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17.3 Heat in Changes of State >17.3 Heat in Changes of State >

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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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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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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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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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• 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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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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• 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


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


1 mol H2O(s)6.01 kJ

Use the thermochemical equation

H2O(s) + 6.01 kJ → H2O(l).



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Express the molar mass of ice as a conversion factor.



18.0 g H2O(s)

1 mol H2O(s)



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Multiply the known enthalpy change by the conversion factors.



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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• 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




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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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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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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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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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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.




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Explain why the evaporation of sweat off your body helps cool you off.


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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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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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?


Using the Heat of Vaporization in Phase-Change Calculations



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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.



18.0 g H2O(l)

1 mol H2O(l)



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Express ΔHvap as a conversion factor.



1 mol H2O(l)

40.7 kJ


H2O(l) + 40.7 kJ → H2O(g).



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Multiply the mass of water in grams by the conversion factors.



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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• 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.




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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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During the formation of a solution, heat is either released or absorbed.



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• The enthalpy change caused by the dissolution of one mole of substance is the molar heat of solution (ΔHsoln).




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CaCl2(s) → Ca2+(aq) + 2Cl–(aq)

ΔHsoln = –82.8 kJ/mol


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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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?


Calculating the Enthalpy Change in Solution Formation



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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.



1 mol NaOH(s)

–44.5 kJ


NaOH(s) → Na+(aq) + OH–(aq) + 44.5 kJ/mol.



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Multiply the number of moles by the conversion factor.



ΔH = 2.50 mol NaOH(s)

= –111 kJ 1 mol NaOH(s)

–44.5 kJ



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• 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.




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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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