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Lecture Presentation
Chapter 5
Thermochemistry
John D. Bookstaver
St. Charles Community College
Cottleville, MO
© 2012 Pearson Education, Inc.
Energy
• Energy is the ability to do work or
transfer heat.
– Energy used to cause an object that has
mass to move is called work.
– Energy used to cause the temperature of
an object to rise is called heat.
Thermochemistry
© 2012 Pearson Education, Inc.
Nature of Energy
• Even though chemistry is the study of
•
•
matter, energy affects matter
Energy is anything that has the capacity to
do work
Work is a force acting over a distance
Energy = Work = Force x Distance
• Heat is the flow of energy caused by a
difference in temperature
• Energy can be exchanged between
objects through contact
collisions
3
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Energy, Heat, and Work
• You can think of energy as a quantity an object
can possess
or collection of objects
• You can think of heat and work as the two
different ways that an object can exchange
energy with other objects
either out of it, or into it
4
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Classification of
Energy
• Kinetic energy is
•
energy of motion or
energy that is being
transferred
Thermal energy is
the energy associated
with temperature
thermal energy is a
form of kinetic energy
5
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Classification of Energy
• Potential energy is energy that is stored in an
object, or energy associated with the
composition and position of the object
energy stored in the structure of a compound is
potential
6
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Kinetic Energy
Kinetic energy is energy an object
possesses by virtue of its motion:
1
Ek =  mv2
2
Thermochemistry
© 2012 Pearson Education, Inc.
Potential Energy
• Potential energy is
energy an object
possesses by virtue of
its position or chemical
composition.
• The most important
form of potential energy
in molecules is
electrostatic potential
energy, Eel:
K Q 1Q 2
Eel =
d
Thermochemistry
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Manifestations of Energy
9
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• Electrical
Some Forms of Energy
 kinetic energy associated with the flow of electrical charge
• Heat or thermal energy
 kinetic energy associated with molecular motion
• Light or radiant energy
 kinetic energy associated with energy transitions in an atom
• Nuclear
 potential energy in the nucleus of atoms
• Chemical
 potential energy due to the structure of the atoms, the
attachment between atoms, the atoms’ positions relative to
each other in the molecule, or the molecules, relative
positions in the structure
10
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Units of Energy
• The SI unit of energy is the joule (J):
kg m2
1 J = 1 
s2
• An older, non-SI unit is still in
widespread use: the calorie (cal):
1 cal = 4.184 J
Thermochemistry
© 2012 Pearson Education, Inc.
Units of Energy
• The amount of kinetic energy an
object has is directly proportional
to its mass and velocity
 KE = ½mv2
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Units of Energy
• joule (J) is the amount of energy needed to
move a 1-kg mass a distance of 1 meter
1 J = 1 N∙m = 1 kg∙m2/s2
• calorie (cal) is the amount of energy needed to
raise the temperature of one gram of water 1°C
kcal = energy needed to raise 1000 g of water 1°C
food Calories = kcals
Energy Conversion Factors
1 calorie (cal) = 4.184 joules (J) (exact)
1 Calorie (Cal) = 1000 cal = 1 kcal = 4184 J
1 kilowatt-hour (kWh) = 3.60 x 106 J
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Energy Use
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System and Surroundings
• We define the system as the material or process
within which we are studying the energy changes
within
• We define the surroundings as everything else
with which the system can exchange energy with
• What we study is the exchange of energy
between the system and the surroundings
Surroundings
Surroundings
System
System
15
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Definitions: System and Surroundings
• The system includes the
molecules we want to
study (here, the hydrogen
and oxygen molecules).
• The surroundings are
everything else (here, the
cylinder and piston).
Thermochemistry
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Comparing the Amount of Energy in the
System and Surroundings During Transfer
• Conservation of energy means that the amount
of energy gained or lost by the system has to
be equal to the amount of energy lost or gained
by the surroundings
17
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Definitions: Work
• Energy used to
move an object over
some distance is
work:
• w=Fd
where w is work, F
is the force, and d is
the distance over
which the force is
exerted.
Thermochemistry
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Heat
• Energy can also be
transferred as heat.
• Heat flows from
warmer objects to
cooler objects.
Thermochemistry
© 2012 Pearson Education, Inc.
Conversion of Energy
• Energy can be converted from one type to
another.
• For example, the cyclist in Figure 5.2 has
potential energy as she sits on top of the hill.
Thermochemistry
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Conversion of Energy
• As she coasts down the hill, her potential
energy is converted to kinetic energy.
• At the bottom, all the potential energy she
had at the top of the hill is now kinetic energy.
Thermochemistry
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First Law of Thermodynamics
• Energy is neither created nor destroyed.
• In other words, the total energy of the universe is
a constant; if the system loses energy, it must be
gained by the surroundings, and vice versa.
