1. Thermochemistry The study of heat transfer in chemical rxns

2. Reading Chapter 13 pages 498 - 506 Chapter 15 page 591- 602 HW due Friday November 10 th Chapter 13 p 535: #43, 47, 49, 51 Chapter 15 p 640: #59 HW For tonight:

3. System That part of nature upon which attention is focused

4. Surroundings That part of nature around the part upon which we focus

5. System + Surroundings = The Universe!

6. Reaction Coordinate graph of energy change vs. time in a chemical reaction

9. Exothermic reaction Releases Gives off Loses heat

10. Endothermic Reaction Absorbs Takes in Gains heat

11. Enthalpy  H= q = mc  T Heat flow/change in a system m is for mass! c is for specific heat! ΔT is for change in temp!

12. Grammar of Thermochemistry Exothermic condensation reaction H 2 O (g)  H 2 O (l) + 44kJ H2O (g)  H2O (l) ΔH o = -44 kJ Endothermic evaporation reaction 2 H 2 O (l) + 88 kJ  2 H 2 O (g) 2 H 2 O (l)  2 H 2 O (g) ΔH o = +88 kJ

13. Specific Heat (capacity) c Ability of a specific quantity (1g) of a substance to store heat as its temp rises by 1 o C units  J g * o C

14. Heat Capacity C Ability of a thing to store heat as its temperature rises units  J o C

15. Calorimeter Device that measures Δ heat It tries to be an adiabatic system In real life, gives experimental yield

16. Adiabatic System Does not lose heat to or take heat from surroundings  H system = 0

17. Calorimetry  H system = 0  H sys =  H cal +  H rxn  H rxn = -  H cal  H rxn = mc  T cal

18. 3.358 kJ of heat added to the 50.0 g water inside a calorimeter. T water increases from 22.34 o C to 36.74 o C. What is the heat capacity of the calorimeter in J/ o C? c water = 4.180 J/g * o C ΔT = (36.74 o C – 22.34 o C) = 14.40 o C 50.00g * (4.184 J/g * o C) * 14.40 o C = 3.012 x 10 3 J 3.012 kJ goes into water 3.358 kJ – 3.012 kJ =.346 kJ absorbed by calorimeter.346 kJ = 346 J ÷ 14.40 o C = 24.0 J/ o C

19. 100.0 g of water at 50.0 o C is added to a calorimeter that already contains 100.0g of water at 30.0 o C. The final temperature is 39 o C. What is the heat capacity of the calorimeter? c water = 4.184 J/g * o C ΔT added water = (50.0 o C – 39.0 o C) = 11 o C 100.0g * (4.184 J/g * o C) * 11 o C = 4.60 x 10 3 J ΔT calorimeter water = (39.0 o C – 30.0 o C) = 9 o C 100.0g * (4.184 J/g * o C) * 9 o C = 3.76 x 10 3 J 4.60 kJ – 3.76 kJ =.834 kJ absorbed by calorimeter.834 kJ = 834 J ÷ 9.0 o C = 93 J/ o C

20. Thermodynamics and Δ of state

21. Heat of Fusion H f Heat req’d to melt 1g of a substance at its MP units  J/g or J/kg

22. Heat of Vaporization H v Heat req’d to boil 1g of a substance at its normal BP units  J/g or J/kg

23. Calculate the amount of heat that must be absorbed by 50.0 grams of ice at -12.0 o C to convert it to water at 20.0 o C. c ice = 2.09 J/g * o C H f for ice = 334 J/g c water = 4.184 J/g * o C Step 1 – warm the ice to 0 o C requires: –(50.0 g) (2.09 J/g * o C) (0 o C – (-12 o C)) = 0.125 x 10 4 J Step 2 – melt the ice with no Δ in temp: –50.0 g * 334J/g = 1.67 x 10 4 J Step 3 – warm the liquid to 20.0 o C requires: –50.0 g * 4.18 J/g * o C * (20 o C - 0 o C) =.418 x 10 4 J

24. Answer = 2.21 x 10 4 J = 22.1 kJ must be absorbed

25. Homework Extra Credit Homework Assignment Due Monday November 13 th –# 55 page 535, chapter 13 Homework due Tuesday November 14 th –Chapter 15, page 641 # 61, 63, 67, 69

26. A certain calorimeter absorbs 20 J/ o C. If 50.0 g of 50 o C water is mixed with 50.0 g of 20 o C water inside the calorimeter, what will be the final temperature of the mixture? Heat lost by the hot water will be gained by the cold water and the calorimeter: ΔH hot water = ΔH cool water + ΔH calorimeter ΔH hot water = (50.0 g) (4.180 J/ o C*g) (50 o C – x) = 209 J/ o C (50 o C – x) ΔH cool water = (50.0 g) (4.180 J/ o C*g) (x – 20 o C) = 209 J/ o C (x – 20 o C ) ΔH calorimeter = 20 J/ o C (x – 20 o C )

