Transcript Ch17 Lesson17_3

17.3 Heat in Changes of State >
Chapter 17
Thermochemistry
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
1
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17.3 Heat in Changes of State >
CHEMISTRY
& YOU
Why does sweating
help cool you off?
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.
2
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17.3 Heat in Changes of State > Heats of Fusion and
Solidification
Heats of Fusion and Solidification
What is the relationship between
molar heat of fusion and molar heat
of solidification?
3
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17.3 Heat in Changes of State > Heats of Fusion and
Solidification
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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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 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
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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17.3 Heat in Changes of State > Heats of Fusion and
Solidification
• 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.
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H2O(s) → H2O(l)
ΔHfus = 6.01 kJ/mol
H2O(l) → H2O(s)
ΔHsolid = –6.01 kJ/mol
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17.3 Heat in Changes of State >
Sample Problem 17.5
Using the Heat of Fusion in PhaseChange Calculations
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
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.
• Convert moles of ice to grams of ice.
KNOWNS
Initial and final
temperature are 0°C
UNKNOWN
mice = ? g
ΔHfus = 6.01 kJ/mol
ΔH = 2.25 kJ
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17.3 Heat in Changes of State >
Sample Problem 17.5
2 Calculate Solve for the unknown.
Start by expressing ΔHfus as a
conversion factor.
1 mol H2O(s)
6.01 kJ
Use the thermochemical equation
H2O(s) + 6.01 kJ → H2O(l).
10
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17.3 Heat in Changes of State >
Sample Problem 17.5
2 Calculate Solve for the unknown.
Express the molar mass of ice as a
conversion factor.
18.0 g H2O(s)
1 mol H2O(s)
11
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17.3 Heat in Changes of State >
Sample Problem 17.5
2 Calculate Solve for the unknown.
Multiply the known enthalpy change by
the conversion factors.
1 mol H2O(s)
18.0 g H2O(s)
mice = 2.25 kJ 

6.01 kJ
1 mol H2O(s)
= 6.74 g H2O(s)
12
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17.3 Heat in Changes of State >
Sample Problem 17.5
3 Evaluate Does the result make sense?
• 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.
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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.
14
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17.3 Heat in Changes of State >
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  32.05 g CH O 
4
= 1.54 kJ
15
3.16 kJ
1 mol
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17.3 Heat in Changes of State > Heats of Vaporization
and Condensation
Heats of Vaporization and Condensation
What is the relationship between molar
heat of vaporization and molar heat of
condensation?
16
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17.3 Heat in Changes of State > Heats of Vaporization
and 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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17.3 Heat in Changes of State >
Interpret Data
This table lists the molar heats of vaporization for
several substances at their normal boiling point.
Heats of Physical Change
Substance
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
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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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.
ΔHvap = –ΔHcond
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17.3 Heat in Changes of State >
CHEMISTRY
& YOU
Explain why the evaporation of sweat
off your body helps cool you off.
21
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17.3 Heat in Changes of State >
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.
22
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17.3 Heat in Changes of State >
Interpret Graphs
A heating curve
graphically
describes the
enthalpy
changes that
take place
during phase
changes.
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
Using the Heat of Vaporization in PhaseChange Calculations
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?
24
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17.3 Heat in Changes of State >
Sample Problem 17.6
1 Analyze List the knowns and the unknown.
• 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.
KNOWNS
Initial and final conditions
are 100°C and 101.3 kPa
UNKNOWN
ΔH = ? kJ
Mass of liquid water converted to steam = 24.8 g
ΔHvap = 40.7 kJ/mol
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17.3 Heat in Changes of State >
Sample Problem 17.6
2 Calculate Solve for the unknown.
Start by expressing the molar mass of
water as a conversion factor.
1 mol H2O(l)
18.0 g H2O(l)
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17.3 Heat in Changes of State >
Sample Problem 17.6
2 Calculate Solve for the unknown.
Express ΔHvap as a conversion factor.
40.7 kJ
1 mol H2O(l)
Use the thermochemical equation
H2O(l) + 40.7 kJ → H2O(g).
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17.3 Heat in Changes of State >
Sample Problem 17.6
2 Calculate Solve for the unknown.
Multiply the mass of water in grams by
the conversion factors.
ΔH = 24.8 g H2O(l) 
1 mol H2O(l)
40.7 kJ

18.0 g H2O(l) 1 mol H2O(l)
= 56.1 kJ
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17.3 Heat in Changes of State >
Sample Problem 17.6
3 Evaluate Does the result make sense?
• 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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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?
A. molar heat of fusion
B. molar heat of vaporization
C. molar heat of solidification
D. molar heat of formation
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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?
A. molar heat of fusion
B. molar heat of vaporization
C. molar heat of solidification
D. molar heat of formation
31
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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?
32
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17.3 Heat in Changes of State > Heat of Solution
During the formation of a solution, heat
is either released or absorbed.
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17.3 Heat in Changes of State > Heat of Solution
During the formation of a solution, heat
is either released or absorbed.
• The enthalpy change caused by the
dissolution of one mole of substance is the
molar heat of solution (ΔHsoln).
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17.3 Heat in Changes of State > Heat 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.
CaCl2(s) → Ca2+(aq) + 2Cl–(aq)
ΔHsoln = –82.8 kJ/mol
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17.3 Heat in Changes of State > Heat 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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17.3 Heat in Changes of State >
Sample Problem 17.7
Calculating the Enthalpy Change in
Solution Formation
How much heat (in kJ) is
released when 2.50 mol
NaOH(s) is dissolved in water?
37
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17.3 Heat in Changes of State >
Sample Problem 17.7
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).
KNOWNS
ΔHsoln = –44.5 kJ/mol
amount of NaOH(s) dissolved = 2.50 mol
UNKNOWN
ΔH = ? kJ
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17.3 Heat in Changes of State >
Sample Problem 17.7
2 Calculate Solve for the unknown.
Start by expressing ΔHsoln as a
conversion factor.
–44.5 kJ
1 mol NaOH(s)
Use the thermochemical equation
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
2 Calculate Solve for the unknown.
Multiply the number of moles by the
conversion factor.
ΔH = 2.50 mol NaOH(s) 
–44.5 kJ
1 mol NaOH(s)
= –111 kJ
40
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17.3 Heat in Changes of State >
Sample Problem 17.7
3 Evaluate Does the result make sense?
• 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.
41
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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 Hsoln = 25.7 kJ/mol?
42
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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 Hsoln = 25.7 kJ/mol?
1 mol
ΔH = 50.0 g NH4NO3  80.04 g NH NO 
4
3
= 16.1 kJ
43
25.7 kJ
1 mol
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17.3 Heat in Changes of State >
Key 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.
44
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17.3 Heat in Changes of State > Glossary 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
45
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17.3 Heat in Changes of State > Glossary 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
46
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17.3 Heat in Changes of State >
END OF 17.3
47
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