Transcript Chapter 6 Thermochemistry - Suffolk County Community College

Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Chapter 6
Thermochemistry
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright © 2011 Pearson Education, Inc.
Chemical Hand Warmers
• Most hand warmers work by using the heat
released from the slow oxidation of iron
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
• The amount your hand temperature rises
depends on several factors
the size of the hand warmer
the size of your glove, etc.
mainly, the amount of heat released by the
reaction
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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
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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
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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
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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
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Manifestations of Energy
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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
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Conservation of Energy
• The Law of Conservation
•
of Energy states that
energy cannot be created
nor destroyed
When energy is transferred
between objects, or
converted from one form
to another, the total
amount of energy present
at the beginning must be
present at the end
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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
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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
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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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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
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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
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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
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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!
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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 ─
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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
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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
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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
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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
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Energy Exchange
• Energy is exchanged between the system and
surroundings through either heat exchange or
work being done
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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
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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
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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
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Example 6.1: If the burning of the fuel in a potato cannon
performs 855 J of work on the potato and
produces 1422 J of heat, what is DE for the burning of the fuel?
Given:
Find:
Concept Plan:
qpotato = 855 J, wpotato = 1422 J
DEfuel, J
q, w
q, w
potato
fuel
q, w
DE
Relationships: q
system = −qsurroundings, wsystem = −wsurroundings, DE = q + w
Solution:
Check:
the unit is correct, the sign make sense as the fuel
should lose energy during the reaction
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Practice – Reacting 50 mL of H2(g) with 50 mL of C2H4(g)
produces 50 mL of C2H6(g) at 1.5 atm. If the reaction
produces 3.1 x 102 J of heat and the decrease in volume
requires the surroundings do 7.6 J of work on the gases, what
is the change in internal energy of the gases?
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If the reaction produces 3.1 x 102 J of heat and the decrease in
volume requires the surroundings do 7.6 J of work on the
gases, what is the change in internal energy of the gases?
Given:
Find:
Concept Plan:
qreaction = −310 J, wsurroundings = −7.6 J
DEgases, J
w
w
surrounding
gases
q, w
DE
Relationships: q
system = −qsurroundings, wsystem = −wsurroundings, DE = q + w
Solution:
Check:
the units are correct, the sign is reasonable as the
amount of heat lost in the reaction is much larger than
the amount of work energy gained
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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
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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
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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
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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
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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
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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)
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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
T1 = −8.0 °C, T2= 37.0 °C, m = 3.10 g
Given:
Information
q, J
Find:
• Strategize
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 (Table 6.4)
Solution:
Check: the unit is correct, the sign is reasonable as
the penny must absorb heat to make its
temperature rise
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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)
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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
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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)
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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
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Example 6.3: 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
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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.
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Practice – Calculate the specific heat and
identify the metal from the data
Given:
Find:
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
Concept Plan:
Relationships:
q
q = m x Cs x DT; qmetal = −qH2O
Solution:
Check:
the units are correct, the number indicates
the metal is copper
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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
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Example 6.4: 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:
V1 = 0.100 L, V2 = 1.85 L, P = 1.00 atm
w, J
Conceptual
Plan:
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
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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)
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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
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Exchanging Energy Between
System and Surroundings
• Exchange of heat energy
•
q = mass x specific heat x DTemperature
Exchange of work
w = −Pressure x DVolume
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Measuring DE,
Calorimetry at Constant Volume
• Because DE = q + w, we can determine DE by measuring q
and w
• In practice, it is easiest to do a process in such a way that
there is no change in volume, so w = 0
 at constant volume, DEsystem = qsystem
• In practice, it is not possible to observe the temperature
changes of the individual chemicals involved in a reaction – so
instead, we measure the temperature change in the
surroundings
 use insulated, controlled surroundings
 qsystem = −qsurroundings
• The surroundings is called a bomb calorimeter and is usually
made of a sealed, insulated container filled with water
qsurroundings = qcalorimeter = ─qsystem
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Bomb Calorimeter
• Used to measure DE
•
because it is a constant
volume system
The heat capacity of the
calorimeter is the amount
of heat absorbed by the
calorimeter for each
degree rise in
temperature and is
called the calorimeter
constant
 Ccal, kJ/ºC
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Example 6.5: 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
Given:
−3 O ,C
2.9506g xC10
O11, DT
Ccal
4.90
1.010
= 22
24.92
°C,=T3.41°C,
°C,= C
4.90 kJ/°C
12H22mol
11 T12
1H
2 = 28.33
cal =kJ/°C
Find:
DErxn
rxn, kJ/mol
Conceptual
Plan:
Ccal, DT
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
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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?
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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?