27. Solve algebraically: 209 (50 – x) = 209 (x – 20) + 24 (x – 20) 209 (50 – x) = 235 (x – 20) 0.889 (50 – x) = x – 20 44 – 0.889x = x – 20 64 = 1.889x x = 33.9 o C = 30 o C

28. A certain calorimeter absorbs 24 J/ o C. If 50.0 g of 52.7 o C water is mixed with 50.0 g of 22.3 o C water inside the calorimeter, what will be the final temperature of the mixture? Heat lost by the hot water will be gained by the cold water and the calorimeter: ΔH hot water = ΔH cool water + ΔH calorimeter ΔH hot water = (50.0 g) (4.180 J/ o C*g) (52.7 o C – x) = 209 J/ o C (52.7 o C – x) ΔH cool water = (50.0 g) (4.180 J/ o C*g) (x – 22.3 o C) = 209 J/ o C (x – 22.3 o C ) ΔH calorimeter = 24 J/ o C (x – 22.3 o C )

29. Solve algebraically: 209 (52.7 – x) = 209 (x – 22.3) + 24 (x – 22.3) 209 (52.7 – x) = 235 (x – 22.3) 0.889 (52.7 – x) = x – 22.3 46.87 – 0.889x = x – 22.3 69.17 = 1.889x x = 36.6 o C = 37 o C

30. Heat of Reaction  H rxn Heat/enthalpy change of a chemical reaction Units  J or kJ Sometimes, units  J/mol rxn

31. Mole of reaction Depends on how it is given in the problem (or how you balance your reaction) Can say that O 2 (g) + 2 H 2 (g)  2 H 2 O (g) + 45 kJ ΔH rxn = 45 kJ/mol rxn You can use the following conversion factors: 1 mol O 2 2 mol H 2 2 mol H 2 O 1 mol rxn 45 kJ 45 kJ 45 kJ 45 kJ

32. When X reacts with water the temp in a 1.5 kg calorimeter containing 2.5 kg water went from 22.5 o C to 26.5 o C. Calculate  H rxn. c water = 4.18 J/g o C c calorimeter = 2.00 J/g o C

33. ΔH rxn = ΔH water + ΔH calorimeter Δ T = 26.5 o C – 22.5 o C = 4 o C Heat absorbed by water: Δ H water = mc ΔT –2.5 kg = 2,500 g –(2,500 g)(4.18J/g* o C)(4 o C) = 41,800 J = 41.8 kJ Heat absorbed by calorimeter: Δ H calorimeter = mc ΔT –1.5 kg = 1,500 g –(1,500 g)(2.00 J/g* o C)(4 o C) = 12,000 J = 12 kJ Total heat added to system = 41.8 + 12 = 53.8 kJ 54 kJ

34. Heat of Solution  H soln The heat or enthalpy change when a substance is dissolved

35. 80 g NaOH is dissolved with 1.40 L of 0.7 M HCl in a calorimeter. HCl solution has a mass of 1.4 kg or 1,400g. C calorimeter = 20 J/ o C  water = 10 o C c HCl same as c water = 4.18 J/g* o C What is the heat released by the solution What is the  H solution for the reaction: –NaOH (s) + HCl (aq)  NaCl (aq) + H 2 O (l)

36. Heat absorbed by calorimeter: –20 J/ o C * 10 o C = 200 J Heat absorbed by HCl solution: –1,400 g * (4.18 J/g* o C) * (10 o C) = 58,520 J H solution = H calorimeter + H HCl solution 200 J + 58,520 J = 58,720 J Heat released by solution = 58,720 J = 59 kJ Go back and see how many moles of NaOH & HCl reacted: 80 g NaOH is 2 moles – therefore you have 2 moles rxn  H solution = 59 kJ/2 mol rxn = 30 kJ/mol rxn

37. Change! To the HW Due Wednesday Chapter 15, page 637: 13 & 15 Due Thursday November 16th Chapter 15, page 637 – 8: 25, 27, 29, 31

38. When 2.61 g of C 2 H 6 O is burned at constant pressure,82.5 kJ of heat is given off. What is ΔH for the reaction: C 2 H 6 O (l) + O 2 (g)  2 CO 2 (g) + 3 H 2 O (l) 82.5 kJ 46.0 g C 2 H 6 O 1 mol C 2 H 6 O 2.61 g C 2 H 6 O mol C 2 H 6 O mol rxn ΔH for the reaction = -1450 kJ/mol rxn

