Chapter 16

1. nonspontaneous spontaneous Chapter 16 • Thermodynamics: Entropy,Free Energy,and Equilibrium In this chapter we will determine the direction of a chemical reaction and calculate equilibrium constant using thermodynamic values: Entropy and Enthalpy.

2. Spontaneous Processes Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence.

3. Universe: System + Surroundings The system is what you observe; surroundings are everything else.

4. Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

5. Laws of Thermodynamics 1) Energy is neither created nor destroyed. 2) In any spontaneous process the total Entropy of a system and its surrounding always increases. We will prove later: ∆G = ∆Hsys – T∆Ssys ∆G < 0 Reaction is spontaneous in forward direction 3) The entropy of a perfect crystalline substance is zero at the absolute zero ( K=0)

6. Enthalpy, Entropy, and Spontaneous Processes State Function: A function or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state. Enthalpy Change (DH): The heat change in a reaction or process at constant pressure. Entropy (DS): The amount of Molecular randomness change in a reaction or process at constant pressure.

7. DH° DH° DH° = +3.88 kJ = -890.3 kJ = +57.1 kJ CH4(g) + 2O2(g) N2O4(g) H2O(s) H2O(l) H2O(l) CO2(g) + 2H2O(l) 2NO2(g) H2O(g) DHfusion DHvap = +6.01 kJ = +40.7 kJ H2O NaCl(s) Na1+(aq) + Cl1-(aq) Enthalpy Change Exothermic: Endothermic:

8. Entropy Change DS = Sfinal - Sinitial

9. Entropy

10. Entropy

11. The sign of entropy change, DS, associated with the boiling of water is _______. • Positive • Negative • Zero

12. Correct Answer: • Positive • Negative • Zero Vaporization of a liquid to a gas leads to a large increase in volume and hence entropy; DS must be positive.

13. Entropy and Temperature 02

14. Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero ( K=0) Ssolid < Sliquid < Sgas

15. What is the sign of the entropy change for the following reaction? 2Zn (s) + O2(g) 2ZnO (s) Entropy Changes in the System (DSsys) When gases are produced (or consumed) • If a reaction produces more gas molecules than it consumes, DS0 > 0. • If the total number of gas molecules diminishes, DS0 < 0. • If there is no net change in the total number of gas molecules, then DS0 may be positive or negative . The total number of gas molecules goes down, DS is negative.

16. 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. Calculated by using S = k ln W , W = Accessible microstates of translational, vibrational and rotational motions

17. ∆G = ∆H – T∆S see page 301 of your book Review of Ch 8: We will show the proof of this formula later in this chapter • Calculating DH° for a reaction: • DH° = DH°f (Products) – DH°f (Reactants) • For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient. • aA + bB cC + dD • DH° = [cDH°f(C) + dDH°f(D)] – [aDH°f(A) + bDH°f(B)] See Appendix B, end of your book

18. The standard entropy change of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C. rxn aS0(A) bS0(B) - [ + ] cS0(C) dS0(D) [ + ] = aA + bB cC + dD - S mS0 (reactants) S nS0 (products) = DS0 DS0 DS0 DS0 rxn rxn rxn rxn What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2(g) 2CO2(g) = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] = 427.2 – [395.2 + 205.0] = -173.0J/K•mol Entropy Changes in the System (DSsys) S0(CO) = 197.6 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol

19. Calculating ∆S for a Reaction ∆So =  So (products) -  So (reactants) Consider 2 H2(g) + O2(g) ---> 2 H2O(liq) ∆So = 2 So (H2O) - [2 So (H2) + So (O2)] ∆So = 2 mol (69.9 J/K•mol) - [2 mol (130.6 J/K•mol) + 1 mol (205.0 J/K•mol)] ∆So = -326.4 J/K Note that there is a decrease in Entropy because 3 mol of gas give 2 mol of liquid.

