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17.3 Heat in Changes of State >
17.3 Heat in Changes of State >
Thermochemistry
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.
17.1 The Flow of Energy
17.2 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 > Heats of Fusion and
Solidification
& YOU
Why does sweating
help cool you off?
Chapter 17
1
CHEMISTRY
2
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17.3 Heat in Changes of State > Heats of Fusion and
Solidification
Heats of Fusion and Solidification
All solids absorb heat as they melt to
become liquids.
What is the relationship between
molar heat of fusion and molar heat
of solidification?
•  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.
3
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17.3 Heat in Changes of State > Heats of Fusion and
Solidification
17.3 Heat in Changes of State > Heats of Fusion and
Solidification
•  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 quantity of heat absorbed by a
melting solid is exactly the same as
the quantity of heat released when
the liquid solidifies.
•  The molar heat of solidification
(ΔHsolid) is the heat lost when one mole
of a liquid substance solidifies at a
constant temperature.
5
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17.3 Heat in Changes of State > Heats of Fusion and
Solidification
ΔHfus = –ΔHsolid
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17.3 Heat in Changes of State >
7
ΔHfus = 6.01 kJ/mol
H2O(l) → H2O(s)
ΔHsolid = –6.01 kJ/mol
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Sample Problem 17.5
Using the Heat of Fusion in PhaseChange Calculations
•  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.
H2O(s) → H2O(l)
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How many grams of ice at 0°C
will melt if 2.25 kJ of heat are
added?
8
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17.3 Heat in Changes of State >
Sample Problem 17.5
17.3 Heat in Changes of State >
1 Analyze List the knowns and the unknown.
•  Find the number of moles of ice that can be
melted by the addition of 2.25 kJ of heat.
2 Calculate Solve for the unknown.
Start by expressing ΔHfus as a
conversion factor.
•  Convert moles of ice to grams of ice.
KNOWNS
Initial and final
temperature are 0°C
1 mol H2O(s)
6.01 kJ
UNKNOWN
mice = ? g
Use the thermochemical equation
ΔHfus = 6.01 kJ/mol
H2O(s) + 6.01 kJ → H2O(l).
ΔH = 2.25 kJ
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17.3 Heat in Changes of State >
Sample Problem 17.5
10
Sample Problem 17.5
2 Calculate Solve for the unknown.
Express the molar mass of ice as a
conversion factor.
Multiply the known enthalpy change by
the conversion factors.
18.0 g H2O(s)
1 mol H2O(s)
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17.3 Heat in Changes of State >
2 Calculate Solve for the unknown.
11
Sample Problem 17.5
1 mol H2O(s)
18.0 g H2O(s)
x
6.01 kJ
1 mol H2O(s)
= 6.74 g H2O(s)
mice = 2.25 kJ x
12
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17.3 Heat in Changes of State >
17.3 Heat in Changes of State >
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.
13
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17.3 Heat in Changes of State > Heats of Vaporization
17.3 Heat in Changes of State >
and Condensation
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.
1 mol
ΔH = 15.6 g CH4O x 32.05 g CH O x
4
= 1.54 kJ
15
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Heats of Vaporization and Condensation
What is the relationship between molar
heat of vaporization and molar heat of
condensation?
3.16 kJ
1 mol
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17.3 Heat in Changes of State > Heats of Vaporization
and Condensation
17.3 Heat in Changes of State >
This table lists the molar heats of vaporization for
several substances at their normal boiling point.
A liquid that absorbs heat at its boiling
point becomes a vapor.
Heats of Physical Change
Substance
•  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).
17
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17.3 Heat in Changes of State > Heats of Vaporization
and Condensation
Interpret Data
18
ΔHfus (kJ/mol)
ΔHvap (kJ/mol)
Ammonia (NH3)
5.66
23.3
Ethanol (C2H6O)
4.93
38.6
Hydrogen (H2)
0.12
Methanol (CH4O)
3.22
Oxygen (O2)
0.44
Water (H2O)
6.01
0.90
35.2
6.82
40.7
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17.3 Heat in Changes of State > Heats of Vaporization
and Condensation
The quantity of heat absorbed by a
vaporizing liquid is exactly the same
as the quantity of heat released when
the vapor condenses.
Condensation is the exact opposite of
vaporization.
•  When a vapor condenses, heat is released.
ΔHvap = –ΔHcond
•  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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17.3 Heat in Changes of State >
CHEMISTRY
& YOU
17.3 Heat in Changes of State >
Explain why the evaporation of sweat
off your body helps cool you off.
CHEMISTRY
& YOU
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.
21
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17.3 Heat in Changes of State >
Interpret Graphs
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17.3 Heat in Changes of State >
Sample Problem 17.6
Using the Heat of Vaporization in PhaseChange Calculations
A heating curve
graphically
describes the
enthalpy
changes that
take place
during phase
changes.
