Chapter
Six
1
Thermochemistry
Prentice Hall © 2005
Chapter Six
2
Energy
• Energy is the capacity to do work (to
displace or move matter).
• Energy literally means “work within”;
however, an object does not contain work.
• Potential energy is energy of position or
composition.
• Kinetic energy is the energy of motion.
Ek = ½ mv2
Energy has the units of joules (J or kg . m2/s2)
Prentice Hall © 2005
Chapter Six
3
Potential Energy and
Kinetic Energy
At what point in each
bounce is the potential
energy of the ball at a
maximum?
Prentice Hall © 2005
Chapter Six
4
Thermochemistry: Basic Terms
• Thermochemistry is the study of energy
changes that occur during chemical
reactions.
• System: the part of the universe being
studied.
• Surroundings: the rest of the universe.
Prentice Hall © 2005
Chapter Six
5
Types of Systems
• Open: energy and
matter can be
exchanged with the
surroundings.
• Closed: energy can
be exchanged with
the surroundings,
matter cannot.
• Isolated: neither
energy nor matter
can be exchanged
with the
surroundings.
Prentice Hall © 2005
A closed system;
energy (not matter)
can be exchanged.
After the lid of the jar
is unscrewed, which
kind of system is it?
Chapter Six
6
Internal Energy (U)
• Internal energy (U) is the total energy
contained within a system
• Part of U is kinetic energy (from molecular
motion)
– Translational motion, rotational motion,
vibrational motion.
– Collectively, these are sometimes called thermal
energy
• Part of U is potential energy
– Intermolecular and intramolecular forces of
attraction, locations of atoms and of bonds.
– Collectively these are sometimes called
chemical energy
Prentice Hall © 2005
Chapter Six
7
Heat (q)
• Technically speaking,
heat is not “energy.”
• Heat is energy transfer
between a system and its
surroundings, caused by
a temperature difference.
• Thermal equilibrium
occurs when the system
and surroundings reach
the same temperature
and heat transfer stops.
Prentice Hall © 2005
More energetic
molecules …
… transfer energy to
less energetic molecules.
How do the root-mean-square
speeds of the Ar atoms and the N2
molecules compare at the point of
thermal equilibrium?
Chapter Six
8
Work (w)
• Like heat, work is an energy transfer between a system and
its surroundings.
• Unlike heat, work is caused by a force moving through a
distance (heat is caused by a temperature difference).
• A negative quantity of work signifies that the system loses
energy.
• A positive quantity of work signifies that the system gains
energy.
• There is no such thing as “negative energy” nor “positive
energy”; the sign of work (or heat) signifies the direction of
energy flow.
Prentice Hall © 2005
Chapter Six
9
Pressure-Volume Work
For now we will consider only
pressure-volume work.
work (w) = –PDV
Prentice Hall © 2005
How would the magnitude of DV
compare to the original gas
volume if the two weights (initial
and final) were identical?
Chapter Six
10
State Functions
• The state of a system: its exact condition at a fixed instant.
• State is determined by the kinds and amounts of matter
present, the structure of this matter at the molecular level,
and the prevailing pressure and temperature.
• A state function is a property that has a unique value that
depends only the present state of a system, and does not
depend on how the state was reached (does not depend on
the history of the system).
• Law of Conservation of Energy – in a physical or
chemical change, energy can be exchanged between a
system and its surroundings, but no energy can be created
or destroyed.
Prentice Hall © 2005
Chapter Six
11
First Law of Thermodynamics
• “Energy cannot be created or destroyed.”
• Inference: the internal energy change of a system
is simply the difference between its final and
initial states:
DU = Ufinal – Uinitial
• Additional inference: if energy change occurs only
as heat (q) and/or work (w), then:
DU = q + w
Prentice Hall © 2005
Chapter Six
12
First Law: Sign Convention
• Energy entering a system carries a positive
sign:
– heat absorbed by the system, or
– work done on the system
• Energy leaving a system carries a negative
sign
– heat given off by the system
– work done by the system
Prentice Hall © 2005
Chapter Six
13
Example 6.1
A gas does 135 J of work while expanding, and at the
same time it absorbs 156 J of heat. What is the change
in internal energy?