3 kJ/mol
107−3
DT T
=22.40
°C,°C,
DE=
−3.23
x 10
kJ/mol
1.22 gxHC
H5mol
O2, HC
T1=20.27
=22.67
DE=
−3.23
x 310
Given: 9.990
7H5O2, °C,
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
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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
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Endothermic and Exothermic Reactions
•
•
•
•
•
•
When DH is ─, heat is being released by the system
Reactions that release heat are called exothermic reactions
When DH is +, heat is being absorbed by the system
Reactions that absorb heat are called endothermic reactions
Chemical heat packs contain iron filings that are oxidized in
an exothermic reaction ─ your hands get warm because the
released heat of the reaction is transferred to your hands
Chemical cold packs contain NH4NO3 that dissolves in water
in an endothermic process ─ your hands get cold because the
pack is absorbing your heat
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Molecular View of Exothermic Reactions
• For an exothermic reaction, the
surrounding’s temperature rises due to
release
of thermal energy by the
reaction
• This extra thermal energy comes
from the conversion of some of the
chemical potential energy in the
reactants into kinetic energy in the
form of heat
• During the course of a reaction,
old bonds are broken and new
bonds are made
• The products of the reaction have less
chemical potential energy than the reactants
• The difference in energy 57is releasedCopyright
as heat
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Molecular View of Endothermic Reactions
• In an endothermic reaction, the surrounding’s
•
•
•
temperature drops due to absorption of some of
its thermal energy by the reaction
During the course of a reaction, old bonds are
broken and new bonds are made
The products of the reaction have more chemical
potential energy than the reactants
To acquire this extra energy, some of the thermal
energy of the surroundings is converted into
chemical potential energy stored in the products
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Enthalpy of Reaction
• The enthalpy change in a chemical reaction is an
extensive property
 the more reactants you use, the larger the enthalpy
change
• By convention, we calculate the enthalpy change
for the number of moles of reactants in the reaction
as written
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
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DH = −2044 kJ
Copyright © 2011 Pearson Education, Inc.
Example 6.7: How much heat is evolved in the
complete combustion of 13.2 kg of C3H8(g)?
Given:
Find:
13.2 kg C3H8,
q, kJ/mol
Conceptual
Plan:
kg
g
mol
kJ
Relationships: 1 kg = 1000 g, 1 mol C3H8 = −2044 kJ, Molar Mass = 44.09 g/mol
Solution:
Check:
the sign is correct and the number is reasonable
because the amount of C3H8 is much more than 1 mole
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Practice – How much heat is evolved
when a 0.483 g diamond is burned?
(DHcombustion = −395.4 kJ/mol C)
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Practice – How much heat is evolved when a
0.483 g diamond is burned?
Given:
Find:
0.483 g C, DH = −395.4 kJ/mol C
q, kJ/mol
Concept Plan:
g
mol
kJ
Relationships: 1 mol C = −395.4 kJ, Molar Mass = 12.01 g/mol
Solution:
Check:
the sign is correct and the number is reasonable
because the amount of diamond is less than 1 mole
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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
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Example 6.8: 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
32.8
°C,s = 4.18 J/g∙°C
6.499
mol Mg,
1.00
102 g,°C,
DTT=
7.2
°C,,C
2=
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
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Practice – What will be the final temperature of the solution in a
coffee cup calorimeter if a 50.00 mL sample of 0.250 M HCl(aq) is
added to a 50.00 mL sample of 0.250 NaOH(aq). The initial
temperature is 19.50 °C and the DHrxn is −57.2 kJ/mol NaOH.
(assume the density of the solution is 1.00 g/mL and the specific
heat of the solution is 4.18 J/g∙°C)
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Practice – What will be the final temperature of the solution in a coffee cup
calorimeter if a 50.00 mL sample of 0.250 M HCl(aq) is added to a 50.00 mL
sample of 0.250 NaOH(aq). The initial temperature is 19.50 °C.
