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