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?
Remember: The temperature of a
substance remains constant during
a change of state.
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17.3 Heat in Changes of State >
Sample Problem 17.6
17.3 Heat in Changes of State >
1 Analyze List the knowns and the unknown.
•  First convert grams of water to moles of water.
2 Calculate Solve for the unknown.
Start by expressing the molar mass of
water as a conversion factor.
•  Then find the amount of heat that is absorbed
when the liquid is converted to steam.
KNOWNS
Initial and final conditions
are 100°C and 101.3 kPa
Sample Problem 17.6
1 mol H2O(l)
18.0 g H2O(l)
UNKNOWN
ΔH = ? kJ
Mass of liquid water converted to steam = 24.8 g
ΔHvap = 40.7 kJ/mol
25
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17.3 Heat in Changes of State >
Sample Problem 17.6
26
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17.3 Heat in Changes of State >
2 Calculate Solve for the unknown.
Sample Problem 17.6
2 Calculate Solve for the unknown.
Express ΔHvap as a conversion factor.
Multiply the mass of water in grams by
the conversion factors.
40.7 kJ
1 mol H2O(l)
ΔH = 24.8 g H2O(l) x
= 56.1 kJ
Use the thermochemical equation
1 mol H2O(l)
40.7 kJ
x
18.0 g H2O(l) 1 mol H2O(l)
H2O(l) + 40.7 kJ → H2O(g).
27
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17.3 Heat in Changes of State >
17.3 Heat in Changes of State >
The molar heat of condensation of a
substance is the same, in magnitude,
as which of the following?
The molar heat of condensation of a
substance is the same, in magnitude,
as which of the following?
A. molar heat of fusion
A. molar heat of fusion
B. molar heat of vaporization
B. molar heat of vaporization
C. molar heat of solidification
C. molar heat of solidification
D. molar heat of formation
D. molar heat of formation
29
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17.3 Heat in Changes of State > Heat of Solution
30
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17.3 Heat in Changes of State > Heat of Solution
Heat of Solution
What thermochemical changes can
occur when a solution forms?
31
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During the formation of a solution, heat
is either released or absorbed.
32
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17.3 Heat in Changes of State > Heat of Solution
17.3 Heat in Changes of State > Heat of Solution
A practical application of an exothermic
dissolution process is a hot pack.
During the formation of a solution, heat
is either released or absorbed.
•  In a hot pack, calcium chloride, CaCl2(s),
mixes with water, producing heat.
•  The enthalpy change caused by the
dissolution of one mole of substance is the
molar heat of solution (ΔHsoln).
CaCl2(s) → Ca2+(aq) + 2Cl–(aq)
ΔHsoln = –82.8 kJ/mol
33
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17.3 Heat in Changes of State > Heat of Solution
34
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17.3 Heat in Changes of State >
The dissolution of ammonium nitrate, NH4NO3(s),
is an example of an endothermic process.
Sample Problem 17.7
Calculating the Enthalpy Change in
Solution Formation
•  The cold pack shown here contains solid ammonium
nitrate crystals and water.
How much heat (in kJ) is
released when 2.50 mol
NaOH(s) is dissolved in 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
35
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17.3 Heat in Changes of State >
Sample Problem 17.7
17.3 Heat in Changes of State >
1 Analyze List the knowns and the unknown.
Use the heat of solution for the dissolution of
NaOH(s) in water to solve for the amount of heat
released (ΔH).
2 Calculate Solve for the unknown.
Start by expressing ΔHsoln as a
conversion factor.
KNOWNS
ΔHsoln = –44.5 kJ/mol
–44.5 kJ
1 mol NaOH(s)
amount of NaOH(s) dissolved = 2.50 mol
Use the thermochemical equation
UNKNOWN
ΔH = ? kJ
37
Sample Problem 17.7
NaOH(s) → Na+(aq) + OH–(aq) + 44.5 kJ/mol.
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17.3 Heat in Changes of State >
Sample Problem 17.7
38
17.3 Heat in Changes of State >
How much heat (in kJ) is absorbed when
50.0 g of NH4NO3(s) are dissolved in
water if ΔHsoln = 25.7 kJ/mol?
2 Calculate Solve for the unknown.
Multiply the number of moles by the
conversion factor.
ΔH = 2.50 mol NaOH(s) x
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–44.5 kJ
1 mol NaOH(s)
= –111 kJ
39
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17.3 Heat in Changes of State >
How much heat (in kJ) is absorbed when
50.0 g of NH4NO3(s) are dissolved in
water if ΔHsoln = 25.7 kJ/mol?
1 mol
25.7 kJ
ΔH = 50.0 g NH4NO3 x 80.0 g NH NO x 1 mol
4
3
= 16.1 kJ
41
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