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Thermochemistry Chapter 17:3 Pages 520-525. A. Heat of Fusion Molar Heat of Fusion ( H fus ) Heat absorbed by one mole of a solid substance when it melts.

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Presentation on theme: "Thermochemistry Chapter 17:3 Pages 520-525. A. Heat of Fusion Molar Heat of Fusion ( H fus ) Heat absorbed by one mole of a solid substance when it melts."— Presentation transcript:

1 Thermochemistry Chapter 17:3 Pages 520-525

2 A. Heat of Fusion Molar Heat of Fusion ( H fus ) Heat absorbed by one mole of a solid substance when it melts to a liquid at a constant temperature Molar Heat of Solidification ( H solid ) Heat lost by one mole of a liquid substance when it solidifies at a constant temperature H fus = - H solid

3 A. Heat of Fusion Molar Heat of Fusion ( H fus ) H 2 O(s) H 2 O(l) H fus = 6.02 kJ/mol Molar Heat of Solidification ( H solid ) H 2 O(l) H 2 O(s) H solid = -6.02 kJ/mol

4 B. Heating Curves Melting - PE Solid - KE Liquid - KE Boiling - PE Gas - KE

5 B. Heating Curves Temperature Change change in KE (molecular motion) depends on heat capacity Phase Change change in PE (molecular arrangement) temp remains constant

6 C. Heat of Vaporization Molar Heat of Vaporization ( H vap ) energy required to boil 1 mole of a substance at its b.p. H vap for water = 40.7 kJ/mol H 2 O(l) H 2 O(g) H vap = 40.7 kJ/mol usually larger than H fus …why? EX: sweating, steam burns

7 How much heat energy is required to melt 25 grams of ice at 0.000 o C to liquid water at a temperature of 0.000 o C? 25 g H 2 O1 mol H 2 O 18.02 g H 2 O = 8.4 kJ 6.02 kJ 1 mol H 2 O ice D. Practice Problems

8 How much heat energy is required to change 500.0 grams of liquid water at 100.0 o C to steam at 100.0 o C? = 1129 kJ 500.0 g H 2 O1 mol H 2 O 18.02 g H 2 O 40.7 kJ 1 mol H 2 O steam

9 D. Practice Problems How many kJ are absorbed when 0.46g of C 2 H 5 Cl vaporizes at its normal boiling point? The molar H vap is 26.4 kJ/mol. = 0.19 kJ 0.46 g C 2 H 5 Cl 1 mol C 2 H 5 Cl 64.52 g C 2 H 5 Cl 26.4 kJ 1 mol C 2 H 5 Cl

10 E. Heat Calculations with State Changes q = mCΔT C H2O(l) = 4.184 J/g o C QTC(l) ΔH fus = 6.02 kJ/mol QPC(f) ΔH vap = 40.7 kJ/mol QPC(v) q = mCΔT C H2O(s) = 2.03 J/g o C QTC(s) QTC(v) q = mCΔT C H2O(g) = 2.02 J/g o C See handout How to Calculate Heat in Changes of State

11 E. Heat Calculations with State Changes q = mCΔT C H2O(l) = 4.184 J/g o C QTC(l) ΔH vap = 40.7 kJ/mol QPC(v) QTC(v) q = mCΔT C H2O(g) = 2.02 J/g o C How much heat energy is required to change 100.0 grams of liquid water at 25.0 o C to steam at 120.0 o C?

12 E. Heat Calculations with State Changes QTC( l ) = (100.0g)(4.184J/g o C)(100-25 o C) = 31380 J = 31.38 kJ QPC( v ) = QTC( v ) = (100.0g)(2.02J/g o C)(120-100 o C) = 4040 J = 40.40 kJ q = mCΔT C H2O(l) = 4.184 J/g o C QTC(l) ΔH vap = 40.7 kJ/mol QPC(v) QTC(v) q = mCΔT C H2O(g) = 2.02 J/g o C = 225.9 kJ 100.0 g H 2 O 1 mol H 2 O 18.02 g H 2 O 40.7 kJ 1 mol H 2 O + + = 298 kJ

13 E. Heat of Solution During the formation of a solution, heat is either released or absorbed Enthalpy change caused by dissolution of 1 mol of a substance is the molar heat of solution H soln Examples: hot packs, cold packs

14 E. Heat of Solution NaOH(s) Na + (aq) + OH - (aq) H soln = -445.1 kJ/mol NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq) H soln = 25.7 kJ/mol

15 F. Practice Problems How much heat (in kJ) is released when 20.0 g of NaOH(s) is dissolved in water? The molar H soln is -445.1 kJ/mol. = 223 kJ 20.0 g NaOH 1 mol NaOH 40.00 g NaOH -445.1 kJ 1 mol NaOH


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