Chemical Energetics and Thermodynamics
A Chem1 Reference Text
Stephen K. Lower
Simon Fraser University
Contents
1 Potential energy and kinetic energy
1.1 Chemical Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
3
2 Energetics of chemical reactions
4
3 The thermodynamic view: system and surroundings
3.1 Heat and work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
5
4 Internal energy
6
4.1
4.2
Changes in internal energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The heat capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
7
4.3
The First Law of thermodynamics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
5 Molecular interpretation of heat capacity
5.1 Polyatomic molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Heat capacities of metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
9
10
6 Pressure-volume work
10
6.1
Reversible processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
7 Heat changes at constant pressure: the enthalpy
13
8 Thermochemical equations and standard states
14
9 Enthalpy of formation
15
10 Hess’ law and thermochemical calculations
16
11 Calorimetry
17
12 Interpretation of reaction enthalpies
12.1 Enthalpy diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.2 Bond enthalpies and bond energies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
21
21
12.3 Energy content of fuels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.4 Energy content of foods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
23
• CONTENTS
13 Variation of the enthalpy with temperature
24
14 The direction of spontaneous change
14.1 A closer look at disorder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
2
15 The entropy
5
15.1 The entropy of the world always increases . . . . . . . . . . . . . . . . . . . . . . . . . . .
15.2 Absolute entropies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15.3 Entropies of substances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
7
8
15.4 Effect of temperature, volume, and concentration on the entropy . . . . . . . . . . . . . .
9
15.5 The second law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
16 The
16.1
16.2
16.3
free energy
The standard Gibbs free energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Temperature dependence of ∆G◦ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
How free energies determine melting and boiling points . . . . . . . . . . . . . . . . . . . .
17 Free energy and concentration
12
12
13
14
14
17.1 The free energy of a gas: Standard states . . . . . . . . . . . . . . . . . . . . . . . . . . .
17.2 The free energy of a solute . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
15
18 Free energy and the equilibrium constant
18.1 The equilibrium constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18.2 Equilibrium and temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18.3 Equilibrium without mixing: coupled reactions . . . . . . . . . . . . . . . . . . . . . . . .
16
18
19
20
19 Coupled reactions
19.1 Extraction of metals from their oxides . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
21
20 Bioenergetics
23
Chem1 General Chemistry
2
Chemical Energetics
• 1 Potential energy and kinetic energy
Part 1:
Introduction to chemical energetics
All chemical changes are accompanied by the absorption or release of heat. The intimate connection
between matter and energy has been a source of wonder and speculation from the most primitive times;
it is no accident that fire was considered one of the four basic elements (along with earth, air, and water)
as early as the fifth century bc. In this unit we will review some of the fundamental concepts of energy
and heat and the relation between them. We will begin the study of thermodynamics, which treats the
energetic aspects of change in general, and will finally apply this specifically to chemical change. Our goal
will be to provide you with the tools to predict the energy changes associated with chemical processes.
This will provide the groundwork for a more ambitious goal: to predict the direction and extent of change
itself.
1
Potential energy and kinetic energy
Ultimately, there are only two kinds of energy: kinetic and potential. Kinetic energy is associated with
the motion of an object; a body with a mass m and moving at a velocity v possesses the kinetic energy
1
2
2 mv . Potential energy is energy a body has by virtue of its location. For example, if an object of mass
m is raised off the floor to a height h, its potential energy increased by gmh, where g is a proportionality
constant known as the acceleration of gravity. This tells us something else about potential energy: it is
energy a body has by virtue of its location in a force field of some kind; a gravitional, electrical, or a
magnetic field.
An interesting thing about energy is that its values are always relative: there is no “zero” of energy.
You might at first think that a book sitting on the table has zero kinetic energy since it is not moving.
You forget, however, that the earth itself is moving; it is spinning on its axis, it is orbiting the sun, and
the sun itself is moving away from the other stars in the general expansion of the universe. Since these
motions are normally of no interest to us, we are free to adopt an arbitrary scale in which the velocity
of the book is measured with respect to the table; on this so-called laboratory coordinate system, the
kinetic energy of the book will be considered zero.
We do the same thing with potential energy. If the book is on the table, its potential energy with
respect to the surface of the table will be zero. If we adopt this as our zero of potential energy, and then
push the book off the table, its potential energy will be negative after it reaches the floor.
1.1
Chemical Energy
When you buy a litre of gasoline for your car, a cubic metre of natural gas to heat your home, or a small
battery for your flashlight, you are purchasing energy in a chemical form. In each case, some kind of
a chemical change will have to occur before this energy can be released and utilized: the fuel must be
burned in the presence of oxygen, or the two poles of the battery must be connected through an external
circuit (thereby initiating a chemical reaction inside the battery). And eventually, when each of these
reactions is complete, our source of energy will be exhausted; the fuel will be used up, or the battery will
be “dead”.
Chemical substances are made of atoms, or more generally, of positively charged nuclei surrounded
by negatively charged electrons. A molecule such as dihydrogen, H2 , is held together by electrostatic
attractions mediated by the shared electrons between the two nuclei. The total potential energy of the
molecule is the sum of the repulsions between like charges and the attractions between electrons and
nuclei:
Chem1 General Chemistry
3
Chemical Energetics
• 2 Energetics of chemical reactions
PEtotal = PEelectron−electron + PEnucleus−nucleus + PEnuclus−electron
In other words, the potential energy of the molecule depends on the relative locations of the two
atoms in the force field of the nuclear charges and the electron cloud. This dependence is expressed
by the familiar potential energy curve which serves as an important description of the chemical bond
between two atoms.
If the H2 is in the gaseous state, the molecules will be moving freely from one location to another; this
is called translational motion, and the molecules therefore possess translational kinetic energy KEtrans =
mv 2 /2, in which v stands for the average velocity of the molecules; you may recall from your study of
gases that v, and therefore KEtrans , depends on the temperature.
In addition to translation, molecules can possess other kinds of motion. Because a chemical bond acts
as a kind of spring, the two nuclei in H2 will have a natural vibrational frequency. In more complicated
molecules, many different modes of vibration become possible, and these all contribute a vibrational term
KEvib to the total kinetic energy. Finally, a molecule can undergo rotational motions which give rise to
a third term KErot . Thus the total kinetic energy of a molecule is the sum
KEtotal = KEtrans + KEvib + KErot
The total energy of the molecule is just the sum
Etotal = KEtotal + PEtotal
(1)
Although this formula is simple and straightforward, it cannot take us very far in understanding and
predicting the behavior of even one molecule, let alone a large number of them. The reason, of course, is
the chaotic and unpredictable nature of molecular motion. Fortunately, the behavior of a large collection
of molecules, like that of a large population of people, can be described by statistical methods.
2
Energetics of chemical reactions
A chemical reaction is defined by its reactants and products, but there is one other property of a reaction
that is not shown in the balanced chemical equation, but is important to know about: the amount of
heat absorbed or liberated when the reaction takes place. If the chemical reaction is serving as a source
of heat, such as in the combustion of a fuel, for example, then these heat effects are of direct and obvious
interest. We will soon see, however, that a study of the energetics of chemical reactions in general can
lead us to a deeper understanding of chemical equilibrium and the basis of chemical change itself. This
more general body of knowledge is known as thermodynamics (from the Greek words referring to the
movement of heat), and this will be the subject of this and the next units of this course.
One of the interesting things about thermodynamics is that although it deals with matter, it makes
no assumptions about the microscopic nature of that matter. Thermodynamics deals with matter in a
macroscopic sense; it would be valid even if the atomic theory of matter were wrong. This is an important
quality, because it means that reasoning based on thermodynamics is unlikely to require alteration as
new facts about atomic structure and atomic interactions come to light.
3
The thermodynamic view: system and surroundings
The thermodynamic view of the world requires that we be very precise about our use of certain words.
The two most important of these are system and surroundings. A thermodynamic system is that part of
Chem1 General Chemistry
4
Chemical Energetics
• Heat and work
the world to which we are directing our attention. Everything that is not a part of the system constitutes
the surroundings. The system and surroundings are separated by a boundary. If our system is one mole
of a gas in a container, then the boundary is simply the container itself. The boundary need not be a
physical barrier; for example, if our system is a factory or a forest, then the boundary can be wherever
we wish to define it. We can even focus our attention on the dissolved ions in an aqueous solution of
a salt, leaving the water molecules as part of the surroundings. The single property that the boundary
must have is that it be clearly defined, so we can unambiguously say whether a given part of the world
is in our system or in the surroundings.
If matter is not able to pass across the boundary, then the system is said to be closed; otherwise, it is
open. A closed system may still exchange energy with the surroundings unless the system is an isolated
one, in which case neither matter nor energy can pass across the boundary. The tea in a closed Thermos
bottle approximates a closed system over a short time interval.
Properties and the state of a system The properties of a system are those quantities such as the
pressure, volume, temperature, and its composition, which are in principle measurable and capable of
assuming definite values. There are of course many properties other than those mentioned above; the
density and thermal conductivity are two examples. However, the values of these latter properties are
determined by the others; thus the density of a gas is determined by its volume and its composition. The
P , V , T (as well as the composition) are known as state properties, because if these have fixed values,
then the values of all the other properties are fixed, and the system is in a definite state.
In the case of chemical substances, we must also be careful to specify the physical state (liquid, solid,
etc.) of the substance in the system.
Change of state In dealing with thermodynamics, we must be able to unambiguously define the change
in the state of a system when it undergoes some process. This is done by specifying changes in the values
of the different state properties as illustrated here for a change in the volume:
∆V = Vfinal − Vinitial
We can compute similar ∆-values for changes in P , T , ni (the number of moles of substance i), and the
other state properties we will meet later.
3.1
Heat and work
Heat and work are both measured in energy units, so they must both represent energy. How do they
differ from each other, and from just plain “energy” itself?
First, recall that energy can take many forms: mechanical, chemical, electrical, radiation (light),
. . .and thermal, or heat. So heat is a form of energy, but it differs from all the others in one crucial
way. All other forms of energy are interconvertible: mechanical energy can be completely converted to
electrical energy, and the latter can be completely converted to heat. However, complete conversion of
heat into other forms of energy is impossible. We will defer a discussion of the profound implications of
this fact until later; for the moment, we will simply state that this places heat in a category of its own
and justifies its special treatment.
There is another special property of heat that you already know about: heat can be transferred from
one body (i.e., one system) to another. We often refer to this as a flow of heat, recalling the 18th-century
notion that heat was an actual substance called “caloric” that could flow like a liquid. Moreover, you
know that heat can only flow from a system at a higher temperature to one at a lower temperature. This
special characteristic is often used to distinguish heat from other modes of transferring energy from one
system to another.
Chem1 General Chemistry
5
Chemical Energetics
• 4 Internal energy
Next, what about work ? Work, like energy, can take various forms: mechanical, electrical, gravitational, etc. All have in common the fact that they depend on two factors. For example, the simplest form
of mechanical work arises when an object moves a certain distance against an opposing force. Electrical
work is done when a body having a certain charge moves through a potential difference. Performance
of work involves a transformation of energy; thus when a book drops to the floor, gravitational work is
done (a mass moving through a gravitational potential difference), and the potential energy the book had
before it was dropped is converted into kinetic energy which ultimately appears as heat.
