Download Document 7997681

2/25/2013
17.1 The Flow of Energy >
17.1 The Flow of Energy >
Thermochemistry
17.2 Measuring and Expressing
Enthalpy Changes
17.3 Heat in Changes of State
17.4 Calculating Heats of Reaction
17.1 The Flow of Energy > Energy Transformations
2
17.1 The Flow of Energy > Energy Transformations
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
3
Every substance has a certain amount of
energy stored inside it.
• When you buy gasoline, you are actually buying
the stored potential energy it contains.
• The controlled explosions of the gasoline in a
car’s engine transform the potential energy into
useful work, which can be
used to propel the car.
• The kinds of atoms and the arrangement of
the atoms in a substance determine the
amount of energy stored in the substance.
5
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Energy Transformations
8
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Energy Transformations
• The energy stored in the chemical bonds of
a substance is called chemical potential
energy.
Energy changes occur as either
heat transfer or work, or a
combination of both.
7
What are the ways in which energy
changes can occur?
Every substance has a certain amount of
energy stored inside it.
• Thermochemistry is the study of energy
changes that occur during chemical reactions
and changes in state.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Energy Transformations
• Energy is the capacity for doing work or
supplying heat.
• Unlike matter, energy has neither mass
nor volume.
• Energy is detected only because of its
effects.
4
Energy Transformations
Lava flowing out of an
erupting volcano is
very hot. As lava flows,
it loses heat and
begins to cool slowly.
The lava may flow into
the ocean, where it
cools more rapidly.
17.1 The Flow of Energy
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Energy Transformations
Why does lava cool faster in water
than in air?
Chapter 17
1
CHEMISTRY & YOU
• Heat is also produced,
making the car’s engine
extremely hot.
6
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Energy Transformations
Energy changes occur as either
heat transfer or work, or a
combination of both.
• Heat flows spontaneously from a warmer
object to a cooler object.
• Heat, represented by q, is energy that
transfers from one object to another
because of a temperature difference
between the objects.
• If two objects remain in contact, heat will
flow from the warmer object to the cooler
object until the temperature of both objects
is the same.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
9
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
1
2/25/2013
17.1 The Flow of Energy >
The energy released when a piece of
wood is burned has been stored in the
wood as
10
Processes
The energy released when a piece of
wood is burned has been stored in the
wood as
A. sunlight.
B. heat.
B. heat.
C. calories.
C. calories.
D. chemical potential energy.
D. chemical potential energy.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Processes
11
Processes
• Chemical reactions and changes in
physical state generally involve either
the absorption or the release of heat.
17.1 The Flow of Energy > Endothermic and Exothermic
Processes
12
14
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Endothermic and Exothermic
Processes
During any chemical or physical
process, the energy of the
universe remains unchanged.
Processes
The law of conservation of energy
states that in any chemical or physical
process, energy is neither created nor
destroyed.
15
Processes
During any chemical or physical
process, the energy of the
universe remains unchanged.
Direction of Heat Flow
The direction of heat flow is given from
the point of view of the system.
• Heat is absorbed from the surroundings in
an endothermic process.
– Heat flowing into a system from its
surroundings is defined as positive; q has
a positive value.
• If the energy of the system decreases during
that process, the energy of the surroundings
must increase by the same amount.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Endothermic and Exothermic
• If the energy of the system increases during
that process, the energy of the surroundings
must decrease by the same amount.
16
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Endothermic and Exothermic
• You can define a system as the part of
the universe on which you focus your
attention.
• Everything else in the universe makes
up the surroundings.
• Together, the system and its
surroundings make up the universe.
What happens to the energy of the
universe during a chemical or
physical process?
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Endothermic and Exothermic
Endothermic and Exothermic Processes
Endothermic and Exothermic Processes
What happens to the energy of the
universe during a chemical or
physical process?
A. sunlight.
17.1 The Flow of Energy > Endothermic and Exothermic
13
17.1 The Flow of Energy > Endothermic and Exothermic
17.1 The Flow of Energy >
18
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
2
2/25/2013
17.1 The Flow of Energy > Endothermic and Exothermic
Processes
17.1 The Flow of Energy > Endothermic and Exothermic
Processes
In an endothermic process,
heat flows into the system
from the surroundings.
Direction of Heat Flow
17.1 The Flow of Energy >
In an exothermic process,
heat flows from the system
to the surroundings.
The direction of heat flow is given from
the point of view of the system.
• An exothermic process is one that
releases heat to its surroundings.
– Heat flowing out of a system into its
surroundings is defined as negative; q
has a negative value.
In both cases, energy is conserved.
19
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
20
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
21
17.1 The Flow of Energy >
17.1 The Flow of Energy >
17.1 The Flow of Energy >
22
23
24
17.1 The Flow of Energy >
17.1 The Flow of Energy >
17.1 The Flow of Energy >
Sample Problem 17.1
Recognizing Endothermic and
Exothermic Processes
On a sunny winter day, the snow on
a rooftop begins to melt. As the
melted water drips from the roof, it
refreezes into icicles. Describe the
direction of heat flow as the water
freezes. Is this process endothermic
or exothermic?
25
26
27
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
3
2/25/2013
17.1 The Flow of Energy >
Sample Problem 17.1
17.1 The Flow of Energy >
1 Analyze Identify the relevant concepts.
Sample Problem 17.1
17.1 The Flow of Energy >
2 Solve Apply concepts to this situation.
• Heat flows from a warmer object to a
cooler object.
2 Solve Apply concepts to this situation.
First identify the system and the surroundings.
Determine the direction of heat flow.
• In order for water to freeze, its
temperature must decrease.
System: water
• An endothermic process absorbs heat
from the surroundings.
Sample Problem 17.1
Surroundings: air
• Heat flows out of the water and into
the air.
• An exothermic process releases heat to
the surroundings.
First identify the system and surroundings.
Then determine the direction of the heat flow.
28
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Sample Problem 17.1
29
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Endothermic and Exothermic
Processes
2 Solve Apply concepts to this situation.
30
17.1 The Flow of Energy > Endothermic and Exothermic
Processes
Units for Measuring Heat Flow
Determine if the process is endothermic or
exothermic.
Heat flow is measured in two common
units:
• Heat is released from the system to
the surroundings.
• the calorie
Units for Measuring Heat Flow
A calorie (cal) is defined as the quantity of
heat needed to raise the temperature of 1 g
of pure water 1 C.
°
• The word calorie is written with a small c except
when referring to the energy contained in food.
• the joule
• The process is exothermic.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
• The dietary Calorie is written with a capital C.
• One dietary Calorie is equal to one kilocalorie,
or 1000 calories.
1 Calorie = 1 kilocalorie = 1000 calories
31
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Endothermic and Exothermic
Processes
32
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Units for Measuring Heat Flow
33
17.1 The Flow of Energy >
Athletes often use instant cold packs to
soothe injuries. Many of these packs use the
dissociation of ammonium nitrate in water to
create a cold-feeling compress. Is this
reaction endothermic or exothermic? Why?
