Fe(s)) + 3 DG°f (CO2(g))] - [1 DG°f (Fe2O3(s)) + 3 DG°f (CO(g))] DG° = [(2 mol)(0 kJ/mol) + (3 mol)(-394.4 kJ/mol)] - [(1 mol)(-742.2 kJ/mol) + (3 mol)(-137.2 kJ/mol)] DG° = -29.4 kJ

26. 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: DG° = [2 DG°f (Fe(s)) + 3 DG°f (CO2(g))] - [1 DG°f (Fe2O3(s)) + 3 DG°f (CO(g))] DG° = [(2 mol)(0 kJ/mol) + (3 mol)(-394.4 kJ/mol)] - [(1 mol)(-742.2 kJ/mol) + (3 mol)(-137.2 kJ/mol)] DG° = -29.4 kJ

27. N2(g) + 3H2(g) 2NH3(g) 2 P NH3 P P H2 N2 Qp = 3 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:

28. 2C(s) + 2H2(g) C2H4(g) P P H2 C2H4 2 0.10 1002 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. Qp = Calculate ln Q: ln = -11.51

29. 1 kJ 1000 J J K mol Free Energy Changes and the Reaction Mixture Calculate DG: DG = DG° + RT ln Q 8.314 (298 K)(-11.51) = 68.1 kJ/mol + DG = 39.6 kJ/mol

30. 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 = DG° + RT ln Q At equilibrium, DG = 0 and Q = K. DG° = -RT ln K

33. CaCO3(s) CaO(s) + CO2(g) Free Energy and Chemical Equilibrium Calculate Kp at 25 °C for the following reaction: Calculate DG°: DG° = [DG°f (CaO(s)) + DG°f (CO2(g))] - [DG°f (CaCO3(s))] = [(1 mol)(-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)] - [(1 mol)(-1128.8.0 kJ/mol)] DG° = +130.4 kJ/mol

34. 1 kJ ln K = 1000 J -DG° J RT K mol Free Energy and Chemical Equilibrium Calculate ln K: DG° = -RT ln K -130.4 kJ/mol = 8.314 (298 K) ln K = -52.63 Calculate K: -52.63 K = e = 1.4 x 10-23