Example 6.2: A Conceptual Example
The internal energy of a fixed quantity of an ideal gas
depends only on its temperature. If a sample of an ideal
gas is allowed to expand against a constant pressure at a
constant temperature,
(a) what is DU for the gas? (b) Does the gas do work? (c)
Is any heat exchanged with the surroundings?
Prentice Hall © 2005
Chapter Six
14
Heats of Reaction (qrxn)
• qrxn is the quantity of heat exchanged between a
reaction system and its surroundings.
• An exothermic reaction gives off heat
– In an isolated system, the temperature increases.
– The system goes from higher to lower energy; qrxn is
negative.
• An endothermic reaction absorbs heat
– In an isolated system, the temperature decreases.
– The system goes from lower to higher energy; qrxn is
positive.
Prentice Hall © 2005
Chapter Six
15
Conceptualizing an Exothermic Reaction
Surroundings are at 25 °C
25 °C
Typical situation:
some heat is released
to the surroundings,
some heat is absorbed
by the solution.
Hypothetical situation: all heat
is instantly released to the
surroundings. Heat = qrxn
Prentice Hall © 2005
32.2 °C
35.4 °C
In an isolated system, all heat is
absorbed by the solution.
Maximum temperature rise.
Chapter Six
16
Internal Energy Change at Constant Volume
• For a system where the
reaction is carried out at
constant volume, DV = 0
and DU = qV.
• All the thermal energy
produced by conversion
from chemical energy is
released as heat; no P-V
work is done.
Prentice Hall © 2005
Chapter Six
17
lnternal Energy Change at Constant Pressure
• For a system where the
reaction is carried out at
constant pressure,
DU = qP – PDV or
DU + PDV = qP
• Most of the thermal energy
is released as heat.
• Some work is done to
expand the system against
the surroundings (push
back the atmosphere).
Prentice Hall © 2005
Chapter Six
18
Enthalpy and Enthalpy Change
Enthalpy is the sum of the
internal energy and the
pressure-volume product
of a system:
H = U + PV
For a process carried out
at constant pressure,
qP = DU + PDV
so
qP = DH
The evolved H2 pushes
back the atmosphere;
work is done at
constant pressure.
Mg + 2 HCl  MgCl2 + H2
Most reactions occur at constant
pressure, so for most reactions, the heat
evolved equals the enthalpy change.
Prentice Hall © 2005
Chapter Six
19
Properties of Enthalpy
• Enthalpy is an extensive property.
– It depends on how much of the
substance is present.
• Since U, P, and V are all state
functions, enthalpy H must be a
state function also.
• Enthalpy changes have unique
values. DH = qP
Prentice Hall © 2005
Two logs on a fire give
off twice as much heat
as does one log.
Enthalpy change depends
only on the initial and
final states. In a chemical
reaction we call the initial
state the ____ and the
final state the ____.
Chapter Six
20
Enthalpy Diagrams
• Values of DH are measured experimentally.
• Negative values indicate exothermic reactions.
• Positive values indicate endothermic reactions.
A decrease in enthalpy
during the reaction; DH
is negative.
Prentice Hall © 2005
An increase in enthalpy
during the reaction; DH
is positive.
Chapter Six
21
Reversing a Reaction
• DH changes sign when a process is reversed.
• Therefore, a cyclic process has the value DH = 0.
Same magnitude; different signs.
Prentice Hall © 2005
Chapter Six
22
Example 6.3
Given the equation
(a) H2(g) + I2(s)  2 HI(g)
DH = +52.96 kJ
calculate DH for the reaction
(b) HI(g)  ½ H2(g) + ½ I2(s).
Example 6.4
The complete combustion of liquid octane, C8H18, to
produce gaseous carbon dioxide and liquid water at 25
°C and at a constant pressure gives off 47.9 kJ of heat
per gram of octane. Write a chemical equation to
represent this information.
Prentice Hall © 2005
Chapter Six
ΔH in Stoichiometric Calculations
23
• For problem-solving, heat evolved (exothermic reaction) can be
thought of as a product. Heat absorbed (endothermic reaction)
can be thought of as a reactant.