Heat and work are best thought of as processes by which energy is exchanged, rather than as energy
itself. That is, heat “exists” only when it is flowing, work “exists” only when it is being done.
When two bodies are placed in thermal contact and energy flows from the warmer
body to the cooler one, we call the process “heat”. A transfer of energy to or
from a system by any means other than heat is called “work”.
4
Internal energy
What is internal energy? It is simply the totality of all forms of kinetic and potential energy of the
system. Thermodynamics makes no distinction between these two forms of energy and it does not
assume the existence of atoms and molecules. But since we are studying thermodynamics in the context
of chemistry, we can allow ourselves to depart from “pure” thermodynamics enough to point out that
the internal energy is the sum of the kinetic energy of motion of the molecules, and the potential energy
represented by the chemical bonds between the atoms and any other intermolecular forces that may be
present. The internal energy of a molecule is given by Eq 1; the internal energy of the system is just the
sum of this quantitity over all the molecules comprising the system.
How can we know how much internal energy a system possesses? The answer is that we cannot, at
least not on an absolute basis; all scales of energy are arbitrary. The best we can do is measure changes in
energy. However, we are perfectly free to define zero energy as the energy of the system in some arbitrary
reference state, and then say that the internal energy of the system in any other state is the difference
∆U between the energies of the system in these two different states.
In chemical thermodynamics, we define the zero of internal energy as the internal energy of the
elements as they exist in their stable forms at 298 ◦ K and 1 atm pressure. Thus the internal energies U
of Xe(g), O2 (g), Br2 (l), Cu(s) and C(diamond) are all zero. When the reaction
H2 (g) + Cl2 (g) −→ 2 HCl(g)
takes place, there is a fall in the internal energy, and the difference ∆U is also the internal energy of two
moles of gaseous HCl. What this represents physically is that the arrangement of electrons and atoms
in the products corresponds to a lower potential energy than in the reactants. This need not always be
the case, however, and it is quite common for the internal energy of the products to exceed those of the
reactants.
In comparing the internal energies of different substances as we have been doing here, it is important
to compare equal numbers of moles, because internal energy is an extensive property of matter. However,
it is commonly expressed on a molar basis, in which case it may be treated as an intensive property.
The internal energy of a system also depends on its temperature; the higher the temperature, the
greater is the average kinetic energy of the molecules comprising the system, and according to Eq 1 this
will lead to an increase in the internal energy.
Chem1 General Chemistry
6
Chemical Energetics
• Changes in internal energy
4.1
Changes in internal energy
We can generalize the foregoing statements that the internal energy of a system depends on its composition
and temperature by saying that the internal energy depends only on the state of the system:
∆U = Ufinal − Uinitial
The word “only” in the preceding statement is important, because it eliminates any reference to how the
system gets from the initial state to the final state. In other words, the value of ∆U for a given change
in state is independent of the pathway of the process. A quantity such as internal energy that depends
only on the state of the system is known as a state function.
To help you appreciate the significance of U being a state function, consider the oxidation of a lump
of sugar to carbon dioxide and water:
C12 H22 O11 + 6 O2 (g) −→ 12 CO2 (g) + 11 H2 O(l)
This process can be carried out in many ways, for example by burning the sugar in air, or by eating the
sugar and letting your body carry out the oxidation. Although the mechanisms of the transformation
are completely different for these two pathways, the overall change in internal energy of the system (the
atoms of carbon, hydrogen and oxygen that were originally in the sugar) will be identical, and can be
calculated simply by looking up the internal energies of the substances and calculating the difference
∆U = U12 CO2 + U11 H2 O − UC12 H22 O11
4.2
The heat capacity
For systems in which no change in composition (chemical reaction) occurs, things are even simpler: to
a very good approximation, the internal energy depends only on the temperature. This means that the
temperature of such a system can serve as a direct measure of its internal energy. The functional relation
between the internal energy and the temperature is given by the heat capacity:
dU
=C
dT
(or ∆U /∆T if you don’t care for calculus!) Heat capacity can be expressed in joules or calories per mole
per degree (molar heat capacity), or in joules or calories per gram per degree; the latter is called the
specific heat capacity or just the specific heat.
If a quantity of heat q crosses the boundaries of a system, the internal energy of the system will change
by an identical amount: ∆U = q. By convention, the signs of q and of ∆U are positive if heat flows
into the system and the internal energy increases; the signs are negative if the system loses heat to the
surroundings.
Remember, however, that a flow of heat into or out of a system is just one way to change the internal
energy. The other way is for the system to do work on the surroundings, or for the surroundings to do
work on the system. The simplest example of work is an expansion of the system against an external
restraining pressure, but there are many other processes that also involve work. If a quantity of work w
is done on the system, its internal energy increases and the sign of w is positive; work done by the system
(and thus on the surroundings) is done at the expense of some internal energy, and its sign is negative.
The distinction between heat and work is quite simple. Heat is the flow of energy into or out of a
system as the result of a temperature difference between the system and surroundings; heat always flows
from higher to lower temperatures. A process that alters the internal energy of a system by any other
means is, by elimination, work.
Chem1 General Chemistry
7
Chemical Energetics
• The First Law of thermodynamics.
4.3
The First Law of thermodynamics.
Since the internal energy of a system is affected by both heat and work, we can write
∆U = q + w
(2)
This expression, a simple sum of two quantities, is a statement of one of the most fundamental laws of
the physical world: it is known as the First Law of thermodynamics. The First Law is also known as the
Law of Conservation of Energy, but in the form given in Eq 2 it is stated more explicitly; it says that
there are two kinds of processes, heat and work, that can lead to a change in the internal energy of a
system. Since both heat and work can be measured and quantified, this is the same as saying that any
change in the energy of a system must result in a corresponding change in the energy of the world outside
the system– in other words, energy cannot be created or destroyed.
The full significance of Eq 2 cannot be grasped without understanding that U is a state function.
This means that a given change in internal energy ∆U can follow an infinite variety of pathways reflected
by all the possible combinations of q and w that can add up to a given value of ∆U .
As a simple example of how this principle can simplify our understanding of change, consider two
identical containers of water initially at the same temperature. We place a flame under one until
its temperature has risen by 1 ◦ C. The water in the other container is stirred vigorously until its
temperature has increased by the same amount. There is now no physical test by which you could
determine which sample of water was warmed by performing work on it, by allowing heat to flow
into it, or by some combination of the two processes. In other words, there is no basis for saying that
one sample of water now contains more “work”, and the other more “heat”. The only thing we can
know for certain is that both samples have undergone identical increases in internal energy, and we
can determine the value of ∆U simply by measuring the increase in the temperature of the water.
5
Molecular interpretation of heat capacity
The heat capacity of a substance is a measure of how sensitively its temperature is affected by a change in
heat content; the greater the heat capacity, the less effect a given change q will have on the temperature.
You will recall that temperature is a measure of the average kinetic energy due to translational motions
of molecules1 . Thus the higher the heat capacity of a substance, the smaller will be the fraction of the
heat it absorbs that gets channeled into translational energy.
Where does the heat go that does not produce more vigorous translational motion? It goes into
vibrational and rotational kinetic energy. These two kinds of motion do not affect the temperature, but
if they are active and can soak up energy, they compete with translation and reduce the effect of a heat
flow q on the temperature.
Vibrational and rotational motions are not possible for monatomic species such as the noble gas
elements, so these substances have the lowest heat capacities. Moreover, as you can see in the leftmost
column of Table 1, their heat capacities are all the same. This reflects the fact that translational motions
are the same for all particles; all such motions can be resolved into three directions in space, each
contributing 12 R to Cv (Cp ’s of gases are always greater by R owing to the work of expansion that occurs
if the gas is heated at constant pressure).
1 Translation
refers to movement of an object as a complete unit. Translational motions of molecules in solids or liquids
are restricted to very short distances, comparable to the dimensions of the molecules themselves, whereas in gases the
molecules travel much greater distances between collisions.
Chem1 General Chemistry
8
Chemical Energetics
• Polyatomic molecules
monatomic
He 20.5
Ne 20.5
Ar 20.5
Kr 20.5
diatomic
CO 29.3
N2
29.3
F2
31.4
Cl2 33.9
triatomic
H2 O 33.5
D2 O 34.3
CO2 37.2
CS2
45.6
Table 1: Molar heat capacities of some gaseous substances at constant pressure (J mol−1 K−1 )
translation
rotation
3/2 R
5.1
1/2 R
vibration
R
Polyatomic molecules
For monatomic molecules such as He or Ar, translation is the only kind of motion allowed. For polyatomic
molecules, two additional kinds of motions become possible. A linear molecule has an axis that defines two
perpendicular directions in which rotations can occur; each represents an additional degree of freedom,
so the two together contribute a total of R to the heat capacity. For a non-linear molecule, rotations
are possible along all three directions of space, so these molecules have a rotational heat capacity of 32 R.
Finally, the individual atoms within a molecule can move relative to each other; this is called vibration.
A molecule consisting of N atoms possesses 3N − 2 degrees of freedom. For mechanical reasons that we
cannot go into here, each vibrational motion contributes R (rather than 12 R to the total heat capacity.
Now we are in a position to understand why more complicated molecules have higher heat capacities.
The total kinetic energy of a molecule is the sum of those due to the various kinds of motions:
KEtotal = KEtrans + KErot + KEvib
When a monatomic gas absorbs heat, all of the energy ends up in translational motion, and thus goes
to increase its temperature. In a polyatomic gas, by contrast, the absorbed energy is partitioned among
the other kinds of motions; since only the translational motions contribute to the temperature, the
temperature rise is smaller, and thus the heat capacity is larger.
There is one very significant complication, however: classical mechanics predicts that the energy
is always partitioned equally between all degrees of freedom. Experiments, however, show that this is
observed only at quite high temperatures. The reason is that these motions are all quantized. This
means that only certain increments of energy are possible for each mode of motion, and unless a certain
minimum amount of energy is availabe, a given mode will not be active at all and will contribute nothing
to the heat capacity.
It turns out that tranlational energy levels are spaced so closely that they these motions are active
almost down to absolute zero, so all gases possess a heat capacity of at least 32 R at all temperatures.
Rotational motions do not get started until intermediate temperatures, typically 300-500K, so within this
rnage heat capacities begin to increase with temperature. Finally, at very high temperaures, vibrations
begin to make a significant contribution to the heat capacity.
The strong intermolecular forces of liquids and many solids allow heat to be channeled into vibrational
motions involving more than a single molecule, further increasing heat capacities. One of the well known
“anomalous” properties of liquid water is its high heat capacity (75 J mol−1 K−1 ) due to intermolecular
Chem1 General Chemistry
9
Chemical Energetics
• 6 Pressure-volume work
4R
H2
30
vibration
3R
20
rotation
2R
C, J KÐ1 molÐ1
10
translation
R
50 100
500 1000
5000
Temperature, K
Figure 1: Heat capacity as a function of temperature for dihydrogen.
hydrogen bonding, which is directly responsible for the moderating influence of large bodies of water on
coastal climates.