Athletes often use instant cold packs to
soothe injuries. Many of these packs use the
dissociation of ammonium nitrate in water to
create a cold-feeling compress. Is this
reaction endothermic or exothermic? Why?
The joule (J) is the SI unit of energy.
• One joule of heat raises the temperature of 1 g
of pure water 0.2390 C.
°
• You can convert between calories and joules
using the following relationships:
1 J = 0.2390 cal
34
The instant cold pack feels cold because it
removes heat from its surroundings. Therefore,
the dissociation of ammonium nitrate in water is
endothermic. The system (the cold pack) gains
heat as the surroundings lose heat.
4.184 J = 1 cal
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
35
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
36
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
4
2/25/2013
17.1 The Flow of Energy > Heat Capacity and Specific
Heat
17.1 The Flow of Energy > Heat Capacity and Specific
Heat
Heat Capacity and Specific Heat
17.1 The Flow of Energy > Heat Capacity and Specific
Heat
The heat capacity of an object
depends on both its mass and its
chemical composition.
Heat Capacity and Specific Heat
On what factors does the heat
capacity of an object depend?
On what factors does the heat
capacity of an object depend?
• The greater the mass of
the object, the greater its
heat capacity.
• The amount of heat needed to increase
the temperature of an object exactly 1oC
is the heat capacity of that object.
37
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
38
17.1 The Flow of Energy > Heat Capacity and Specific
Interpret Data
Heat
The specific heat capacity, or simply the specific
heat, of a substance is the amount of heat it takes to
raise the temperature of 1 g of the substance 1 C.
°
• Water has a
very high
specific heat
compared with
the other
substances.
• Metals
generally have
low specific
heats.
40
Specific heat
J/(g·
°C)
cal/(g·
°C)
Liquid water
4.18
1.00
Ethanol
2.4
0.58
Ice
2.1
0.50
Steam
1.9
0.45
Chloroform
0.96
0.23
Aluminum
0.90
0.21
Iron
0.46
0.11
Silver
0.24
0.057
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Heat Capacity and Specific
Heat
Specific Heat of Water
When a freshly baked apple pie comes out of the
oven, both the filling and the crust are at the same
temperature.
• The filling, which is mostly water,
has a higher specific heat than
the crust.
39
Heat
Specific Heat of Water
Water in lakes and oceans absorbs heat
from the air on hot days
and releases it back into
the air on cool days.
• This property of water is
responsible for
moderate climates in
coastal areas.
41
17.1 The Flow of Energy >
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
CHEMISTRY & YOU
Heat will flow from the lava to the surroundings
until the lava and surroundings are at the same
temperature. Air has a smaller specific heat than
water. Why would lava then cool more quickly in
water than in air?
42
CHEMISTRY & YOU
Heat will flow from the lava to the surroundings
until the lava and surroundings are at the same
temperature. Air has a smaller specific heat than
water. Why would lava then cool more quickly in
water than in air?
Water requires more energy to raise its
temperature than air.
Therefore, lava in contact with water loses more
heat energy than lava in contact with air, allowing it
to cool more quickly.
• This release of heat is why you
have to be careful not to burn
your tongue when eating hot
apple pie.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
• In order to cool down, the filling
must give off a lot of heat.
43
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Heat Capacity and Specific
Specific Heat of Water
Just as it takes a lot of heat to raise the
temperature of water, water also releases a
lot of heat as it cools.
Specific Heats of Some Common Substances
Substance
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
• A massive steel cable
requires more heat to raise
its temperature by 1oC
than a steel nail does.
44
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
45
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
5
2/25/2013
17.1 The Flow of Energy > Heat Capacity and Specific
Heat
17.1 The Flow of Energy > Heat Capacity and Specific
Heat
Calculating Specific Heat
Calculating Specific Heat
To calculate the specific heat (C) of a
substance, you divide the heat input by the
mass of the substance times the temperature
change.
q
C=
=
m × ∆T
q
C=
m × ∆T
=
17.1 The Flow of Energy >
Calculating the Specific Heat of a Substance
heat (J or cal)
mass (g) × change in temperature
( C)
The temperature of a 95.4-g piece
of copper increases from 25.0 C
to 48.0 C when the copper
absorbs 849 J of heat. What is the
specific heat of copper?
°
• m is mass.
• ∆T is the change in temperature.
∆T = Tf – Ti
°
°
• q is heat, expressed in terms of joules or calories.
heat (J or cal)
mass (g) × change in temperature (oC)
Sample Problem 17.2
°
• The units of specific heat are either J/(g· C) or
cal/(g· C).
46
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Sample Problem 17.2
47
°
Sample Problem 17.2
48
• Substitute the known quantities into
the equation to calculate the unknown
value CCu.
q
CCu = m × ∆T
Cu
∆T = (48.0 C – 48.0 C) = 23.0 C
q = 849 J
Sample Problem 17.2
2 Calculate Solve for the unknown.
• Start with the equation for specific
heat.
°
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
2 Calculate Solve for the unknown.
Use the known values and the definition of
specific heat.
°
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
1 Analyze List the knowns and the unknown.
KNOWNS
mCu = 95.4 g
°
849 J
CCu =
95.4 g × 23.0oC
J/(g· C)
°
= 0.387
UNKNOWN
°
C = ? J/(g· C)
49
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Sample Problem 17.2
50
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
51
17.1 The Flow of Energy >
°
°
The specific heat of ethanol is 2.4 J/(g· C). A
sample of ethanol absorbs 676 J of heat, and
the temperature rises from 22 C to 64 C.
What is the mass of ethanol in the sample?
3 Evaluate Does the result make sense?
°
• Remember that liquid water has a specific
heat of 4.18 J/(g· C).
°
°
• Metals have specific heats lower than
water.
C=
q
m × ∆T
• Thus, the calculated value of 0.387
J/(g· C) seems reasonable.
m=
676 J
2.4 J/(g· C) × (64 C –
22 C)
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
53
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
°
°
The specific heat of ethanol is 2.4 J/(g· C). A
sample of ethanol absorbs 676 J of heat, and
the temperature rises from 22 C to 64 C.
What is the mass of ethanol in the sample?
°
52
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
54
m=
°
°
°
q
C × ∆T
= 6.7 g ethanol
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
6
2/25/2013
17.1 The Flow of Energy > Key Concepts & Key Equation
17.1 The Flow of Energy > Glossary Terms
• thermochemistry: the study of energy changes that
occur during chemical reactions and changes in state
Energy changes occur as either heat
transfer or work, or a combination of both.
q
m × ∆T
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
BIG IDEA
• endothermic process: a process that absorbs heat
from the surroundings
• heat (q): energy that transfers from one object to
another because of a temperature difference between
the objects
The heat capacity of an object depends on
both its mass and its chemical
composition.
C=
• law of conservation of energy: in any chemical or
physical process, energy is neither created nor
destroyed
• chemical potential energy: energy stored in
chemical bonds
During any chemical or physical process,
the energy of the universe remains
unchanged.