• We can generate conversion factors involving DH.
• For example, the reaction:
H2(g) + Cl2(g)  2 HCl(g)
DH = –184.6 kJ
can be used to write:
–184.6 kJ
————
1 mol H2
Prentice Hall © 2005
–184.6 kJ
————
1 mol Cl2
–184.6 kJ
————
2 mol HCl
Chapter Six
24
Example 6.5
What is the enthalpy change associated with the
formation of 5.67 mol HCl(g) in this reaction?
H2(g) + Cl2(g)  2 HCl(g)
Prentice Hall © 2005
DH = –184.6 kJ
Chapter Six
25
Calorimetry
• We measure heat flow using calorimetry.
• A calorimeter is a device used to make this
measurement.
• A “coffee cup” calorimeter may be used for
measuring heat involving solutions.
A “bomb” calorimeter is used to
find heat of combustion; the
“bomb” contains oxygen and a
sample of the material to be burned.
Prentice Hall © 2005
Chapter Six
26
Calorimetry, Heat Capacity, Specific Heat
• Heat evolved in a reaction is absorbed by the calorimeter
and its contents.
• In a calorimeter we measure the temperature change of
water or a solution to determine the heat absorbed or
evolved by a reaction.
• The heat capacity (C) of a system is the quantity of heat
required to change the temperature of the system by 1 °C.
C = q/DT (units are J/°C)
• Molar heat capacity is the heat capacity of one mole of a
substance.
• The specific heat (s) is the heat capacity of one gram of a
pure substance (or homogeneous mixture).
s = C/m = q/(mDT)
q = s m DT
Prentice Hall © 2005
Chapter Six
27
Heat Capacity: A Thought Experiment
• Place an empty iron pot weighing 5 lb on the
burner of a stove.
• Place an iron pot weighing 1 lb and containing 4
lb water on a second identical burner (same total
mass).
• Turn on both burners. Wait five minutes.
• Which pot handle can you grab with your bare
hand?
• Iron has a lower specific heat than does water. It
takes less heat to “warm up” iron than it does
water.
Prentice Hall © 2005
Chapter Six
28
Example 6.6
Calculate the heat capacity of an aluminum block
that must absorb 629 J of heat from its
surroundings in order for its temperature to rise
from 22 °C to 145 °C.
Prentice Hall © 2005
Chapter Six
29
More on Specific Heat
•
•
•
•
•
•
q = mass x specific heat x DT
If DT is positive (temperature increases), q is
positive and heat is gained by the system.
If DT is negative (temperature decreases), q is
negative and heat is lost by the system.
The calorie, while not an SI unit, is still used to
some extent.
Water has a specific heat of 1 cal/(g oC).
4.184 J = 1 cal
One food calorie (Cal or kcal) is actually equal to
1000 cal.
Prentice Hall © 2005
Chapter Six
30
Many metals have
low specific heats.
The specific heat of
water is higher than
that of almost any
other substance.
Prentice Hall © 2005
Chapter Six
31
Example 6.7
How much heat, in joules and in kilojoules, does
it take to raise the temperature of 225 g of water
from 25.0 to 100.0 °C?
Example 6.8
What will be the final temperature if a 5.00-g
silver ring at 37.0 °C gives off 25.0 J of heat to its
surroundings? Use the specific heat of silver
listed in Table 6.1.
Prentice Hall © 2005
Chapter Six
32
Example 6.9
A 15.5-g sample of a metal alloy is heated to 98.9 °C and
then dropped into 25.0 g of water in a calorimeter. The
temperature of the water rises from 22.5 to 25.7 °C.
Calculate the specific heat of the alloy.
Example 6.10: An Estimation Example
Without doing detailed calculations, determine which of
the following is a likely approximate final temperature
when 100 g of iron at 100 °C is added to 100 g of 20 °C
water in a calorimeter:
(a) 20 °C
Prentice Hall © 2005
(b) 30 °C
(c) 60 °C
(d) 70 °C.