Heat capacities of metals
Metallic solids are a rather special case. In metals, the atoms oscillate about their equilibrium positions
in a rather uniform way which is essentially the same for all metals, so they should all have about the
same heat capacity. That this is indeed the case is embodied in the Law of Dulong and Petit. In the 19th
century these workers discovered that the molar heat capacities of all the metallic elements they studied
were around to 25 J mol−1 K−1 , which is close to what classical physics predicts for crystalline metals.
This observation played an important role in characterizing new elements, for it provided a means of
estimating their molar masses by a simple heat capacity measurement.
6
Pressure-volume work
Before we can explore the chemical applications of the First Law, we need to examine the nature of work
more closely. Although work can take many forms, they all have in common the feature that they are
the product of two terms:
type of work
mechanical
gravitational
electrical
intensity term
force
gravitational potential
potential difference
capacity term
change in distance
mass
quantity of charge
formula
f ∆x
mg∆h
Q∆E
The kind of work we are most concerned with in chemistry is connected with changes in the volume
that a system undergoes as the result of some physical or chemical process. This is sometimes called
expansion work or PV-work, and it can most easily be understood by reference to the simplest form of
matter we can deal with, the ideal gas. Fig. 2 shows a quantity of gas confined in a cylinder by means
Chem1 General Chemistry
10
Chemical Energetics
• 6 Pressure-volume work
P
P
external
AAAA
AAAA
force of w eights acting on area
a of piston produces external
p ressure P = f/a
external
AAAA
expansion
initial state:
P1,V1,T1
final state:
P2,V2,T2
Figure 2: Expansion of a gas subjected to an external pressure.
of a moveable piston. Weights placed on top of the piston exert a force f over the cross-section area A,
producing a pressure P = f /A which is exactly countered by the pressure of the gas, so that the piston
remains stationary. Now suppose that we heat the gas slightly; according to Charles’ law, this will cause
the gas to expand, so the piston will be forced upward by a distance ∆x. Since this motion is opposed
by the force f , a quantity of work f ∆x will be done by the gas on the piston. By convention, work done
by the system (in this case, the gas) on the surroundings is negative, so the work is given by
w = −f ∆x
When dealing with a gas, it is more convenient to think in terms of the more relevant quantities
pressure and volume rather than force and distance. We can accomplish this by multiplying the second
term by A/A which of course leaves it unchanged:
w = −f · (∆x) ·
A
A
By grouping the terms differently, but still not changing anything, we obtain
w=−
f
· (∆x) · A
A
Since pressure is force per unit area and the product of the length ∆x and the area has the dimensions
of volume, this expression becomes
w = −P · ∆V
(3)
Eq 3 is a good illustration of how a non-state function like the work depends on the path by which a
given change is carried out. In this case the path is governed by the external pressure P .
Problem Example 1
Find the amount of work done on the surroundings when 1 litre of an ideal gas, initially at a pressure
of 10 atm, is allowed to expand at constant temperature to 10 litres by a) reducing the external
pressure to 1 atm in a single step, b) reducing P first to 5 atm, and then to 1 atm, c) allowing the
gas to expand into an evacuated space so its total volume is 10 litres.
Solution. First, note that ∆V , which is a state function, is the same for each path: V2 =
(10/1) × 1 L = 10 L, so ∆V = 9 L.
For path (a), w = −(1 atm) × 9 L = −9 `-atm.
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Chemical Energetics
• Reversible processes
Compression
Expansion
1 step
Ð15 L-atm
1 step
60 L-atm
2 steps
Ð20 L-atm
2 steps
47.5 L-atm
3 steps
37.5 L-atm
3 steps
Ð22 L-atm
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P
Reversible
expansion
Ð31 L-atm
(maximum work)
Reversible
compression
31L-atm
(minimum work)
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P
V
V
The shaded areas are proportional to the amount of work that must be done on the gas to compress it, or that
is done on the surroundings when the gas expands. In a cyclic process in which the gas is returned to its initial
state, ∆P = ∆V = 0, but the net work is zero only in the unattainable limit of the reversible pathway. All real
cyclic processes are accompanied by a net decrease in the amount of work that can be done in the world.
Figure 3: Work done in reversible and irreversible P V changes.
For path (b), the work is calculated for each stage separately:
w = −(5 atm) × (2 − 1)L − (1 atm) × (10 − 2)L = −13 `-atm
For path (c) would be carried out by removing all weights from the piston in Fig. 2 so that the
gas expands to 10 L against zero external pressure. In this case, w = (0 atm) × 9 L = 0; no work is
done.
6.1
Reversible processes
The preceding example shows clearly how the work, a non-state function, depends on the manner in
which a process is carried out.
Evidently the minimum work that can be obtained from the expansion of a gas is zero. What is the
maximum work? To answer this, notice that more work was done when the process was carried out in
two stages than in one stage, and a simple calculation will show that even more work can be obtained by
increasing the number of stages. In order to extract the maximum work from the process, the expansion
would have to be carried out in an infinite sequence of infinitessimal steps. Each step yields an increment
Chem1 General Chemistry
12
Chemical Energetics
• 7 Heat changes at constant pressure: the enthalpy
of work P dV which can be expressed as (RT /V ) dV and integrated:
Z
V2
w=
V1
RT
P2
dV = RT ln
V
P1
Although such a path (which corresponds to what is called a reversible process) cannot be realized in
practice, it can be approximated as closely as desired.
Even though no real process can take place reversibly (it would take an infinitely long time!), reversible
processes play an essential role in thermodynamics. The main reason for this is that qrev and wrev
are state functions which are important and are easily calculated. Moreover, many real processes
take place sufficiently gradually that they can be treated as approximately reversible processes for
easier calculation.
7
Heat changes at constant pressure: the enthalpy
Most chemical processes are accompanied by changes in the volume of the system, and therefore involve
both heat and work terms. If the process takes place at a constant pressure, then the work is given by
Eq 3 and the change in internal energy will be
∆U = q − P ∆V
(4)
Since most changes that occur in the laboratory, on the surface of the earth, and in organisms are
subjected to a constant pressure of one atmosphere, Eq 4 is the form of the First Law that is of greatest
interest to most chemists.
Problem Example 2
Hydrogen chloride gas readily dissolves in water, releasing 75.3 kJ/mol of heat in the process. If one
mole of HCl at 298 ◦ K and 1 atm pressure occupies 24.5 litres, find the ∆U for the system when one
mole of HCl dissolves in water under these conditions.
Solution: In this process the volume of liquid remains practically unchanged, so ∆V = −24.5 L.
The work done is
w = −P · ∆V = −(1 atm)(−24.5 L) = 24.5 `-atm
Using the conversion factor 1 atm = 101.33 J mol−1 and substituting in Eq 4, we obtain
∆U = q + P ∆V = −75300 J − [(101.33 J/`-atm) × −24.5 `-atm] = −72.82 kJ
Notice that since the system undergoes a decrease in volume, work is done on the system and this
tends to increase the internal energy of the system.
From a practical standpoint we are usually most interested in the quantity of heat q associated with a
change. This is because q can be easily measured, and its sign determines whether a reaction is classified
as endothermic or exothermic. Solving Eq 4 for q, we obtain
q = ∆U + P ∆V
(5)
Notice that the ∆U + P ∆V term in this expression is composed entirely of state functions. This means
that U + P V is itself a state function which we refer to as the enthalpy H. In terms of H, the First Law
becomes
(6)
qP = ∆H
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13
Chemical Energetics
• 8 Thermochemical equations and standard states
Remember that q is not a state function, and its value will depend on the particular pathway by which a
process is carried out. The subscript on q in the above expression means that we are referring to the heat
associated with only those pathways in which the pressure remains constant. The equation says that the
heat associated with any process that is carried out at a constant pressure depends only on the difference
between the enthalpies of the final and initial states of the system.
Because of the importance of constant-pressure processes (usually at 1 atm), enthalpies are much
more commonly used than are internal energies. About the only people who widely use ∆U values are
chemical engineers, who typically work with processes that take place in enclosed containers of constant
volume. For the rest of us, the “heat” of a reaction normally means ∆H.
It is worth noting that the difference between ∆U and ∆H depends on the change in volume. ∆V is
only significant in processes that involve changes in the number of moles of gas. In fact, if the gases are
assumed to behave ideally, ∆V = ∆ng RT where ng refers to the change in the number of moles of gas.
Thermochemistry The heat that flows across the boundaries of a system undergoing a change is a
fundamental property that characterizes the change. It is easily measured, and if the process is carried out
at constant pressure, it can also be predicted from the difference between the enthalpies of the products
and reactants. The quantitative study and measurement of heat and enthalphy changes is known as
thermochemistry.
8
Thermochemical equations and standard states
In order to define the thermochemical properties of a process, it is first necessary to write a thermochemical
equation that defines the system (thus allowing it to be distinghished from the surroundings). This
equation must also define the initial and final states of the system in terms of the states of its various
components.
To take a very simple example, here is the complete thermochemical equation for the vaporization of
water at its normal boiling point:
H2 O(l, 373 K, 1 atm) −→ H2 O(g, 373 K, 1 atm)
∆H = 40.7 kJ mol−1
The quantity 40.7 kJ mol−1 is known as the enthalpy of vaporization (“heat of vaporization”) of liquid
water.
For the neutralization of an acid by a strong base in aqueous solution, we must also specify the
concentrations of the various solutes:
H+ (aq, 1 M , 298K, 1 atm) + OH− (aq, 1 M , 298K, 1 atm)
−→ H2 O(l, 298K, 1 atm)
∆H = −56.9 kJ mol−1
Since most thermochemical equations are written for the standard conditions of 298 ◦ K and 1 atm pressure, we can leave these quantities out if these conditions apply both before and after the reaction. If,
under these same conditions, the substance is in its preferred (most stable) physical state, then the substance is said to be in its standard state. Thus the standard state of water at 1 atm is the solid below
0 ◦ C, and the gas above 100 ◦ C. A thermochemical quantity such as ∆H that refers to reactants and
products in their standard states is denoted by ∆H ◦ .
In the case of dissolved substances, the standard state of a solute is that in which the “effective
concentration”, known as the activity, is unity. This will be close to an actual concentration of 1 M for a
non-ionic substance, but for an ionic solute it will be less than 1 M and will depend on the particular ion
and on the other components of the solution.
Chem1 General Chemistry
14
Chemical Energetics
• 9 Enthalpy of formation
Any thermodynamic quantity such as ∆H or ∆V that is associated with a thermochemical equation
always refers to the number of moles of substances explicitly shown in the equation. Thus for the synthesis
of water we can write
2 H2 (g) + O2 (g) −→ 2 H2 O(l)
or
H2 (g) +
1
2 O2 (g)
−→
1
2 H2 O(l)
∆H ◦ = −572 kJ mol−1
∆H ◦ = −286 kJ mol−1
When a solute dissolves in a liquid, the resulting heat change is known as the enthalpy of solution.