Thermochemistry
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The First Law of Thermodynamics
Law of Conservation of Energy
• Thermodynamics is the study of energy and
•
•
•
its interconversions
The First Law of Thermodynamics is the Law of
Conservation of Energy
This means that the total amount of energy in
the universe is constant
You can therefore never design a system that
will continue to produce energy without some
source of energy
23
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Energy Flow and
Conservation of Energy
• Conservation of energy requires that the sum of the
energy changes in the system and the surroundings
must be zero
 DEnergyuniverse = 0 = DEnergysystem + DEnergysurroundings
 D Is the symbol that is
used to mean change
 final amount – initial amount
24
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Internal Energy
The internal energy of a system is the sum of all
kinetic and potential energies of all components
of the system; we call it E.
Thermochemistry
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Internal Energy
By definition, the change in internal energy, DE,
is the final energy of the system minus the initial
energy of the system:
DE = Efinal − Einitial
Thermochemistry
© 2012 Pearson Education, Inc.
Internal Energy
• The internal energy is the sum of the kinetic
•
and potential energies of all of the particles that
compose the system
The change in the internal energy of a system
only depends on the amount of energy in the
system at the beginning and end
a state function is a mathematical function whose
result only depends on the initial and final
conditions, not on the process used
DE = Efinal – Einitial
DEreaction = Eproducts − Ereactants
27
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State Function
You can travel either of
two trails to reach the
top of the mountain.
One is long and
winding, the other is
shorter but steep.
Regardless of which
trail you take, when you
reach the top you will be
10,000 ft above the
base.
The distance from the base to the peak of the mountain is a
state function. It depends only on the difference in elevation
between the base and the peak, not on how you arrive there!
28
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“graphical” way of showing
the direction of energy flow
during a process
• If the final condition has a
larger amount of internal
energy than the initial
condition, the change in the
internal energy will be +
• If the final condition has a
smaller amount of internal
energy than the initial
condition, the change in the
internal energy will be ─
29
Internal Energy
• Energy diagrams are a
Internal Energy
Energy Diagrams
final
initial
energy added
DE = +
initial
final
energy removed
DE = ─
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Energy Flow
• When energy flows out of a
•
•
•
system, it must all flow into
the surroundings
When energy flows out of a
system, DEsystem is ─
When energy flows into the
surroundings, DEsurroundings
is +
Therefore:
─ DEsystem= DEsurroundings
30
Surroundings
DE +
System
DE ─
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Energy Flow
• When energy flows into a
•
•
•
system, it must all come
from the surroundings
When energy flows into a
system, DEsystem is +
When energy flows out
of the surroundings,
DEsurroundings is ─
Therefore:
DEsystem= ─ DEsurroundings
31
Surroundings
DE ─
System
DE +
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Energy Flow in a Chemical Reaction
in 1mol of C(s) and 1 mole of O2(g)
is greater than the internal energy in
1 mole of CO2(g)
 at the same temperature and pressure
• In the reaction C(s) + O2(g) → CO2(g),
there will be a net release of energy
into the surroundings
 −DEreaction = DEsurroundings
Internal Energy
• The total amount of internal energy
C(s), O2(g)
CO2(g)
energy
energy
released
absorbed
DE
DErxn
+
rxn==─
Surroundings
• In the reaction CO2(g) → C(s) + O2(g),
there will be an absorption of
energy from the surroundings into
the reaction
System
CO
O22
C
+ 2O→C
2 →+CO
 DEreaction = − DEsurroundings
32
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Energy Exchange
• Energy is exchanged between the system and
surroundings through heat and work
 q = heat (thermal) energy
 w = work energy
 q and w are NOT state functions, their value depends
on the process
DE = q + w
33
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Energy Exchange
• Energy is exchanged between the system and
surroundings through either heat exchange or
work being done
34
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Heat & Work
• The white ball has an initial amount of 5.0 J of
•
•
kinetic energy
As it rolls on the table, some of the energy is
converted to heat by friction
The rest of the kinetic energy is transferred to the
purple ball by collision
35
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Heat & Work
On a rough
smoothtable,
table,most
mostofofthe
thekinetic
kineticenergy
energy
of
the white ball
is lost
friction
less
is transferred
from
the through
white ball
to the–purple
than
is transferred
to the
purplefriction
ball
– withhalf
a small
amount lost
through
energy change for the white
ball, DE = KEfinal − KEinitial =
0 J − 5.0 J = −5.0 J
kinetic energy transferred to
to
purple
ball,
= −3.0
purple
ball,
w =w −4.5
J J
kinetic energy lost as heat,
q = −2.0
−0.5 J
q + w = (−2.0
(−0.5 J) + (−3.0
(−4.5 J) =
=
−5.0
−5.0
J =J =
DEDE
36
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Heat, Work, and Internal Energy
• In the previous billiard ball example, the DE of the white
ball is the same for both cases, but q and w are not
• On the rougher table, the heat loss, q, is greater
 q is a more negative number
• But on the rougher table, less kinetic energy is
transferred to the purple ball, so the work done by the
white ball, w, is less
 w is a less negative number
• The DE is a state function and depends only on the
velocity of the white ball before and after the collision
 in both cases it started with 5.0 kJ of kinetic energy and ended
with 0 kJ because it stopped
 q + w is the same for both tables, even though the values of q
and w are different
37
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Heat Exchange
• Heat is the exchange of thermal energy
•
•
•
between the system and surroundings
Heat exchange occurs when system and
surroundings have a difference in temperature
Temperature is the measure of the amount of
thermal energy within a sample of matter
Heat flows from matter with high temperature
to matter with low temperature until both
objects reach the same temperature
thermal equilibrium
38
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Changes in Internal Energy
• If DE > 0, Efinal > Einitial
– Therefore, the system absorbed energy
from the surroundings.
