Calculations in
Chapter 10
Molar Enthalpy of Fusion
• Used when melting or freezing
•
= ___energy ____
mol of substance
• Can be arranged to find any of the three
• Units for molar enthalpy of fusion will be
kJ/mol and units for energy is kJ or J,
(kilojoules or joules)
Example Problem
• The molar heat of fusion for water is 6.009
kJ/mol. How much energy is needed to
convert 60.0 g of ice at 0°C to liquid water
at 0°C?
• Rearrange the formula for energy
• Energy =
 mol of substance
•
•
•
•
•
Have to convert the 60.0 g of ice to mol
Divide by the molar mass of water
60.0 g/ 18.016 g/mol =
3.330373002 mol H2O(s)
Energy = 6.009 kJ/mol  3.330373002 mol
H2O(s) =
• 20.01221137 kJ
• 20.0 kJ needed to convert 60.0 g of ice at
0°C to liquid water at 0°C
Molar Enthalpy of Vaporization
• Used when boiling or condensing
•
=
energy
mol of substance
• Can be arranged to find any of the three
• Boiling has energy required, condensing
gives off energy
Example Problem
• How much heat is evolved when 275 g of
ammonia gas condenses to a liquid at its
boiling point? The molar enthalpy of
vaporization for ammonia is 23.3 kJ/mol
• Have to convert the g of ammonia to mol
of ammonia
• 275 g/ 17.034 g/mol = 16.14418222mol
•
•
•
•
•
Rearrange the formula for energy evolved
Energy =
 mol of substance
Energy = 23.3 kj/mol  16.14418222mol
Energy = 376.1594458 kJ
Energy = 376 kJ
Another Problem
• What mass of steam is required to release
4.97 x 105 kJ of heat energy on
condensation? The molar enthalpy of
vaporization for water is 40.79 kJ/mol
• First rearrange to solve for mol
• Solve for mol
• Multiply by the molar mass to find grams
• Answer: 2.20 x 105 g H2O