39. When Al metal is exposed to O 2 it is oxidized to form Al 2 O 3. How much heat is released by the complete oxidation of 24.2 g of Al at 25 o C and 1 atm? 4 Al (s) + 3 O 2 (g)  2 Al 2 O 3 (s) ΔH = -3352 kJ/mol rxn 24.2 g Al 1 mol Al 1 mol rxn -3352 kJ 27 g Al 4 mol Al mol rxn -751 kJ = 751 kJ of heat are released

40. Heat of Combustion  H combustion The heat or enthalpy change when a substance is burned

41. Heat of Formation  H f o The heat req’d to form 1 mol of a compound from pure elements units  kJ/mole

42. Gibb’s Free Energy  G o Energy of a system that can be converted to work Determines spontaneity

43. Energy of Formation  G f o The energy req’d to form 1 mol of a compound from pure elements units  kJ/mole

44. Exergonic Reaction A reaction in which free energy is given off  G < 0

45. Endergonic Reaction A reaction in which free energy is absorbed  G > 0

46. Reaction at Equilibrium  G = 0

47. Entropy A measure of disorder  S o

48. Entropy of Formation The entropy change when one mole of a substance is formed S f o (J/mole o K)

49. Thermochemical Equation An equation that shows changes in heat, energy, etc

50. Thermochemical Equation  H o rxn   H f o products   H f o reactants

51. Thermochemical Equation  G o rxn   G f o products   G f o reactants

52. Thermochemical Equation  S o rxn  S f o products  S f o reactants

53. Thermochemical Equation Stoichiometry of heat change Solves theoretical yield

54. Interrelating Equation  G  H  S

55. Calculate  H,  G, &  S when 19.7 kg of BaCO 3 is decomposed into BaO + CO 2 Cmpd BaCO 3 CO 2. BaO  H f o -1216.3 -393.5 -553.5  G f o -1137.6 -394.4 -525.1 S f o 112.1 213.6 70.4

56. Calculate  H,  G, &  S when 13.6 g of CaSO 4 is changed into CaO + SO 2 + O 2 at 27 o C Cmpd CaSO 4 SO 2 CaO  H f o -1434.1 -296.8 -635.1  G f o -1321.8 -300.2 -604.0

57. Lab Results: Cup H 2 O NaOH Thermo 5.0 g 50.0 g 4.0 g 15.0 g T i = 22.0 o CT f = 27.0 o C Cmpd NaOH Na + OH -  H f o -425.6 -240.1 -230.0 Determine: theoretical and experimental heat changes, &

58. Calculate the potential  H,  G, &  S for the reaction & S f o for O 2 when burning 8.8 kg of C 3 H 8 Cpd C 3 H 8 CO 2 H 2 O  H f o -103.8 -393.5-241.8  G f o - 23.5 -394.4-228.6 S f o 269.9 213.6 188.7

59. Bond Energy The energy change when one mole of bonds are broken  H o bond

60. Bond Equation  H bond o rxn   H bond o products  H bond o reactants

61. Bond Energies (kJ/mole) C-C347 C-H414 O-H464 C=O715

62. Calculate  H,  G, &  S in the production of 831ML ammonia at 227 o C under 125 kPa pressure Compd NH 3  H f o -46.1  G f o -16.5

63. 1st Law Thermodynamics Total energy change = heat + work  E = q + W

64. Work W = Fd P = F/A V = Ad W = P  V =  nRT

65. 2nd Law Thermodynamics Total entropy in a system always increases assuming no energy is added to the system

66. Thermodynamic Rxns are State Rxns

67. State Reaction Reactions that are independent of the path; thus not dependent on intermediates

68. Calculate  H o,  G o, &  S when A + BC  AC 2 + B at -23 o C & T eq Compd BC AC 2  H f o (kJ/mole) -150-250  G f o (kJ/mole)-125-225

69. Write TE for the process 2 A + BC + D C + AH D + B 2 K H + KM + B K + MProduct

70. Write TE for the process 2 A + 2 BC + D C + A2 H D + B 2 K H + KP + B

71. Write TE for the process 2 A + 2 BC + D C + A3 H D + B 2 K H + KP + Q

72. Review

73. Calculate  H total, when 40.0 g of H 2 O is changed from - 25 o C to 125 o C. FP w = 0.0 o C BP w = 100.0 o C H v = 2260 J/g C ice = 2.06 (J/g K) H f = 330 J/g C water = 4.18 (J/g K) C steam = 2.02 (J/g K)

74. Calculate  H o,  G o, &  S for AD 2 + BCAC 2 + BD at (-23 o C) Cpd BC AD 2 AC 2 BD  H f o -150 -250 -300 -175  G f o -125 -225 -250 -150 S f o 75 50 80 ? Determine S f o BD

75. Calculate  H o,  G o, &  S o for PbO 2 + COCO 2 + Pb Cpd PbO 2 CO CO 2  H f o -277.4 -110.5 -393.5  G f o -217.4 -137.2 -394.4 Calculate: T eq &  H of 48 g PbO 2