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Energy and Chemical Reactions: Understanding Kinetic and Potential Energy

Learn about the different forms of energy, including kinetic and potential energy, and how they relate to chemical reactions. Explore concepts such as temperature, heat transfer, and the first law of thermodynamics. Discover the specific heat capacity and molar heat capacity of substances, and understand the processes of phase change and heat flow.

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Energy and Chemical Reactions: Understanding Kinetic and Potential Energy

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  1. Chapter 6 Energy and Chemical Reactions

  2. Kinetic Energy energy that something has because it is moving Potential Energy energy that something has because of its position Molecular Kinetic Energy energy of motion of molecules = temperature Chemical Potential Energy reactive molecules have the potential to form stable ones and release energy Macroscale Nanoscale

  3. 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

  4. First Law of Thermodynamics • 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

  5. Measuring Temperature K.E. of motion  Temperature

  6. Energy Transfer Energy is always transferred from the hotter to the cooler sample

  7. Heat Transfer Energy is always transferred from the hotter to the cooler sample

  8. Internal Energy E • the sum of the individual energies of all nanoscale particles (atoms, ions, or molecules) in that sample • including all chemical bonds and intermolecular interactions • we cannot measure the internal energy accurately • but chemistry depends only on changes in internal energy E

  9. Thermochemistry Divide universe into two parts system  that part of the universe under investigation surroundings  the rest of the universe universe = system + surroundings consider everything from the system’s viewpoint

  10. Internal Energy

  11. First Law of Thermodynamics System can exchange energy with surroundings via heat or work heat  q work  w internal energy change  DE DE = q + w

  12. 130 10 0 0 Combustion of gasoline in your car engine results in the production of 124 kJ of energy. If the engine does 31 kJ of work to move the car how much heat must be released by the radiator? • 31 kJ • 124 kJ • 93 kJ • 155 kJ

  13. Internal Energy, Heat, and Work

  14. 130 8 0 0 Which of these changes always results in a decrease in the internal energy of the system? • The system absorbs heat and does work on the surroundings. • The system releases heat and does work on the surroundings. • The system absorbs heat and work is done on it by the surroundings. • The system releases heat and work is done on it by the surroundings.

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

  16. Specific Heat Capacity

  17. Heat q = m  c DT where q  heat, J m  mass, g c  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 • cm • metals: cm≈ 25 J/mol.C

  19. Hot and Cold Iron

  20. Freezing and Melting

  21. Heat Transfer Two substances in an isolated system – no energy exchange with the surroundings E = 0, no work qlost + qgained = E = 0, qlost = -qgained (m  c DT)lost = - (m  c DT)gained DT = final temperature – initial temperature DT = Tf – Ti Tf is same for both substances

  22. 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

  23. 130 9 0 0 At 25ºC, the same quantity of heat is added to 1.00-g samples of gold and iron.Metal Specific Heat Capacity (J g-1ºC-1) Gold 0.128Iron 0.451Identify the correct statement. • The temperatures of both will remain the same. • The temperatures of both will increase by the same amount. • The temperatures of both will decrease by the same amount. • The temperature of the gold sample will rise higher than that of iron. • The temperature of the iron sample will rise higher than that of gold.

  24. Heating, Temperature Change, and Phase Change

  25. Vaporization and Condensation

  26. 130 10 0 0 Which of the following is an exothermic process? • Water boiling • Ice melting • Heating soup • Steam condensing

  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  c DTice cice = 2.09J/goC qwater =m  c DTwater cwater = 4.18J/goC qsteam = m  c Dtsteam csteam = 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 + 3.33×103 + 4.18×103 + 2.26×104 + 548)J = 23.3 kJ

  29. Heat Flow in Reactions exothermic –reaction that gives off energy q < 0 isolated system E=0 heat released by reaction raises the temperature of the system constant T, heat is released to the surroundings endothermic – reaction that absorbs energy q > 0

  30. Expansion Type Work w = -PDV system does work DV = Vfinal - Vinitial P V P Vinitial qp = +2kJ

  31. 130 10 0 0 Do 250 J of work to compress a gas, 180 J of heat are released by the gasWhat is E for the gas? • 430 J • 70 J • -70 J • -180 J • -250 J

  32. Enthalpy H DE = q + w at constant V, wexpansion = 0 DE = qv at constant P, wexpansion = -PDV DE = qp - PDV DefineH=E + (PV) = E + PV DH = qp

  33. 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

  34. State Functions • H and E along with P, T, V are state functions. They are the same no matter what path we take for the change. • q and w are not state functions, they depend on which path we take between two points. q initial E=Efinal-Einitialq and w can be anything w q E w final

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