50.00
mL−2ofmol
0.250
M HCl,
0.250
M°C,
NaOH T1 = 19.50 °C,
x 10
NaOH,
1.0050.00
x 102mL
g, Tof1 =
19.50
Given: 1.25
d ==−57.2
1.00 g/mL,
Cs C
=s4.18
J/g∙°C,
D H = −57.2 kJ/mol
DH
kJ/mol,
= 4.18
J/g∙°C
°C
Find: TT22,,°C
Conceptual
Plan:
Relationships:
DH
qrxn
qsol’n
qrxn  mol  DH qsol'n   qrxn
DT
Cs, qsol’n, m
DT 
qsol'n
m  Cs
qsol’n = m x Cs, sol’n x DT = −qrxn, 0.250 mol NaOH = 1 L
DH 
T2
qrxn
mol
Solution:
Check: the sign is correct and the value is reasonable, because the reaction is
exothermic we expect the final temperature to be higher than the initial
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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)
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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
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Example: Hess’s Law
Given the following information:
2 NO(g) + O2(g)  2 NO2(g)
2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq)
N2(g) + O2(g)  2 NO(g)
DH° = −116 kJ
DH° = −256 kJ
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
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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
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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)
2 Cu(s) + Cl2(g)  2 CuCl(s)
Cu(s) + CuCl2(s)  2 CuCl(s)
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DH° = +206 kJ
DH° = − 36 kJ
DH° = +170. kJ
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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
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Standard Enthalpies of Formation
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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 IIB that has a
DHf° = 0
because the definition requires 1 mole of compound
be made, the coefficients of the reactants may be
fractions
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Writing Formation Reactions
Write the Formation Reaction for CO(g)
• The formation reaction is the reaction between the
•
elements in the compound, which are C and O
C + O → CO(g)
The elements must be in their standard state
 there are several forms of solid C, but the one with
DHf° = 0 is graphite
 oxygen’s standard state is the diatomic gas
•
C(s, graphite) + O2(g) → CO(g)
The equation must be balanced, but the coefficient
of the product compound must be 1
 use whatever coefficient in front of the reactants is
necessary to make the atoms on both sides equal
without changing the product coefficient
C(s, graphite) + ½ O2(g) → CO(g)
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Write the formation reactions for the following:
CO2(g)
C(s, graphite) + O2(g)  CO2(g)
Al2(SO4)3(s)
2 Al(s) + 3/8 S8(s, rhombic) + 6 O2(g)  Al2(SO4)3(s)
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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 = S n DHf°(products) − S n DHf°(reactants)
S means sum
n is the coefficient of the reaction
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CH4(g)+ 2 O2(g)→ CO2(g) + H2O(g)
DH°4graphite)
= [(DH
(g)H
+2(g)
2∙DH
H
)− (DHf° CH
+ −2∙DH
CH
(g)
→ C(s,
22O(g)
H42(g)
(g)
D4(g)
H°
+74.6
74.6
kJ)] CH4
C(s,
+ graphite)
→+f°CH
DH
kJ/mol
22
2(g)
f° CO
f°O
f°= =
DH° graphite)
= [((−393.5
kJ)+
2(−241.8 kJ)− ((−74.6
kJ)+
2(0 kJ))]
C(s,
+O
DHf°=
−393.5
kJ/mol CO2
2(g) → CO2(g)
= −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
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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
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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)
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DH = −2600.4 kJ
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Example: Calculate the enthalpy change in
the reaction
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l)
DH°reaction = S n DHf°(products) − S 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
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Example 6.12: 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
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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)
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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
DH°rxn = +572.7 kJ
DH°rxn per mol CaCO3
mol CaCO3
g CaCO3
Relationships: MMlimestone = 100.09 g/mol,
Solution:
Check:
kJ
from
above
CaCO3(s) → CaO(s) + O2(g)
the units and sign are correct
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Energy Use and the Environment
• In the United States, each person uses over 105 kWh of
energy per year
• Most comes from the combustion of fossil fuels
 combustible materials that originate from ancient life
C(s) + O2(g) → CO2(g)
DH°rxn = −393.5 kJ
CH4(g) +2 O2(g) → CO2(g) + 2 H2O(g)
DH°rxn = −802.3 kJ
C8H18(g) +12.5 O2(g) → 8 CO2(g) + 9 H2O(g)
DH°rxn = −5074.1
kJ
• Fossil fuels cannot be replenished
• At current rates of consumption, oil and natural gas
supplies will be depleted in 50–100 yrs
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Energy Consumption
• The increase in
energy consumption
in the United States
• The distribution of
energy consumption
in the United States
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The Effect of Combustion Products
on Our Environment
• Because of additives and impurities in the fossil
•
fuel, incomplete combustion, and side
reactions, harmful materials are added to the
atmosphere when fossil fuels are burned for
energy
Therefore fossil fuel emissions contribute to air
pollution, acid rain, and global warming
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Global Warming
• CO2 is a greenhouse gas
 it allows light from the sun to reach the earth, but does
not allow the heat (infrared light) reflected off the Earth to
escape into outer space
 it acts like a blanket
• CO2 levels in the atmosphere have been steadily
•
•
•
increasing
Current observations suggest that the average
global air temperature has risen 0.6 °C in the past
100 yrs.
Atmospheric models suggest that the warming
effect could worsen if CO2 levels are not curbed
Some models predict that the result will be more
severe storms, more floods and droughts, shifts in
agricultural zones, rising sea levels, and changes in
habitats
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Renewable Energy
• Our greatest unlimited supply of energy is the
•
sun
New technologies are being developed to
capture the energy of sunlight
parabolic troughs, solar power towers, and dish
engines concentrate the sun’s light to generate
electricity
solar energy used to decompose water into H2(g) and
O2(g); the H2 can then be used by fuel cells to
generate electricity
H2(g) + ½ O2(g) → H2O(l) DH°rxn = −285.8 kJ
• Hydroelectric power
• Wind power
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