Packet 17: Free Energy and Thermodynamics

1. Packet 17:Free Energy and Thermodynamics

2. Concept Area I: Terminology spontaneous entropy, S second law of thermodynamics Gibbs free energy, G standard entropy change, ΔSºrxn standard Molar entropies third law of thermodynamics standard free energy change, ΔGºrxn standard free energy of formation, ΔGfº reversible reaction irreversible reaction free energy change, ΔGrxn

3. Concept Area II: Entropy • You should understand that entropy is a measure of disorder. • You should recognize that entropy can be determined experimentally as the heat change of a reversible process. • You should know how to calculate entropy changes. • You should be able to identify common processes that are entropy-favored.

4. “DE = q + w: Energy can neither be created nor destroyed.” Consequence Review:The First Law of Thermodynamics New stuff now…

5. Entropy, S • S = is used to quantify the extent of disorder resulting from dispersal of energy and matter • DS = is the change in entropy So, for a reaction DS = Sproducts – Sreactants • The bigger and more positive the DS, the more entropy we have and the more likely the reaction will go as written – spontaneous.

6. Dispersal of Matter Which of these situations is the most likely after we open the valve? Why? Tro page 645

7. Probabilities can help us understand why disorder is favored over order. Do mixtures ever spontaneously separate into the chemicals they consist of? Dispersal of Matter

8. “DSuniv = DSsys + DSsurr. So, in a reversible process, DSuniv = 0; in an irreversible (spontaneous) process, DSuniv > 0.” Consequence The Second Law of Thermodynamics

9. Experimentally determining ΔS • To determine the entropy change experimentally, heat transfer must be measured for a reversible process. • A reversible process is any process that exists in equilibrium. (top picture) • In an irreversible process, the surroundings are changed to get the system back to the original state. (bottom picture)

10. Why must we have a reversible process to experimentally measure ΔS? • Well, remember DE = q + w? • If the work term, w, is zero (as it is in a reversible process), then ΔE is equal to q, the heat! • And to determine the entropy change experimentally, heat transfer must be measured for a reversible process.

11. “The entropy of a pure crystalline solid at 0 K is zero.” Consequence The Third Law of Thermodynamics A perfect crystal at 0 K has only one possible way to arrange its components. Tro page 656

12. Standard Molar Entropies can be found in Table 17.2 or in Appendix II-B Tro page 657

13. So is this reaction favored according to entropy? Let’s calculate a DS! Does the following chemical reaction become more or less ordered?N2(g) + O2(g)→ 2 NO(g) Entropy, S (J/K·mol) N2(g) 191.6 O2(g) 205.2 NO(g) 210.8

14. Identifying entropy-favored processes: • Great! We can calculate the ΔS for a reaction. • But, what if we just need to have a general idea and don’t need a specific value? • What processes are generally entropy favored?

15. Entropy Rules of Thumb What has more entropy? • gases > liquids > solids • larger molecules/atoms > smaller molecules/atoms • complex molecules > simple molecules • higher temperature > lower temperature • larger volume > smaller volume • more moles > less moles Tro page 659

16. Predicting Relative Entropy For the pairs listed below, determine which one has the higher entropy. Defend your choice. • 1 mol of SO2(g) or 1 mol of SO3(g) • 1 mol of CO2(g) or 1 mol of CO2(s) • 3 mol of O2(g) or 2 mol of O3(g) • 1 mol of KBr(s) or 1 mol of KBr(aq) • seawater at 2ºC or seawater at 23ºC • 1 mol of CF4(g) or 1 mol of CCl4(g)

17. Your turn! Balance the following equations, then predict the sign of ΔSºrxn if possible. Finally, calculate its value at 25ºC using the entropy values in Appendix II-B. • NaOH(s) + CO2(g)  Na2CO3(s) + H2O(ℓ) • Fe(s) + H2O(g)  Fe2O3(s) + H2(g)

18. NaOH(s) + CO2(g)  Na2CO3(s) + H2O(ℓ) • Fe(s) + H2O(g)  Fe2O3(s) + H2(g)

19. Concept Area III: Will a process be spontaneous? • You should be able to use entropy and enthalpy changes to predict whether a reaction is spontaneous or not. • You should understand the roll of temperature on whether a reaction is spontaneous or not.

