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Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.3 Heat in Changes of State >17.3 Heat in Changes of State >
1
Chapter 17Thermochemistry
17.1 The Flow of Energy17.2 Measuring and Expressing Enthalpy Changes
17.3 Heat in Changes of State
17.4 Calculating Heats of Reaction
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17.3 Heat in Changes of State >17.3 Heat in Changes of State >
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Why does sweating help cool you off?
CHEMISTRY & YOUCHEMISTRY & YOU
When your body heats up, you start to sweat. The evaporation of sweat is your body’s way of cooling itself to a normal temperature.
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17.3 Heat in Changes of State >17.3 Heat in Changes of State >
3
Heats of Fusion and Heats of Fusion and SolidificationSolidification
Heats of Fusion and Solidification
What is the relationship between molar heat of fusion and molar heat of solidification?
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Heats of Fusion and Heats of Fusion and SolidificationSolidification
All solids absorb heat as they melt to become liquids.
• The gain of heat causes a change of state instead of a change in temperature.
• The temperature of the substance undergoing the change remains constant.
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Heats of Fusion and Heats of Fusion and SolidificationSolidification
• The heat absorbed by one mole of a solid substance as it melts to a liquid at constant temperature is the molar heat of fusion (ΔHfus).
• The molar heat of solidification (ΔHsolid) is the heat lost when one mole of a liquid substance solidifies at a constant temperature.
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Heats of Fusion and Heats of Fusion and SolidificationSolidification
The quantity of heat absorbed by a melting solid is exactly the same as the quantity of heat released when the liquid solidifies.
ΔHfus = –ΔHsolid
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Heats of Fusion and Heats of Fusion and SolidificationSolidification
• The melting of 1 mol of ice at 0°C to 1 mol of liquid water at 0°C requires the absorption of 6.01 kJ of heat.
• The conversion of 1 mol of liquid water at 0°C to 1 mol of ice at 0°C releases 6.01 kJ of heat.
ΔHfus = 6.01 kJ/mol
ΔHsolid = –6.01 kJ/mol
H2O(s) → H2O(l)
H2O(l) → H2O(s)
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How many grams of ice at 0°C will melt if 2.25 kJ of heat are added?
Sample Problem 17.5Sample Problem 17.5
Using the Heat of Fusion in Phase-Change Calculations
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Analyze List the knowns and the unknown.1
Sample Problem 17.5Sample Problem 17.5
KNOWNS UNKNOWNInitial and final temperature are 0°C
ΔHfus = 6.01 kJ/mol
ΔH = 2.25 kJ
mice = ? g
• Find the number of moles of ice that can be melted by the addition of 2.25 kJ of heat.
• Convert moles of ice to grams of ice.
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Start by expressing ΔHfus as a conversion factor.
Calculate Solve for the unknown.2
Sample Problem 17.5Sample Problem 17.5
1 mol H2O(s)6.01 kJ
Use the thermochemical equation
H2O(s) + 6.01 kJ → H2O(l).
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17.3 Heat in Changes of State >17.3 Heat in Changes of State >
11
Express the molar mass of ice as a conversion factor.
Calculate Solve for the unknown.2
Sample Problem 17.5Sample Problem 17.5
18.0 g H2O(s)
1 mol H2O(s)
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Multiply the known enthalpy change by the conversion factors.
Calculate Solve for the unknown.2
Sample Problem 17.5Sample Problem 17.5
mice = 2.25 kJ
= 6.74 g H2O(s)
1 mol H2O(s)6.01 kJ
18.0 g H2O(s)
1 mol H2O(s)
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17.3 Heat in Changes of State >17.3 Heat in Changes of State >
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• To melt 1 mol of ice, 6.01 kJ of energy is required.
• Only about one-third of this amount of heat (roughly 2 kJ) is available.
• So, only about one-third mol of ice, or 18.0 g/3 = 6 g, should melt.
• This estimate is close to the calculated answer.
Evaluate Does the result make sense?3
Sample Problem 17.5Sample Problem 17.5
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Calculate the amount of heat absorbed to liquefy 15.0 g of methanol (CH4O) at its melting point. The molar heat of fusion for methanol is 3.16 kJ/mol.
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Calculate the amount of heat absorbed to liquefy 15.6 g of methanol (CH4O) at its melting point. The molar heat of fusion for methanol is 3.16 kJ/mol.