55
17.1 The Flow of Energy > Glossary Terms
• exothermic process: a process that releases heat to
its surroundings
• system: a part of the universe on which you focus
your attention
• heat capacity: the amount of heat needed to increase
the temperature of an object exactly 1 C
• surroundings: everything in the universe outside the
system
• specific heat: the amount of heat needed to increase
the temperature of 1 g of a substance 1 C; also
called specific heat capacity
56
17.1 The Flow of Energy >
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
BIG IDEA
°
°
57
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Matter and Energy
• During a chemical or physical process, the
energy of the universe is conserved.
– If energy is absorbed by the system in a
chemical or physical process, the same
amount of energy is released by the
surroundings.
– Conversely, if energy is released by the
system, the same amount of energy is
absorbed by the surroundings.
58
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Solar hot water system installed on low cost housing in the
Kouga Local Municipality, South Africa
59
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
60
17.1 The Flow of Energy >
Thermochemistry
Remember: The
concept of specific
heat allows you to
measure heat flow in
chemical and physical
processes.
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
61
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
62
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
CHEMISTRY & YOU
How can you measure the amount of
heat released when a match burns?
Chapter 17
END OF 17.1
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
63
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
7
2/25/2013
17.1 The Flow of Energy >
Calorimetry
17.1 The Flow of Energy > Calorimetry
17.1 The Flow of Energy > Calorimetry
Calorimetry is the measurement of the
heat flow into or out of a system for
chemical and physical processes.
Calorimetry
How can you measure the change in
enthalpy of a reaction?
Calorimetry is the measurement of the
heat flow into or out of a system for
chemical and physical processes.
• In a calorimetry experiment involving an
endothermic process, the heat absorbed by
the system is equal to the heat released by
its surroundings.
• In an exothermic process, the heat released
by the system is equal to the heat absorbed
by its surroundings.
64
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Calorimetry
65
17.1 The Flow of Energy > Calorimetry
Calorimetry is the measurement of the
heat flow into or out of a system for
chemical and physical processes.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Calorimetry
66
Constant-Pressure Calorimeters
The enthalpy (H) of a system accounts
for the heat flow of the system at
constant pressure.
Foam cups can be used as simple
calorimeters because they do not let
much heat in or out.
• The heat absorbed or released by a reaction
at constant pressure is the same as the
change in enthalpy, symbolized as ∆H.
• Most chemical reactions and physical
changes carried out in the laboratory are
open to the atmosphere and thus occur at
constant pressure.
68
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Calorimetry
Constant-Pressure Calorimeters
The value of ∆H of a reaction can be
determined by measuring the heat flow
of the reaction at constant pressure.
• In this textbook, the terms heat and
enthalpy change are used
interchangeably.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Calorimetry
Constant-Pressure Calorimeters
• The insulated device used to measure the
absorption or release of heat in chemical or
physical processes is called a calorimeter.
67
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
69
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Calorimetry
Constant-Pressure Calorimeters
Constant-Pressure Calorimeters
• To measure the enthalpy
change for a reaction in
aqueous solution in a
foam cup calorimeter,
dissolve the reacting
chemicals (the system) in
known volumes of water
(the surroundings).
• Measure the initial
temperature of each
solution, and mix the
solutions in the foam cup.
• After the reaction is
complete, measure the
final temperature of the
mixed solutions.
• In other words, q = ∆H.
70
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
71
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
72
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
8
2/25/2013
17.1 The Flow of Energy > Calorimetry
17.1 The Flow of Energy > Calorimetry
Constant-Pressure Calorimeters
Constant-Pressure Calorimeters
You can calculate the heat
absorbed or released by the
surroundings (qsurr) using the
formula for the specific heat,
the initial and final
temperatures, and the heat
capacity of water.
Constant-Pressure Calorimeters
The heat absorbed by
the surroundings is
equal to, but has the
opposite sign of, the
heat released by the
system.
qsurr = m × C × ∆T
• m is the mass of the
water.
• C is the specific heat
of water.
qsurr = m × C × ∆T
73
17.1 The Flow of Energy > Calorimetry
qsurr = –qsys
• ∆T = Tf – Ti
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Calorimetry
74
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Calorimetry
Constant-Pressure Calorimeters
Constant-Volume Calorimeters
The enthalpy change for the reaction
(∆H) can be written as follows:
Calorimetry experiments can also be
performed at a constant volume using a
device called a bomb calorimeter.
75
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Calorimetry
Constant-Volume Calorimeters
In a bomb
calorimeter, a
sample of a
compound is
burned in a
constant-volume
chamber in the
presence of
oxygen at high
pressure.
qsys = ∆H = –qsurr = –m × C × ∆T
• The sign of ∆H is positive for an
endothermic reaction and negative for an
exothermic reaction.
76
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Calorimetry
77
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
78
17.1 The Flow of Energy >
17.1 The Flow of Energy >
80
81
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Constant-Volume Calorimeters
• The heat that is
released warms the
water surrounding the
chamber.
• By measuring the
temperature increase
of the water, it is
possible to calculate
the quantity of heat
released during the
combustion reaction.
79
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
9
2/25/2013
17.1 The Flow of Energy >
17.1 The Flow of Energy >
17.1 The Flow of Energy >
82
83
84
17.1 The Flow of Energy >
CHEMISTRY & YOU
17.1 The Flow of Energy >
What type of calorimeter would you use to
measure the heat released when a match
burns? Describe the experiment and how you
would calculate the heat released.
CHEMISTRY & YOU
17.1 The Flow of Energy >
Enthalpy Change in a Calorimetry
Experiment
What type of calorimeter would you use to
measure the heat released when a match
burns? Describe the experiment and how you
would calculate the heat released.
When 25.0 mL of water containing 0.025 mol HCl
at 25.0 C is added to 25.0 mL of water
containing 0.025 mol NaOH at 25.0 C in a
foam-cup calorimeter, a reaction occurs.
Calculate the enthalpy change (in kJ) during this
reaction if the highest temperature observed is
32.0 C. Assume that the densities of the
solutions are 1.00 g/mL and the volume of the
final solution is equal to the sum of the volumes
of the reacting solutions.
°
A constant-volume, or bomb, calorimeter would be used
to measure the heat released when a match burns. The
match would be ignited in the chamber. By measuring
the temperature increase in the water and using the
equation q = –m × C × ∆T , the heat released, q, can be
calculated.
85
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Sample Problem 17.3
86
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
1 Analyze List the knowns and the unknown.
Sample Problem 17.3
KNOWNS
87
°
Sample Problem 17.3
2 Calculate Solve for the unknown.
First calculate the total mass of the
water.
∆H = ? kJ
Vfinal = VHCl + VNaOH
• Use ∆H = –qsurr = –m × C × ∆T to solve for ∆H.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
UNKNOWN
Cwater = 4.18 J/(g· C)
• You must also calculate ∆T.
°
°
1 Analyze List the knowns and the unknown.
• Use dimensional analysis to determine the
mass of the water.