Since this will depend on the amount of solvent as well as that of the solute, the equation must show
these explicitly:
HCl(g) + 50 H2 O(l) −→ HCl · H2 O(soln)
9
Enthalpy of formation
The enthalpy change for a chemical reaction is the difference
∆H = Hproducts − Hreactants
. . . but how can we evaluate the terms on the right side? We take advantage of the fact that energies are
only relative; that is, we can define the zero of energy arbitrarily. If we arbitrarily set the enthalpy and
internal energy of an element to zero, then the enthalpy of a compound such as H2 O is defined by the
thermochemical equation
H2 (g) + 12 O2 (g) −→ H2 O(l)
When this reaction is carried out at constant pressure, the quantity of heat released is found to be
−268 kJ mol−1 , so we can write
qP = ∆H = Hf◦ (H2 O) − [Hf◦ ( 12 O2 ) + Hf◦ (H2 )] = −268 kJ mol−1
Hf◦ (H2 O) = −268 kJ mol−1
Hf◦ (H2 O) is the standard enthalpy of formation of water.
The standard enthalpy of formation of a compound is defined as the heat associated with the formation of one mole of the compound from its elements in their
standard states.
The following examples illustrate some important aspects of the standard enthalpy of formation of
substances.
• The thermochemical equation definining ∆H ◦ must always be written in terms of one mole of the
substance in question:
1
2
N2 (g) +
3
2
H2 (g) −→ NH3 (g)
∆H ◦ = −46.1 kJ mol−1
• A number of elements, of which sulfur and carbon are well-known examples, can exist in more
then one solid crystalline form. The standard heat of formation of a compound is always taken in
reference to the form that is most stable at 25 ◦ C and 1 atm pressure. In the case of carbon, this
is the graphite, rather than the diamond form:
C(graphite) + O2 (g) −→ CO2 (g)
∆H ◦ = −393.5 kJ mol−1
C(diamond) + O2 (g) −→ CO2 (g)
∆H ◦ = −395.8 kJ mol−1
Chem1 General Chemistry
15
Chemical Energetics
• 10 Hess’ law and thermochemical calculations
• In contrast, the product can be in any physical state, even if it is not the most stable one at 25 ◦ C
and 1 atm pressure:
H2 (g) +
1
2
O2 (g) −→ H2 O(l)
∆H ◦ = −285.8 kJ mol−1
H2 (g) +
1
2
O2 (g) −→ H2 O(g)
∆H ◦ = −241.8 kJ mol−1
Notice that the difference between these two ∆H ◦ values is just the heat of vaporization of water.
• Although the formation of most molecules from their elements is an exothermic process, positive
heats of formation are possible:
1
2
N2 (g) + O2 (g) −→ NO2 (g)
∆H ◦ = +33.2 kJ mol−1
A positive heat of formation usually means that the molecule will be unstable (tend to decompose
into its elements) at room temperature. In many cases, however, the rate of this decomposition
is essentially zero, so it is still possible for the substance to exist. In this connection, it is worth
noting that all molecules will become unstable at higher temperatures.
• The thermochemical reactions that define the heats of formation of most compounds cannot actually
take place; for example, the direct synthesis of methane from its elements
C(graphite) + 2 H2 (g) −→ CH4 (g)
cannot be carried out in the laboratory. All this means is that the ∆H ◦ for such a reaction must be
found indirectly from other Hf◦ data, using Hess’s law as explained in the next section.
• The standard enthalpy of formation of gaseous atoms from the element is known as the heat of
atomization. Heats of atomization are always positive, and are important in the calculation of bond
energies.
∆H ◦ = 417 kJ mol−1
Fe(s) −→ Fe(g)
• The standard heat of formation of a dissolved ion such as Cl− (aq) cannot be determined with
respect to the element, since it is impossible to have a solution containing a single kind of ion. For
this reason, ionic enthalpies are expressed on a separate scale on which Hf◦ of the hydrogen ion at
unit activity (1 M effective concentration) is defined as zero. Thus for Ca2+ , Hf◦ = −248 kJ mol−1 ;
this means that the reaction
Ca(s) −→ Ca2+ (aq) + 2 e− (aq)
would release 248 kJ mol−1 more heat than the reaction
1
2 H2 (g)
−→ H+ (aq) + e− (aq)
The standard enthalpy of formation the hydrogen ion is defined as zero.
10
Hess’ law and thermochemical calculations
You probably know that two or more chemical equations can be combined algebraically to give a new
equation. Even before the science of thermodynamics developed in the late nineteenth century, it was
observed by Germaine Hess (1802-1850) that the heats associated with chemical reactions can be combined
in the same way to yield the heat of another reaction. For example, the standard enthalpy changes for
Chem1 General Chemistry
16
Chemical Energetics
• 11 Calorimetry
the oxidation of graphite and diamond can be combined to obtain ∆H ◦ for the transformation between
these two forms of solid carbon– a reaction that cannot be studied experimentally.
C(graphite) + O2 (g) −→ CO2 (g)
∆H ◦ = −393.51 kJ mol−1
C(diamond) + O2 (g) −→ CO2 (g)
∆H ◦ = −395.40 kJ mol−1
Subtraction of the second reaction from the first (i.e., writing the second equation in reverse and adding
it to the first one) yields
C(graphite) −→ C(diamond)
∆H ◦ = 1.89 kJ mol−1
(7)
This principle, known as Hess’ law of independent heat summation is a direct consequence of the
enthalpy being a state function. ∆H ◦ for Eq 7 is determined only by the properties of graphite and
diamond, and is independent of the pathway by which the transformation is carried out. Hess’ law
is one of the most powerful tools of chemistry, for it allows the change in the enthalpy (and in other
thermodynamic functions) of huge numbers of chemical reactions to be predicted from a relatively small
base of experimental data.
Of the various kinds of information in the world’s thermochemical database, the most important are
the standard heats of formation, for these lead directly to the ∆H ◦ values for the chemical transformation
of one substance to another. You will find tables of thermochemical data near the back of most Chemistry
textbooks.
As important as these quantities are, heats of formation are rarely measured directly. The reason is
that most substances cannot be prepared directly from their elements. Instead, Hess’ law is employed
to calculate enthalpies of formation from more accessible data. The most important of these are the
standard enthalpies of combustion. Most elements and compounds combine with oxygen, and many of
these oxidations are highly exothermic, making the measurement of their heats relatively easy.
For example, by combining the heats of combustion of carbon, hydrogen, and methane, we obtain the
standard enthalpy of formation of methane, which as we noted above, cannot be determined directly:
C(graphite) + O2 (g) −→ CO2 (g)
2 H2 (g) + O2 (g) −→ 2 H2 O(l)
CO2 (g) + 2 H2 O(l) −→ CH4 (g) + O2 (g)
C(graphite) + 2 H2 (g) −→ CH4 (g)
−393.7 kJ
−571.5 kJ
+784.3 kJ
–74.4 kJ
Tables of heats of formation, atomization, and combustion can be found in most textbooks.
11
Calorimetry
How are enthalpy changes determined experimentally? First, you must understand that the only thermal
quantity that can be observed directly is the heat q that flows into or out of a reaction vessel, and that
q is numerically equal to ∆H only under the special conditions of constant pressure. Moreover, q is
equal to the standard enthalpy change ∆H ◦ only when the reactants and products are both at the same
temperature, normally 25 ◦ C.
The measurement of q is generally known as calorimetry. A very simple calorimetric determination
of the standard enthalpy of the reaction
H+ (aq) + OH− (aq) −→ H2 O(l)
could be carried out by combining equal volumes of 0.1 M solutions of HCl and of NaOH initially at
25 ◦ C. Since this reaction is exothermic, a quantity of heat q will be released. If the reaction is carried out
Chem1 General Chemistry
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Chemical Energetics
• 11 Calorimetry
ignition wires
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thermometer
stirrer
insulation
bomb
sample holder
water in inner container
Figure 4: Bomb calorimeter for measurement of heat of combustion.
in an insulated vessel such as a Thermos container, this heat will be absorbed by the solution, raising its
temperature. What we actually measure is this temperature rise; if we multiply ∆T by the specific heat
capacity of the solution, which will be close to that of pure water (4.184 J/g-K), we obtain the number
of joules of heat released into each gram of the solution, and q can then be calculated from the mass of
the solution. Since the entire process is carried out at constant pressure, we have ∆H ◦ = q.
For reactions that cannot be carried out in dilute aqueous solution, the reaction vessel is commonly
placed within a larger insulated container of water. During the reaction, heat passes between the inner
and outer containers until their temperatures become identical. Again, the temperature change of the
water is observed, but in this case the value of q cannot be found just from the mass and the specific
heat capacity of the water, for we now have to allow for the absorption of some of the heat by the walls
of the inner vessel. Instead, the calorimeter is “calibrated” by measuring the temperature change that
results from the introduction of a known quantity of heat. The resulting calorimeter constant, expressed
in J K−1 , can be regarded as the “heat capacity of the calorimeter”. The known source of heat is usually
produced by passing an electric current through a resistor within the calorimeter.
Probably the most frequent kind of calorimetric measurement is the determination of heats of combustion. Such measurements present special problems owing to the gaseous nature of O2 and CO2 , and
the necessity of igniting the mixture in order to start the reaction. Since the gases must be confined, heat
of combustion determinations are carried out under constant volume, rather than constant pressure conditions, so what is actually measured in the internal energy change. From the formula of the compound
the change ∆ng in the number of moles of gas is calculated, and this is used to find ∆H:
∆H = ∆U + P ∆V = qV + ∆ng RT
(8)
Since the process takes place at constant volume, the reaction vessel must be constructed to withstand the
high pressure resulting from the combustion process, which amounts to a confined explosion. The vessel
is usually called a “bomb”, and the technique is known as bomb calorimetry. In order to ensure complete
combustion, the bomb is initially charged with pure oxygen above atmospheric pressure. The reaction
is initiated by discharging a capacitor through a thin wire which ignites the mixture. The product of
the capacitance and the voltage gives the quantity of charge, and thus the energy introduced into the
Chem1 General Chemistry
18
Chemical Energetics
• 11 Calorimetry
calorimeter by the spark; this value is subtracted from that calculated from the observed temperature rise
of the calorimeter to yield the heat of the combustion reaction. A more common practice is to determine
the temperature rise produced by combustion, under identical conditions, of a substance whose heat of
combustion is accurately known. This yields a calorimeter constant that is usually expressed in Joules
per degree.
Problem Example 3
A sample of biphenyl (C6 H5 )2 weighing 0.526 g was ignited in a bomb calorimeter initially at 25 ◦ C,
producing a temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid,
C6 H5 COOH, weighing 0.825 g was ignited under identical conditions and produced a temperature
rise of 1.94 K. For benzoic acid, ∆U ◦ comb is known to be −3226 kJ mol−1 . Use this information to
determine the standard enthalpy of combusion of biphenyl.