20. Quick Review! (already covered) (haven’t covered yet) (finish covering now) • How do we decide the extent of a reaction? • Equilibrium! • How do we predict how fast a reaction will go? • Kinetics! • How do we predict if a reaction will go or is spontaneous if given enough time? • Thermodynamics!

21. Thermodynamics & Kinetics • We will finish learning how to calculate if a reaction is spontaneous or not. • Usually to be spontaneous, a reaction must be thermodynamically favored or ____________. • However, just because a reaction is spontaneous does not mean it happens quickly! For that we need kinetics.

22. Thermodynamics & Kinetics • Let’s look at the kinetics for this spontaneous reaction for instance…does it happen very quickly? _________ C(diamond)  C(graphite) DHrxn= –1.8 kJ/mol Tro page 426

23. Thermodynamics & Kinetics • Can we convert a nonspontaneous reaction to a spontaneous one using a catalyst or other means? • Can we speed up a really slow reaction with a catalyst or other means? • So if something is non-spontaneous , does that mean it will neverhappen? Tro page 643

24. Spontaneous? • We’ve been seeing the word “spontaneous” an awful lot this chapter. What does it mean?

25. When is a reaction thermodynamically spontaneous?

26. Review! If ΔHrxn = +182 kJ, so is that exo- or endothermic?

27. So, is the reaction thermodynamically favored? • For the reaction: N2(g) + O2(g)→ 2 NO(g) • The ΔHrxn = +182 kJ (read off graph on previous slide), which is not spontaneous for enthalpy. • The DSrxn = +24.8 J/K (we calculated on slide 13), which is spontaneous for entropy. • So, is this reaction spontaneous overall?

28. Thermodynamics 2 Handout Do you get it? Let’s see! • Calculate ΔH°rxn and ΔSºrxn for part “a” on the Thermodynamics 2 worksheet.

29. Concept Area IV: Gibbs Free Energy • You should know how entropy changes, enthalpy changes and temperature can be used to calculate the Gibbs free energy for a process. • You should understand how a change in temperature changes the Gibbs free energy of a process. • You should be able to calculate the standard free energy of formation from standard entropy and enthalpy values or from standard free energy values.

30. How can we tell if a reaction will go when “it depends”? Gibbs Free Energy DGrxn = DHrxn – TDSrxn • If DGrxn is … • negative, the reaction is spontaneous. • zero, the reaction is at equilibrium. • positive, the reaction is not spontaneous.

31. Spotlight on J. Willard Gibbs • A little info on Gibbs: • lived 1839-1903 • earned Ph.D. from Yale in 1863, first Ph.D. in science awarded in U.S. • pretty much founded the science of chemical thermodynamics and also made large contributions to equilibria, and electrochemistry • elected to the Hall of Fame of Distinguished Americans in 1950 (took that long to receive enough votes!)

32. What is “free energy”? • Free energy simply means the energy available to do work. • It might seem that all energy in a chemical process is able to do work. • However, if the products are more ordered than the reactants, some of the energy must go to ordering the products. • Thus, ΔG is the total energy given off or needed for a chemical reaction.

33. Let’s calculate a DG! Will the following reaction go spontaneously at 298 K?N2(g) + O2(g)→ 2 NO(g) Previously calculated values: DH +182 kJ DS +24.8 J/K

34. Another DG calculation! Approximately what temperature will this reaction proceed spontaneously?N2(g) + O2(g)→ 2 NO(g) Previously calculated values: DH +182 kJ DS +24.8 J/K Make sense?

