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Chapter 6

Chapter 6. Thermochemistry. Thermite Reaction. Thermite Reaction. Terminology. Energy capacity to do work Kinetic Energy energy that something has because it is moving Potential Energy energy that something has because of its position. Kinetic Energy. Chemical Potential Energy.

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Chapter 6

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  1. Chapter 6 Thermochemistry Dr. S. M. Condren

  2. Thermite Reaction Dr. S. M. Condren

  3. Thermite Reaction Dr. S. M. Condren

  4. Terminology Energy • capacity to do work Kinetic Energy • energy that something has because it is moving Potential Energy • energy that something has because of its position Dr. S. M. Condren

  5. Kinetic Energy Dr. S. M. Condren

  6. Chemical Potential Energy Dr. S. M. Condren

  7. Chemical Potential Energy Dr. S. M. Condren

  8. Internal Energy • The sum of the individual energies of all nanoscale particles (atoms, ions, or molecules) in that sample. • E = 1/2mc2 • The total internal energy of a sample of matter depends on temperature, the type of particles, and how many of them there are in the sample. Dr. S. M. Condren

  9. Energy Units • calorie - energy required to heat 1-g of water 1oC • Calorie - unit of food energy; 1 Cal = 1-kcal = 1000-cal • Joule - 1-cal = 4.184 J = 1-kg*m2/sec2 Dr. S. M. Condren

  10. Law of Conservation of Energy • energy can neither be created nor destroyed • the total amount of energy in the universe is a constant • energy can be transformed from one form to another Dr. S. M. Condren

  11. First Law of Thermodynamics • the amount of heat transferred into a system plus the amount of work done on the system must result in a corresponding increase of internal energy in the system Dr. S. M. Condren

  12. Thermochemistry Terminology system => that part of the universe under investigation surroundings => the rest of the universe universe = system + surroundings Dr. S. M. Condren

  13. Thermodynamic System Dr. S. M. Condren

  14. Energy Transfer Energy is always transferred from the hotter to the cooler sample Heat – the energy that flows into or out of a system because of a difference in temperature between the thermodynamic system and its surroundings Dr. S. M. Condren

  15. Thermochemistry Terminology state properties => properties which depend only on the initial and final states => properties which are path independent non-state properties => properties which are path dependent state properties => E non-state properties => q & w Dr. S. M. Condren

  16. Thermochemistry Terminology exothermic - reaction that gives off energy endothermic - reaction that absorbs energy chemical energy - energy associated with a chemical reaction thermochemistry - the quantitative study of the heat changes accompanying chemical reactions thermodynamics - the study of energy and its transformations Dr. S. M. Condren

  17. Energy & Chemistry 2 H2(g) + O2(g) --> 2 H2O(g) + heat and light This can be set up to provide ELECTRIC ENERGY in a fuel cell. Oxidation: 2 H2 ---> 4 H+ + 4 e- Reduction: 4 e- + O2 + 2 H2O ---> 4 OH- Dr. S. M. Condren

  18. Energy & Chemistry Dr. S. M. Condren

  19. Enthalpy • heat at constant pressure qp = DH = Hproducts - Hreactants Exothermic Reaction DH = (Hproducts - Hreactants) < 0 H2O(l) -----> H2O(s)DH < 0 Endothermic Reaction DH = (Hproducts - Hreactants) > 0 H2O(l) -----> H2O(g)DH > 0 Dr. S. M. Condren

  20. Enthalpy H = E + PV DH = DE + PDV DE = DH – PDV Where text uses U for internal energy Dr. S. M. Condren

  21. Pressure-Volume Work Dr. S. M. Condren

  22. First Law of Thermodynamics heat => q internal energy => E internal energy change =>DE work => w DE = q - w (Engineering convention) Dr. S. M. Condren

  23. Specific Heat • the amount of heat necessary to raise the temperature of 1 gram of the substance 1oC • independent of mass • substance dependent • s.h. • Specific Heat of Water = 4.184 J/goC Dr. S. M. Condren

  24. Heat q = m * s.h. * Dt where q => heat, J m => mass, g s.h. => specific heat, J/g*oC Dt = change in temperature, oC Dr. S. M. Condren

  25. Molar Heat Capacity • the heat necessary to raise the temperature of one mole of substance by 1oC • substance dependent • C Dr. S. M. Condren

  26. Heat Capacity • the heat necessary to raise the temperature 1oC • mass dependent • substance dependent • C Dr. S. M. Condren

  27. Heat Capacity C = m X s.h. where C => heat capacity, J/oC m => mass, g s.h. => specific heat, J/goC Dr. S. M. Condren

  28. Plotted are graphs of heat absorbed versus temperature for two systems. Which system has the larger heat capacity? A, B Dr. S. M. Condren

  29. Heat Transfer qlost = - qgained (m X s.h. X Dt)lost = - (m X s.h. X Dt)gained Dr. S. M. Condren

  30. EXAMPLEIf 100. g of iron at 100.0oC is placed in 200. g of water at 20.0oC in an insulated container, what will the temperature, oC, of the iron and water when both are at the same temperature? The specific heat of iron is 0.106 cal/goC. (100.g*0.106cal/goC*(Tf - 100.)oC) = qlost - qgained = (200.g*1.00cal/goC*(Tf - 20.0)oC) 10.6(Tf - 100.oC) = - 200.(Tf - 20.0oC) 10.6Tf - 1060oC = - 200.Tf + 4000oC (10.6 + 200.)Tf = (1060 + 4000)oC Tf = (5060/211.)oC = 24.0oC Dr. S. M. Condren