Chapter Six
33
Measuring Enthalpy Changes for
Chemical Reactions
For a reaction carried out in a calorimeter, the heat evolved by
a reaction is absorbed by the calorimeter and its contents.
qrxn = – qcalorimeter
qcalorimeter = mass x specific heat x DT
By measuring the temperature change that occurs in a
calorimeter, and using the specific heat and mass of the
contents, the heat evolved (or absorbed) by a reaction can be
determined and the enthalpy change calculated.
Prentice Hall © 2005
Chapter Six
34
Example 6.11
A 50.0-mL sample of 0.250 M HCl at 19.50 °C is added to
50.0 mL of 0.250 M NaOH, also at 19.50 °C, in a
calorimeter. After mixing, the solution temperature rises to
21.21 °C. Calculate the heat of this reaction.
Example 6.12
Express the result of Example 6.11 for molar amounts of
the reactants and products. That is, determine the value
of DH that should be written in the equation for the
neutralization reaction:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Prentice Hall © 2005
DH = ?
Chapter Six
Bomb Calorimetry:
Reactions at Constant
Volume
35
• Some reactions, such as combustion,
cannot be carried out in a coffee-cup
calorimeter.
• In a bomb calorimeter, a sample of
known mass is placed in a heavywalled “bomb,” which is then
pressurized with oxygen.
• Since the reaction is carried out at
constant volume,
–qrxn = qcalorimeter = DU
… but in many cases the value of DU
is a good approximation of DH.
Prentice Hall © 2005
Chapter Six
36
Hess’s Law of Constant
Heat Summation
• Some reactions cannot be carried out “as written.”
• Consider the reaction:
C(graphite) + ½ O2(g)  CO(g).
• If we burned 1 mol C in ½ mol O2, both CO and CO2
would probably form. Some C might be left over.
However …
Prentice Hall © 2005
Chapter Six
37
Hess’s Law of Constant
Heat Summation
• … enthalpy change is a state function.
• The enthalpy change of a reaction is the same
whether the reaction is carried out in one step or
through a number of steps.
• Hess’s Law: If an equation can be expressed as the
sum of two or more other equations, the enthalpy
change for the desired equation is the sum of the
enthalpy changes of the other equations.
Prentice Hall © 2005
Chapter Six
38
Hess’s Law: An Enthalpy Diagram
We can find DH(a) by
subtracting DH(b) from DH(c)
Prentice Hall © 2005
Chapter Six
39
Example 6.14
Calculate the enthalpy change for reaction (a) given the data
in equations (b), (c), and (d).
(a) 2 C(graphite) + 2 H2(g)  C2H4(g)
DH = ?
(b) C(graphite) + O2(g)  CO2(g)
DH = –393.5 kJ
(c) C2H4(g) + 3 O2  2 CO2(g) + 2 H2O(l)
DH = –1410.9 kJ
(d) H2(g) + ½ O2  H2O(l)
DH = –285.8 kJ
Prentice Hall © 2005
Chapter Six
40
Standard Enthalpies of Formation
• It would be convenient to be able to use the simple relationship
ΔH = Hproducts – Hreactants
to determine enthalpy changes.
• Although we don’t know absolute values of enthalpy, we don’t
need them; we can use a relative scale.
• We define the standard state of a substance as the state of the pure
substance at 1 atm pressure and the temperature of interest (usually
25 °C).
• The standard enthalpy change (ΔH°) for a reaction is the enthalpy
change in which reactants and products are in their standard states.
• The standard enthalpy of formation (ΔHf°) for a reaction is the
enthalpy change that occurs when 1 mol of a substance is formed
from its component elements in their standard states.
Prentice Hall © 2005
Chapter Six
41
Standard Enthalpy of Formation
When we say “The standard enthalpy of formation of
CH3OH(l) is –238.7 kJ”, we are saying that the reaction:
C(graphite) + 2 H2(g) + ½ O2(g)  CH3OH(l)
has a value of ΔH of –238.7 kJ.
We can treat ΔHf° values as though they were absolute
enthalpies, to determine enthalpy changes for reactions.