Solution. The calorimeter constant is given by
(3226 kJ mol−1 )(.825 g)
= 11.1 kJ/K
(1.94 K)(123 g/mol)
The heat released by the combustion of the biphenyl is then
qV =
(11.1 kJ/K)(1.91 K)(154 g/mol)
= −6240 kJ mol−1
.526 g
(The negative sign indicates that heat is released in this process.) From the reaction equation
(C6 H5 )2 (s) +
we have ∆ng = 12 −
19
2
19
O2 (g) −→ 12 CO2 (g) + 5 H2 O(l)
2
= − 52 . Substituting into Eq 8, we have
∆H ◦ = qp = ∆U ◦ −
5
(.008314 J mol−1 K−1 )(298 K) = −6246 kJ mol−1
2
This is the amount of heat that must be lost by the system if reaction takes place at constant pressure
and the temperature is restored to its initial value.
Although calorimetry is simple in principle, its practice is a highly exacting art, especially when
applied to processes which take place slowly or involve very small heat changes.
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Chemical Energetics
• 11 Calorimetry
sample tube
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weighing container
mercury
water
ice-water mixture
sample
The ice calorimeter is an important tool for measuring the heat capacities of liquids and solids, as well as the
heats of certain reactions. This simple yet ingenious apparatus is essentially a device for measuring the change
in volume due to melting of ice. To measure a heat capacity, a warm sample is placed in the inner compartment,
which is surrounded by a mixture of ice and water. The heat withdrawn from the sample as it cools causes some
of this ice to melt. Since ice is less dense than water, the volume of water in the insulated chamber decreases.
This causes an equivalent volume of mercury to be sucked into the inner reservoir from the outside container.
The loss in weight of this container gives the decrease in volume of the water, and thus the mass of ice melted.
This, combined with the heat of fusion of ice, gives the quantity of heat lost by the sample as it cools to 0 ◦ C.
Figure 5: The ice calorimeter
Chem1 General Chemistry
20
Chemical Energetics
• 12 Interpretation of reaction enthalpies
C + O2(g)
0
reference level
110 kJ
Ð100
CO(g)
kJ molÐ1
394kJ
Ð200
284 kJ
Ð300
CO2(g)
Ð400
Figure 6: Enthalpy diagram for the carbon-oxygen system
12
12.1
Interpretation of reaction enthalpies
Enthalpy diagrams
Comparison and interpretation of enthalpy changes is materially aided by a graphical construction in
which the relative enthalpies of various substances are represented by horizontal lines on a vertical energy
scale. The zero of the scale can be placed anywere, since energies are always arbitrary anyway. An
enthalpy diagram for carbon and oxygen and its two stable oxides is shown in Fig. 6. In this diagram,
we follow the convention of assigning zero enthalpy to graphite, the most stable state of the elements at
298 ◦ K and 1 atm pressure. The labeled arrows show the changes in enthalpy associated with the various
reactions this system can undergo. Notice how Hess’ law is implicit in this diagram; we can calculate the
enthalpy change for the combustion of carbon monoxide to carbon dioxide, for example, by subtraction
of the appropriate arrow lengths without writing out the thermochemical equations in a formal way.
Enthalpy diagrams are especially useful for comparing groups of substances having some common
feature. In Fig. 7, the molar enthalpies of species relating to two hydrogen halides are shown with
respect to those of the elements. From this diagram we can see at a glance that the formation of HF
from the elements is considerably more exothermic than the corresponding formation of HCl. The upper
part of this diagram shows the gaseous atoms at positive enthalpies with respect to the elements. The
endothermic processes in which the H2 and the dihalogen are dissociated into atoms can be imagined as
taking place in two stages, also shown. From the enthalpy change associated with the dissociation of H2
(218 kJ mol−1 ), the dissociation enthalpies of F2 and Cl2 can be calculated and placed on the diagram.
12.2
Bond enthalpies and bond energies
The energy change associated with the reaction
HI(g) −→ H(g) + I(g)
is the heat of dissociation of the HI molecule; it is also the bond energy of the hydrogen-iodine bond in
◦
◦
this molecule. Under the usual standard conditions, it would be expressed as HHI(298)
or UHI(298)
; in
this case they differ from each other by ∆P V = RT . Since this reaction cannot be observed under these
conditions, the H–I bond enthalpy is calculated from the appropriate standard enthalpies of formation:
Chem1 General Chemistry
21
Chemical Energetics
• Energy content of fuels
300
Cl(g) + H(g)
F(g) + H(g)
+121
+79
H(g)
200
100
kJ
Ð564
molÐ1
Ð431
+218
0
Ð92
Ð100
Ð269
H2, Cl2, F2
HCl(g)
Ð200
HF(g)
Ð300
Figure 7: Enthalpy diagram comparing HF and HCl
1
2 H2 (g) −→ H(g)
1
2 I2 (g) −→ I(g)
1
1
2 H2 (g) + 2 I2 (g)
+218 kJ
+107 kJ
−→ HI(g)
HI(g) −→ H(g) + I(g)
+26 kJ
299 kJ
Bond energies and enthalpies are important properties of chemical bonds, and it is very important
to be able to estimate their values from other thermochemical data. The total bond enthalpy of a more
complex molecule such as ethane can be found from
C2 H6 (g) −→ 2 C(graphite) + 3 H2 (g)
3 H2 (g) −→ 6 H(g)
2 C(graphite) −→ 2 C(g)
C2 H6 (g) −→ 2 C(g) + 6 H(g)
84.7 kJ
1308 kJ
1430 kJ
2823 kJ
The total bond energy of a molecule can be thought of as the sum of the energies of the individual
bonds. This principle, known as Pauling’s Rule, is only an approximation, because the energy of a given
type of bond is not really a constant, but depends somewhat on the particular chemical environment of
the two atoms. In other words, all we can really talk about is the average C–Cl bond energy, for example,
the average being taken over a representative sample of compounds containing this type of bond.
Despite the lack of strict additivity of bond energies, Pauling’s Rule is extremely useful because it
allows one to estimate the heats of formation of compounds that have not been studied, or have not even
been prepared. Thus in the foregoing example, if we know the enthalpies of the C–C and C–H bonds
from other data, we could estimate the total bond enthalpy of ethane, and then work back to get some
other quantity of interest, such as ethane’s enthalpy of formation.
12.3
Energy content of fuels
The enthalpy of combustion is obviously an important criterion for a substance’s suitability as a fuel,
but it is not the only one; a useful fuel must also be easily ignited, and in the case of a fuel intended
Chem1 General Chemistry
22
Chemical Energetics
• Energy content of foods
for self-powered vehicles, its energy density (kJ/m3 ) must be reasonably large. Thus substances such as
methane and propane which are gases at 1 atm must be stored as pressurized liquids for transportation
and portable applications.
Owing to its low molar mass and high heat of combustion, dihydrogen possesses an extraordinarily
high energy density, and would be an ideal fuel if its critical temperature (33 ◦ K, the temperature above
which it cannot exist as a liquid) were not so low. The potential benefits of using hydrogen as a fuel
have motivated a great deal of research into other methods of getting a large amount of H2 into a small
volume of space. Simply compressing the gas to a very high pressure is not practical because the weight
of the heavy-walled steel vessel required to withstand the pressure would increase the effective weight of
the fuel to an unacceptably large value. One scheme that has shown some promise exploits the ability
of H2 to “dissolve” in certain transition metals. The hydrogen can be recovered from the resulting solid
solution (actually a loosely-bound compound) by heating.
12.4
Energy content of foods
What, exactly, is meant by the statement that a particular food “contains 1200 calories” per serving?
This simply refers to the standard enthalpy of combustion of the foodstuff, as measured in a bomb
calorimeter. Note, however, that in nutritional usage, the calorie is really a kilocalorie (sometimes called
“large calories”), that is, 4184 J. Although this unit is still employed in the popular literature, the SI
unit is now commonly used in the scientific and clinical literature, in which energy contents of foods are
usually quoted in megaJoules (MJ) or kJ per 1-, 100-, or 1000 grams.
Although the mechanisms of oxidation of a carbohydrate such as glucose to carbon dioxide and water
in a bomb calorimeter and in the body are complex and completely different, the net reaction involves the
same initial and final states, and ∆H, being a state function, must be the same for any possible pathway.
C6 H12 O6 + 6 O2 −→ 6 CO2 + 6 H2 O
∆H ◦ = 20.8 kJ mol−1
(9)
Glucose is a sugar, a breakdown product of starch, and is the most important energy source at the cellular
level; fats, proteins, and other sugars are readily converted to glucose. By writing balanced equations for
the combustion of sugars, fats, and proteins, a comparison of their relative energy contents can be made.
The stoichiometry of each reaction gives the amounts of oxygen taken up and CO2 released when a given
amount of each kind of food is oxidized; these gas volumes are often taken as indirect measures of energy
consumption and metabolic activity.
For some components of food, oxidation may not always be complete in the body, so the energy that
is actually available will be smaller than that given by the heat of combustion. Mammals, for example,
are unable to break down cellulose (a polymer of sugar) at all; animals that derive a major part of their
nutrition from grass and leaves must rely on the action of symbiotic bacteria which colonize their digestive
tracts. The amount of energy available from a food can be found by measuring the heat of combustion of
the waste products excreted by an organism that has been restricted to a controlled diet, and subtracting
this from the heat of combustion of the food.
The amount of energy a person requires depends on the age, sex, surface area of the body, and of
course on the amount of physical activity. The rate at which energy is expended is expressed in watts:
1 W = 1 J/s. For humans, this value varies from about 200–800 W. This translates into daily food intakes
having energy equivalents of about 10–15 MJ for most working adults. In order to just maintain weight
in the absence of any physical activity, about 6 MJ per day is required.
Chem1 General Chemistry
23
Chemical Energetics
• 13 Variation of the enthalpy with temperature
type of food
Protein
meat
egg
Fat
butter
animal fat
Carbohydrate
starch
glucose
ethanol
∆Hcomb , kJ/g
percent availability
22.4
23.4
92
38.2
39.2
95
17.2
15.5
29.7
99
100
Table 2: Energy content and availability of the major food components
13
Variation of the enthalpy with temperature
The enthalpy of a system increases with the temperature. This is mainly due to the similar dependence
of the internal energy on the temperature. The defining relation
∆H = ∆U + P ∆V
tells us that the enthalpy is dominated by the internal energy, subject to a slight correction for pressurevolume work. Heating a substance causes it to expand, making ∆V positive and causing the enthalpy to
increase slightly more than the internal energy. Physically, what this means is that if the temperature
is increased while holding the pressure constant, some extra energy must be expended to push back the
external atmosphere while the system expands. The difference between the dependence of U and H on
temperature is only really significant for gases, since the coefficients of thermal expansion of liquids and
solids are very small.