Sample Problem 17.3
mwater = 50.0 mL ×
= 25.0 mL + 25.0 mL = 50.0 mL
°
°
1.00 g
= 50.0 g
1 mL
Ti = 25.0 C
Tf = 32.0 C
Assume that the densities of the
solutions are 1.00 g/mL to find the
total mass of the water.
densitysolution = 1.00 g/mL
88
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
89
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
90
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
10
2/25/2013
17.1 The Flow of Energy >
Sample Problem 17.3
17.1 The Flow of Energy >
2 Calculate Solve for the unknown.
Now calculate ∆T.
°
°
Sample Problem 17.3
17.1 The Flow of Energy >
2 Calculate Solve for the unknown.
°
∆T = Tf – Ti = 32.0 C – 25.0 C = 7.0 C
3 Evaluate Does the result make sense?
Use the values for mwater, Cwater,
and ∆T to calculate ∆H.
• The temperature of the solution increases, which
means that the reaction is exothermic, and thus the
sign of ∆H should be negative.
∆H = –qsurr = –mwater × Cwater × ∆T
• About 4 J of heat raises the temperature of 1 g of
water 1 C, so 200 J of heat is required to raise 50
g of water 1 C. Raising the temperature of 50 g of
water 7 C requires about 1400 J, or 1.4 kJ.
°
°
°
°
= –(50.0 g)(4.18J/(g·oC))(7.0 C)
= –1500 J = –1.5 kJ
91
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
92
°
Use the relationship
1 kJ = 1000 J to
convert your answer
from J to kJ.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
The initial temperature of the water in a
constant-pressure calorimeter is 24 C. A
reaction takes place in the calorimeter, and the
temperature rises to 87 C. The calorimeter
contains 367 g of water, which has a specific
heat of 4.18 J/(g· C). Calculate the enthalpy
change during this reaction.
°
• This estimated answer is very close to the
calculated value of ∆H.
93
°
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Thermochemical Equations
The initial temperature of the water in a
constant-pressure calorimeter is 24 C. A
reaction takes place in the calorimeter, and the
temperature rises to 87 C. The calorimeter
contains 367 g of water, which has a specific
heat of 4.18 J/(g· C). Calculate the enthalpy
change during this reaction.
°
Sample Problem 17.3
°
°
∆H = –m × C × ∆T
°
Thermochemical Equations
How can you express the enthalpy
change for a reaction in a chemical
equation?
°
= –367 g × 4.18 J/(g· C) × (87 C –
24 C)
°
94
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Thermochemical Equations
95
= –97000 J = –97 kJ
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Thermochemical Equations
96
17.1 The Flow of Energy > Thermochemical Equations
In the equation describing the exothermic
reaction of calcium oxide and water, the
enthalpy change can be considered a product.
In a chemical equation, the enthalpy
change for the reaction can be written
as either a reactant or a product.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
A chemical equation that includes the enthalpy
change is called a thermochemical equation.
CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ
CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ
Calcium oxide is
one of the
components of
cement.
97
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
98
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
99
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
11
2/25/2013
17.1 The Flow of Energy > Thermochemical Equations
17.1 The Flow of Energy > Thermochemical Equations
Heats of Reaction
The heat of reaction is the enthalpy change
for the chemical equation exactly as it is
written.
Heats of Reaction
Each mole of calcium oxide and water
that reacts to form calcium hydroxide
produces 65.2 kJ of heat.
• Heats of reaction are reported as ∆H.
CaO(s) + H2O(l) → Ca(OH)2(s)
• The physical state of the reactants and
products must also be given.
°
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Thermochemical Equations
101
Baking soda (sodium bicarbonate) decomposes
when it is heated. This process is endothermic.
2NaHCO3(s) + 85 kJ → Na2CO3(s) + H2O(l) + CO2(g)
The carbon
dioxide released
in the reaction
causes muffins
to rise while
baking.
∆H = –65.2 kJ
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Thermochemical Equations
Heats of Reaction
2NaHCO3(s) + 85 kJ → Na2CO3(s) + H2O(l) + CO2(g)
Remember that ∆H is positive for endothermic
reactions. Therefore, you can write the reaction
as follows:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
Heats of Reaction
• In exothermic processes, the chemical
potential energy of the reactants is higher than
the chemical potential energy of the products.
• The standard conditions are that the reaction is
carried out at 101.3 kPa (1 atm) and 25 C.
100
17.1 The Flow of Energy > Thermochemical Equations
102
17.1 The Flow of Energy > Thermochemical Equations
Heats of Reaction
Heats of Reaction
The amount of heat released or absorbed
during a reaction depends on the number of
moles of the reactant involved.
To see why the physical state of the reactants and
products must be stated, compare the following
two equations.
H2O(l) → H2(g) +
• The decomposition of 2 mol of sodium
bicarbonate requires 85 kJ of heat.
∆H = 85 kJ
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
H2O(g) → H2(g) +
1
2
1
2
O2(g)
O2(g)
• Therefore, the decomposition of 4 mol of the
same substance would require twice as much
heat, or 170 kJ.
103
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Thermochemical Equations
104
17.1 The Flow of Energy >
Heats of Reaction
To see why the physical state of the reactants and
products must be stated, compare the following
two equations.
H2O(l) → H2(g) +
H2O(g) → H2(g) +
1
2
1
2
O2(g)
O2(g)
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Sample Problem 17.4
105
Calculate the amount of heat (in kJ)
required to decompose 2.24 mol
NaHCO3(s).
∆H = 85 kJ for 2 mol NaHCO3
°
106
UNKNOWN
∆H = ? kJ for 2.24 mol NaHCO3
∆H = 44.0 kJ
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Sample Problem 17.4
KNOWNS
amount of NaHCO3(s) that decomposes = 2.24 mol
• The vaporization of 1 mol of liquid water to water
vapor at 25 C requires 44.0 kJ of heat.
H2O(l) → H2O(g)
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
1 Analyze List the knowns and the unknown.
Use the thermochemical equation to write a
conversion factor relating kJ of heat and moles of
NaHCO3. Then use the conversion factor to
determine ∆H for 2.24 mol NaHCO3.
2NaHCO3(s) + 85 kJ → Na2CO3(s) + H2O(l) CO2(g)
∆H = 285.8 kJ
∆H = 241.8 kJ
difference = 44.0 kJ
17.1 The Flow of Energy >
Using the Heat of Reaction to Calculate
Enthalpy Change
∆H = 241.8 kJ
difference = 44.0 kJ
∆H = 285.8 kJ
107
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
108
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
12
2/25/2013
17.1 The Flow of Energy >
Sample Problem 17.4
17.1 The Flow of Energy >
2 Calculate Solve for the unknown.
Sample Problem 17.4
17.1 The Flow of Energy >
2 Calculate Solve for the unknown.
Write the conversion factor relating kJ of
heat and moles of NaHCO3.
3 Evaluate Does the result make sense?
Using dimensional analysis, solve for ∆H.