– This energy change is called endergonic.
Thermochemistry
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Changes in Internal Energy
• If DE < 0, Efinal < Einitial
– Therefore, the system released energy to
the surroundings.
– This energy change is called exergonic.
Thermochemistry
© 2012 Pearson Education, Inc.
Changes in Internal Energy
• When energy is
exchanged between
the system and the
surroundings, it is
exchanged as either
heat (q) or work (w).
• That is, DE = q + w.
Thermochemistry
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DE, q, w, and Their Signs
Thermochemistry
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Exchange of Heat between System and
Surroundings
• When heat is absorbed by the system from the
surroundings, the process is endothermic.
Thermochemistry
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Exchange of Heat between System and
Surroundings
• When heat is absorbed by the system from the
surroundings, the process is endothermic.
• When heat is released by the system into the
surroundings, the process is exothermic.
Thermochemistry
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State Functions
Usually we have no way of knowing the
internal energy of a system; finding that value
is simply too complex a problem.
Thermochemistry
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State Functions
• However, we do know that the internal energy
of a system is independent of the path by
which the system achieved that state.
– In the system depicted in Figure below, the water
could have reached room temperature from either
direction.
Thermochemistry
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State Functions
• Therefore, internal energy is a state function.
• It depends only on the present state of the
system, not on the path by which the system
arrived at that state.
• And so, DE depends only on Einitial and Efinal.
Thermochemistry
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State Functions
• However, q and w are
not state functions.
• Whether the battery is
shorted out or is
discharged by running
the fan, its DE is the
same.
– But q and w are different
in the two cases.
Thermochemistry
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Work
Usually in an open
container the only work
done is by a gas
pushing on the
surroundings (or by
the surroundings
pushing on the gas).
Thermochemistry
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Work
We can measure the work done by the gas if
the reaction is done in a vessel that has been
fitted with a piston:
w = −PDV
Thermochemistry
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Enthalpy
• If a process takes place at constant
pressure (as the majority of processes we
study do) and the only work done is this
pressure–volume work, we can account
for heat flow during the process by
measuring the enthalpy of the system.
• Enthalpy is the internal energy plus the
product of pressure and volume:
H = E + PV
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Thermochemistry
Enthalpy
• When the system changes at constant
pressure, the change in enthalpy, DH, is
DH = D(E + PV)
• This can be written
DH = DE + PDV
Thermochemistry
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Enthalpy
• Since DE = q + w and w = −PDV, we
can substitute these into the enthalpy
expression:
DH = DE + PDV
DH = (q + w) − w
DH = q
• So, at constant pressure, the change in
enthalpy is the heat gained or lost.
Thermochemistry
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Enthalpy
• The enthalpy, H, of a system is the sum of the
internal energy of the system and the product of
pressure and volume
 H is a state function
•
•
H = E + PV
The enthalpy change, DH, of a reaction is the
heat evolved in a reaction at constant pressure
DHreaction = qreaction at constant pressure
Usually DH and DE are similar in value, the
difference is largest for reactions that produce or
use large quantities of gas
54
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Endothermicity and Exothermicity
• A process is
endothermic
when DH is
positive.
Thermochemistry
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Endothermicity and Exothermicity
• A process is
endothermic
when DH is
positive.
• A process is
exothermic when
DH is negative.
Thermochemistry
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Enthalpy of Reaction
The change in
enthalpy, DH, is the
enthalpy of the
products minus the
enthalpy of the
reactants:
DH = Hproducts − Hreactants
Thermochemistry
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Enthalpy of Reaction
This quantity, DH, is called the enthalpy of
reaction, or the heat of reaction.
Thermochemistry
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The Truth about Enthalpy
1. Enthalpy is an extensive property.
2. DH for a reaction in the forward
direction is equal in size, but opposite
in sign, to DH for the reverse reaction.
