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1. Chapter 19: Thermodynamics Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop

2. Thermodynamics • Study of energy changes and flow of energy • Answers several fundamental questions: • Is it possible for a given reaction to occur? • Will the reaction occur spontaneously (without outside interference) at a given T? • Will reaction release or absorb heat? • Tells us nothing about time frame of reaction • Kinetics • Two major considerations • Enthalpy changes, H (heats of reaction) • Heat exchange between system and surroundings • Nature's trend to randomness or disorder • Entropy

3. Review of First Law of Thermodynamics • Internal energy, E • System's total energy • Sum of KE and PE of all particles in system • or for chemical reaction • E + energy into system • E– energy out of system

4. Two Methods of Energy Exchange Between System and Surroundings • Heat qWork w • E = q + w • Conventions of heat and work

5. First Law of Thermodynamics • Energy can neither be created nor destroyed • It can only be converted from one form to another • Kinetic  Potential • Chemical  Electrical • Electrical  Mechanical • E is a state function • E is a change in a state function • Path independent • E = q + w

6. Work in Chemical Systems • Electrical • Pressure-volume orPV • w = –PV • Where P = external pressure • If PV only work in chemical system, then

7. Heat at Constant Volume • Reaction done at constantV • V = 0 • PV = 0, so • E = qV • Entire energy change due to heat absorbed or lost • Rarely done, not too useful

8. Heat at Constant Pressure • More common • Reactions open to atmosphere • Constant P • Enthalpy • H = E + PV • Enthalpy change • H = E + PV • Substituting in first law for E gives • H = (q –PV)+ PV = qP • H = qP • Heat of reaction at constant pressure

9. Converting Between E and H For Chemical Reactions • H E • Differ byH– E = PV • Only differ significantly when gases formed or consumed • Assume gases are ideal • Since P and T are constant

10. Converting Between E and H For Chemical Reactions • When reaction occurs • V caused by n of gas • Not all reactants and products are gases • So redefine as ngas • Where ngas= (ngas)products– (ngas)reactants • Substituting into H = E + PV gives • or

11. Ex. 1 What is the difference between H and E for the following reaction at 25 °C? 2 N2O5(g) 4 NO2(g) + O2(g) What is the % difference between H and E? Step 1: Calculate H using data (Table 7.2) Recall H° = (4 mol)(33.8 kJ/mol) + (1 mol)(0.0 kJ/mol) – (2 mol)(11 kJ/mol) H° = 113 kJ

12. Ex. 1. H and E (cont.) Step 2: Calculate ngas ngas = (ngas)products–(ngas)reactants ngas = (4 + 1 – 2) mol = 3 mol Step 3: Calculate E using R = 8.31451 J/K mol T = 298 K E = 113 kJ – (3 mol)(8.314 J/K mol)(298 K)(1 kJ/1000 J) E = 113 kJ – 7.43 kJ = 106 kJ

13. Ex. 1. H and E (cont.) Step 4: Calculate percent difference Bigger than most, but still small Note: Assumes that volumes of solids and liquids are negligible VsolidVliquid << Vgas

14. Is Assumption that VsolidVliquid << Vgas Justified? • Consider CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O+ CO2(g) 37.0 mL 2×18.0 mL 18 mL 18 mL 24.4 L • Volumes assuming each coefficient equal number of moles • So V = Vprod– Vreac = 24.363 L  24.4 L • Yes, assumption is justified Note: If no gases are present reduces to

15. Learning Check • Consider the following reaction for picric acid: 8O2(g) + 2C6H2(NO2)3OH(l)→ 3 N2(g) + 12CO2(g) + 6H2O(l) • Calculate Η°, Ε° H° = 12 mol(–393.5 kJ/mol) + 6 mol(–241.83 kJ/mol) + 6 mol(0.00 kJ/mol) – 8 mol(0.00 kJ/mol) – 2 mol(3862.94 kJ/mol) ΔH° = –13,898.9 kJ Ε° = H° – ngasRT = H° – (15 – 8) mol × 298 K × 8.314 × 10–3 kJ/(mol K) Ε° = –13,898.9 kJ – 29.0 kJ =–13,927.9 kJ

16. Your Turn! Given the following: 3H2(g) + N2(g) → 2NH3(g) Η°= –46.19 kJ mol–1 Determine E for the reaction. A. –51.14 kJ mol–1 B. –41.23 kJ mol–1 C. –46.19 kJ mol–1 D. –46.60 kJ mol–1 • Η = E + nRTE = Η – nRT • E = –46.19 kJ mol – (–2 mol)(8.314 J K–1mol–1)(298 K)(1 kJ/1000 J) • E = –51.14 kJ mol–1