35. Your turn! Potassium chlorate, a common oxidizing agent in fireworks and match heads, undergoes a solid-state disproportion-ation reaction when heated: 4 KClO3(s)  3 KClO4(s) + KCl(s) Use ΔHfº and Sº values to calculate ΔGºrxn at 25ºC for this reaction.

36. 4 KClO3(s)  3 KClO4(s) + KCl(s) at 25ºC • First, calculate ΔHºrxn • Next, calculate ΔSºrxn • Finally, calculate ΔGºrxn at 25º

37. Calculating ΔGºrxn another way… • We can also calculate ΔGºrxn from ΔGºf values. • What are ΔGºf values? • Thus it can be calculated with the following equation:

38. ΔGºf Values (also found in Appendix II-B in our text) Tro page 663

39. Let’s calculate a DG! Use ΔGfº values to calculate ΔGºrxn at 25ºC for this reaction.N2(g) + O2(g)→ 2 NO(g) ΔGfº (kJ/mol) N2(g) 0 O2(g) 0 NO(g) 87.6 Let’s compare the answer to what we got before using DGrxn=DHrxn–TDSrxn which was 175 kJ.Are our answers similar?

40. Your turn! Potassium chlorate, a common oxidizing agent in fireworks and match heads, undergoes a solid-state disproportion-ation reaction when heated: 4 KClO3(s)  3 KClO4(s) + KCl(s) Use ΔGfº values to calculate ΔGºrxn at 25ºC for this reaction.

41. 4 KClO3(s)  3 KClO4(s) + KCl(s) • So, calculate ΔGºrxn • Let’s compare the answer to what we got before using DGrxn=DHrxn–TDSrxnwhich was – 133.1 kJ. Are our answers similar?

42. Concept Area V: Application of Thermodynamics to Equilibrium • You should be able to describe the relationship between the free energy change and equilibrium constants. • You should be able to calculate equilibrium constants from DGºrxn. • You should be able to calculate DGrxn at nonstandard conditions.

43. Free Energy & Equilibrium • Remember, if: • K > Q (Q/K < 1) – reaction proceeds to the right • K = Q (Q/K = 1) – reaction is at equilibrium • K < Q (Q/K > 1) – reaction proceeds to the left • And, • ΔG < 0 – reaction spontaneous towards the right • ΔG = 0 – reaction is at equilibrium • ΔG > 0 – reaction spontaneous towards the left • What if we take the natural log of Q/K? Tro page 653

44. Free Energy & Equilibrium • Remember, if: • ln Q/K < 0 – reaction proceeds to the right • ln Q/K = 0 – reaction is at equilibrium • ln Q/K > 0 – reaction proceeds to the left • And, • ΔG < 0 – reaction spontaneous towards the right • ΔG = 0 – reaction is at equilibrium • ΔG > 0 – reaction spontaneous towards the left • Notice some similarities?

45. Free Energy & Equilibrium • So, we can take this information and eventually derive the following useful equations: • We can rearrange the second equation to: to solve for an equilibrium constant! • Now, does everyone remember what R is?

46. forward reaction reverse reaction The relationship Between ΔGº and K at 298 K

47. Let’s calculate a K using ΔG! What is the equilibrium constant at 25ºC for the following reaction?N2(g) + O2(g)→ 2 NO(g) Previously calculated values: DHºrxn +182 kJ DSºrxn +24.8 J/K DGºrxn +175kJ

48. Why not the same?!?! • We just got: N2(g) + O2(g)→ 2 NO(g) Keq= 2.11×10–31 • But, the previous text reported: ½ N2(g) + ½ O2(g)→ NO(g) Keq= 4.59×10–16 • Why the difference? Or, are they really? • Your mission, for next class or later in this class, is to prove that these answers are really not different from each other!

49. Worksheet Time, again! • Let’s finish the Thermodynamics II worksheet.

50. Worksheet Time, again! • Let’s finish the Thermodynamics II worksheet.