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CHAPTER 6

CHAPTER 6. All Bold Numbered Problems. Chapter 6 Outline. Energy -vs- Heat Specific Heat First Law of Thermodynamics q (Heat) Hess’s Law. THERMOCHEMISTRY or Thermodynamics. 3. Burning peanuts supply sufficient energy to boil a cup of water.

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CHAPTER 6

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  1. CHAPTER 6 All Bold Numbered Problems

  2. Chapter 6 Outline • Energy -vs- Heat • Specific Heat • First Law of Thermodynamics • q (Heat) • Hess’s Law

  3. THERMOCHEMISTRYorThermodynamics 3

  4. Burning peanuts supply sufficient energy to boil a cup of water. Burning sugar (sugar reacts with KClO3, a strong oxidizing agent) Energy & Chemistry

  5. Energy and 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-

  6. Energy and Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 samples because of their difference in temperature. Other types of energy • light • electrical • kinetic and potential

  7. Kinetic and Potential Energy Potential energy: energy stored in chemical due to its structure. The energy of a motionless body due to its position.

  8. Kinetic and Potential Energy Kinetic energy: energy of motion. • Translational • Rotational • Vibrational (IR)

  9. Thermodynamics Thermodynamics is the science of heat (energy) transfer. Heat energy is associated with molecular motions.

  10. Energy and Chemistry All of thermodynamics depends on the law of THE CONSERVATION OF ENERGY. • The total energy of a system is constant.

  11. UNITS OF ENERGY 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC. 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food “calorie”) The S.I sytem uses the unit called the JOULE 1 cal = 4.18 Joules James Joule 1818-1889

  12. Specific heat capacity = heat lost or gained by substance (J) (mass, g)(T change, K) Specific Heat Capacity Thermochemistry is the science of heat (energy) flow. A difference in temperature leads to energy transfer. The heat “lost” or “gained” is related to a) sample mass b) change in T c) specific heat capacity

  13. Specific Heat Capacity Substance Spec. Heat (J/g•K) H2O 4.18 Al 0.902 glass 0.84 Aluminum

  14. CHEMICAL REACTIVITY What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS and the second by KINETICS. Have already seen a number of “driving forces” for reactions that are PRODUCT-FAVORED. • formation of a precipitate • gas formation • H2O formation (acid-base reaction) • electron transfer in a battery

  15. CHEMICAL REACTIVITY But ENERGY TRANSFER also allows us to predict reactivity. In general, reactions that transfer energy to their surroundings are product-favored. So, let us consider heat transfer in chemical processes.

  16. Heat Energy Transfer in Chemical Processes CO2 (s, -78 oC) ---> CO2 (g, -78 oC)

  17. work done (by the surroundings) energy change FIRST LAW OF THERMODYNAMICS DE = q + w heat energy transferred (to the system) Energy is conserved!

  18. The First Law of Thermodynamics is the law of conservation of energy. DE = q + w DE is the change in Energy, q is heat, w is work • Enthalpy (H), defined H = E + PV. DH = DE + PDVat constant pressure(only V and T change). • If w = - PDV (because V does not change in a typical lab experiment, only T changes) and DH = q + w + PDV DH = qp(for chemist in a lab which means only T changes)

  19. ENTHALPY Most chemical reactions occur at constant P. qp = DH DH = change in heat content of the system DH = Hfinal - Hinitial How do we measure q in the lab?

  20. Specific Heat Capacity Substance Spec. Heat (J/g•K) H2O 4.18 Al 0.902 glass 0.84 If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? Specific heat capacity = heat lost or gained by substance (J) or g)(T change, (mass, K) C = Specific Heat, units of J/(gK)

  21. heat gain/lost = q = (C)(mass)(DT) Specific Heat Capacity If 25.0 g of Al cool from 310. oC to 37 oC, how many joules of heat energy are lost by the Al? where DT = Tfinal - Tinitial q = (0.902 J/g•K)(25.0 g)(37 - 310.)K q = - 6160 J

  22. Heat Energy Transfer in Chemical Processes Heat flows FROM the SYSTEM into the SURROUNDINGSis an EXOTHERMICprocess. q is “-” The reaction feels hot. Energy given Off!

  23. Heat Energy Transfer in Chemical Processes Heat flows INTO the SYSTEMfrom the SURROUNDINGSis an ENDOTHERMICprocess. q is “+” The reaction feels cold. Energy Added!

  24. Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? q = - 6160 J q = - 6.16 kJ Notice that the negative sign on q signals heat “lost by” or transferred out of Al. Is this Exo or Endothermic?

  25. Specific Heat Capacity A 500.0 g piece of metal is heated to 75.0 0C. It is placed into 300. mL of water at 22.0 0C and the final temperature is 31.0 0C. Calculate the specific heat of the metal. qwater + qmetal= 0 qwater = -qmetal (300. g)(4.18 J/gK)(9.0oC) = -(500.0 g)(c)(-44.0oC) c = 0.51 J/gK

  26. Specific Heat Capacity To 150 g of water at 25oC is added 45 g of Al at 115oC, what will be the final temperature? Al 0.902

  27. Heat Transfer and Changes of State Changes of state involve energy Ice -----> Water 333 J/g (heat of fusion) + energy Exo or Endo? Sign of q?