3. DH for a reaction depends on the state
of the products and the state of the
reactants.
Thermochemistry
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Calorimetry
Since we cannot
know the exact
enthalpy of the
reactants and
products, we
measure DH through
calorimetry, the
measurement of
heat flow.
Thermochemistry
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Heat Capacity and Specific Heat
The amount of energy required to raise the
temperature of a substance by 1 K (1C) is its
heat capacity.
Thermochemistry
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Heat Capacity and Specific Heat
We define specific
heat capacity (or
simply specific heat)
as the amount of
energy required to
raise the temperature
of 1 g of a substance
by 1 K (or 1 C).
Thermochemistry
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Heat Capacity and Specific Heat
Specific heat, then, is
Specific heat =
heat transferred
mass  temperature change
s=
q
m  DT
Thermochemistry
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Quantity of Heat Energy Absorbed:
Heat Capacity
• When a system absorbs heat, its
•
•
temperature increases
The increase in temperature is directly
proportional to the amount of heat absorbed
The proportionality constant is called the
heat capacity, C
units of C are J/°C or J/K
q = C x DT
• The larger the heat capacity of the object
being studied, the smaller the temperature
rise will be for a given amount of heat
Tro: Chemistry: A Molecular Approach, 2/e
64
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Factors Affecting Heat Capacity
• The heat capacity of an object depends on its
amount of matter
usually measured by its mass
200 g of water requires twice as much heat to raise
its temperature by 1°C as does 100 g of water
• The heat capacity of an object depends on the
type of material
1000 J of heat energy will raise the temperature of
100 g of sand 12 °C, but only raise the temperature
of 100 g of water by 2.4 °C
Tro: Chemistry: A Molecular Approach, 2/e
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Specific Heat Capacity
• Measure of a substance’s
•
intrinsic ability to absorb heat
The specific heat capacity is the
amount of heat energy required to
raise the temperature of one gram
of a substance 1°C
 Cs
 units are J/(g∙°C)
• The molar heat capacity is the
amount of heat energy required to
raise the temperature of one mole
of a substance 1°C
Tro: Chemistry: A Molecular Approach, 2/e
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Specific Heat of Water
• The rather high specific heat of water allows water to
absorb a lot of heat energy without a large increase in
its temperature
• The large amount of water absorbing heat from the air
keeps beaches cool in the summer
 without water, the Earth’s temperature would be about the
same as the moon’s temperature on the side that is facing
the sun (average 107 °C or 225 °F)
• Water is commonly used as a coolant because it can
absorb a lot of heat and remove it from the important
mechanical parts to keep them from overheating
 or even melting
 it can also be used to transfer the heat to something else
because it is a fluid
67
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Quantifying Heat Energy
• The heat capacity of an object is proportional to
•
its mass and the specific heat of the material
So we can calculate the quantity of heat
absorbed by an object if we know the mass,
the specific heat, and the temperature change
of the object
Heat = (mass) x (specific heat) x (temp. change)
q = (m) x (Cs) x (DT)
68
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Example 6.2: How much heat is absorbed by a copper
penny with mass 3.10 g whose temperature rises from
−8.0 °C to 37.0 °C?
• Sort
Information
• Strategize
Given: T1 = −8.0 °C, T2= 37.0 °C, m = 3.10 g
Find: q, J
Cs m, DT
Conceptual
Plan:
q
Relationships: q = m ∙ Cs ∙ DT
• Follow the
conceptual
plan to
solve the
problem
• Check
Cs = 0.385 J/g•ºC (Given)
Solution:
Check: the unit is correct, the sign is reasonable as
the penny must absorb heat to make its
temperature rise
69
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Practice – Calculate the amount of heat released
when 7.40 g of water cools from 49° to 29 °C
(water’s specific heat is 4.18 J/gºC)
70
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Practice – Calculate the amount of heat released
when 7.40 g of water cools from 49° to 29 °C
• Sort
Given:
Information
Find:
• Strategize
q, J
Cs m, DT
Conceptual
Plan:
Relationships:
• Follow the
T1= 49 °C, T2 = 29 °C, m = 7.40 g
q
q = m ∙ Cs ∙ DT
Cs = 4.18 J/gC (Table 6.4)
Solution:
concept plan
to solve the
problem
• Check
Check: the unit is correct, the sign is reasonable as
the water must release heat to make its
temperature fall
71
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Heat Transfer & Final Temperature
• When two objects at different temperatures are
•
•
placed in contact, heat flows from the material at
the higher temperature to the material at the lower
temperature
Heat flows until both materials reach the same
final temperature
The amount of heat energy lost by the hot material
equals the amount of heat gained by the cold
material
qhot = −qcold
mhotCs,hotDThot = −(mcoldCs,coldDTcold)
72
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A piece of metal at 85 °C is added to water
at 25 °C, the final temperature of the both
metal and water is 30 °C. Which of the
following is true?