ΔH = 15.6 g CH4O
= 1.54 kJ
32.05 g CH4O1 mol 3.16 kJ
1 mol
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17.3 Heat in Changes of State >17.3 Heat in Changes of State >
16
Heats of Vaporization Heats of Vaporization and Condensationand Condensation
Heats of Vaporization and Condensation
What is the relationship between molar heat of vaporization and molar heat of condensation?
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Heats of Vaporization Heats of Vaporization and Condensationand Condensation
A liquid that absorbs heat at its boiling point becomes a vapor.
• The amount of heat required to vaporize one mole of a given liquid at a constant temperature is called its molar heat of vaporization (ΔHvap).
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This table lists the molar heats of vaporization for several substances at their normal boiling point.
Heats of Physical Change
Substance ΔHfus (kJ/mol) ΔHvap (kJ/mol)
Ammonia (NH3) 5.66 23.3
Ethanol (C2H6O) 4.93 38.6
Hydrogen (H2) 0.12 0.90
Methanol (CH4O) 3.22 35.2
Oxygen (O2) 0.44 6.82
Water (H2O) 6.01 40.7
Interpret DataInterpret Data
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Heats of Vaporization Heats of Vaporization and Condensationand Condensation
Condensation is the exact opposite of vaporization.
• When a vapor condenses, heat is released.
• The molar heat of condensation (ΔHcond) is the amount of heat released when one mole of vapor condenses at its normal boiling point.
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Heats of Vaporization Heats of Vaporization and Condensationand Condensation
The quantity of heat absorbed by a vaporizing liquid is exactly the same as the quantity of heat released when the vapor condenses.
ΔHvap = –ΔHcond
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Explain why the evaporation of sweat off your body helps cool you off.
CHEMISTRY & YOUCHEMISTRY & YOU
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Explain why the evaporation of sweat off your body helps cool you off.
CHEMISTRY & YOUCHEMISTRY & YOU
Energy is required to vaporize (or evaporate) a liquid into a gas. When liquid sweat absorbs energy from your skin, the temperature of your skin decreases.
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17.3 Heat in Changes of State >17.3 Heat in Changes of State >
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A heating curve graphically describes the enthalpy changes that take place during phase changes.
Interpret GraphsInterpret Graphs
Remember: The temperature of a substance remains constant during a change of state.
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17.3 Heat in Changes of State >17.3 Heat in Changes of State >
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How much heat (in kJ) is absorbed when 24.8 g H2O(l) at 100°C and 101.3 kPa is converted to H2O(g) at 100°C?
Sample Problem 17.6Sample Problem 17.6
Using the Heat of Vaporization in Phase-Change Calculations
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Analyze List the knowns and the unknown.1
Sample Problem 17.6Sample Problem 17.6
KNOWNS UNKNOWNInitial and final conditions are 100°C and 101.3 kPa
Mass of liquid water converted to steam = 24.8 g
ΔHvap = 40.7 kJ/mol
ΔH = ? kJ
• First convert grams of water to moles of water.
• Then find the amount of heat that is absorbed when the liquid is converted to steam.
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Start by expressing the molar mass of water as a conversion factor.
Calculate Solve for the unknown.2
Sample Problem 17.6Sample Problem 17.6
18.0 g H2O(l)
1 mol H2O(l)
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Express ΔHvap as a conversion factor.
Calculate Solve for the unknown.2
Sample Problem 17.6Sample Problem 17.6
1 mol H2O(l)
40.7 kJ
Use the thermochemical equation
H2O(l) + 40.7 kJ → H2O(g).
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Multiply the mass of water in grams by the conversion factors.
Calculate Solve for the unknown.2
Sample Problem 17.6Sample Problem 17.6
ΔH = 24.8 g H2O(l)
= 56.1 kJ 18.0 g H2O(l)
1 mol H2O(l)
1 mol H2O(l)
40.7 kJ
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Evaluate Does the result make sense?3
• Knowing that the molar mass of water is 18.0 g/mol, 24.8 g H2O(l) can be estimated
to be somewhat less than 1.5 mol H2O.
• The calculated enthalpy change should be a little less than 1.5 mol 40 kJ/mol = 60 kJ, and it is.
Sample Problem 17.6Sample Problem 17.6
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The molar heat of condensation of a substance is the same, in magnitude, as which of the following?