• The 85 kJ in the thermochemical equation
refers to the decomposition of 2 mol
NaHCO3(s).
85 kJ
∆H = 2.24 mol NaHCO3(s) × 2 mol NaHCO (s)
3
= 95 kJ
85 kJ
2 mol NaHCO3(s)
• Therefore, the decomposition of 2.24 mol
should absorb more heat than 85 kJ.
The thermochemical equation
indicates that 85 kJ are needed
to decompose 2 mol NaHCO3(s).
109
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Thermochemical Equations
• The answer of 95 kJ is consistent with this
estimate.
110
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Thermochemical Equations
111
°C
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) + 890 kJ
• This is an exothermic reaction.
• The heat of combustion (∆H)
for this reaction is –890 kJ per
mole of methane burned.
17.1 The Flow of Energy >
113
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Which of the following thermochemical
equations represents an endothermic
reaction?
A. Cgraphite(s) + 2 kJ
A. Cgraphite(s) + 2 kJ
B. 2H2(g) + O2(g)
Cdiamond(s)
B. 2H2(g) + O2(g)
2H2O + 483.6 kJ
Substance
Formula
Hydrogen
H2(g)
Carbon
C(s,
graphite)
∆H (kJ/mol)
–286
–394
Methane
CH4(g)
Acetylene
C2H2(g)
–1300
Ethanol
C2H6O(l)
–1368
–890
Propane
C3H8(g)
Glucose
C6H12O6(s)
–2808
Octane
C8H18(l)
–5471
Sucrose
C12H22O11(s)
–5645
–2220
114
Like other heats of
reaction, heats of
combustion are
reported as the
enthalpy changes
when the reactions
are carried out at
101.3 kPa and
25 C.
°
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Which of the following thermochemical
equations represents an endothermic
reaction?
Interpret Data
Heats of Combustion at 25
• Burning 1 mol of methane
releases 890 kJ of heat.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Heats of Combustion
Small amounts of natural gas within crude
oil are burned off at oil refineries.
Heats of Combustion
The heat of combustion is the heat of
reaction for the complete burning of one mole
of a substance.
112
Sample Problem 17.4
Key Concepts & Key
Equation
The value of ∆H of a reaction can be
determined by measuring the heat flow of
the reaction at a constant pressure.
In a chemical equation, the enthalpy
change for the reaction can be written as
either a reaction or a product.
Cdiamond(s)
2H2O + 483.6 kJ
qsys = ∆H = –qsurr = –m × C × ∆T
115
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
116
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
117
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
13
2/25/2013
17.1 The Flow of Energy >
Glossary Terms
17.1 The Flow of Energy >
The heat of reaction or process can be
determined experimentally through
calorimetry.
• heat of combustion: the heat of reaction for
the complete burning of one mole of a
substance
• enthalpy (H): the heat content of a system at
constant pressure
17.1 The Flow of Energy >
119
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
120
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
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Heats of Fusion and
17.1 The Flow of Energy > Solidification
122
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Heats of Fusion and
17.1 The Flow of Energy > Solidification
Heats of Fusion and Solidification
123
Heats of Fusion and
• 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 (∆Hfus).
• The gain of heat causes a change of state
instead of a change in temperature.
• The molar heat of solidification
(∆Hsolid) is the heat lost when one mole
of a liquid substance solidifies at a
constant temperature.
• The temperature of the substance
undergoing the change remains constant.
124
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
125
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > 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?
CHEMISTRY & YOU
Why does sweating
help cool you off?
Thermochemistry
END OF 17.2
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Chapter 17
121
BIG IDEA
Matter and Energy
• heat of reaction: the enthalpy change for a
chemical equation exactly as it is written
• calorimeter: an insulated device used to
measure the absorption or release of heat in
chemical or physical processes
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
• thermochemical equation: a chemical
equation that includes the enthalpy change
• calorimetry: the precise measurement of heat
flow out of a system for chemical and physical
processes
118
Glossary Terms
126
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
14
2/25/2013
Heats of Fusion and
17.1 The Flow of Energy > Solidification
Heats of Fusion and
17.1 The Flow of Energy > Solidification
°
°
∆Hfus = –∆Hsolid
Sample Problem 17.5
°
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Sample Problem 17.5
129
Sample Problem 17.5
2 Calculate Solve for the unknown.
Start by expressing ∆Hfus as a
conversion factor.
Express the molar mass of ice as a
conversion factor.
1 mol H2O(s)
6.01 kJ
UNKNOWN
mice = ? g
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
2 Calculate Solve for the unknown.
• Convert moles of ice to grams of ice.
KNOWNS
Initial and final
temperature are 0 C
°
∆Hsolid = –6.01 kJ/mol
128
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.
How many grams of ice at
0 C will melt if 2.25 kJ of heat
are added?
∆Hfus = 6.01 kJ/mol
H2O(l) → H2O(s)
17.1 The Flow of Energy >
Sample Problem 17.5
Using the Heat of Fusion in PhaseChange Calculations
°
H2O(s) → H2O(l)
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
°
• 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.
The quantity of heat absorbed by a
melting solid is exactly the same as
the quantity of heat released when
the liquid solidifies.
127
17.1 The Flow of Energy >
18.0 g H2O(s)
1 mol H2O(s)
Use the thermochemical equation
∆Hfus = 6.01 kJ/mol
H2O(s) + 6.01 kJ → H2O(l).
∆H = 2.25 kJ
130
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Sample Problem 17.5
131
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
2 Calculate Solve for the unknown.
Sample Problem 17.5
132
17.1 The Flow of Energy >
3 Evaluate Does the result make sense?
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.
• To melt 1 mol of ice, 6.01 kJ of energy is
required.
Multiply the known enthalpy change by
the conversion factors.
• Only about one-third of this amount of
heat (roughly 2 kJ) is available.
1 mol H2O(s)
18.0 g H2O(s)
×
6.01 kJ
1 mol H2O(s)
= 6.74 g H2O(s)
mice = 2.25 kJ ×
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
• 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.
133
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
134
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
135
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
15
2/25/2013
Heats of Vaporization and
17.1 The Flow of Energy >
17.1 The Flow of Energy > 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.
Heats of Vaporization and Condensation
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Interpret Data
137
Heats of Vaporization and
∆Hfus (kJ/mol)
∆Hvap (kJ/mol)
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
139
138
Heats of Vaporization and
The quantity of heat absorbed by a
vaporizing liquid is exactly the same
as the quantity of heat released when
the vapor condenses.
∆Hvap = –∆Hcond
• The molar heat of condensation (∆Hcond) is
the amount of heat released when one mole
of vapor condenses at its normal boiling point.
0.90
6.82
40.7
CHEMISTRY & YOU
140
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Explain why the evaporation of sweat
off your body helps cool you off.
CHEMISTRY & YOU
141
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Interpret Graphs
A heating curve
graphically
describes the
enthalpy
changes that
take place
during phase
changes.
Explain why the evaporation of sweat
off your body helps cool you off.
