1 / 75

Thermochemistry

Thermochemistry. The study of heat transfer in chemical rxns. 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:. System. That part of nature upon which attention is focused.

livia
Download Presentation

Thermochemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related