Question: What is ΔHf° for an element in its standard
state [such as O2(g)]? Hint: since the reactants are
the same as the products …
Prentice Hall © 2005
Chapter Six
42
Calculations Based on
Standard Enthalpies of Formation
•
•
•
DH°rxn = Snp x DHf°(products) – Snr x DHf°(reactants)
The symbol S signifies the summation of several terms.
The symbol n signifies the stoichiometric coefficient used
in front of a chemical symbol or formula.
In other words …
1. Add all of the values for DHf° of the products.
2. Add all of the values for DHf° of the reactants.
3. Subtract #2 from #1
(This is usually much easier than using Hess’s Law!)
Prentice Hall © 2005
Chapter Six
43
Example 6.15
Synthesis gas is a mixture of carbon monoxide and hydrogen that is
used to synthesize a variety of organic compounds. One reaction for
producing synthesis gas is
3 CH4(g) + 2 H2O(l) + CO2(g)  4 CO(g) + 8 H2(g) ΔH° = ?
Use standard enthalpies of formation from Table 6.2 to calculate the
standard enthalpy change for this reaction.
Example 6.16
The combustion of isopropyl alcohol, common rubbing alcohol, is
represented by the equation
2 (CH3)2CHOH(l) + 9 O2(g)  6 CO2(g) + 8 H2O(l) ΔH° = –4011 kJ
Use this equation and data from Table 6.2 to establish the standard
enthalpy of formation for isopropyl alcohol.
Example 6.17: A Conceptual Example
Without performing a calculation, determine which of these two
substances should yield the greater quantity of heat per mole upon
complete combustion: ethane, C2H6(g), or ethanol, CH3CH2OH(l).
Prentice Hall © 2005
Chapter Six
44
Ionic Reactions
in Solution
• We can apply
thermochemical concepts
to reactions in ionic
solution by arbitrarily
assigning an enthalpy of
formation of zero to
H+(aq).
Prentice Hall © 2005
Chapter Six
45
Example 6.18
H+(aq) + OH–(aq)  H2O(l)
ΔH° = –55.8 kJ
Use the net ionic equation just given, together with ΔHf°
= 0 for H+(aq), to obtain ΔHf° for OH–(aq).
Prentice Hall © 2005
Chapter Six
46
Looking Ahead
• A reaction that occurs (by itself) when the reactants are
brought together under the appropriate conditions is
said to be spontaneous.
• A discussion of entropy is needed to fully understand
the concept of spontaneity, and will be discussed in
Chapter 17.
• A spontaneous reaction isn’t necessarily fast (rusting;
diamond  graphite; etc. are slow).
• The difference between the tendency of a reaction to
occur and the rate at which a reaction occurs will be
discussed in Chapter 13.
Prentice Hall © 2005
Chapter Six
47
Combustion and Respiration:
Fuels and Foods
• Fossil Fuels: Coal, Natural Gas, and Petroleum
– A fuel is a substance that burns with the release of
heat.
– These fossil fuels were formed over a period of
millions of years from organic matter that became
buried and compressed under mud and water.
• Foods: Fuels for the Body
– The three principal classes of foods are
carbohydrates, fats, and proteins.
– 1 Food Calorie (Cal) is equal to 1000 cal (or 1
kcal).
Prentice Hall © 2005
Chapter Six
48
Lower in carbon,
higher in moisture, less
effective for heating.
“Hard” coal is
highest in fuel value,
most expensive.
Prentice Hall © 2005
Chapter Six
49
Glyceryl
Trilaurate:
A Typical Fat
Fats are esters.
This particular fat is
saturated; all the C–C
bonds are single bonds.
Prentice Hall © 2005
Chapter Six
50
CUMULATIVE EXAMPLE
The human body is about 67% water by mass. What
mass of sucrose, C12H22O11(s), for which ΔHf° = –2225
kJ/mol, must be metabolized by a 55-kg person with
hypothermia (low body temperature) to raise the
temperature of the body water from 33.5 °C to the normal
body temperature of 37.0 °C? Assume the products of
metabolism are in their most stable states at 25 °C.
What volume of air at 37 °C having a partial pressure of
O2 of 151 Torr is required for the metabolism?
Prentice Hall © 2005
Chapter Six