17. Enthalpy Changes and Spontaneity • What are relationships among factors that influence spontaneity? • Spontaneous Change • Occurs by itself • Without outside assistance until finished • e.g. • Water flowing over waterfall • Melting of ice cubes in glass on warm day

18. Nonspontaneous Change • Occurs only with outside assistance • Never occurs by itself: • Room gets straightened up • Pile of bricks turns into a brick wall • Decomposition of H2O by electrolysis • Continues only as long as outside assistance occurs: • Person does work to clean up room • Bricklayer layers mortar and bricks • Electric current passed through H2O

19. Nonspontaneous Change • Occur only when accompanied by some spontaneous change • You consume food, spontaneous biochemical reactions occur to supply muscle power • to tidy up room or • to build wall • Spontaneous mechanical or chemical change to generate electricity

20. Direction of Spontaneous Change • Many reactions which occur spontaneously are exothermic: • Iron rusting • Fuel burning • H and E are negative • Heat given off • Energy leaving system • Thus, H is one factor that influences spontaneity

21. Direction of Spontaneous Change • Some endothermicreactions occur spontaneously: • Ice melting • Evaporation of water • Expansion of CO2 gas into vacuum • H and E are positive • Heat absorbed • Energy entering system • Clearly other factors influence spontaneity

22. Your Turn! We can expect the combustion of propane to be: A. spontaneous B. non-spontaneous C. neither

23. Entropy (Symbol S) • Thermodynamic quantity • Describes number of equivalent ways that energy can be distributed • Quantity that describes randomness of system • Greater statistical probability of particular state means greater the entropy! • Larger S, means more possible ways to distribute energy and that it is a more probable result

24. Fig. 19.6 - Entropy Distribution • Low Entropy – (a) • A absorbs E in units of 10 • Few ways to distribute E • ●represent E’s of molecules of A • High Entropy – (b) • More ways to distribute E • B absorbs E in units of 5 • ●represent E’s of molecules of B

25. Entropy • If Energy = money • Entropy (S) describes number of different ways of counting it

26. Examples of Spontaneity • Spontaneous reactions • Things get rusty spontaneously • Don't get shiny again • Sugar dissolves in coffee • Stir more—it doesn't undissolve • Ice  liquid water at RT • Opposite does NOT occur • Fire burns wood, smoke goes up chimney • Can't regenerate wood • Common factor in all of these: • Increase in randomness and disorder of system • Something that brings about randomness more likely to occur than something that brings order

27. Entropy, S • State function • Independent of path • S = Change in entropy • For chemical reactions or physical processes

28. Effect of Volume on Entropy • For gases, entropy increases as volume increases • Gas separated from vacuum by partition • Partition removed, more ways to distribute energy • Gas expands to achieve more probable particle distribution • More random, higher probability, more positive S

29. Effect of Temperature on Entropy • As Tincreases, entropy increases (a) T = 0 K, particles in equilibrium lattice positions and S relatively low (b) T> 0 K, molecules vibrate, Sincreases (c) Tincreases further, more violent vibrations occur and Shigher than in (b)

30. Effect of Physical State on Entropy • Crystalline solid very low entropy • Liquid higher entropy,molecules can move freely • More ways to distribute KE among them • Gas highest entropy,particles randomly distributed throughout container • Many, many ways to distribute KE

31. Entropy Affected by Number of Particles • Adding particles to system • Increase number of ways energy can be distributed in system • So all other things being equal • Reaction that produces more particles will have positive S

32. Your Turn! Which represents an increase in entropy? A. Water vapor condensing to liquid B. Carbon dioxide subliming C. Liquefying helium gas D. Proteins forming from amino acids

33. Entropy Changes in Chemical Reactions Reactions without gases • Calculate number of mole molecules n = nproducts – nreactants • If n is positive, entropy increases • More molecules, means more disorder • Usually the side with more molecules, has less complex molecules Reactions involving gases • Calculate change in number of moles of gas, ngas • If ngas is positive , S is positive • ngas is more important than nmolecules

34. Entropy Changes in Chemical Reactions Ex. N2(g) + 3H2(g) 2NH3(g) nreactant = 4 nproduct = 2 n = 2 – 4 = –2 Predict Srxn < 0 Lower positional probability Higher positional probability