  31. EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC? q = DHice + DHfusion + DHwater + DHboil. + DHsteam q = DHice + DHfusion + DHwater + DHboil. + DHsteam Dr. S. M. Condren

  32. Heat Transfer Dr. S. M. Condren

  33. EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC?q = DHice + DHfusion + DHwater + DHboil. + DHsteam q = (10.0g*2.09J/goC*((0.0 – (-15.0))oC)) { specific heat of ice Mass of the ice Temperature change Dr. S. M. Condren

  34. EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC?q = DHice + DHfusion + DHwater + DHboil. + DHsteam q = (10.0g*2.09J/goC*15.0oC) + (10.0g*333J/g) Melting of ice occurs at a constant temperature Mass of ice Heat of fusion Dr. S. M. Condren

  35. EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC?q = DHice + DHfusion + DHwater + DHboil. + DHsteam q = (10.0g*2.09J/goC*15.0oC) + (10.0g*333J/g) + (10.0g*4.18J/goC*((100.0-0.00)oC)) Mass of water Specific heat of liquid water Temperature change of the liquid water Dr. S. M. Condren

  36. EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC?q = DHice + DHfusion + DHwater + DHboil. + DHsteam q = (10.0g*2.09J/goC*15.0oC) + (10.0g*333J/g) + (10.0g*4.18J/goC*100.0oC) + (10.0g*2260J/g) Boiling of water occurs at a constant temperature Mass of water Heat of vaporization Dr. S. M. Condren

  37. EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC?q = DHice + DHfusion + DHwater + DHboil. + DHsteam q = (10.0g*2.09J/goC*15.0oC) + (10.0g*333J/g) + (10.0g*4.18J/goC*100.0oC) + (10.0g*2260J/g) + (10.0g*2.03J/goC*((127.0-100.0)oC)) Specific heatof steam Temperature change for the steam Mass of steam Dr. S. M. Condren

  38. EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC?q = DHice + DHfusion + DHwater + DHboil. + DHsteam q = (10.0g*2.09J/goC*15.0oC) + (10.0g*333J/g) + (10.0g*4.18J/goC*100.0oC) + (10.0g*2260J/g) + (10.0g*2.03J/goC*27.0oC) q = (314 )J + 3.33X103 + 4.18X103 + 2.26X104 + 548 = 30.96 kJ Dr. S. M. Condren

  39. Spreadsheet of Previous Problem Dr. S. M. Condren

  40. Parr Bomb Calorimeter calorimeter Dr. S. M. Condren

  41. EXAMPLE A 1.000g sample of a particular compound produced 11.0 kJ of heat. The temperature of the calorimeter and 3000 g of water was raised 0.629oC. How much heat is gained by the calorimeter? heat gained = - heat lost heatcalorimeter + heatwater = heatreaction heatcalorimeter = heatreaction - heatwater Dr. S. M. Condren

  42. EXAMPLE A 1.000g sample of a particular compound produced 11.0 kJ of heat. The temperature of the calorimeter and 3000 g of water was raised 0.629oC. How much heat is gained by the calorimeter? heatcalorimeter = heatreaction - heatwater heat = 11.0 kJ - ((3.00kg)(0.629oC)(4.184kJ/kgoC)) = 3.1 kJ Dr. S. M. Condren

  43. Example What is the mass of water equivalent of the heat absorbed by the calorimeter? #g = (3.1 kJ/0.629oC)(1.00kg*oC/4.184kJ) = 6.5 x 102 g Dr. S. M. Condren

  44. Example A 1.000 g sample of ethanol was burned in the sealed bomb calorimeter described above. The temperature of the water rose from 24.284oC to 26.225oC. Determine the heat for the reaction. m = (3000 + "647")g H2O q = m X s.h. X Dt = (3647g)(4.184J/goC)(1.941oC) = 29.61 kJ Dr. S. M. Condren

  45. When graphite is burned to yield CO2, 394 kJ of energy are released per mole of C atoms burned. When C60 is burned to yield CO2 approximately 435 kJ of energy is released per mole of carbon atoms burned. Would the buckyball-to-graphite conversion be exothermic or endothermic? exothermic, endothermic Dr. S. M. Condren

  46. Laws of Thermochemistry 1. The magnitude of DH is directly proportional to the amount of reactant or product. s --> l DH => heat of fusion l --> g DH => heat of vaporization Dr. S. M. Condren

  47. Laws of Thermochemistry 2. DH for a reaction is equal in magnitude but opposite in sign to DH for the reverse reaction. H2O(l) -----> H2O(s)DH < 0 H2O(s) -----> H2O(l)DH > 0 Dr. S. M. Condren

  48. Laws of Thermochemistry 3. The value of H for the reaction is the same whether it occurs directly or in a series of steps. DHoverall = DH1 + DH2 + DH3 + · · · Dr. S. M. Condren

  49. Hess' Law • a relation stating that the heat flow in a reaction which is the sum of a series of reactions is equal to the sum of the heat flows in those reactions Dr. S. M. Condren

  50. EXAMPLECH4(g) + 2 O2(g) -----> CO2(g) + 2 H2O(l) CH4(g) -----> C(s) + 2 H2(g)DH1 2 O2(g) -----> 2 O2(g)DH2 C(s) + O2(g) -----> CO2(g)DH3 2 H2(g) + O2(g) -----> 2 H2O(l)DH4 --------------------------------------------- CH4(g) + 2 O2(g) -----> CO2(g) + 2 H2O(l) DHoverall = DH1 + DH2 + DH3 + DH4 Dr. S. M. Condren

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