Molar Enthalpy of Vaporization The amount of energy as heat that is needed to vaporize 1 mole of liquid(at the liquids boiling point at constant pressure) The symbol used is Δ Hv Given in kJ/mol J is Joules which is the SI unit of energy

Molar Enthalpy of Fusion The amount of energy as heat that is required to melt 1 mole of solid. (at the solid’s melting point) Given in kJ/mol Symbol Δ Hf

Example 1 – page 351 How much energy is absorbed when 47.0 g of ice (H2O) melts? Given that ΔHf= 6.009 kJ/mol

To Solve Since the given molar enthalpies are energy per mole we need to convert 47.0 grams of H2O to moles. Remember how? We need the molar mass of H2O Molar mass of H2O is 1.01 + 1.01 + 16.00 = 18.02 (from periodic table) Divide the given grams by the molar mass of H2O

47.0 g × 1mol/18.02 g = 2.61 mol H2O Then multiply the moles by the given molar enthalpy. 2.61 mol H2O × 6.009 kJ/mol = 15.7 kJ energy absorbed when 47 grams of water melts.

Example 2 Calculate the molar enthalpy of vaporization of a substance given that 0.433 mol of the substance absorbs 36.5 kJ of energy when it is vaporized?

To solve We know that molar enthalpy of vaporization is - how much energy it takes (kJ) to vaporize 1 mole of substance so…. If it takes 36.5 kJ to vaporize 0.433 mol, just divide 36.5kJ ÷ 0.433 mol = 84.3 kJ/mol