2 A. Heat of FusionMolar Heat of Fusion (Hfus)Heat absorbed by one mole of a solid substance when it melts to a liquid at a constant temperatureMolar Heat of Solidification (Hsolid)Heat lost by one mole of a liquid substance when it solidifies at a constant temperatureHfus = - Hsolid
3 A. Heat of Fusion H2O(l) H2O(s) Hsolid = -6.02 kJ/mol Molar Heat of Fusion (Hfus)H2O(s) H2O(l) Hfus = 6.02 kJ/molMolar Heat of Solidification (Hsolid)H2O(l) H2O(s) Hsolid = kJ/mol
4 B. Heating Curves Gas - KE Boiling - PE Liquid - KE Melting - PE Solid - KE
5 B. Heating Curves Temperature Change change in KE (molecular motion) depends on heat capacityPhase Changechange in PE (molecular arrangement)temp remains constant
6 C. Heat of Vaporization Molar Heat of Vaporization (Hvap) energy required to boil 1 mole of a substance at its b.p.Hvap for water = 40.7 kJ/molH2O(l) H2O(g) Hvap = 40.7 kJ/molusually larger than Hfus…why?EX: sweating, steam burns
7 D. Practice ProblemsHow much heat energy is required to melt 25 grams of ice at 0.000oC to liquid water at a temperature of 0.000oC?ice6.02 kJ1 mol H2O25 g H2O1 mol H2O18.02 g H2O= 8.4 kJ
8 D. Practice ProblemsHow much heat energy is required to change grams of liquid water at 100.0oC to steam at 100.0oC?steam40.7 kJ1 mol H2O500.0 g H2O1 mol H2O18.02 g H2O= 1129 kJ
9 D. Practice ProblemsHow many kJ are absorbed when 0.46g of C2H5Cl vaporizes at its normal boiling point? The molar Hvap is 26.4 kJ/mol.26.4 kJ1 mol C2H5Cl0.46 g C2H5Cl1 mol C2H5Cl64.52 g C2H5Cl= 0.19 kJ
10 E. Heat Calculations with State Changes ΔHvap = 40.7 kJ/molQTC(v)QPC(v)QTC(l)q = mCΔTCH2O(g) = 2.02 J/goCq = mCΔTCH2O(l) = J/goCQPC(f)See handout “How to Calculate Heat in Changes of State”QTC(s)ΔHfus = 6.02 kJ/molq = mCΔTCH2O(s) = 2.03 J/goC
11 E. Heat Calculations with State Changes ΔHvap = 40.7 kJ/molQTC(v)QPC(v)QTC(l)q = mCΔTCH2O(g) = 2.02 J/goCq = mCΔTCH2O(l) = J/goCHow much heat energy is required to change grams of liquid water at 25.0oC to steam at 120.0oC?
12 E. Heat Calculations with State Changes q = mCΔTCH2O(l) = J/goCQTC(l)QTC(l) = (100.0g)(4.184J/goC)(100-25oC)= J = kJQPC(v) =QTC(v) = (100.0g)(2.02J/goC)( oC)= 4040 J = kJ+QPC(v)ΔHvap = 40.7 kJ/mol100.0 g H2O1 mol H2O18.02 g H2O40.7 kJ1 mol H2O=225.9 kJQTC(v)q = mCΔTCH2O(g) = 2.02 J/goC+= 298 kJ
13 E. Heat of SolutionDuring the formation of a solution, heat is either released or absorbedEnthalpy change caused by dissolution of 1 mol of a substance is the molar heat of solution HsolnExamples: hot packs, cold packs
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