20. Calculate DS° for the equation below using the standard entropy data given: 2 NO(g) + O2(g) 2 NO2(g) DS° values (J/mol-K): NO2(g) = 240, NO(g) = 211, O2(g) = 205. 147 J/mol 76 J/mol +176 J/mol +147 J/mol

21. Correct Answer: 147 J/K 76 J/K +176 J/K +147 J/K ( ) ( )     D  = D  D S S products S reactants DS° = 2(240)  [2(211) + 205] DS° = 480  [627] DS° =  147

22. Thermite Reaction

23. Spontaneous Processesand Entropy 07 • Consider the gas phase reaction of A2 molecules (red) with B atoms (blue). • (a) Write a balanced equation for the reaction. • (b) Predict the sign of ∆S for the reaction.

24. Second Law Of Thermodynamic: In any spontaneous process the total Entropy of a system and its surrounding always increases. Entropy Changes in the Surroundings (DSsurr) Endothermic Process DSsurr < 0 Exothermic Process DSsurr > 0

25. 1 T - DH Dssurr = T Entropy and the Second Law of Thermodynamics Dssurra ( - DH) Dssurra ( ) See example of tossing a rock into a calm waters vs. rough waters Page 661 of your book

26. 2nd Law of Thermodynamics 01 • The total entropy increases in a spontaneous process and remains unchanged in an equilibrium process. • Spontaneous: ∆Stotal = ∆Ssys + ∆Ssur > 0 • Equilibrium: ∆Stotal = ∆Ssys + ∆Ssur = 0 • The system is what you observe; surroundings are everything else.

27. 2nd Law of Thermodynamics H -D sys S = D surr T Calculate change of entropy of surrounding in the following reaction: 2 H2(g) + O2(g) ---> 2 H2O(liq) ∆H ° = -571.7 kJ ∆Sosurroundings = +1917 J/K

28. H H -D -D sys sys S = D sur T T -D H sys T -T (∆Stotal = ∆Ssys + ( ) ) < 0 • ∆Stotal = ∆Ssys + ( ) > 0 Spontaneous Reactions 01 • The 2nd law tells us a process will be spontaneous if ∆Stotal > 0 which requires a knowledge of ∆Ssurr.spontaneous: ∆Stotal = ∆Ssys + ∆Ssur > 0

29. -T . ∆Stotal = -T. ∆Ssys + ΔHsys < 0 Gibbs Free Energy 02 • The expression –T∆Stotal is equated asGibbs free energy change (∆G)*, or simply free energy change: • ∆G = ∆Hsys – T∆Ssys • ∆G < 0 Reaction is spontaneous in forward direction.∆G = 0 Reaction is at equilibrium.∆G > 0 Reaction is spontaneous in reverse direction. –T∆Stotal = ΔG *Driving force of a reaction, Maximum work you can get from a system.

30. Calculate ∆Go rxn for the following: C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g) Use enthalpies of formation to calculate ∆Horxn = -1256 kJ Use standard molar entropies to calculate ∆Sorxn ( see page 658, appendix A-10) ∆Sorxn = -97.4 J/K or -0.0974 kJ/K ∆Gorxn = -1256 kJ - (298 K)(-0.0974 kJ/K) = -1227 kJ Reaction is product-favored in spite of negative ∆Sorxn. Reaction is “enthalpy driven”

31. Calculate DG° for the equation below using the thermodynamic data given: N2(g) + 3 H2(g) 2 NH3(g) DHf° (NH3) = 46 kJ; S° values (J/mol-K) are: NH3 = 192.5, N2 = 191.5, H2 = 130.6. +33 kJ +66 kJ 66 kJ 33 kJ

32. Correct Answer: +33 kJ +66 kJ 66 kJ 33 kJ DH° = 92 kJ DS° = 198.3J/mol . K DG° = DH° TDS° DG° = (92)  (298)(0.1983) DG° = (92) + (59.2) = 32.9 KJ

33. Gibbs Free Energy 03 Using ∆G = ∆H – T∆S, we can predict the sign of ∆G from the sign of ∆H and ∆S. 1) If both ∆H and ∆S are positive, ∆G will be negative only when the temperature value is large. Therefore, the reaction is spontaneous only at high temperature. 2) If ∆H is positive and ∆S is negative, ∆G will always be positive. Therefore, the reaction is not spontaneous

34. Gibbs Free Energy: 04 ∆G = ∆H – T∆S 3) If ∆H is negative and ∆S is positive, ∆G will always be negative. Therefore, the reaction is spontaneous 4) If both ∆H and ∆S are negative, ∆G will be negative only when the temperature value is small. Therefore, the reaction is spontaneous only at Low temperatures.