You will recall that the proportionality constant that relates the internal energy to the temperature
is called the heat capacity. Since U and H increase with temperature by slightly different amounts, we
can define two heat capacities, one for processes which take place at constant volume
Cv =
∆U
∆T
(10)
Cp =
∆H
∆T
(11)
and one for processes at constant pressure
A plot of the enthalpy of a system as a function of its temperature is called an enthalpy diagram. The
slope of the line is given by Cp . The enthalpy diagram of a pure substance such as water shows that this
plot is not uniform, but is interrupted by sharp breaks at which the value of Cp is apparently infinite,
meaning that the substance can absorb or lose heat without undergoing any change in temperature at
all. This, of course, is exactly what happens when a substance undergoes a phase change; you already
know that the temperature the water boiling in a kettle can never exceed 100 ◦ C until all the liquid has
evaporated, at which point the temperature of the steam will rise as more heat flows into the system.
Fusion and boiling are not the only kinds of phase changes that matter can undergo. Most solids can
exist in different structural modifications at different temperatures, and the resulting solid-solid phase
changes produce similar discontinuities in the heat capacity. Enthalpy diagrams are easily determined
Chem1 General Chemistry
24
Chemical Energetics
• 13 Variation of the enthalpy with temperature
by following the temperature of a sample as heat flows into our out of the substance at a constant rate.
The resulting diagrams are widely used in materials science and forensic investigations to characterize
complex and unknown substances.
Chem1 General Chemistry
25
Chemical Energetics
• 13 Variation of the enthalpy with temperature
100
CCl4
80
DHvaporization
60
H, kJ molÐ1
40
D Hfusion
DHphase transition
20
100
200
300
400
T, ¡K
Figure 8: Enthalpy of carbon tetrachloride as a function of temperature at 1 atm
Chem1 General Chemistry
26
Chemical Energetics
• 14 The direction of spontaneous change
Part 2: The approach to equilibrium
The greater part of what we call chemistry is concerned with the different kinds of reactions that
substances can undergo. The statement that “hydrogen fluoride is a stable molecule” is really a way of
saying that the reaction HF −→ 12 H2 + 12 F2 has an overwhelming tendency to occur in the reverse
direction, and a negligible tendency to occur in the forward direction. More generally, we can predict
how the composition of an arbitrary mixture of H2 , F2 , and HF will tend to change by comparing the
values of the equilibrium constant K and the equilibrium quotient Q; your study of equilibrium, you will
recall that if Q/K > 1 the reaction will proceed to the left, whereas if Q/K < 1 it will proceed to the
right. In either case, the system will undergo a change in composition until it reaches the equilibrium
state where Q = K.
Clearly, the value of K is the crucial quantity that characterizes a chemical reaction, but what factors
affect its value? In particular, is there any way that we can predict the value of the equilibrium constant of
a reaction solely from information about the products and reactants themselves, without any knowledge
at all about the mechanism or other details of the reaction? The answer is yes, and this turns out to be
the central purpose of chemical thermodynamics:
The purpose of thermodynamics is to predict the equilibrium state
of a system from the properties of its components.
Don’t let the significance of this pass you by; it means that we can say with complete certainty whether
or not a given change is possible, and if it is possible, to what extent it will occur, without the need to
study the particular reaction in question. To a large extent, this is what makes chemistry a science, rather
than a mere cataloging of facts.
14
The direction of spontaneous change
Drop a teabag into a pot of hot water, and you will see the tea diffuse into the water until it is uniformly
distributed throughout the water. What you will never see is the reverse of this process, in which the tea
would be re-absorbed by the teabag. The making of tea, like all changes that take place in the world,
possesses a “natural” direction. Here are a few other examples:
• A stack of one hundred coins is thrown onto the floor, and the numbers that land “heads up” and
“tails up” are noted.
• One mole of gas, initially at 300 ◦ K and 2 atm pressure, is allowed to expand to double its volume,
keeping the temperature constant.
• A raindrop, supercooled to −1 ◦ C at 1 atm, changes into ice (“freezing rain”) as it strikes the
windshield of your car.
• One mole of N2 O4 gas is placed in a container and allowed to come to equilibrium with NO2 .
All of these changes take place spontaneously, meaning that once they are started, they will proceed
to the finish without any outside intervention. It would be inconceiveable that any of these changes could
occur in the reverse direction (that is, be undone) without changing the conditions or actively disturbing
the system in some way.
What determines the direction in which spontaneous change will occur? Your first thought might be
that the direction of change will be the one that leads to a lower energy, but this is not generally correct.
As a matter of fact, in all of the examples cited above, the energy of the system in its final state is either
the same or greater that that of the initial state.
Chem1 General Chemistry
1
Chemical Energetics
• A closer look at disorder
This might seem strange to you; after all, you know that a simple mechanical system will always
change in a way that leads to a lower potential energy, so why should a more complex system made
up of many molecules not do the same? When you drop a book onto the floor there is no question
about what will happen, because the initial potential energy of the book is entirely converted into
kinetic energy of random thermal motion (heat) at the moment of impact, and this has no effect on
the book other than to warm it up a bit.
Now imagine that instead of a heavy book, we observe a single molecule as it collides with a
surface. Because its mass is very small, its kinetic energy is also small and comparable to that of
the molecules making up the surface; on colliding with the surface it could just as easily gain kinetic
energy as lose it. The average gas molecule in a container will simply bounce off the wall with about
the same kinetic energy as it has before the collision, but some will acquire more, and others will
suffer a net loss.
The short (and admittedly incomplete) answer to the above question is that in an exothermic
change, energy is never lost, but simply exchanged between system and surroundings, and when the
packets of energy are of sufficiently small magnitude (as in a collection of molecules that are free to
act individually), this exchange can operate in both directions.
After any of the processes listed on the preceding page is completed, the system may have gained or
lost energy, but the total energy of the system and the surroundings (and thus, of the world) is unchanged.
However, there is something about the world that has changed, and this is its degree of randomness, or
disorder.
This can be seen most easily in the case of the coins. The process
100 heads −→ 50 heads + 50 tails
which expresses the state of maximum disorder of 100 coins is a far more likely outcome of random tosses
than
50 heads + 50 tails −→ 100 heads
which, if not impossible, is so unlikely that we would have grave doubts about the fairness of the coins if
we were to witness such an outcome.
Similarly, the molecules of a gas can occupy a larger number of possible positions in space if the
volume is larger, so the expansion of a gas is similarly accompanied by an increase in randomness, or
disorder. And looking at the dissociation reaction of dinitrogen tetroxide
∆H ◦ = +13.9 kJ
N2 O4 (g) −→ 2 NO2 (g)
it is apparent that the number of molecules increases, and this also leads to an increase in disorder.
But what about the freezing of water? It might appear that this process leads to a decrease in disorder,
since ice is certainly a more ordered state of matter than is liquid water. Furthermore, if you think about
it, a sample of pure gaseous NO2 will tend to undergo the reverse of the above dissociation reaction,
since the equilibrium composition must be independent of the direction in which it is approached. And
if we increase the pressure on a gas, it will spontaneously contract, leading to a decrease rather than
an increase in disorder of the gas. Clearly, the relation of randomness to similar processes and to most
chemical reactions, requires some further refinement of our thinking.
14.1
A closer look at disorder
First, how can we express disorder quantitatively? From the example of the coins, you can probably see
that simple statistics plays a role: the probability of obtaining three heads and seven tails after tossing
ten coins is just the ratio of the number of ways that ten different coins can be arranged in this way, to
the number of all possible arrangements of ten coins.
Chem1 General Chemistry
2
Chemical Energetics
• A closer look at disorder
macrostate
0 heads
1 head
2 heads
3 heads
4 heads
ways
1
4
6
4
1
probability
1/16
4/16 = 1/4
6/16 = 3/8
4/16 = 1/4
1/16
microscopic states
↓↓↓↓
↑↓↓↓ ↓↑↓↓ ↓↓↑↓
↑↑↓↓ ↑↓↑↓ ↑↓↓↑
↑↑↑↓ ↑↓↑↑ ↑↑↓↑
↑↑↑↑
↓↓↓↑
↓↑↑↓
↓↑↑↑
↓↓↑↑
↓↑↓↑
The probability of a given macrostate is proportional to the number of ways of realizing that state,
and thus to the number of microstates that correspond to the macrostate. For the general case of n
coins being in a macrostate with h heads-up, this number (in the second column of the table) is given
by n!/(h!)2 . The probability of a given macrostate is just the ratio of the number of corresponding
microstates to the total number of microstates. This assumes, of course, that all microstates are
equally probable.
Table 3: Macroscopic and microscopic states of a set of four coins
Using the language of molecular statistics, we say that a collection of coins in which 30% of its
members are heads-up constitutes a macroscopic state of the system. Since we don’t care which coins are
heads-up, there are clearly various configurations of coins which can result in this “macrostate”. Each of
these configurations specifies a microscopic state of the system.
The greater the number of microstates that correspond to a given macrostate, the greater the probability of that macrostate. To see what this means, study Table 3, which shows the possible outcomes of
a toss of four coins.
In applying these same ideas to molecules, we need to distinguish between two kinds of randomness,
or disorder. The examples we have just given illustrate what might be called positional disorder. We can
quantify positional randomness in a monatomic gas by dividing a container up into imaginary compartments, and counting the number of different ways that the gas molecules could be distributed amongst
these different compartments. The greater the volume occupied by the gas, the more configurations
(microstates) that are consistent with this volume. If the number of molecules is large, the number of
microstates in which the gas occupies the entire volume of the container is much, much greater than the
number that would correspond to a crowding of the molecules into a volume that is smaller by even a tiny
amount; any of the macrostates corresponding to smaller volumes are so improbable statistically that we
can call them “impossible”.
Energy disorder. There is another kind of randomness that is generally more important than
positional randomness. This relates to the number of ways in which energy can be distributed in a
collection of molecules. At the atomic and molecular level, all energy is quantized; each particle possesses
discrete states of kinetic energy and is able to accept thermal energy only in packets whose values
correspond to the energies of one or more of these states.
Suppose that we have three molecules and enough kinetic energy to excite three energy states (Fig. 9).
We can give all the kinetic energy to one molecule, leaving the others with nothing, we can give two units
to one molecule and one unit to another, or we share out the energy equally and give one unit to each
molecule. All told, there are ten possible ways of distributing three units of energy among three identical
molecules.
However, if the allowed energy levels of the molecule are closer together so that the same amount of
energy can be accepted in smaller packets, then the number of possible distributions is greatly increased,
and so is the “energy randomness” of the system. The spacing of these energy states becomes closer as
the mass and number of bonds in the molecule increases, so we can generally say that the more complex
the molecule, the greater the amount of energy-disorder it has.
Chem1 General Chemistry
3
Chemical Energetics
• A closer look at disorder
5
4
3
2
1
0
3
2
available
energy
1
0
The quantity of energy indicated by the vertical
arrow can be distributed in three ways in a collection
of molecules whose allowed energy levels are
spaced as shown here.
quantized
energy state
( denotes
excited)
In molecules with more widelyspaced energy levels, there are
fewer ways of distributing the same
total energy.
3
2
1
molecul
e
microstat
e
0
a b c
a b c
a b c
a b c
a b c
a b c
a b c
a b c
a b c
1
10
2
3
4
5
6
7
8
9
If the allowed energy levels are very close (more likely in molecules having stronger bonds and heavier
atoms), then the available energy can be distributed equally among them, greatlly increasing the
energy randomness and thus the probability of the system. This example shows how one macroscopic
state (3 units of energy distributed amongst three molecules) can be realized in ten different ways
(different microscopic states).
