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Section 11.3 HEAT IN CHANGES OF STATE. After reading Section 11.3, you should know: The difference between fusion, solidification, vaporization and condensation.

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Presentation on theme: "Section 11.3 HEAT IN CHANGES OF STATE. After reading Section 11.3, you should know: The difference between fusion, solidification, vaporization and condensation."— Presentation transcript:

1 Section 11.3 HEAT IN CHANGES OF STATE

2 After reading Section 11.3, you should know: The difference between fusion, solidification, vaporization and condensation How to calculate the heat changes for each type of reaction

3 Definitions Molar heat of fusion (H fus ) – the heat absorbed by one mole of a substance in melting from a solid to a liquid at constant temperature. Table 11.5 (pg 308) Molar heat of solidification (H solid ) – the heat released when one mole of a liquid solidifies at constant temperature. Determined from Table 11.5 (pg 308)

4 Definitions Molar heat of vaporization (H vap ) – the amount of heat absorbed when one mole of a given liquid is vaporized. Table 11.5 (pg 308) Molar heat of condensation (H cond ) – the amount of heat released when one mole of a vapor condenses. Determined from Table 11.5 (pg 308)

5 Definitions Molar heat of solution (H soln ) – the heat change caused by the dissolution of one mole of a substance; can be absorbed or released. Molar heat of combustion ( Δ H comb ) – heat of reaction for the complete burning of one mole of a substance. Table 11.4 (pg 305)

6 How they all relate (values on pg 308) H solid and H fus are opposites H solid = - H vap H 2 O (s) --> H 2 O (l) H fus = 6.01 kJ / mol H 2 O (l) --> H 2 O (s) H solid = kJ / mol H vap and H cond are opposites, too! H vap = - H cond H 2 O (l) --> H 2 O (g) H vap = 40.7 kJ / mol H 2 O (g) --> H 2 O (l) H cond = kJ / mol

7 Heating Curve for Water H vap H fus

8 Constants for H 2 0 H fus = 6.01 kJ/mol H vap = 40.7 kJ/mol H solid = 6.01 kJ/mol H cond = 40.7 kJ/mol C p (ice) = 2.06 J/goC C p (water) = 4.18 J/goC C p (steam) = 2.02 J/goC

9 Sample Problem How many grams of ice at 0 o C and kPa could be melted by the addition of 2.25 kJ of heat? Known Values:Unknowns: Melting = H fusion grams of ice = ? H fusion (H 2 O) = 6.01 kJ/mol 2.25 kJ of heat are absorbed ? Grams = 2.25 kJ1 mol 18.0 grams = 6.74 grams 6.01 kJ 1 mol

10 After reading Section 11.3, you should know: The difference between fusion, solidification, vaporization and condensation How to calculate the heat changes for each type of reaction


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