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Thermochemistry

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  1. Thermochemistry The study of heat transfer in chemical rxns

  2. ReadingChapter 13 pages 498 - 506Chapter 15 page 591- 602HW due Friday November 10thChapter 13 p 535: #43, 47, 49, 51Chapter 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

  7. Exothermic reaction Releases Gives off Loses heat

  8. Endothermic Reaction Absorbs Takes in Gains heat

  9. EnthalpyDH= q = mcDT m is for mass! Heat flow/change in a system c is for specific heat! ΔT is for change in temp!

  10. Grammar of Thermochemistry Exothermic condensation reaction H2O (g)  H2O (l) + 44kJ H2O (g)  H2O (l) ΔHo = -44 kJ Endothermic evaporation reaction 2 H2O (l) + 88 kJ  2 H2O (g) 2 H2O (l)  2 H2O (g) ΔHo = +88 kJ

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

  12. Heat CapacityC Ability of a thing to store heat as its temperature rises units  J oC

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

  14. Adiabatic System Does not lose heat to or take heat from surroundings DHsystem = 0

  15. Calorimetry DHsystem = 0 DHsys = DHcal + DHrxn DHrxn = -DHcal DHrxn = mcDTcal

  16. 3.358 kJ of heat added to the 50.0 g water inside a calorimeter. Twater increases from 22.34oC to 36.74oC. What is the heat capacity of the calorimeter in J/oC? • cwater = 4.180 J/g * oC • ΔT = (36.74oC – 22.34oC) = 14.40oC • 50.00g * (4.184 J/g * oC) * 14.40oC = 3.012 x 103 J • 3.012 kJ goes into water • 3.358 kJ – 3.012 kJ = .346 kJ absorbed by calorimeter • .346 kJ = 346 J ÷ 14.40oC = 24.0 J/oC

  17. 100.0 g of water at 50.0oC is added to a calorimeter that already contains 100.0g of water at 30.0oC. The final temperature is 39oC. What is the heat capacity of the calorimeter? • cwater = 4.184 J/g * oC • ΔTadded water = (50.0oC – 39.0oC) = 11oC • 100.0g * (4.184 J/g * oC) * 11oC = 4.60 x 103 J • ΔTcalorimeter water = (39.0oC – 30.0oC) = 9oC • 100.0g * (4.184 J/g * oC) * 9oC = 3.76 x 103 J • 4.60 kJ – 3.76 kJ = .834 kJ absorbed by calorimeter • .834 kJ = 834 J ÷ 9.0oC = 93 J/oC

  18. Thermodynamics and Δ of state

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

  20. Heat of VaporizationHv Heat req’d to boil 1g of a substance at its normal BP units  J/g or J/kg

  21. Calculate the amount of heat that must be absorbed by 50.0 grams of ice at -12.0oC to convert it to water at 20.0oC. • cice = 2.09 J/g * oC Hf for ice = 334 J/g • cwater = 4.184 J/g * oC • Step 1 – warm the ice to 0oC requires: • (50.0 g) (2.09 J/g * oC) (0oC – (-12oC)) = 0.125 x 104 J • Step 2 – melt the ice with no Δin temp: • 50.0 g * 334J/g = 1.67 x 104 J • Step 3 – warm the liquid to 20.0oC requires: • 50.0 g * 4.18 J/g * oC * (20 oC - 0 oC) = .418 x 104 J

  22. Answer = 2.21 x 104 J = 22.1 kJ must be absorbed

  23. Homework • Extra Credit Homework AssignmentDue Monday November 13th • # 55 page 535, chapter 13 • Homework due Tuesday November 14th • Chapter 15, page 641 # 61, 63, 67, 69

  24. A certain calorimeter absorbs 20 J/oC. If 50.0 g of 50oC water is mixed with 50.0 g of 20oC 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: ΔHhot water = ΔHcool water + ΔHcalorimeter ΔHhot water = (50.0 g) (4.180 J/oC*g) (50oC – x) = 209J/oC (50oC – x) ΔHcool water = (50.0 g) (4.180 J/oC*g) (x – 20oC) =209 J/oC (x – 20oC) ΔHcalorimeter= 20 J/oC (x – 20oC)