Figure 9: Energy randomness at the molecular level
Chem1 General Chemistry
4
Chemical Energetics
a b c
• 15 The entropy
Three kinds of states associated with kinetic energy are possible. A monatomic molecule such as
helium can absorb energy only by moving faster; the resulting energy is known as translational
kinetic energy. You will recall from the kinetic theory of gases that the average translational kinetic
energy of a gas is related to its temperature. In polyatomic molecules, energy can also be taken up
in the form of vibrational and rotational motions; the number of such motions, and hence the degree
of randomness in which energy can be distributed in a collection of molecules, increases rapidly as
the molecules become more complicated in structure. Because only a portion of the energy absorbed
by such molecules goes into increasing the translational motion (and thus the temperature), larger
molecules have larger heat capacities; that is, their temperatures are less affected by absorption or
loss of a given quantity of heat.
15
The entropy
How can we relate disorder to measurable thermodynamic quantities? It turns out that the quantity
of importance is the ratio q/T . If you stop to think about the meaning of this ratio, it makes sense.
Absorption of a quantity of heat q will always lead to in increase in disorder by providing more packets of
energy that can be distributed over the various modes of motion characteristic of the particular molecule.
To understand why we have to divide the quantity of heat by the temperature, consider the effect of very
large and very small values of T in the denominator. If a substance is initially at a very low temperature,
it possesses relatively little kinetic energy and what there is cannot be divided up in very many ways. If it
now absorbs a quantity of heat, there will be a relatively large increase in energy randomness. If, however,
the temperature is initially large, there will already be a great amount of disorder and absorption of the
same increase in energy as before will have relatively little effect on the randomness of the system.
The only problem with using q/T as a measure of the change in randomness is that q is not a
thermodynamic state function; that is, its value is dependent on the pathway, or manner, by which a
process is carried out. This means, of course, that the quotient q/T cannot be a state function either, so
we are unable to use it to get differences between reactants and products as we do with the other state
functions. The way around this is to restrict our consideration to a special class of pathways that are
designated reversible. A change is said to occur reversibly when it can be undone without any exchange
of heat or work with the surroundings. In effect, this means that the process is carried out in a series
of infinitesimal steps. Of course, such a process would take infinitely long to occur, so thermodynamic
reversibility is an idealization that is never achieved in real processes, except when the system is already
at equilibrium, in which case no change will occur anyway!
Why is the concept of a reversible process so important? The answer can be seen by recalling that
the change in the internal energy that characterizes any process can be distributed in an infinity of ways
between heat flow across the boundaries of the system and work done on or by the system, as expressed
by the First Law ∆U = q + w. Each combination of q and w represents a different pathway between the
initial and final states. It can be shown that as a process, such as the expansion of a gas, is carried out
in successively longer series of smaller steps, the absolute value of q approaches a minimum, and that of
w approaches a maximum that is characteristic of the particular process. Thus when a process is carried
out reversibly, the w-term in the First Law expression has its greatest possible value, and the q-term is
at its smallest. These special quantities wmax and qmin (which we denote as qrev or “q reversible”) have
unique values 12 for any given process and are therefore state functions.
Since qrev is a state function, so is qrev /T . This quotient is one of the most important quantities
in thermodynamics, because it is expresses the change in disorder that accompanies a process. Note
carefully that the change in disorder is always related to the limiting value qrev /T even when the process
is carried out irreversibly and the actual value of q/T is different.
Being a state function, qrev /T deserves a name and symbol of its own; it is called the entropy,
Chem1 General Chemistry
5
Chemical Energetics
• The entropy of the world always increases
designated by S. Since qrev /T describes a change in state, we write the definition
∆S = qrev /T
15.1
(12)
The entropy of the world always increases
All natural processes that involve more than a few particles2 occur in a direction that leads to a more
random distribution of matter and energy; that is, they lead to an increase in the entropy of the world.
Once initiated, this process will tend to continue until any further change would produce a decrease in
disorder, and thus in the entropy. At this point, no further change will occur, and the system will be in
its equilibrium state.
If this is the case, how can we explain the occurrence of processes that seem at first glance to result
in a decrease in disorder? Examples are the freezing of a liquid, the formation of a precipitate, and the
growth of an organism.
It is important to understand that the criterion for spontaneous change is the entropy change of the
system and the surroundings– that is, of the universe:
∆S total = ∆S system + ∆S surroundings
The only way the entropy of the surroundings can be affected is by exchange of heat with the system;
if the system absorbs a quantity of heat q, then ∆S surroundings = −q/T . Note that it does not matter
whether or not the change in the system occurs reversibly or irreversibly. As mentioned previously, it is
always possible to define an alternative pathway in which the same amount of heat is exchanged with the
surroundings reversibly, yielding the same value of ∆S.
Examples of processes for which q = 0 are the free expansion of an ideal gas into a vacuum, and the
mixing of two ideal gases. In practice, almost all processes involving mixing and diffusion can be regarded
as driven exclusively by the entropy increase of the system. Most processes involving chemical and phase
changes involve the exchange of heat with the surroundings, so their tendency to occur cannot always
be predicted by focussing attention on the randomness of the system alone; further, this tendency will
always depend on the temperature.
As a specific example, let us consider the freezing of water. We know that liquid water will
spontaneously change into ice when the temperature drops below 0 ◦ C at 1 atm pressure. Since the
water molecules are far more ordered in ice than they are in the liquid, the entropy of the water (the
system here) is decreasing. The amount of decrease is found by dividing the heat of fusion of ice by
the temperature for the reversible pathway, which occurs at the normal freezing point:
∆S system =
−6000 J
= −21.978 J K−1 mol−1
273 K
If the process actually occurs at 0 ◦ K , then the heat of fusion is transferred to the surroundings at
the same temperature, and the entropy of the surroundings increases by
∆S surroundings =
6000 J
= +21.978 J K−1 mol−1
273 K
2 If you throw only two coins, the probability that both will land heads up 25%. This should warn you that arguments
based on randomness are basically statistical arguments, and are only valid for large quantities of things. Given the numbers
of molecules we normally deal with in Chemistry, this rarely presents a problem.
The point being made here is that the laws of probability have meaningful application only to systems made up of large
numbers of independent actors. For example, you can rest comfortably in certainty that the spontaneous movement of half
the molecules of the air to one side of the room you now occupy will not occur, even though the molecules are moving
randomly. On the other hand, if we consider a box whose dimensions are only a few molecular diameters, then we would
expect that the random displacement of the small number of particles it contains to one side of the box would occur quite
frequently. This is, in fact, the cause of the blueness of the sky: random fluctuations in the air density over tiny volumes of
space whose dimensions are comparable with the wavelength of light results in selective scattering of the shorter wavelengths,
so that blue light is scattered out, leaving the red light for the enjoyment of sunset-watchers to the east.
Chem1 General Chemistry
6
Chemical Energetics
• Absolute entropies
so that ∆S total = 0. Under these conditions, the ice and water are in equilibrium, and no net change
(no additional freezing) will occur.
Suppose now that the water is supercooled to −1 ◦ C before it freezes. The entropy change of the
water still corresponds to the reversible value qrev /T = −6000 J/273 K, or −21.978 J K−1 mol−1 .
The entropy change of the surroundings, however, is now given by
∆S surroundings =
6000 J
= +22.059 J K−1 mol−1
272 K
(we can use 272 ◦ K here because it is possible to formulate a reversible pathway by which heat can
be transferred to the surroundings at any temperature). The entropy change of the world is now
∆S total == −21.978 J K−1 mol−1 + 22.059 J K−1 mol−1 = .081 J K−1 mol−1
Freezing of supercooled water is of course an irreversible process (once it starts, it cannot be stopped
except by raising the temperature), and the positive value of ∆S total tells us that this process will occur
spontaneously at temperatures below 273 ◦ K. Under these conditions, the process is driven by the entropy
increase of the surroundings resulting from heat flow into the surroundings from the system.
In any spontaneous macroscopic change, the entropy of the world increases. If a process is endothermic
or exothermic, heat is exchanged with the surroundings, and we have to consider the entropy change of
the surroundings as well as that of the system in order to predict what the direction of change will be.
Failure to take the surroundings into account is what leads to the apparent paradox that freezing of a
liquid, compression of a gas, and growth of an organism are all processes in which entropy decreases. It
is only the entropy of the system that undergoes a decrease when these processes occur.
15.2
Absolute entropies
Energy values, as you know, are all relative, and must be defined on a scale that is completely arbitrary;
there is no such thing as the absolute energy of a substance, so we can arbitrarily define the enthalpy or
internal energy of an element in its most stable form at 298 ◦ K and 1 atm pressure as zero.
The same is not true of the entropy; since entropy is a measure of disorder, we would expect a
perfectly-ordered system to have zero entropy. But what do we mean by “perfect order” as applied to a
chemical substance? Recalling that disorder has both positional and energetic aspects, it is easy to see
that the most highly ordered state of a substance must be the solid, and it must be at a low temperature
so that all particles are in their lowest energy states.
This reasoning is supported by experiments which show that differences between the entropies of
different substances tend to disappear as the temperature is reduced; all entropies converge to zero as the
temperature approaches absolute zero. This principle is the basis of the Third law of thermodynamics,
which states that the entropy of a perfect crystal at 0 ◦ K is zero.
The absolute entropy of a substance at any temperature above 0 ◦ K must be determined by calculating
the increments of heat q required to bring the substance from 0 ◦ K to the temperature of interest, and
then summing the ratios q/T . Two kinds of experimental measurements are needed:
• The enthalpies associated with any phase changes the substance may undergo within the temperature range of interest. In addition to heats of fusion and vaporization, these often include solid-solid
phase changes that must be found by careful observation. The entropy increase associated with
melting, for example, is just ∆H fusion /Tm .
• The heat capacity Cp of a phase expresses the quantity of heat required to change the temperature by
a small amount ∆T , or more precisely, by an infinitessimal amount dT . Thus the entropy increase
brought about by warming a substance over a range of temperatures that does not encompass a
Chem1 General Chemistry
7
Chemical Energetics
• Entropies of substances
50
gas
Entrop y, calÐ1 molÐ1
40
DS vap orization
30
20
d
liqui
10
DS fusion
solid
0
0
100
200
300
temp erature, ¡K
400
Figure 10: Entropy of water as a function of temperature
phase transition is given by the sum of the quantities Cp dT /T for each increment of temperature
dT . This is of course just the integral
Z
S0◦ K → T ◦ K =
T ◦K
0 ◦K
Cp
dT
T
Because the heat capacity is itself slightly temperature dependent, the most precise determinations
of absolute entropies require that the functional dependence of Cp on T be used in the above integral
in place of a constant Cp . For rough determinations, one can simply plot the enthalpy increment
Cp ∆T as a function of temperature and measure the area under the curve.