35. Gibbs Free Energy 04

36. Gibbs Free Energy 06 • What are the signs (+, –, or 0) of ∆H, ∆S, and ∆G for the following spontaneous reaction of A atoms (red) and B atoms (blue)?

37. -T . ∆Stotal = -T. ∆Ssys + ΔHsys < 0 Gibbs Free Energy 02 • The expression –T∆Stotal is equated asGibbs free energy change (∆G)*, or simply free energy change: • ∆G = ∆Hsys – T∆Ssys • ∆G < 0 Reaction is spontaneous in forward direction.∆G = 0 Reaction is at equilibrium.∆G > 0 Reaction is spontaneous in reverse direction. –T∆Stotal = ΔG *Driving force of a reaction, Maximum work you can get from a system.

38. 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

39. Gibbs Free Energy 05 • Iron metal can be produced by reducing iron(III) oxide with hydrogen: Fe2O3(s) + 3 H2(g)  2 Fe(s) + 3 H2O(g) ∆H° = +98.8 kJ; ∆S° = +141.5 J/K • Is this reaction spontaneous at 25°C? • At what temperature will the reaction become spontaneous?

40. Decomposition of CaCO3 has a H° = 178.3 kJ/mol and S° = 159 J/mol K. At what temperature does this become spontaneous? 121°C 395°C 848°C 1121°C

41. Correct Answer: 121°C 395°C 848°C 1121°C T = H°/S° T = 178.3 kJ/mol/0.159 kJ/mol K T = 1121 K T (°C) = 1121 – 273 = 848

42. 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

43. Standard free energy of formation (DG0f) is the free-energy change that occurs when1 mole of the compound is formed from its elements in their standard ( 1 atm) states.

44. The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn aA + bB cC + dD - [ + ] [ + ] = - mDG0 (reactants) S S = f DG0 DG0 rxn rxn DG0 of any element in its stable form is zero. f dDG0 (D) nDG0 (products) bDG0 (B) cDG0 (C) aDG0 (A) f f f f f --

45. Calculate DG° for the equation below using the Gibbs free energy data given: 2 SO2(g) + O2(g) 2 SO3(g) DGf° values (kJ): SO2(g) = 300.4, SO3(g) = 370.4 +70 kJ +140 kJ 140 kJ 70 kJ

46. Correct Answer: +70 kJ +140 kJ 140 kJ 70 kJ ( ) ( )   ° °  ° =   G G products G reactants  f f DG° = (2  370.4)  [(2  300.4) + 0] DG° = 740.8  [600.8] = 140

47. Gibbs Free Energy 07

48. Calculation of Nonstandard ∆G • The sign of ∆G° tells the direction of spontaneous reaction when both reactants and products are present at standard state conditions. • Under nonstandard( condition where pressure is not 1atm or concentrations of solutions are not 1M) conditions, ∆G˚ becomes ∆G.∆G = ∆G˚ + RT lnQ • The reaction quotient is obtained in the same way as an equilibrium expression.

49. 2C(s) + 2H2(g) C2H4(g) P P H2 C2H4 2 ln = -11.51 0.10 1002 Free Energy Changes and the Reaction Mixture DG = DG° + RT ln Q 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 Qp:

50. 1 kJ 1000 J J K mol Free Energy Changes and the Reaction Mixture Calculate DG: DG = DG° + RT ln Q 8.314 = 68.1 kJ/mol + (298 K)(-11.51) ( see page A-13 of your book, page 667) DG = 39.6 kJ/mol