  25. 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 = 33.9oC = 30oC

  26. A certain calorimeter absorbs 24 J/oC. If 50.0 g of 52.7oC water is mixed with 50.0 g of 22.3oC 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: ΔHhot water = ΔHcool water + ΔHcalorimeter ΔHhot water = (50.0 g) (4.180 J/oC*g) (52.7oC – x) = 209J/oC (52.7oC – x) ΔHcool water = (50.0 g) (4.180 J/oC*g) (x – 22.3oC) =209 J/oC (x – 22.3oC) ΔHcalorimeter= 24 J/oC (x – 22.3oC)

  27. 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 = 36.6oC = 37oC

  28. Heat of ReactionDHrxn Heat/enthalpy change of a chemical reaction Units  J or kJ Sometimes, units  J/mol rxn

  29. Mole of reaction • Depends on how it is given in the problem (or how you balance your reaction) • Can say that O2 (g) + 2 H2 (g)  2 H2O (g) + 45 kJ • ΔHrxn = 45 kJ/mol rxn • You can use the following conversion factors: 1 mol O2 2 mol H22 mol H2O1 mol rxn 45 kJ 45 kJ 45 kJ 45 kJ

  30. When X reacts with water the temp in a 1.5 kg calorimeter containing 2.5 kg water went from 22.5oC to 26.5oC. Calculate DHrxn. cwater = 4.18 J/g oC ccalorimeter = 2.00 J/g oC

  31. ΔHrxn = ΔHwater + ΔHcalorimeter Δ T = 26.5oC – 22.5oC = 4oC Heat absorbed by water: Δ Hwater = mc ΔT • 2.5 kg = 2,500 g • (2,500 g)(4.18J/g*oC)(4oC) = 41,800 J = 41.8 kJ Heat absorbed by calorimeter: Δ Hcalorimeter = mc ΔT • 1.5 kg = 1,500 g • (1,500 g)(2.00 J/g*oC)(4oC) = 12,000 J = 12 kJ Total heat added to system = 41.8 + 12 = 53.8 kJ 54 kJ

  32. Heat of SolutionDHsoln • The heat or enthalpy change when a substance is dissolved

  33. 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. • Ccalorimeter = 20 J/oCDTwater = 10oC • cHCl same as cwater = 4.18 J/g*oC • What is the heat released by the solution • What is the DHsolution for the reaction: • NaOH (s) + HCl (aq)  NaCl (aq) + H2O (l)

  34. Heat absorbed by calorimeter: • 20 J/oC * 10oC = 200 J • Heat absorbed by HCl solution: • 1,400 g * (4.18 J/g*oC) * (10oC) = 58,520 J • Hsolution = Hcalorimeter + HHCl 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 • DHsolution = 59 kJ/2 mol rxn = 30 kJ/mol rxn

  35. 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

  36. When 2.61 g of C2H6O is burned at constant pressure,82.5 kJ of heat is given off. What is ΔH for the reaction: C2H6O (l)+ O2 (g) 2 CO2 (g) + 3 H2O (l) • 82.5 kJ 46.0 g C2H6O1 mol C2H6O 2.61 g C2H6Omol C2H6O mol rxn • ΔH for the reaction = -1450 kJ/mol rxn

  37. When Al metal is exposed to O2 it is oxidized to form Al2O3. How much heat is released by the complete oxidation of 24.2 g of Al at 25oC and 1 atm? 4 Al (s) + 3 O2 (g) 2 Al2O3(s)ΔH = -3352 kJ/mol rxn • 24.2 g Al 1 mol Al1 mol rxn-3352 kJ 27 g Al 4 mol Al mol rxn • -751 kJ = 751 kJ of heat are released

  38. Heat of CombustionDHcombustion • The heat or enthalpy change when a substance is burned

  39. Heat of FormationDHfo • The heat req’d to form 1 mol of a compound from pure elements units  kJ/mole

  40. Gibb’s Free EnergyDGo • Energy of a system that can be converted to work • Determines spontaneity

  41. Energy of FormationDGfo The energy req’d to form 1 mol of a compound from pure elements units kJ/mole

  42. ExergonicReaction • A reaction in which free energy is given off DG< 0

  43. Endergonic Reaction • A reaction in which free energy is absorbed DG> 0

  44. Reaction at Equilibrium DG= 0

  45. Entropy • A measure of disorder DSo

  46. Entropy of Formation • The entropy change when one mole of a substance is formed • Sfo (J/moleoK)

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

  48. Thermochemical Equation DHorxn = • SDHfoproducts - • SDHforeactants