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Molar Enthalpy of Vaporization

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Presentation on theme: "Molar Enthalpy of Vaporization"— Presentation transcript:

1 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

2 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

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

4 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 = (from periodic table) Divide the given grams by the molar mass of H2O

5 47.0 g × 1mol/18.02 g = 2.61 mol H2O Then multiply the moles by the given molar enthalpy. 2.61 mol H2O × kJ/mol = 15.7 kJ energy absorbed when 47 grams of water melts.

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

7 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 mol, just divide 36.5kJ ÷ mol = 84.3 kJ/mol


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