143
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
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.
142
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Condensation
• When a vapor condenses, heat is released.
35.2
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
• The amount of heat required to vaporize one
mole of a given liquid at a constant
temperature is called its molar heat of
vaporization (∆Hvap).
Condensation is the exact opposite of
vaporization.
Heats of Physical Change
Ammonia (NH3)
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Condensation
This table lists the molar heats of vaporization for
several substances at their normal boiling point.
Substance
A liquid that absorbs heat at its boiling
point becomes a vapor.
What is the relationship between molar
heat of vaporization and molar heat of
condensation?
1 mol
3.16 kJ
∆H = 15.6 g CH4O × 32.05 g CH O × 1 mol
4
= 1.54 kJ
136
Heats of Vaporization and
17.1 The Flow of Energy > Condensation
Remember: The temperature of a
substance remains constant during
a change of state.
144
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
16
2/25/2013
17.1 The Flow of Energy >
Sample Problem 17.6
17.1 The Flow of Energy >
Using the Heat of Vaporization in PhaseChange Calculations
17.1 The Flow of Energy >
1 Analyze List the knowns and the unknown.
• First convert grams of water to moles of water.
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 17.6
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.
°
Sample Problem 17.6
1 mol H2O(l)
18.0 g H2O(l)
KNOWNS
UNKNOWN
Initial and final conditions
∆H = ? kJ
are 100 C and 101.3 kPa
°
Mass of liquid water converted to steam = 24.8 g
∆Hvap = 40.7 kJ/mol
145
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Sample Problem 17.6
146
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
2 Calculate Solve for the unknown.
Sample Problem 17.6
147
17.1 The Flow of Energy >
2 Calculate Solve for the unknown.
Express ∆Hvap as a conversion factor.
∆H = 24.8 g H2O(l) ×
• 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.
1 mol H2O(l)
40.7 kJ
×
18.0 g H2O(l) 1 mol H2O(l)
• The calculated enthalpy change should be a
little less than 1.5 mol × 40 kJ/mol = 60 kJ,
and it is.
= 56.1 kJ
Use the thermochemical equation
Sample Problem 17.6
3 Evaluate Does the result make sense?
Multiply the mass of water in grams by
the conversion factors.
40.7 kJ
1 mol H2O(l)
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
H2O(l) + 40.7 kJ → H2O(g).
148
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
149
17.1 The Flow of Energy >
The molar heat of condensation of a
substance is the same, in magnitude,
as which of the following?
151
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
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
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
152
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Heat of Solution
A. molar heat of fusion
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
150
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Heat of Solution
What thermochemical changes can
occur when a solution forms?
153
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17
2/25/2013
17.1 The Flow of Energy > Heat of Solution
17.1 The Flow of Energy > Heat of Solution
During the formation of a solution, heat
is either released or absorbed.
17.1 The Flow of Energy > 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 (∆Hsoln).
CaCl2(s) → Ca2+(aq) + 2Cl–(aq)
∆Hsoln = –82.8 kJ/mol
154
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Heat of Solution
155
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
The dissolution of ammonium nitrate, NH4NO3(s),
is an example of an endothermic process.
Sample Problem 17.7
156
17.1 The Flow of Energy >
Calculating the Enthalpy Change in
Solution Formation
• The cold pack shown here contains solid ammonium
nitrate crystals and water.
• The solution process absorbs energy from the
surroundings.
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 (∆H).
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.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
KNOWNS
∆Hsoln = –44.5 kJ/mol
NH4NO3(s) → NH4+(aq) + NO3–(aq)
amount of NaOH(s) dissolved = 2.50 mol
∆Hsoln = 25.7 kJ/mol
UNKNOWN
∆H = ? kJ
157
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Sample Problem 17.7
158
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
2 Calculate Solve for the unknown.
Sample Problem 17.7
159
17.1 The Flow of Energy >
2 Calculate Solve for the unknown.
Start by expressing ∆Hsoln as a
conversion factor.
∆H = 2.50 mol NaOH(s) ×
Sample Problem 17.7
3 Evaluate Does the result make sense?
• ∆H is 2.5 times greater than ∆Hsoln, as it should
be.
Multiply the number of moles by the
conversion factor.
–44.5 kJ
1 mol NaOH(s)
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
• Also, ∆H should be negative, as the dissolution
of NaOH(s) in water is exothermic.
–44.5 kJ
1 mol NaOH(s)
= –111 kJ
Use the thermochemical equation
NaOH(s) → Na+(aq) + OH–(aq) + 44.5 kJ/mol.
160
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
161
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
162
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
18
2/25/2013
17.1 The Flow of Energy >
17.1 The Flow of Energy >
How much heat (in kJ) is absorbed when
50.0 g of NH4NO3(s) are dissolved in
water if ∆Hsoln = 25.7 kJ/mol?
17.1 The Flow of Energy >
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
∆H = 50.0 g NH4NO3 × 80.04 g NH NO ×
4
3
= 16.1 kJ
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, ∆Hfus = –∆Hsolid.
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, ∆Hvap = –∆Hcond.
25.7 kJ
1 mol
During the formation of a solution, heat is
either released or absorbed.
163
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Glossary Terms
164
17.1 The Flow of Energy >
• molar heat of fusion (∆Hfus): the amount of
heat absorbed by one mole of a solid substance
as it melts to a liquid at a constant temperature
17.1 The Flow of Energy >
167
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Thermochemistry
END OF 17.3
CHEMISTRY & YOU
168
17.4 Calculating Heats of
Reaction
170
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Hess’s Law
Hess’s Law
How can you calculate the heat of
reaction when it cannot be directly
measured?
Diamonds are gemstones
composed of carbon.
Over a time period of
millions and millions of
years, diamond will break
down into graphite, which
is another form of carbon.
17.1 The Flow of Energy
17.2 Measuring and Expressing
Enthalpy Changes
17.3 Heat in Changes of State
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
How much heat is released when a
diamond changes into graphite?
Chapter 17
169
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
• molar heat of solution (∆Hsoln): the enthalpy
change caused by the dissolution of one mole of
a substance
• molar heat of vaporization (∆Hvap): the
amount of heat absorbed by one mole of a liquid
as it vaporizes at a constant temperature
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Glossary Terms
165
• molar heat of condensation (∆Hcond): the
amount of heat released by one mole of a vapor
as it condenses to a liquid at a constant
temperature
• molar heat of solidification (∆Hsolid): the
amount of heat lost by one mole of a liquid as it
solidifies at a constant temperature
166
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
171
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
19
2/25/2013
17.1 The Flow of Energy > Hess’s Law
17.1 The Flow of Energy > Hess’s Law
17.1 The Flow of Energy > Hess’s Law
C(s, diamond) → C(s, graphite)
Hess’s law of heat summation states that
if you add two or more thermochemical
equations to give a final equation, then you
can also add the heats of reaction to give
the final heat of reaction.
172
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Hess’s Law
Hess’s law allows you to determine
the heat of reaction indirectly by using
the known heats of reaction of two or
more thermochemical equations.
