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Heat in changes of state
Section 11.3
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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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Definitions Molar heat of fusion (∆Hfus) – 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 (∆Hsolid) – the heat released when one mole of a liquid solidifies at constant temperature. Determined from Table 11.5 (pg 308)
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Definitions Molar heat of vaporization (∆Hvap) – the amount of heat absorbed when one mole of a given liquid is vaporized. Table 11.5 (pg 308) Molar heat of condensation (∆Hcond) – the amount of heat released when one mole of a vapor condenses. Determined from Table 11.5 (pg 308)
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Definitions Molar heat of solution (∆Hsoln) – the heat change caused by the dissolution of one mole of a substance; can be absorbed or released. Molar heat of combustion (ΔHcomb) – heat of reaction for the complete burning of one mole of a substance. Table 11.4 (pg 305)
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How they all relate (values on pg 308)
Hsolid and Hfus are opposites Hsolid = - Hvap H2O(s) --> H2O(l) Hfus = 6.01 kJ / mol H2O(l) --> H2O(s) Hsolid = kJ / mol Hvap and Hcond are opposites, too! Hvap = - Hcond H2O(l) --> H2O(g) Hvap = 40.7 kJ / mol H2O(g) --> H2O(l) Hcond = kJ / mol
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Heating Curve for Water
∆Hvap ∆Hfus
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Constants for H20 ∆Hfus = 6.01 kJ/mol ∆Hvap = 40.7 kJ/mol
∆Hsolid = − 6.01 kJ/mol ∆Hcond = −40.7 kJ/mol Cp (ice) = J/goC Cp (water) = J/goC Cp (steam) = J/goC
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Sample Problem How many grams of ice at 0oC and kPa could be melted by the addition of 2.25 kJ of heat? Known Values: Unknowns: Melting = ∆Hfusion grams of ice = ? ∆Hfusion (H2O) = 6.01 kJ/mol 2.25 kJ of heat are absorbed ? Grams = 2.25 kJ 1 mol grams = grams 6.01 kJ mol
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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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