1. Heat lost by the metal > heat gained by water
2. Heat gained by water > heat lost by the metal
3. Heat lost by the metal > heat lost by the water
4. Heat lost by the metal = heat gained by water
5. More information is required
73
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Example: A 32.5-g cube of aluminum initially at 45.8 °C is
submerged into 105.3 g of water at 15.4 °C. What is the final
temperature of both substances at thermal equilibrium?
Given:
Find:
Conceptual
Plan:
Relationships:
mAl = 32.5 g, Tal = 45.8 °C, mH20 = 105.3 g, TH2O = 15.4 °C
Tfinal, °C
Cs, Al mAl, Cs, H2O mqH2O
DTAl = kDTH2O
Al = −qH2O
Tfinal
q = m ∙ Cs ∙ DT
Cs, Al = 0.903 J/g•ºC, Cs, H2O = 4.18 J/g•ºC(Table 6.4)
Solution:
Check:
the unit is correct, the number is reasonable as the final
temperature should be between the two initial temperatures
74
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Practice – A hot piece of metal weighing
350.0 g is heated to 100.0 °C. It is then
placed into a coffee cup calorimeter
containing 160.0 g of water at 22.4 °C.
The water warms and the copper cools
until the final temperature is 35.2 °C.
Calculate the specific heat of the metal
and identify the metal.
75
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Practice – Calculate the specific heat and
identify the metal from the data
Given:
Find:
Concept Plan:
Relationships:
metal: 350.0 g, T1 = 100.0 °C, T2 = 35.2 °C
H2O: 160.0 g, T1 = 22.4 °C, T2 = 35.2°C, Cs = 4.18 J/g °C
Cs , metal, J/gºC
m, Cs, DT
q
q = m x Cs x DT; qmetal = −qH2O
Solution:
Check:
the units are correct, the number indicates
the metal is copper
76
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Pressure –Volume Work
• PV work is work caused by a volume change against
•
•
an external pressure
When gases expand, DV is +, but the system is doing
work on the surroundings, so wgas is ─
As long as the external pressure is kept constant
─ Workgas = External Pressure x Change in
Volumegas
w = ─PDV
 to convert the units to joules use 101.3 J = 1 atm∙L
77
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Example : If a balloon is inflated from 0.100 L to
1.85 L against an external pressure of 1.00 atm,
how much work is done?
Given:
Find:
Conceptual
Plan:
V1 = 0.100 L, V2 = 1.85 L, P = 1.00 atm
w, J
P, DV
w
Relationships: 101.3 J = 1 atm∙L
Solution:
Check:
the unit is correct; the sign is reasonable because when a gas
expands it does work on the surroundings and loses energy
78
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Practice – A certain process results in a gas system releasing
68.3 kJ of energy. During the process, 15.8 kcal of heat is
released by the system. If the external pressure is kept
constant at 1.00 atm and the initial volume of the gas is 10.0 L,
what is the final volume of the gas?
(1 cal = 4.18 J, 101.3 J = 1.00 atm•L)
79
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Practice – A certain process results in a gas system releasing 68.3 kJ of
energy. During the process, 15.8 kcal of heat is released by the system.
If the external pressure is kept constant at 1.00 atm and the initial volume
of the gas is 10.0 L, what is the final volume of the gas?
Given:
Find:
Conceptual
Plan:
DE == −6.83
−68.3xkJ,
kcal,
V41J,= V10.0
L, P =P 1.00
atm
104qJ,=q−15.8
= −6.604
x 10
= 1.00
atm
1 = 10.0L,
V22,, LL
V
q, DE
w
V2
P, V1
Relationships: DE = q + w, w = −PDV, 1 kJ = 1000 J, 1 cal = 4.18 J, 101.3 J = 1 atm∙L
Solution:
Check:
so both DE and q are −, and DE > q, w must be −; and
when w is − the system is expanding, so V2 should be
greater than V1; and it is
80
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Constant Pressure Calorimetry
By carrying out a
reaction in aqueous
solution in a simple
calorimeter such as this
one, one can indirectly
measure the heat
change for the system
by measuring the heat
change for the water in
the calorimeter.
Thermochemistry
© 2012 Pearson Education, Inc.
Constant Pressure Calorimetry
Because the specific
heat for water is well
known (4.184 J/g-K), we
can measure DH for the
reaction with this
equation:
q = m  s  DT
Thermochemistry
© 2012 Pearson Education, Inc.
Measuring DH
Calorimetry at Constant Pressure
• Reactions done in aqueous
solution are at constant pressure
open to the atmosphere
• The calorimeter is often nested
foam cups containing the solution
qreaction = ─ qsolution
= ─(masssolution x Cs, solution x DT)
 DHreaction = qconstant pressure = qreaction
to get DHreaction per mol, divide by the
number of moles
83
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Example : What is DHrxn/mol Mg for the reaction
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacts in
100.0 mL of solution and changes the temperature from 25.6 °C to 32.8 °C?