173
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Hess’s Law
C(s, diamond) → C(s, graphite)
a. C(s, graphite) + O2(g) → CO2(g)
∆H = –393.5 kJ
b. C(s, diamond) + O2(g) → CO2(g)
∆H = –395.4 kJ
174
∆H = –393.5 kJ
b. C(s, diamond) + O2(g) → CO2(g)
∆H = –395.4 kJ
b. C(s, diamond) + O2(g) → CO2(g)
c. CO2(g) → C(s, graphite) + O2(g)
C(s, diamond) → C(s, graphite)
∆H = –395.4 kJ
∆H = 393.5 kJ
If you also add the values of ∆H for equations b
and c, you get the heat of reaction for this
conversion.
If you add equations b and c, you get the equation
for the conversion of diamond to graphite.
Write equation a in reverse to give:
∆H = 393.5 kJ
When you reverse a reaction, you must also
change the sign of ∆H.
C(s, diamond) + O2(g) → CO2(g)
∆H = –395.4 kJ
C(s, diamond) + O2(g) → CO2(g)
∆H = –395.4 kJ
CO2(g) → C(s, graphite) + O2(g)
∆H =
CO2(g) → C(s, graphite) + O2(g)
∆H =
393.5 kJ
C(s, diamond) → C(s, graphite)
∆H =
–1.9 kJ
393.5 kJ
C(s, diamond) → C(s, graphite)
175
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Hess’s Law
176
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
C(s, diamond) + O2(g) → CO2(g)
∆H = –395.4 kJ
CO2(g) → C(s, graphite) + O2(g)
∆H =
393.5 kJ
C(s, diamond) → C(s, graphite)
∆H =
–1.9 kJ
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy > Hess’s Law
C(s, diamond) → C(s, graphite)
a. C(s, graphite) + O2(g) → CO2(g)
c. CO2(g) → C(s, graphite) + O2(g)
Although the enthalpy change for this reaction
cannot be measured directly, you can use Hess’s
law to find the enthalpy change for the conversion
of diamond to graphite by using the following
combustion reactions.
CHEMISTRY & YOU
177
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
How can you determine ∆H for the
conversion of diamond to graphite
without performing the reaction?
CHEMISTRY & YOU
How can you determine ∆H for the
conversion of diamond to graphite
without performing the reaction?
You can use Hess’s law by adding
thermochemical equations in which the
enthalpy changes are known and whose sum
will result in an equation for the conversion of
diamond to graphite.
178
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
179
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
180
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
20
2/25/2013
17.1 The Flow of Energy > Hess’s Law
17.1 The Flow of Energy > Hess’s Law
17.1 The Flow of Energy > Hess’s Law
C(s, graphite) + O2(g) → CO2(g)
Another case where Hess’s law is useful is
when reactions yield products in addition to
the product of interest.
You can calculate the desired enthalpy
change by using Hess’s law and the
following two reactions that can be carried
out in the laboratory:
• Suppose you want to determine the enthalpy
change for the formation of carbon monoxide
from its elements.
C(s, graphite)+ 12 O2(g) → CO(g) ∆H = ?
• Carrying out the reaction in the laboratory as
written is virtually impossible.
181
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
C(s, graphite) + O2(g) → CO2(g)
∆H = –393.5 kJ
CO2(g) → CO(g) + 12 O2(g)
∆H =
C(s, graphite)+ 21 O2(g) → CO(g)
∆H = –110.5 kJ
182
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
A. heats of fusion for each of the compounds in
the reaction
A. heats of fusion for each of the compounds in
the reaction
B. two other reactions with known heats of
reaction
B. two other reactions with known heats of
reaction
C. specific heat capacities for each compound in
the reaction
C. specific heat capacities for each compound in
the reaction
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Standard Heats
of Formation
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Standard Heats
of Formation
°
The standard heat of formation (∆Hf ) of
a compound is the change in enthalpy that
accompanies the formation of one mole of a
compound from its elements with all
substances in their standard states.
• Scientists specify a common set of conditions
as a reference point.
°
• Thus, ∆H ° = 0 for the diatomic molecules
• These conditions, called the standard state,
refer to the stable form of a substance at
25 C and 101.3 kPa.
• The ∆Hf of a free element in its standard
state is arbitrarily set at zero.
°
f
H2(g), N2(g), O2(g), F2(g), Cl2(g), Br2(l), and
I2(s).
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
283.0 kJ
183
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Standard Heats
of Formation
Standard Heats of Formation
How can you calculate the heat of
reaction when it cannot be directly
measured?
D. density for each compound in the reaction
185
Enthalpy changes generally depend on the
conditions of the process.
187
∆H = –110.5 kJ
17.1 The Flow of Energy >
According to Hess’s law, it is possible to
calculate an unknown heat of reaction
by using which of the following?
D. density for each compound in the reaction
∆H =
C(s, graphite)+ 12 O2(g) → CO(g)
283.0 kJ
According to Hess’s law, it is possible to
calculate an unknown heat of reaction
by using which of the following?
184
∆H = –393.5 kJ
CO2(g) → CO(g) + 12 O2(g)
188
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
186
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Interpret Data
Standard Heats of Formation (∆Hf
°
Substance
∆Hf
(kJ/mol)
Substance
Al2O3(s)
–1676.0
F2(g)
Br2(g)
30.91
Fe(s)
Br2(l)
0.0
Fe2O3(s)
C(s, diamond)
1.9
H2(g)
C(s, graphite)
CH4(g)
NO2(g)
NaCl(s)
0.0
0.0
O3(g)
P(s, white)
–187.8
I2(g)
–1207.0
I2(s)
–635.1
N2(g)
0.0
NH3(g)
62.4
0.0
33.85
–411.2
142.0
–241.8
–285.8
H2O2(l)
90.37
O2(g)
H2O(g)
–110.5
189
0.0
–822.1
H2O(l)
–393.5
Cl2(g)
f
(kJ/mol)
NO(g)
0.0
CO2(g)
CaO(s)
0.0
–74.86
CO(g)
CaCO3(s)
°) at 25°C and 101.3 kPa
° Substance ∆H °
∆Hf
(kJ/mol)
P(s, red)
S(s, rhombic)
S(s, monoclinic)
0.0
–18.4
0.0
0.30
0.0
SO2(g)
–296.8
–46.19
SO3(g)
–395.7
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
21
2/25/2013
Standard Heats
of Formation
17.1 The Flow of Energy >
Standard Heats
of Formation
17.1 The Flow of Energy >
For a reaction that occurs at standard
conditions, you can calculate the heat
of reaction by using standard heats of
formation.
Standard Heats
of Formation
17.1 The Flow of Energy >
For a reaction that occurs at standard
conditions, you can calculate the heat
of reaction by using standard heats of
formation.
For a reaction that occurs at standard
conditions, you can calculate the heat
of reaction by using standard heats of
formation.
• Such an enthalpy change is called
the standard heat of reaction (∆H ).
