1. Thermochemistry
Chapter 17:3
Pages

2. 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

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

7. 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

8. D. Practice Problems
How much heat energy is required to change 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

9. 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

10. 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) = 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

11. 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) = J/goC
How 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 = mCΔT
CH2O(l) = J/goC
QTC(l)
QTC(l) = (100.0g)(4.184J/goC)(100-25oC)
= J = kJ
QPC(v) =
QTC(v) = (100.0g)(2.02J/goC)( oC)
= 4040 J = 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

13. 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

14. 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

15. 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 kJ/mol. kJ
1 mol NaOH
20.0 g NaOH
1 mol NaOH
40.00 g NaOH
= 223 kJ