Given:
Find:
Conceptual
Plan:
−3 100.0
0.158 xg 10
Mg,
mL,
T1 =x 25.6
T27.2
= 32.8
°C,
6.499
mol Mg,
1.00
102 g,°C,
DT =
°C,,C
s = 4.18 J/g∙°C
assume d = 1.00 g/mL, Cs = 4.18 J/g∙°C
H, kJ/mol
DH,
kJ/mol
Cs, DT, m
qsol’n
qrxn
qrxn, mol
DH
Relationships: qsol’n = m x Cs, sol’n x DT = −qrxn, MM Mg= 24.31 g/mol
Solution:
Check:
the sign is correct and the value is reasonable
because the reaction is exothermic
84
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Bomb Calorimetry
• Reactions can be
carried out in a sealed
“bomb” such as this
one.
• The heat absorbed
(or released) by the
water is a very good
approximation of the
enthalpy change for
the reaction.
Thermochemistry
© 2012 Pearson Education, Inc.
Bomb Calorimetry
• Because the volume
in the bomb
calorimeter is
constant, what is
measured is really the
change in internal
energy, DE, not DH.
• For most reactions,
the difference is very
small.
Thermochemistry
© 2012 Pearson Education, Inc.
Example : When 1.010 g of sugar is burned in a bomb
calorimeter, the temperature rises from 24.92 °C to 28.33 °C.
If Ccal = 4.90 kJ/°C, find DE for burning 1 mole
−3 O ,C
Given: 2.9506
O11, DT
= 3.41°C,
Ccal
= 4.90
1.010 gxC10
=2224.92
°C,
T2 = 28.33
°C,
Ccal kJ/°C
= 4.90 kJ/°C
12H22mol
11 T12
1H
Find: DErxn, kJ/mol
Conceptual
Ccal, DT
Plan:
Relationships:
qcal
qrxn
qrxn, mol
DE
qcal = Ccal x DT = −qrxn
MM C12H22O11 = 342.3 g/mol
Solution:
Check:
the units are correct, the sign is as
expected for combustion
87
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Practice – When 1.22 g of HC7H5O2 (MM 122.12) is burned in
a bomb calorimeter, the temperature rises from 20.27 °C to
22.67 °C. If DE for the reaction is −3.23 x 103 kJ/mol, what is
the heat capacity of the calorimeter in kJ/°C?
88
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Practice – When 1.22 g of HC7H5O2 is burned in a bomb
calorimeter, the temperature rises from 20.27 °C to 22.67 °C.
If D E for the reaction is −3.23 x 103 kJ/mol, what is the heat
capacity of the calorimeter in kJ/°C?
107−3
DT =T2.40
°C,°C,
DE=DE=
−3.23
x 10x3 kJ/mol
1.22 gxHC
H5mol
O2, HC
T1=20.27
−3.23
103 kJ/mol
Given: 9.990
7H5O2, °C,
2=22.67
Find: Ccal, kJ/°C
Conceptual
Plan:
Relationships:
mol, DE
qrxn
qcal, DT
qcal
Ccal
qcal = Ccal x DT = −qrxn
MM HC7H5O2 = 122.12 g/mol
Solution:
Check:
the units and sign are correct
89
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Relationships Involving DHrxn
• When reaction is multiplied by a factor, DHrxn is
multiplied by that factor
because DHrxn is extensive
C(s) + O2(g) → CO2(g)
DH = −393.5 kJ
2 C(s) + 2 O2(g) → 2 CO2(g) DH = 2(−393.5 kJ) = −787.0 kJ
• If a reaction is reversed, then the sign of DH is
changed
DH = +393.5 kJ
CO2(g) → C(s) + O2(g)
90
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Hess’s Law
 DH is well known for many reactions,
and it is inconvenient to measure DH
for every reaction in which we are
interested.
• However, we can estimate DH using
published DH values and the
properties of enthalpy.
Thermochemistry
© 2012 Pearson Education, Inc.
Hess’s Law
Hess’s law states that
“[i]f a reaction is
carried out in a series
of steps, DH for the
overall reaction will be
equal to the sum of
the enthalpy changes
for the individual
steps.”
Thermochemistry
© 2012 Pearson Education, Inc.
Hess’s Law
Because DH is a state
function, the total
enthalpy change
depends only on the
initial state of the
reactants and the final
state of the products.
Thermochemistry
© 2012 Pearson Education, Inc.