A. molar heat of fusion
B. molar heat of vaporization
C. molar heat of solidification
D. molar heat of formation
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The molar heat of condensation of a substance is the same, in magnitude, as which of the following?
A. molar heat of fusion
B. molar heat of vaporization
C. molar heat of solidification
D. molar heat of formation
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Heat of SolutionHeat of Solution
Heat of Solution
What thermochemical changes can occur when a solution forms?
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Heat of SolutionHeat of Solution
During the formation of a solution, heat is either released or absorbed.
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Heat of SolutionHeat of Solution
• The enthalpy change caused by the dissolution of one mole of substance is the molar heat of solution (ΔHsoln).
During the formation of a solution, heat is either released or absorbed.
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CaCl2(s) → Ca2+(aq) + 2Cl–(aq)
ΔHsoln = –82.8 kJ/mol
Heat of SolutionHeat of Solution
A practical application of an exothermic dissolution process is a hot pack.
• In a hot pack, calcium chloride, CaCl2(s), mixes with water, producing heat.
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Heat of SolutionHeat of Solution
The dissolution of ammonium nitrate, NH4NO3(s), is an example of an endothermic process.• The cold pack shown here contains solid ammonium
nitrate crystals and water.
• Once the solute dissolves, the pack becomes cold.
• The solution process absorbs energy from the surroundings.
NH4NO3(s) → NH4+(aq) + NO3
–(aq)
ΔHsoln = 25.7 kJ/mol
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How much heat (in kJ) is released when 2.50 mol NaOH(s) is dissolved in water?
Sample Problem 17.7Sample Problem 17.7
Calculating the Enthalpy Change in Solution Formation
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Analyze List the knowns and the unknown.1
Sample Problem 17.7Sample Problem 17.7
KNOWNS
UNKNOWN
ΔHsoln = –44.5 kJ/mol
amount of NaOH(s) dissolved = 2.50 mol
ΔH = ? kJ
Use the heat of solution for the dissolution of NaOH(s) in water to solve for the amount of heat released (ΔH).
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Start by expressing ΔHsoln as a conversion factor.
Calculate Solve for the unknown.2
Sample Problem 17.7Sample Problem 17.7
1 mol NaOH(s)
–44.5 kJ
Use the thermochemical equation
NaOH(s) → Na+(aq) + OH–(aq) + 44.5 kJ/mol.
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Multiply the number of moles by the conversion factor.
Calculate Solve for the unknown.2
Sample Problem 17.7Sample Problem 17.7
ΔH = 2.50 mol NaOH(s)
= –111 kJ 1 mol NaOH(s)
–44.5 kJ
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Evaluate Does the result make sense?3
Sample Problem 17.7Sample Problem 17.7
• ΔH is 2.5 times greater than ΔHsoln, as it should
be.
• Also, ΔH should be negative, as the dissolution of NaOH(s) in water is exothermic.
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How much heat (in kJ) is absorbed when 50.0 g of NH4NO3(s) are dissolved in water if Hsoln = 25.7 kJ/mol?
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How much heat (in kJ) is absorbed when 50.0 g of NH4NO3(s) are dissolved in water if Hsoln = 25.7 kJ/mol?
ΔH = 50.0 g NH4NO3
= 16.1 kJ
80.04 g NH4NO3
1 mol 25.7 kJ1 mol
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Key ConceptsKey Concepts
The quantity of heat absorbed by a melting solid is exactly the same as the quantity of heat released when the liquid solidifies; that is, ΔHfus = –ΔHsolid.
The quantity of heat absorbed by a vaporizing liquid is exactly the same as the quantity of heat released when the vapor condenses; that is, ΔHvap = –ΔHcond.
During the formation of a solution, heat is either released or absorbed.
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Glossary TermsGlossary Terms
• molar heat of fusion (ΔHfus): the amount of heat absorbed by one mole of a solid substance as it melts to a liquid at a constant temperature
• molar heat of solidification (ΔHsolid): the amount of heat lost by one mole of a liquid as it solidifies at a constant temperature
• molar heat of vaporization (ΔHvap): the amount of heat absorbed by one mole of a liquid as it vaporizes at a constant temperature
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Glossary TermsGlossary Terms
• molar heat of condensation (ΔHcond): the amount of heat released by one mole of a vapor as it condenses to a liquid at a constant temperature
• molar heat of solution (ΔHsoln): the enthalpy change caused by the dissolution of one mole of a substance
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END OF 17.3END OF 17.3