  28. Heat Transfer and Changes of State Liquid ---> Vapor Requires energy (heat). This is the reason a) you cool down after swimming b) you use water to put out a fire. + energy

  29. 29 Heat and Changes of State What quantity of heat is required to change 10.0 g of ice at -50.0 oC to steam at 200.0 oC? Endo or Exo? Sign of q? Heat of fusion of ice = 333 J/g Heat of vaporization = 2260 J/g +333 J/g +2260 J/g

  30. Heating/Cooling Curve for Water 5 4 3 2 1

  31. Heat and Changes of State 1. To heat ice q = (10.0 g)(2.09 J/g•K)(50.0K) = 1050J 2. To melt ice q = (10.0 g)(333 J/g) = 3330J 3. To heat water q = (10.0 g)(4.18 J/g•K)(100.K) = 4180J 4. To evaporate water q = (10.0 g)(2260 J/g) = 22600J 5. To heat steam q = (10.0 g)(2.03 J/g•K)(100.0K) = 2030J Total heat energy = 33190.J = 33.190kJ

  32. Heat and Changes of State Calculate the amount of heat energy necessary to change 25.0 g of copper solid at 925o C to liquid at 1083o C. Melting point = 1083o C, csolid = 0.382 J/g.K, Heat of fusion = 205 J/g. q1 = (25.0 g)(.382 J/gK)(1083oC-925oC)= 1510 J q2 = (25.0 g)(205 J/g) = 5120 J qT = q1 + q2 = 6630 J

  33. Endo- and Exothermic qsystem > 0 qsystem < 0 T(system) goes down T(system) goes up EXOTHERMIC Heat leaves the system ENDOTHERMIC Heat goes in the system

  34. USING ENTHALPY Consider the decomposition of water H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g) Endo or Exo? Endothermic reaction Heat is a “reactant” DH = + 242 kJ

  35. USING ENTHALPY Making H2 from H2O involves two steps. Each step requires energy. Liquid H2O H2 + O2 gas H2O vapor

  36. USING ENTHALPY Making H2 from H2O involves two steps. H2O(liq) + 44 kJ ---> H2O(g) H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g) ----------------------------------------------------------------------- H2O(liq) + 286 kJ --> H2(g) + 1/2 O2(g) Example of HESS’S LAW If a rxn. is the sum of 2 or more others, the net DH is the sum of the DH’s of the other rxns.

  37. USING ENTHALPY Calc. DH for: S(s) + 3/2 O2(g) --> SO3(g) Given: S(s) + O2(g) --> SO2(g)DH1 = -320.5 kJ SO2(g) + 1/2 O2(g) --> SO3(g)DH2 = -75.2 kJ The two equations add up to give the desired equation, so - DHnet = DH1 + DH2 = -395.7 kJ

  38. ENERGY S solid +O 2 direct path DH = 1 -320.5 kJ + 3/2 O 2 DH = SO gas 2 -395.7 kJ + 1/2 O 2 SO gas 3 DH = -75.2 kJ 2 DH along one path = DH along another path

  39. DH along one path = DH along another path • This equation is valid because DH is a STATE FUNCTION • State functions depend only on the state of the system and not how it got there. • Examples: V, T, P, energy — and your bank account! • Unlike V, T, and P, one cannot measure absolute H. We can only measure DH.

  40. Standard Enthalpy Values Most DH values are labeled ΔHo Measured under standard conditions P = 1 atmosphere or 1 bar, (105Pascals) Concentration = 1 mol/L T = 25 oC with all species in standard states e.g., C = graphite and O2 = gas, etc.

  41. Standard Enthalpy Values NIST (National Institute for Standards and Technology) gives values of DHof = standard molar enthalpy of formation This is the enthalpy change when 1 mole of compound is formed from elements under standard conditions. See Table 6.2 and Appendix L

  42. DHof, standard molar enthalpy of formation H2(g) + 1/2 O2(g) --> H2O(g) DHof = -241.8 kJ/mol By definition, DHof = 0 for elements in their standard states.

  43. Using Standard Enthalpy Values Use DH°f’s to calculate enthalpy change for: H2O(g) + C(graphite) --> H2(g) + CO(g) (product is called “water gas”)

  44. Using Standard Enthalpy Values H2O(g) + C(graphite) --> H2(g) + CO(g) From reference books we find: H2(g) + 1/2 O2(g) --> H2O(g) DH°f of H2O vapor = - 241.8 kJ/mol C(s) + 1/2 O2(g) --> CO(g) DH°f of CO = - 110.5 kJ/mol

  45. Using Standard Enthalpy Values H2O(g) --> H2(g) + 1/2 O2(g) DHo = +241.8kJ C(s) + 1/2 O2(g) --> CO(g)DHo = -110.5 kJ -------------------------------------------------------------------------------- H2O(g) + C(graphite) --> H2(g) + CO(g) DHonet = +131.3 kJ To convert 1 mol of water to 1 mol each of H2 and CO requires 131.3 kJ of energy. The “water gas” reaction is ENDOthermic.

  46. Using Standard Enthalpy Values Calculate D H of reaction? In general, when ALL enthalpies of formation are known, DHorxn = DHof (products) - DHof (reactants)

  47. Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., DHorxn for: CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) DHorxn = DHof (prod) - DHof (react) Use Appendix L

  48. Using Standard Enthalpy Values CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) DHorxn = DHof (prod) - DHof (react) DHorxn = {DHof (CO2) + 2 DHof (H2O)} - { 3/2 DHof (O2) + DHof (CH3OH) } = {(-393.5 kJ) + 2 (-241.8 kJ)} - { 0 + (-200.7 kJ) } DHorxn = - 676.4 kJ per mol of methanol

  49. Using Standard Enthalpy Values OF2(g) + H2O(g) --> O2(g) + 2 HF(g) DHorxn =- 318.0kJ DHof(H2O) = - 241.8 kJ and DHof (HF) = - 271.1 kJ Calculate DHof (OF2) DHorxn = {DHof (O2) + 2 DHof (HF) } - { DHof (OF2) + DHof (H2O) } -318.0 kJ = {( 0 ) + 2 (-271.1 kJ) } - {DHof (OF2) + (-241.8 kJ) } DHof (OF2) = 17.6 kJ /mole

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