Relationships Involving DHrxn
Hess’s Law
• If a reaction can be
expressed as a
series of steps, then
the DHrxn for the
overall reaction is
the sum of the heats
of reaction for each
step
94
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Example: Hess’s Law
Given the following information:
2 NO(g) + O2(g)  2 NO2(g)
DH° = −116 kJ
2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq)DH° = −256 kJ
N2(g) + O2(g)  2 NO(g)
DH° = +183 kJ
Calculate the DH° for the reaction below:
3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g)
DH° = ?
[2
1.5
DH°
[3NO
NO22(g)
(g)
23NO(g)
NO(g)++O1.5
O2x(g)]
DH°==1.5(+116
(+174 kJ)kJ)
2(g)]
[2
O2O
(g)2(g)
+ 2+H12O(l)

4 HNO
x 0.5 DH°
[1NN22(g)
(g)++52.5
H2O(l)
2 HNO
DH°==0.5(−256
(−128 kJ)kJ)
3(aq)]
3(aq)]
[2
DH°
[2NO(g)
NO(g)
NN22(g)
(g)++OO22(g)]
(g)]
DH°==−183
(−183kJkJ)
3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) DH° = − 137 kJ
95
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Practice – Hess’s Law
Given the following information:
Cu(s) + Cl2(g)  CuCl2(s)
DH° = −206 kJ
2 Cu(s) + Cl2(g)  2 CuCl(s) DH° = − 36 kJ
Calculate the DH° for the reaction below:
Cu(s) + CuCl2(s)  2 CuCl(s)
DH° = ? kJ
96
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Practice – Hess’s Law
Given the following information:
Cu(s) + Cl2(g)  CuCl2(s)
2 Cu(s) + Cl2(g)  2 CuCl(s)
DH° = −206 kJ
DH° = − 36 kJ
Calculate the DH° for the reaction below:
Cu(s) + CuCl2(s)  2 CuCl(s)
DH° = ?
kJ
CuCl2(s)  Cu(s) + Cl2(g)
DH° = +206 kJ
2 Cu(s) + Cl2(g)  2 CuCl(s)
DH° = − 36 kJ
Cu(s) + CuCl2(s)  2 CuCl(s)
DH° = +170. kJ
97
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Enthalpies of Formation
An enthalpy of formation, DHf, is defined
as the enthalpy change for the reaction
in which a compound is made from its
constituent elements in their elemental
forms.
Thermochemistry
© 2012 Pearson Education, Inc.
Standard Conditions
• The standard state is the state of a material at a defined
set of conditions
 pure gas at exactly 1 atm pressure
 pure solid or liquid in its most stable form at exactly 1 atm pressure
and temperature of interest
 usually 25 °C
 substance in a solution with concentration 1 M
• The standard enthalpy change, DH°, is the enthalpy
•
change when all reactants and products are in their
standard states
The standard enthalpy of formation, DHf°, is the
enthalpy change for the reaction forming 1 mole of a pure
compound from its constituent elements
 the elements must be in their standard states
 the DHf° for a pure element in its standard state = 0 kJ/mol
 by definition
99
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Standard Enthalpies of Formation
Standard enthalpies of formation, DHf°, are
measured under standard conditions (25 °C
and 1.00 atm pressure).
Thermochemistry
© 2012 Pearson Education, Inc.
Standard Enthalpies of Formation
101
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Formation Reactions
• Reactions of elements in their standard state to
form 1 mole of a pure compound
if you are not sure what the standard state of an
element is, find the form in Appendix that has a
DHf° = 0
because the definition requires 1 mole of compound
be made, the coefficients of the reactants may be
fractions
102
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Calculation of DH
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
• Imagine this as occurring
in three steps:
C3H8(g)  3C(graphite) + 4H2(g)
Thermochemistry
© 2012 Pearson Education, Inc.
Calculation of DH
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
• Imagine this as occurring
in three steps:
C3H8(g)  3C(graphite) + 4H2(g)
3C(graphite) + 3O2(g) 3CO2(g)
Thermochemistry
© 2012 Pearson Education, Inc.
Calculation of DH
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
• Imagine this as occurring
in three steps:
C3H8(g)  3C(graphite) + 4H2(g)
3C(graphite) + 3O2(g) 3CO2(g)
4H2(g) + 2O2(g)  4H2O(l)
Thermochemistry
© 2012 Pearson Education, Inc.
Calculation of DH
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
• The sum of these
equations is
C3H8(g)  3C(graphite) + 4H2(g)
3C(graphite) + 3O2(g) 3CO2(g)
4H2(g) + 2O2(g)  4H2O(l)
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
Thermochemistry
© 2012 Pearson Education, Inc.
Calculation of DH
We can use Hess’s law in this way:
DH = nDHf,products – mDHf°,reactants
where n and m are the stoichiometric
coefficients.
Thermochemistry
© 2012 Pearson Education, Inc.
