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Terminology

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. Energy Units. 1J = 1 kg m 2 /sec 2 1 cal = 4.184 J 1kcal = 1 Cal thus 1 Cal = 1 kcal = 1000 cal = 4.184 kJ = 4184 J.

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Terminology

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  1. 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

  2. Energy Units 1J = 1 kg m2/sec2 1 cal = 4.184 J 1kcal = 1 Cal thus 1 Cal = 1 kcal = 1000 cal = 4.184 kJ = 4184 J

  3. 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

  4. First Law of Thermodynamics • the total amount of energy in the universe is a constant • 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

  5. Energy Transformations

  6. Thermal Energy

  7. Measuring Temperature

  8. Heat TransferEnergy is always transferred from the hotter to the cooler sample

  9. 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

  10. Thermochemistry Terminology 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

  11. Thermochemistry Terminology system  that part of the universe under investigation surroundings  the rest of the universe universe = system + surroundings

  12. Internal Energy

  13. First Law of Thermodynamics heat  q internal energy  E internal energy change  DE work  w DE = q + w

  14. Internal Energy, Heat, and Work

  15. Specific Heat • the amount of heat necessary to raise the temperature of 1 gram of the substance 1°C • independent of mass • substance dependent • s.h. • Specific Heat of Water = 4.184 J/g°C

  16. Specific Heat Capacity

  17. Heat q = m  s.h. Dt where q  heat, J m  mass, g s.h.  specific heat, J/g  C Dt = change in temperature, C

  18. Molar Heat Capacity • the heat necessary to raise the temperature of one mole of substance by 1C • substance dependent • C

  19. Heat Capacity • the heat necessary to raise the temperature 1C • mass dependent • substance dependent • C

  20. Heat Capacity C = m  s.h. where C  heat capacity, J/C m  mass, g s.h.  specific heat, J/gC

  21. Hot and Cold Iron

  22. Freezing and Melting

  23. Heat Transfer qlost = - qgained (m  s.h. Dt)lost = - (m  s.h. Dt)gained Dt = final temperature – initial temperature Dt = tf - ti

  24. Heating, Temperature Change, and Phase Change

  25. 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

  26. Vaporization and Condensation

  27. EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC? q ice = m  s.h. Dtice 2.09J/goC qwater =m  s.h. Dtwater 4.18J/goC qsteam = m  s.h. Dtsteam 2.03J/goC qfusion = m  heat of fusion333J/g qboil. = m  heat of vaporization 2260J/g qoverall = qice + qfusion + qwater + qboil. + qsteam

  28. EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC?qoverall = qice + qfusion + qwater + qboil. + qsteam 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 + 3.33X103 + 4.18X103 + 2.26X104 + 548)J = 23.3 kJ

  29. Heat Flow in Reactions exothermic –reaction that gives off energy q < 0 endothermic – reaction that absorbs energy q > 0

  30. Expansion Type Work w = PDV DV = Vinitial + Vincrease P Vincrease P Vinitial qp = +2kJ

  31. First Law of Thermodynamics DE = q + w at constant V, wexpansion = 0 DE = qv + 0 at constant P, wexpansion = PDV DE = qp + PDV qp = DH = DE - PDV

  32. Enthalpy 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

  33. Standard State Enthalpy • enthalpy at thermodynamic standard conditions of 298 K, 1 atm, and 1molar for solutions

  34. 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

  35. Path Independent Energy Changes

  36. Enthalpy Diagram

  37. Bond Energy • the energy associated with holding 2 atoms together

  38. Stepwise Energy Changesin Reactions

  39. Bomb Calorimeter

  40. EXAMPLE A 1.000g sample of a compound was burned in an oxygen bomb calorimeter. It produced 12.0 kJ of heat. The temperature of the calorimeter and 2000 g of water was raised 4.645oC. How much heat is gained by the calorimeter? heat gained = – heat lost heatcalorimeter + heatwater = heatreaction heatcalorimeter = heatreaction - heatwater

  41. EXAMPLE A 1.000g sample of a compound was burned in an oxygen bomb calorimeter. It produced 42.0 kJ of heat. The temperature of the calorimeter and 2000 g of water was raised 4.645oC. How much heat is gained by the calorimeter? heatcalorimeter = heatreaction - heatwater heat = 42.0 kJ - ((2.000kg)(4.645oC)(4.184kJ/kgoC)) = 3.13 kJ

  42. Example What is the mass of water equivalent of the heat absorbed by the calorimeter? #g = (3.13 kJ/4.645oC)(1.00kg  C/4.184kJ) = 1.61 x 102 g

  43. Example A 1.000 g sample of ethanol (MM 46.07) was burned in the sealed bomb calorimeter described above. The temperature of the water rose from 24.284oC to 27.559oC. Determine the heat for the reaction. m = (2000 + ”161")g H2O Dt = 27.559oC - 24.284oC = 3.275oC q = m  s.h. Dt q= (2161g)(4.184J/goC)(3.275oC) = 29.61 kJ q = (29.61 kJ/1.000g)(46.07 g/mol) = 1364 kJ/mol

  44. “Coffee Cup” Calorimeter Photo by George Lisensky

  45. 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

  46. 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

  47. 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 + · · · also called Hess’ Law

  48. 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

  49. Standard Enthalpy of Formation the enthalpy associated with the formation of a substance from its constituent elements under standard state conditions

  50. Calculation of DHo DHo = Sc  DHfoproducts – Sc  DHforeactants

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