1. Heat in Changes of State

2. What happens when you place an ice cube on a table in a warm room? Molar Heat of Fusion (ΔH fus ): heat absorbed by one mole of a substance in melting from a solid to a liquid at a constant temperature Molar heat of Solidification ( ΔH solid ): heat lost when one mole of a liquid solidifies at a constant temperature ΔH fus = - ΔH solid

3. Molar Heat of Vaporization (ΔH vap ): amount of heat necessary to vaporize one mole of a given liquid H 2 O (l)  H 2 O (g) ΔH vap = 40.7 kJ/mol Molar Heat of Condensation (ΔH cond ): amount of heat released when 1 mol of vapor condenses ΔH vap = - Δ H cond

4. Vapor Liquid Solid vaporization condensation fusionsolidification Δ H vap -ΔH fus -ΔH cond -ΔH solid High Enthalpy Low Enthalpy

5. ΔH vap ΔH fus The Heating Curve of Water http://netcamp.prn.bc.ca/nuggets/heatingcurve.swf

6. Molar Heat of Solution (ΔH soln ): heat change caused by dissolution of one mole of substance Exothermic Reaction: CaCl 2 (s)  Ca 2+ (aq) + 2Cl - (aq) ΔH soln = -445.1 kJ/mo l Endothermic Reaction: NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq) ΔH soln = 25.7 kJ/mol H 2 O(l)  H soln = H soln - H components

7. How much heat (in kJ) is released when 2.500 mol NaOH (s) is dissovled in water? ΔH soln = -445.1 kJ/mol List Knowns and Unknowns Knowns: ΔH soln = -445.1 kJ/mol Amount of NaOH(s): 2.500 mol Unknown: ΔH = ?kJ Solve: ΔH = 2.500 mol NaOH (s) x -445.1 kJ = -1113 kJ 1 mol NaOH

8. Hess’s Law Hess’s Law of heat summation: If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction. Example: C (diamond)  C (graphite)

9. Use Hess’s Law to find the enthalpy changes for the conversion of diamond to graphite by using the following combustion rxns C (s, graphite) + O 2 (g)  CO 2 (g) ΔH = -393.5 kJ C (s, diamond) + O 2 (g)  CO 2 (g) ΔH = -395.4 kJ 1 st Step: Write equation the first equation in reverse CO 2 (g)  C (s, graphite) + O2 (g) ΔH = 393.5 kJ **when you write an equation in reverse, change sign** 2 nd Step: Add the two equations together CO 2 (g)  C (s, graphite) + O 2 (g) ΔH = 393.5 kJ C (s, diamond) + O 2 (g)  CO 2 (g) ΔH = -395.4 kJ C (s, diamond)  C (s, graphite) ΔH = -1.9 kJ

10. Use Hess’s Law to find the enthalpy change for the formation of carbon monoxide from its elements. C (s, graphite) + O 2 (g)  CO 2 (g) ΔH = -393.5 kJ CO (g) + ½ O 2 (g)  CO 2 (g) ΔH = -283.0 kJ 1 st Step: Write the first equation in reverse CO 2 (g)  CO (g) + ½ O 2 (g) ΔH = 283.0 kJ 2 nd step: Add the equations together CO 2 (g)  CO (g) + ½ O 2 (g) ΔH = 283.0 kJ C (s, graphite) + O 2 (g)  CO 2 (g) ΔH = -393.5 kJ C (s, graphite) + ½ O 2 (g)  CO (g) ΔH = -110.5 kJ

11. Standard Heat of Formation (ΔH f 0 ): the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states at 25°C ΔH f 0 of a free element in its standard state is arbitrarily set at 0. ΔH f 0 = 0 for diatomic molecules Standard Heat of Reaction (ΔH 0 rxn ):heats of reaction at standard conditions ΔH 0 rxn = ΔH f 0 (products) - ΔH f 0 (reactants)