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Heats of Vaporization and Condensation

Heats of Vaporization and Condensation. By Reid and Shad. Overview. Vaporization – Process of phase transition from liquid to gas. A Heating process And ABSORBS energy. Condensation – A process of phase change from gas to liquid. A Cooling process and RELEASES energy.

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Heats of Vaporization and Condensation

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  1. Heats of Vaporization and Condensation By Reid and Shad

  2. Overview Vaporization – Process of phase transition from liquid to gas. A Heating process And ABSORBS energy

  3. Condensation – A process of phase change from gas to liquid. A Cooling process and RELEASES energy

  4. General Rule: The temperature remains CONSTANT during phase changes.

  5. Molar Heat of Vaporization: The amount of heat needed to vaporizeone mole of a given substance. Symbol: Units: kJ/mol Molar Heat of Vaporization: The amount of heat needed to vaporizeone mole of a given substance. Symbol: Units: kJ/mol Molar Heat of Vaporization: The amount of heat needed to vaporizeone mole of a given substance. Symbol: Units: kJ/mol Molar Heat of Vaporization: The amount of heat needed to vaporizeone mole of a given substance. Symbol: Units: kJ/mol Molar Heat of Vaporization: The amount of heat needed to vaporizeone mole of a given substance. Symbol: Units: kJ/mol Molar Heat of Condensation: The amount of heat released to condense 1 mol of vapor of a given substance. Units: kJ/mol Molar Heat of Condensation: The amount of heat released to condense 1 mol of vapor of a given substance. Units: kJ/mol For any given substance, the molar heat of vaporization is equal and OPPOSITE to the molar heat of condensation. Molar ΔHvap= - Molar ΔHcond （What does it mean to add negative heat?）

  6. ONE MOLE of water Two moles? Three moles? WATER Molar Heat of Vaporization = 40.7kJ/mol Molar Heat of Condensation = -40.7kJ/mol Note: heat of vaporization is basically negative heat of condensation.

  7. How much heat is absorbed when 24.8g H2O(l) at 100℃ and 101.3kPa when it is converted to steam converted to steam at 100℃? Knowns Unknown • Mass of H2O = 24.8g • Initial and final conditions are 100 ℃. And 101.3kPa • Δhvap for water = 40.7kJ/mol Δhvap = ? kJ General Formula: ΔHvap = Molar heat of vaporization * # of moles ΔHcond = Molar heat of condensation * # of moles -------# moles = mass / molar mass = 24.8gH2O * (1mol/18.0gH2O) = 1.38 moles -------Molar heat of vaporization = 40.7kJ/mol ΔHvap = 40.7kJ/mol * (1.38mol) = 56.1 kJ

  8. See Shad’s explanation

  9. Everything in this lesson only concerns the boiling point of the substance and NOTHING ELSE.

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