Calculating Standard Enthalpy
Change for a Reaction
• Any reaction can be written as the sum of
•
formation reactions (or the reverse of formation
reactions) for the reactants and products
The DH° for the reaction is then the sum of the
DHf° for the component reactions
DH°reaction =  n DHf°(products) −  n DHf°(reactants)
 means sum
n is the coefficient of the reaction
Tro: Chemistry: A Molecular Approach, 2/e
108
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Calculation of DH
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
DH = [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(−103.85 kJ) + 5(0 kJ)]
= [(−1180.5 kJ) + (−1143.2 kJ)] – [(−103.85 kJ) + (0 kJ)]
= (−2323.7 kJ) – (−103.85 kJ) = −2219.9 kJ
Thermochemistry
© 2012 Pearson Education, Inc.
CH4(g)+ 2 O2(g)→ CO2(g) + H2O(g)
DH°
[(DH
+ 2∙DH
H(g)
CH4−
(g)+
+74.6
2∙DH
)] 4
CH
(g)=→
C(s,
+ CH
2f°
H42(g)
Df°
H°
kJ
C(s,
+ graphite)
2COH2(g)
DH
74.6
kJ/mol
CH
2O(g))− (DH
f°
f°O2(g)
4graphite)
2(g) →
f°= =
DH°graphite)
= [((−393.5
2(−241.8
kJ)+
2(0 kJ))]
C(s,
+ O2kJ)+
(g) →
CO2(g) kJ)− ((−74.6
DHf°=
−393.5
kJ/mol CO2
= −802.5 kJ
2H2H(g)
DH°
−483.6
kJ H2O
+ +½OO2(g)
→2HH2O(g)
DH
−241.8
kJ/mol
2(g)
2O(g)
2(g)→
f°= =
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) DH° = −802.5 kJ
110
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Example: Calculate the enthalpy
change in the reaction
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l)
1. Write formation reactions for each compound
and determine the DHf° for each
2 C(s, gr) + H2(g)  C2H2(g)
DHf° = +227.4 kJ/mol
C(s, gr) + O2(g)  CO2(g)
DHf° = −393.5 kJ/mol
H2(g) + ½ O2(g)  H2O(l)
DHf° = −285.8 kJ/mol
111
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Example: Calculate the enthalpy change
in the reaction
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l)
2. Arrange equations so they add up to desired reaction
2 C2H2(g)  4 C(s) + 2 H2(g)
DH° = 2(−227.4) kJ
4 C(s) + 4 O2(g)  4CO2(g)
DH° = 4(−393.5) kJ
2 H2(g) + O2(g)  2 H2O(l)
DH° = 2(−285.8) kJ
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l)
112
DH = −2600.4 kJ
Copyright © 2011 Pearson Education, Inc.
Example: Calculate the enthalpy change in
the reaction
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l)
DH°reaction =  n DHf°(products) −  n DHf°(reactants)
DHrxn = [(4•DHCO2 + 2•DHH2O) – (2•DHC2H2 + 5•DHO2)]
DHrxn = [(4•(−393.5) + 2•(−285.8)) – (2•(+227.4) + 5•(0))]
DHrxn = −2600.4 kJ
113
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Example : How many kg of octane must be
combusted to supply 1.0 x 1011 kJ of energy?
Given:
Find:
Conceptual
Plan:
DH°rxn = −5074.1 kJ
1.0 x 1011 kJ
mass octane, kg
Write the balanced equation per mole of octane
DHf°’s
DHrxn°
kJ
mol C8H18
from
above
g C8H18
kg C8H18
Relationships: MMoctane = 114.2 g/mol, 1 kg = 1000 g
Solution:
C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g)
Look up the DHf°
for each material
in Appendix IIB
Check:
the units and sign are correct,
the large value is expected
Tro: Chemistry: A Molecular Approach, 2/e
114
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Practice – Calculate the DH° for decomposing 10.0 g
of limestone, CaCO3, under standard conditions.
CaCO3(s) → CaO(s) + O2(g)
115
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Practice – Calculate the DH° for decomposing 10.0 g of
limestone, CaCO3(s) → CaO(s) + O2(g)
Given:
Find:
Conceptual
Plan:
10.0 g CaCO3
DH°, kJ
DHf°’s
g CaCO3
DH°rxn = +572.7 kJ
DH°rxn per mol CaCO3
mol CaCO3
Relationships: MMlimestone = 100.09 g/mol,
Solution:
Check:
kJ
from
above
CaCO3(s) → CaO(s) + O2(g)
the units and sign are correct
116
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Energy in Foods
Most of the fuel in the food we eat comes
from carbohydrates and fats.
Thermochemistry
© 2012 Pearson Education, Inc.
Energy in Fuels
The vast
majority of the
energy
consumed in
this country
comes from
fossil fuels.
Thermochemistry
© 2012 Pearson Education, Inc.