• The standard heat of reaction is the
difference between the standard
heats of formation of all the reactants
and products.
°
° = ∆H °(products) – ∆H °(reactants)
∆H
190
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Standard Heats
of Formation
17.1 The Flow of Energy >
191
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Sample Problem 17.8
Balance the equation of the reaction of CO(g)
with O2(g) to form CO2(g). Then determine
∆H using the standard heats of formation of
the reactants and products.
°
KNOWNS
∆Hf CO(g) = –110.5 kJ/mol
°
∆H °O (g) = 0 kJ/mol (free
element)
∆H °CO (g) = –393.5 kJ/mol
f
• Notice that water has a lower
enthalpy than the elements
from which it is formed.
193
17.1 The Flow of Energy >
Sample Problem 17.8
194
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
2 Calculate Solve for the unknown.
° of the product in a similar way.
Find ∆Hf
°(reactants) = 2 mol CO(g) × ∆H °CO(g) + 1 mol O (g) ×
°O (g)
–110.5 kJ
2
= 2 mol CO(g) ×
∆Hf
∆Hf
2
0 kJ
2 mol CO(g)+ 1 mol O2(g) × 1 mol O2(g)
= –221.0 kJ
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
197
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
°(products) = 2 mol CO (g) ×
°CO (g)
2
2
–393.5 kJ
= 2 mol CO2(g) × 1 mol CO (g)
2
= –787.0 kJ
Remember to take into
account the number of moles
of each reactant and product.
196
Sample Problem 17.8
2 Calculate Solve for the unknown.
° of all the reactants.
f
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Find and add ∆Hf
∆Hf
∆Hf
°
2
195
2 Calculate Solve for the unknown.
First write the balanced equation.
2CO(g) + O2(g) → 2CO2(g)
Sample Problem 17.8
UNKNOWN
∆H = ? kJ
2
f
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Sample Problem 17.8
1 Analyze List the knowns and the unknown.
°
• The enthalpy difference
between the reactants and
products, –285.8 kJ/mol, is
the standard heat of formation
of liquid water from the gases
hydrogen and oxygen.
f
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Calculating the Standard Heat
of Reaction
What is the standard heat of
reaction (∆H ) for the reaction
of CO(g) with O2(g) to form
CO2(g)?
This enthalpy diagram shows the
standard heat of formation of water.
f
192
Remember to take into
account the number of moles
of each reactant and product.
198
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
22
2/25/2013
17.1 The Flow of Energy >
Sample Problem 17.8
17.1 The Flow of Energy >
2 Calculate Solve for the unknown.
Calculate ∆H
° for the reaction.
f
Standard heats of formation are used to
calculate the enthalpy change for the
reaction of carbon monoxide and oxygen.
°
• The ∆H is negative, so the reaction is
exothermic.
• 2CO(g) + O2(g) → 2CO2(g)
f
• This outcome makes sense because
combustion reactions always release heat.
= (–787.0 kJ) – (–221.0 kJ)
• The diagram shows the
difference between
∆Hf (product) and
∆Hf (reactants) after taking
into account the number of
moles of each.
°
°
= –566.0 kJ
199
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
200
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Calculate the standard heat of reaction
for the following:
CH4(g) + Cl2(g) → C(s, diamond) + 4HCl(g)
CH4(g) + Cl2(g) → C(s, diamond) + 4HCl(g)
°(CH (g)) = –74.86 kJ/mol
°(C(s, diamond)) = 1.9 kJ/mol
∆H °(HCl(g)) = –92.3 kJ/mol
°(CH (g)) = –74.86 kJ/mol
°(C(s, diamond)) = 1.9 kJ/mol
∆H °(HCl(g)) = –92.3 kJ/mol
∆Hf
4
∆Hf
201
4
f
∆Hf
°(reactants) = [1 mol CH (g) × ∆H °CH (g)] + [1 mol Cl
∆Hf
∆Hf
°(products) = [1 mol C(s) × ∆H °C(s, diamond)] + [4 mol HCl ×
°HCl(g)]
4
f
4
2
× ∆Hf
For a reaction that occurs at standard
conditions, you can calculate the heat of
reaction by using standard heats of
formation.
°Cl (g)]
2
= –74.86 kJ + 0.0 kJ = –74.86 kJ
202
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Glossary Terms
°
f
°(products) – ∆H °(reactants) = –367.3 kJ – (–74.86 kJ) = –
f
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
BIG IDEA
° = ∆H °(products) –
∆H °(reactants)
∆H
= 1.9 kJ + (4 × –92.3 kJ) = –367.3 kJ
∆H = ∆Hf
292.4 kJ
203
Key Concepts &
Key Equation
Hess’s law allows you to determine the
heat of reaction indirectly by using the
known heats of reaction of two or more
thermochemical equations.
∆Hf
f
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Calculate the standard heat of reaction
for the following:
∆Hf
Standard Heats
of Formation
17.1 The Flow of Energy >
3 Evaluate Does the result make sense?
° = ∆H °(products) – ∆H °(reactants)
∆H
Sample Problem 17.8
f
f
204
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
17.1 The Flow of Energy >
Matter and Energy
• Hess’s law of heat summation: if you add
two or more thermochemical equations to give
a final equation, then you also add the heats of
reaction to give the final heat of reaction
The heat of reaction can be calculated by
using the known heats of reaction of two
or more thermochemical equations or by
using standard heats of formation.
°
• standard heat of formation (∆Hf ): the
change in enthalpy that accompanies the
formation of one mole of a compound from its
elements with all substances in their standard
states at 25 C
END OF 17.4
°
205
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
206
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
207
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
23
2/25/2013
17.1 The Flow of Energy >
17.1 The Flow of Energy >
17.1 The Flow of Energy >
208
209
210
17.1 The Flow of Energy >
17.1 The Flow of Energy >
17.1 The Flow of Energy >
211
212
213
17.1 The Flow of Energy >
17.1 The Flow of Energy >
17.1 The Flow of Energy >
214
215
216
24
2/25/2013
17.1 The Flow of Energy >
17.1 The Flow of Energy >
17.1 The Flow of Energy >
217
218
219
17.1 The Flow of Energy >
17.1 The Flow of Energy >
17.1 The Flow of Energy >
220
221
222
17.1 The Flow of Energy >
17.1 The Flow of Energy >
17.1 The Flow of Energy >
223
224
225
25
2/25/2013
17.1 The Flow of Energy >
17.1 The Flow of Energy >
17.1 The Flow of Energy >
226
227
228
17.1 The Flow of Energy >
17.1 The Flow of Energy >
17.1 The Flow of Energy >
229
230
231
17.1 The Flow of Energy >
17.1 The Flow of Energy >
17.1 The Flow of Energy >
232
233
234
26
2/25/2013
17.1 The Flow of Energy >
17.1 The Flow of Energy >
17.1 The Flow of Energy >
235
236
237
17.1 The Flow of Energy >
17.1 The Flow of Energy >
238
239
27