35. Ex. 2 Predict Sign of S for Following Reactions CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O + CO2(g) • ngas = 1 mol – 0 mol = 1 mol • sincengas is positive, S is positive 2 N2O5(g) 4 NO2(g) + O2(g) • ngas = 4 mol + 1 mol – 2 mol = 3 mol • sincengas is positive, S is positive OH–(aq) + H+(aq) H2O • ngas = 0 mol • n = 1 mol – 2 mol = –1 mol • sincengas is negative, S is negative

36. Predict Sign of S in the Following: Dry ice → carbon dioxide gas Moisture condenses on a cool window AB → A + B A drop of food coloring added to a glass of water disperses 2Al(s) + 3Br2(l) → 2AlBr3(s) CO2(s) → CO2(g) positive H2O(g) → H2O(l) negative positive positive negative

37. Your Turn! Which of the following has the most entropy at standard conditions? • H2O(l) • NaCl(aq) • AlCl3(s) • Can’t tell from the information

38. Your Turn! Which reaction would have a negative entropy? A. Ag+(aq) + Cl–(aq) → AgCl(s) B. N2O4(g) → 2NO2(g) C. C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(g) D. CaCO3(s) → CaO(s) + CO2(g)

39. Both Entropy and Enthalpy Affect Reaction Spontaneity • Sometimes they work together • Building collapses • PE decreases His negative • Stones disordered S is positive • Sometimes work against each other • Ice melting (ice/water mix) • Endothermic • H is positive nonspontaneous • Increase in disorder of molecules • S is positive spontaneous

40. Which Prevails? • Hard to tell—depends on temperature! • At 25 °C, ice melts • At –25 °C, water freezes • So three factors affect spontaneity: • H • S • T • Next few slides will develop the relationship between H, S,and T that defines a spontaneous process

41. Second Law of Thermodynamics • When a spontaneous event occurs, total entropy of universe increases • (Stotal > 0) • In a spontaneous process, Ssystem can decrease as long as total entropy of universe increases • Stotal = Ssystem + Ssurroundings • It can be shown that

42. Spontaneous Reactions (cont.) Law of Conservation of Energy • Says q lost by system must be gained by surroundings • qsurroundings= –qsystem • If system at constant P, then • qsystem = H • So • qsurroundings = –Hsystem • and

43. Thus Entropy for Entire Universe is Multiplying both sides by T we get TStotal = TSsystem – Hsystem or TStotal = – (Hsystem– TSsystem) • For reaction to be spontaneous • TStotal > 0 (entropy must increase) So, (Hsystem– TSsystem) < 0 must be negative for reaction to be spontaneous

44. Gibbs Free Energy • Would like one quantity that includes all three factors that affect spontaneity of a reaction • Define new state function, G • Gibbs Free Energy • Maximum energy in reaction that is "free" or available to do useful work GH – TS • At constant P and T, changes in free energy G = H – TS

45. G = H – TS G  state function Made up of T, H and S = state functions Has units of energy Extensive property G = Gfinal– Ginitial Gibbs Free Energy

46. Criteria for Spontaneity? • At constant P and T, process spontaneous only if it is accompanied by decrease in free energy of system

47. Summary • When H and S have same sign, T determines whether spontaneous or nonspontaneous • Temperature-controlledreactions are spontaneous at one temperature and not at another

48. Your Turn! At what temperature (K) will a reaction become nonspontaneous when H = –50.2 kJ mol–1 and S = +20.5 J K–1 mol–1? A. 298 K B. 1200 K C. 2448 K D. The reaction cannot become non-spontaneous at any temperature

49. Third Law of Thermodynamics • At absolute zero (0 K) • Entropy of perfectly ordered, pure crystalline substance is zero • S = 0 at T = 0 K • Since S = 0 at T = 0 K • Define absolute entropy of substance at higher temperatures • Standard entropy, S° • Entropy of 1 mole of substance at 298 K (25 °C) and 1 atm pressure • S° = S for warming substance from 0 K to 298 K (25 °C)

50. Consequences of Third Law • All substances have positive entropies as they are more disordered than at 0 K • Heating increases randomness • S° is biggest for gases—most disordered • For elements in their standard states • S°  0 (but Hf° = 0) • Units of S°  J/(mol K) Standard Entropy Change • To calculate S° for reaction, do Hess's Law type calculation • Use S° rather than entropies of formation