1. NOTES: 17.3 – Heat in Changes of State

2. RECALL… ● when a substance changes state (i.e. melts, freezes, vaporizes, condenses) it does not change temperature ● however, it does absorb or release heat during the phase change ● melting and vaporizing: ENDOTHERMIC ● freezing and condensing: EXOTHERMIC

3. Heat of Fusion and Solidification: ● Solid ↔ Liquid ● molar heat of fusion (∆H fus ) = heat absorbed when one mole of a solid substance is melted ● molar heat of solidification (∆H solid ) = heat released when one mole of a liquid substance solidifies (freezes) ● ∆H fus = -∆H solid ● Examples: H 2 O (s)  H 2 O (l) ∆H fus = 6.01 kJ/mol H 2 O (l)  H 2 O (s) ∆H solid = -6.01 kJ/mol

4. Example #1: How many grams of ice at 0 ° C could be melted by the addition of 2.25 kJ of heat?

5. Example #1: How many grams of ice at 0 ° C could be melted by the addition of 2.25 kJ of heat?

6. Example #2: Sometimes heat of fusion is given in J/g: for ice, H fusion = 333 J/g. How much energy (in J) is required to melt 17.75 g of ice at 0 ° C?

7. Example #2: Sometimes heat of fusion is given in J/g: for ice, H fusion = 333 J/g. How much energy (in J) is required to melt 17.75 g of ice at 0 ° C?

8. Heat of Vaporization / Condensation: ● Liquid ↔ Gas ● molar heat of vaporization (∆H vap ) = heat absorbed when one mole of a liquid substance is vaporized ● molar heat of condensation (∆H cond ) = heat released when one mole of a gaseous substance condenses ● ∆H vap = -∆H cond ● Examples: H 2 O (l)  H 2 O (g) ∆H vap = 40.7 kJ/mol H 2 O (g)  H 2 O (l) ∆H cond = -40.7 kJ/mol

9. Example #3: How much heat (in kJ) is absorbed when 24.8 g of liquid H 2 O at 100 ° C is converted to steam at 100 ° C?

10. Example #3: How much heat (in kJ) is absorbed when 24.8 g of liquid H 2 O at 100 ° C is converted to steam at 100 ° C?

11. Example #4: Sometimes heat of vaporization is given in J/g: for water, H vapor. = 2260 J/g. How much energy (in J) is required to vaporize 113.2 g of water at 100 ° C?

12. Example #4: Sometimes heat of vaporization is given in J/g: for water, H vapor. = 2260 J/g. How much energy (in J) is required to vaporize 113.2 g of water at 100 ° C?

13. Heat of Solution: ● Solid  Solution ● molar heat of solution (∆H soln ) = heat change caused by the dissolution (dissolving) of one mole of a substance ● Example: NaOH (s) + H 2 O (l)  Na + (aq) + OH - (aq) ∆H soln = -445.1 kJ/mol

14. Example #5: NaOH (s) + H 2 O (l)  Na + (aq) + OH - (aq) ∆H soln = -445.1 kJ/mol How much heat (in kJ) is released when 11.5 g of NaOH (s) is dissolved in water?

15. Example #5: NaOH (s) + H 2 O (l)  Na + (aq) + OH - (aq) ∆H soln = -445.1 kJ/mol How much heat (in kJ) is released when 11.5 g of NaOH (s) is dissolved in water?

16. ExampleExample #6: How much heat energy is necessary to turn 10.0 grams of ice at -15.0 o C into water vapor at 120.0 o C? **Hint: heat ice, melt ice, heat water, vaporize water, heat vapor ● Given: H 2 O(s)  H 2 O(l)∆H fus = 333 J/g H 2 O(l)  H 2 O(g)∆H vap = 2260 J/g C ice = 2.01 J/(g o C) C water = 4.18 J/(g o C) C vapor = 2.03 J/(g o C)

17. ExampleExample #6: Step 1: heat up the ice to its melting point: q = mcΔT q = (10.0 g)(2.01 J/g˚C)(15.0˚C) q = 301.5 J

18. ExampleExample #6: Step 2: melt the ice q = (mass)(H fusion ) q = (10.0 g)(333 J/g) q = 3330 J

19. ExampleExample #6: Step 3: heat up the liquid water to its boiling point: q = mcΔT q = (10.0 g)(4.18 J/g˚C)(100.0˚C) q = 4180 J

20. ExampleExample #6: Step 4: vaporize the liquid water q = (mass)(H vapor ) q = (10.0 g)(2260 J/g) q = 22600 J

21. ExampleExample #6: Step 5: heat up the steam to its final temp. q = mcΔT q = (10.0 g)(2.03 J/g˚C)(20.0˚C) q = 406 J

22. ExampleExample #6: Finally…add up all steps! 301.5 J + 3330 J + 4180 J + 22600 J + 406 J = 30,818 J =30.8 kJ