Thermochemistry Chapter 17:3 Pages 520-525

A. Heat of Fusion Molar Heat of Fusion (Hfus) Heat absorbed by one mole of a solid substance when it melts to a liquid at a constant temperature Molar Heat of Solidification (Hsolid) Heat lost by one mole of a liquid substance when it solidifies at a constant temperature Hfus = - Hsolid

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/mol Molar Heat of Solidification (Hsolid) H2O(l)  H2O(s) Hsolid = -6.02 kJ/mol

B. Heating Curves Gas - KE  Boiling - PE  Liquid - KE  Melting - PE  Solid - KE 

B. Heating Curves Temperature Change change in KE (molecular motion) depends on heat capacity Phase Change change in PE (molecular arrangement) temp remains constant

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/mol H2O(l)  H2O(g) Hvap = 40.7 kJ/mol usually larger than Hfus…why? EX: sweating, steam burns

D. Practice Problems How much heat energy is required to melt 25 grams of ice at 0.000oC to liquid water at a temperature of 0.000oC? ice 6.02 kJ 1 mol H2O 25 g H2O 1 mol H2O 18.02 g H2O = 8.4 kJ

D. Practice Problems How much heat energy is required to change 500.0 grams of liquid water at 100.0oC to steam at 100.0oC? steam 40.7 kJ 1 mol H2O 500.0 g H2O 1 mol H2O 18.02 g H2O = 1129 kJ

D. Practice Problems How 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 kJ 1 mol C2H5Cl 0.46 g C2H5Cl 1 mol C2H5Cl 64.52 g C2H5Cl = 0.19 kJ

E. Heat Calculations with State Changes ΔHvap = 40.7 kJ/mol QTC(v) QPC(v) QTC(l) q = mCΔT CH2O(g) = 2.02 J/goC q = mCΔT CH2O(l) = 4.184 J/goC QPC(f) See handout “How to Calculate Heat in Changes of State” QTC(s) ΔHfus = 6.02 kJ/mol q = mCΔT CH2O(s) = 2.03 J/goC

E. Heat Calculations with State Changes ΔHvap = 40.7 kJ/mol QTC(v) QPC(v) QTC(l) q = mCΔT CH2O(g) = 2.02 J/goC q = mCΔT CH2O(l) = 4.184 J/goC How much heat energy is required to change 100.0 grams of liquid water at 25.0oC to steam at 120.0oC? 

E. Heat Calculations with State Changes q = mCΔT CH2O(l) = 4.184 J/goC QTC(l) QTC(l) = (100.0g)(4.184J/goC)(100-25oC) = 31380 J = 31.38 kJ QPC(v) = QTC(v) = (100.0g)(2.02J/goC)(120-100oC) = 4040 J = 40.40 kJ + QPC(v) ΔHvap = 40.7 kJ/mol 100.0 g H2O 1 mol H2O 18.02 g H2O 40.7 kJ 1 mol H2O = 225.9 kJ QTC(v) q = mCΔT CH2O(g) = 2.02 J/goC + = 298 kJ

E. Heat of Solution During the formation of a solution, heat is either released or absorbed Enthalpy change caused by dissolution of 1 mol of a substance is the molar heat of solution Hsoln Examples: hot packs, cold packs

E. Heat of Solution NaOH(s) → Na+(aq) + OH-(aq) Hsoln = -445.1 kJ/mol NH4NO3(s) → NH4+(aq) + NO3-(aq) Hsoln = 25.7 kJ/mol

F. Practice Problems How much heat (in kJ) is released when 20.0 g of NaOH(s) is dissolved in water? The molar Hsoln is -445.1 kJ/mol. -445.1 kJ 1 mol NaOH 20.0 g NaOH 1 mol NaOH 40.00 g NaOH = 223 kJ