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LESSON 3 Heat in Changes of State

LESSON 3 Heat in Changes of State. Heats of Fusion and Solidification. Molar Heat of Fusion ( H fus ) - the heat absorbed by one mole of a substance in melting from a solid to a liquid Molar Heat of Solidification ( H solid ) heat lost when one mole of liquid solidifies.

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LESSON 3 Heat in Changes of State

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  1. LESSON 3Heat in Changes of State

  2. Heats of Fusion and Solidification • Molar Heat of Fusion (Hfus) - the heat absorbed by one mole of a substance in melting from a solid to a liquid • Molar Heat of Solidification(Hsolid) heat lost when one mole of liquid solidifies

  3. Heats of Fusion and Solidification • Heat absorbed by a melting solid is equal to heat lost when a liquid solidifies • Thus, Hfus = -Hsolid

  4. Heats of Fusion and Solidification • H20(s) H20(l) Hfus = 6.01kJ/mol • H20(l) H20(s) Hsolid = -6.01kJ/mol

  5. Using the Heat of Fusion in Phase-Change Calculations How many kilojoules of heat are required to melt a 10.0g popsicle at 0°C? Assume the popsicle has the same molar mass and heat of fusion of water. ∆Hfus= 6.01kJ/mol

  6. Using the Heat of Fusion in Phase-Change Calculations How many grams of ice at 0°C could be melted by the addition of 0.400 kJ of heat?

  7. Heats of Vaporization and Condensation • When liquids absorb heat at their boiling points, they become vapors. • Molar Heat of Vaporization(Hvap) the amount of heat necessary to vaporize one mole of a given liquid.

  8. Heats of Vaporization and Condensation • Condensation is the opposite of vaporization. • Molar Heat of Condensation(Hcond) amount of heat released when one mole of vapor condenses • Hvap = - Hcond

  9. Heats of Vaporization and Condensation • H20(g) H20(l) Hcond = - 40.7kJ/mol • H20(l) H20(g) Hvap = 40.7kJ/mol

  10. Using the Heat of Vaporization in Phase-Change Calculations How much heat is absorbed when 63.7g of liquid water at 100°C and 101.3Kpa is converted to steam at 100°C? ∆Hvap = 40.7 kJ/mol

  11. Using the Heat of Vaporization in Phase-Change Calculations How many kilojoules of heat are absorbed when 0.46 g of Chloroethane (C2H5Cl) vaporizes at its normal boiling point? The molar heat of vaporization of chloroethane is 26.4kJ/mol.

  12. Heating Curve of Water

  13. Heating Curve of Water vaporization/melting = endothermic H positive (energy absorbed to get hotter). Solidification/condensation = exothermic H negative (energy released as it gets cooler).

  14. Common Heat of Physical Change

  15. Heat of Solution • Heat changes can also occur when a solute dissolves in a solvent. • Molar Heat of Solution (Hsoln) heat change caused by dissolution of one mole of substance • Sodium hydroxide provides a good example of an exothermic molar heat of solution:

  16. Heat of Solution NaOH(s)  Na1+(aq) + OH1-(aq) Hsoln = - 445.1 kJ/mol • The heat is released as the ions separate and interact with water, releasing 445.1 kJ of heat as Hsoln thus becoming so hot it steams! H2O(l)

  17. Heat of Solution NaOH(s)  Na1+(aq) + OH1-(aq) Hsoln = - 445.1 kJ/mol • How much heat (in kJ) is released when 0.677 mol of NaOH(s) is dissolved in water? H2O(l)

  18. Heat of Solution NaOH(s)  Na1+(aq) + OH1-(aq) Hsoln = - 445.1 kJ/mol • How many kJ of heat are released when 25.0 g of NaOH are dissolved in water? H2O(l)

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