ThermochemistryKinetic EnergyHeat1st Law of ThermodynamicsSlide 5Slide 6Slide 7Slide 8E = Efinal  EinitialState FunctionSlide 11Slide 12Slide 13EnthalpySlide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22EnthalpySlide 24Slide 25Slide 26Slide 27Slide 28Determine the H for the sublimation of ice to water vapor at 0oC.Slide 30Slide 31Slide 32Standard Heat of FormationSlide 34Slide 35Slide 36Slide 37Enthalpies from Bond EnergiesSlide 39Slide 40Energy UnitsSlide 42Stoichiometry using EnthalpySlide 44EntropyGibbs Free Energy, GG = H  TSSlide 48Slide 49Slide 50ThermochemistryChapter 6Kinetic Energy•energy of motion•thermal energy is kinetic• •1/2 mv2Heat•energy transferred between two objects as a result of the temperature difference between them.Temperature•A measure of kinetic energy1st Law of Thermodynamics•The energy of the universe is constant.•i.e. the energy of the universe is conservedWhich of the pictures at the right represents the energy gauge for the system above when the Esystem is negative?E = Efinal  Einitial E if energy leaves system•+ E if energy enters system•Note the E of a system doesn’t depend on how system got there -- i.e. it is a state functionState Function•A function or property whose value depends only on the present state (condition) of the system, not on the path used to arrive at that condition.Energy can be either work or heat•E = q + wHeat gain or lossWork done = -PVMatches our earlier convention that Ein is + and Eout is –E = q + w = q - PVq = E + PV (note that at constant V, qV = E)EnthalpyH = qP = E + PVH = Hfinal  Hinitial • = Hproducts  HreactantsTmq(C)Heat Specificor Capacity Heat SpecificTells how much heat is required to change the temp of a substance. Some specific heats areAl 0.902 J/g oKCu 0.385 J/g oKH2O 4.184 J/g oK Quantity of heat suppliedTemperature change (always Tf-Ti)•A 55.0 g piece of metal was heated in boiling water to a temperature of 99.8oC and dropped into an insulated beaker with 225 mL of water (d = 1.00 g/ml) at 21.0 oC. The final temperature of the metal and water is 23.1oC. Calculate the specific heat of the metal assuming that no heat was lost to the surroundings.•Octane, C8H18, a primary constituent of gasoline, burns in air. •C8H18(l) + 25/2 O2(g)  8 CO2(g) + 9 H2O(l)•Suppose that a 1.00 g sample of octane is burned in a calorimeter that contains 1.20 kg of water. The temperature of the water and the bomb rises from 25.00oC to 33.20oC. If the specific heat of the bomb, Cbomb, is known to be 837 J/oC, calculate the molar heat of reaction of C8H18.A quantity of ice at 0oC is added to 90.0 g of water at 80oC. After the ice melted, the temperature of the water was 25oC. How much ice was added?•specific heat of ice 2.06 J/goC 0.91 kJ/moloC•specific heat of water 4.184 J/goC 7.54 kJ/moloC •specific heat of steam 2.0 J/goC 0.92 kJ/moloC•heat of fusion 333 J/g 6.01 kJ/mol•heat of vaporization 2226 J/g 40.67 kJ/mol•50.0 g of ice at -20.0 oC are added to 342.0 g of water at 86.0 oC. What will be the final temperature of the sample?•specific heat of ice 2.06 J/goC 0.91 kJ/moloC•specific heat of water 4.184 J/goC 7.54 kJ/moloC •specific heat of steam 2.0 J/goC 0.92 kJ/moloC•heat of fusion 333 J/g 6.01 kJ/mol•heat of vaporization 2226 J/g 40.67 kJ/mol•A 33.14 g sample of copper and aluminum was heated to 119.25oC and dropped into a calorimeter containing 250.0 g of water at 21.00oC. The temperature rose to 23.05oC. Assuming no heat was lost to the surroundings, what is the percent copper in the sample?Enthalpy•Enthalpy transferred out of reactants  exothermic  H = •Enthalpy transferred into products  endothermic  H = +EnthalpyHforward = Hreverse (For reversible reactions)•H2O(g)  H2(g) + 1/2 O2(g)H = +241.8 kJ•H2(g) + 1/2 O2(g)   H2O(g)H = 241.8 kJEnthalpy•The H is proportional to the amount of substance undergoing change.•H2O(g)  H2(g) + 1/2 O2(g)H = +241.8 kJ•2 H2O(g)  2 H2(g) + 1 O2(g) H = +483.6 kJEnthalpy•The physical state of reactants and products is important.•H2O(g)  H2(g) + 1/2 O2(g)H = +241.8 kJ•H2O(l)  H2(g) + 1/2 O2(g) H = +285.8 kJEnthalpy•Enthalpy is a state function -- it doesn’t matter how you go from one place to another -- enthalpy and enthalpy changes are the same!!•The H value is the same no matter how you get from A BDetermine the H for the sublimation of ice to water vapor at 0oC.•H2O(s)  H2O(l)H = 6.02 kJ/reaction•H2O(l)  H2O(g) H = 40.7 kJ/reaction•-----------------------------------------------------•H2O(s)  H2O(g) H = 46.7 kJ/reaction•Calculate the enthalpy change for the formation of methane, CH4, from solid carbon (as graphite) and hydrogen gas.•C(s) + 2 H2(g)  CH4(g)•The enthalpies for the combustion of graphite, hydrogen gas and methane are given.•C(s) + O2(g)  CO2(g) 393.5 kJ•H2(g) + ½ O2(g)  H2O(l) 285.8 kJ•CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l) 890.3 kJ•Calculate the enthalpy change for the reaction•S(s) + O2(g)  SO2(g)•given•2 SO2(g) + O2(g)  2 SO3(g) H = 196 kJ•2 S(s) + 3 O2(g)  2 SO3(g) H = 790 kJStandard Heat of Formation•The enthalpy change, Hfo, for the formation of 1 mol of a substance in the standard state from the most stable forms of its constituent elements in their standard states.ΔHfosuperscript o means standard state25oC and 1 atm pressuresubscript f means formation from most stable elements•Benzene, C6H6, is an important hydrocarbon. Calculate its enthalpy of combustion; that is, find the value of Ho for the following reaction.•C6H6(l)+15/2 O2(g)  6 CO2(g)+3 H2O(l)•Given•Hfo [C6H6(l)] = +49.0 kJ/mol•Hfo [CO2(g)] = 393.5 kJ/mol•Hfo [H2O(l)] = 285.8 kJ/mol•Nitroglycerin is a powerful explosive, giving four different gases when detonated.•2 C3H5(NO3)3(l)  3 N2(g) + ½ O2(g) + 6 CO2(g) + 5 H2O(g)•Given the enthalpy of formation of nitroglycerin, Hfo, is 364 kJ/mol, calculate the energy liberated when 10.0 g of nitroglycerin is detonated.Enthalpies from Bond Energies-Calculate