15.3
Entropies of substances
The standard entropy of a substance is its entropy at 1 atm pressure. The values found in tables are
normally those for 298 ◦ K , and are expressed in J K−1 mol−1 or, more commonly, on a molar basis as
J mol−1 K−1 . Remember that since entropies are expressed on an absolute scale, the standard entropies
of the elements are not zero, and the values given for various substances are absolute entropies, not
entropies of formation.
Examination of some typical entropy values such as those listed in Table 4 will give you a feeling for
the way that the entropy of a substance is affected by its chemical nature. In general, we can expect that
• Gases have higher entropies than liquids, and these in turn have higher entropies than solids. This
can be clearly seen by comparing the entropies of substances of similar molar mass in the three
sections of Table 4. An especially interesting comparison can be made by measuring the temperature
dependence of the entropies of water vapor and ice, and then extrapolating the values to 298 ◦ K
so that the effect of temperature is eliminated. The resulting values are 41 (ice), 70 (liquid), and
186 J K−1 mol−1 (vapor).
Chem1 General Chemistry
8
Chemical Energetics
• Effect of temperature, volume, and concentration on the entropy
C(diamond)
C(graphite)
Si
Fe
S(rhombic)
Cu
Ag
Hg
2.5
5.7
18.9
27.1
32.0
33.3
43.7
76.0
CaO
CuO
ZnO
ZnS
NaF
NaCl
39.7
43.5
43.9
57.7
58.6
58.6
Ca(OH)2
CaCO3
CuSO4
BaSO4
PbSO4
72.8
92.9
113
132
147
Br2
152
He
Ne
Ar
Kr
Xe
126
146
155
164
170
H2
N2
CO
F2
O2
Cl2
n-CH3 CH2 CH2 CH2 CH3
131
192
197
203
205
223
349
H2 O
CH3 OH
CH3 CH2 OH
n-CH3 CH2 CH2 CH2 CH2 CH3
CH4
H2 O(g)
CO2
CH3 CH3
n-CH3 CH2 CH3
n-CH3 CH2 CH2 CH3
70.0
127
177
296
186
187
213
229
270
310
Table 4: Standard entropies of selected substances at 298 ◦ K (J K−1 mol−1 )
• harder substance have smaller entropies than softer substances. Compare the entropies of the
diamond and graphite forms of carbon in the table, or the values for iron (27.1), aluminum (28.5),
and sodium (51.0 J K−1 mol−1 ).
• Complex substances have higher entropies than simpler substances of the same general character. Note, for example, the trend in the entropies of the aliphatic hydrocarbons methane through
pentane.
Inspection of Table 4 shows that the entropy of the noble gas elements increases with atomic weight. In what
way is a heavier atom more “disordered” than a lighter one? The answer lies in the fact that the kinetic energies
of individual atoms are subject to the laws of quantum mechanics, and that only certain discrete energy levels
associated with translational, vibrational, and rotational motions are allowed. Further, the spacing between the
allowed energy levels decreases as the particles become heavier. (This is consistent with the fact that we do not
observe quantization of kinetic energies in macroscopic objects; the energies are still quantized, but the levels are
so closely spaced that they appear to be continuous). Thus neon, being heavier then helium, possesses a greater
number of allowed levels of translational kinetic energy within a given range of energies than does helium. As
a consequence, the available thermal energy, given by RT per mole, can be distributed over a larger number of
energy levels in neon, thus contributing to “energy disorder”.
15.4
Effect of temperature, volume, and concentration on the entropy
As discussed previously (see Fig. 10), the entropy of a substance always increases with temperature. This
is due mainly to the increased number of ways that the thermal energy can be distributed amongst the
allowed energy levels as this energy is raised. The rate of increase dS/dT is just the ratio of the heat
capacity to the temperature C/T . When integrated over a range of temperatures, this yields (for an ideal
gas confined to a fixed volume)
µ ¶
T2
∆S = Cv ln
T1
Chem1 General Chemistry
9
Chemical Energetics
• The second law of thermodynamics
In general, a larger volume also leads to increased entropy, in this case reflecting increased positional
disorder. For an ideal gas that expands at a constant temperature (meaning that it absorbs heat from
the surroundings to compensate for the work it does during the expansion), the increase in entropy is
given by
µ ¶
V2
∆S = R ln
(13)
V1
The dilution of a solute by a solvent can be regarded in very much the same was as the expansion
of a gas at constant temperature, and in well-behaved solutions (that is, where no special effects such
as hydrogen bonding or changes in intermolecular interactions are significant) the consequent change in
entropy follows essentially the same law:
µ ¶
C1
∆S = R ln
(14)
C2
(The subscripts of C are opposite to those of V because the concentration is inversely proportional to
the volume). We will see later that the entropy of mixing that occurs when products and reactants are
both present is an important factor in determining the equilibrium constant of a reaction.
15.5
The second law of thermodynamics
You will recall that the first law of thermodynamics, expressed as ∆U = q + w, is essentially a statement
of the law of conservation of energy. The significance of this law is that it tells us that any proposed
process that would violate this law can be dismissed as impossible, without even inquiring further into
the details of the process.
The second law of thermodynamics goes beyond this by saying, in effect, that the extent to which any
natural process can occur is limited by the increase in disorder that accompanies it, and once the change
has occurred, it can never be un-done without creating even more disorder. For example, you can restore
a pile of randomly-ordered coins to a stack of all-heads, but in doing so your own metabolism will have
created more disorder than you removed.
Historically, the second law was first applied to the problem of converting heat into work in steam
engines. In this context, it says that there is an upper limit to the fraction of heat input that an engine can
extract as work. This limit has nothing to do with the mechanics of the engine, but is determined solely by
the increase in disorder that is possible. This, in turn, is a function of the input and exhaust temperatures.
If they are the same, then no work at all can be extracted, and the engine will have an efficiency of 0%.
Thus a device which would drive a ship across the ocean by extracting the thermal energy from seawater
would be prohibited by the Second Law unless you could find some lower-temperature reservoir for the
heat to flow into.
Chem1 General Chemistry
10
Chemical Energetics
• The second law of thermodynamics
T
high-temperature
source
T
H
AAA
H
q
a
H
engine
work
out
b
T
L
q
L
low-temperature
sink
T
L
e = a/b
AAA
The fall of heat. A heat engine is a device into
which a quantity of heat qH , extracted from a source
at temperature TH , is partly converted into work w.
The remainder of the heat qL is exhausted at a lower
temperature TL . The fraction of the heat that can
be converted into work qH /w is known as the efficiency of the engine, ². According to the Second
Law, ² is proportional to the “distance” in temperature that the heat can fall as it passes through the
engine:
TL
²=1−
TH
If the heat can flow into an exhaust at 0 ◦ K, the efficiency could be 100%, but no such exhaust reservoir
could exist.
For all practical heat engines, TL is just the temperature of the environment, normally around 300 ◦ K,
so the only way to improve efficiency is to make
TH as high as possible. The upper limit of around
600 ◦ K is determined largely by the ability of materials such as turbine rotors to withstand the high
temperatures.
Figure 11: The efficiency of a heat engine
Chem1 General Chemistry
11
Chemical Energetics
• 16 The free energy
16
The free energy
In the previous section we saw that it is the sum of the entropy changes of the system and surroundings
that determines whether a process will occur spontaneously. In chemical thermodynamics we prefer
to focus our attention on the system rather than the surroundings, and would like to avoid having to
calculate the entropy change of the surroundings explicity. The key to doing this is to define a new state
function known as the Gibbs free energy
G = H − TS
(15)
Since H, T and S are all state functions, G is a function of state. For a process that takes place reversibly,
we can write
qsys
qsurr
+
∆S total =
T
T
Multiplying through by −T , we obtain
qsys
= Hsys − T ∆S sys
−T ∆S total = −qsurr − T
T
which expresses the entropy change of the world in terms of thermodynamic properties of the system
exclusively. If −T ∆S total is denoted by ∆G, then we have
∆G = ∆H − T ∆S
(16)
which defines the Gibbs free energy change for the process.
From the foregoing, you should convince yourself that G will decrease in any process occurring at
constant temperature and pressure which is accompanied by an overall increase in the entropy. (The
constant temperature and pressure are a consequence of the temperature and the enthalpy appearing in
Eq 15.) Since most chemical and phase changes of interest to chemists take place under such conditions,
the Gibbs function (as it is often called) is the most useful of all the thermodynamic properties of a
substance, and it is closely linked to the equilibrium constant.
You may recall (page ) that the two quantities qrev (qmin ) and wmax associated with reversible
processes are state functions. We gave the quotient qrev /T a new name, the entropy. What we did
not say is that wmax also has its own name, the free energy. Thus the Gibbs free energy is the
maximum useful work (excluding P V work associated with volume changes of the system) that a
system can do on the surroundings when the process occurs reversibly at constant temperature and
pressure. This work is done at the expense of the internal energy of the system, and whatever part
of ∆U that is not extracted as work is exchanged with the surroundings as heat; this latter quantity
will have the value −T ∆S.
16.1
The standard Gibbs free energy
In order to make use of free energies to predict chemical changes, we need to know the free energies of
the individual components of the reaction. For this purpose we can combine the standard enthalpy of
formation and the standard entropy of a substance according to Eq 15 to get its standard free energy
of formation for a given temperature. As with standard heats of formation, the ∆G◦ f of a substance
represents the free energy change associated with the formation of the substance from the elements in
their most stable forms as they exist under the standard conditions of 1 atm pressure and 298 ◦ K.
Standard Gibbs free energies of formation are normally found directly from tables. Once the ∆G◦
values for all the components of a reaction are known, the standard Gibbs free energy change ∆G◦ for
the reaction is found in the normal way.
The interpretation of ∆G◦ for a process is very simple. For a reaction A −→ B,
∆G◦ < 0
∆G◦ > 0
∆G◦ = 0
the process will tend to occur to the right, producing more A
the process will tend to occur to the left, producing more B
A and B are in equilibrium; there will be no net change
Chem1 General Chemistry
12
Chemical Energetics
• Temperature dependence of ∆G◦
DGû>0
DGû<0
DHû
kJ
DGû
100
TDSû
200
300
400
temperature ûK
For the vaporization of water, both ∆H ◦ and ∆S ◦ are positive. At higher temperatures, T ∆H ◦ will eventually
overtake ∆H ◦ , driving ∆G◦ negative; all reactions of this kind will therefore become spontaneous at higher
temperatures. At 373K, the free energies of liquid water and water vapor at 1 atm partial pressure are identical,
thus establishing the normal boiling temperature.
Figure 12: Thermodynamics of the vaporization of water
16.2
Temperature dependence of ∆G◦
From Eq 16, it is apparent that the temperature dependence of ∆G◦ depends almost entirely on the
entropy change associated with the process. (We say almost because the values of ∆H ◦ and ∆S ◦ are
themselves slightly temperature dependent; both gradually increase with temperature). In particular,
notice that the sign of the entropy change determines whether the reaction becomes more or less spontaneous as the temperature is raised. Since the signs of both ∆H ◦ and ∆S ◦ can be positive or negative,
we can identify four possibilities as outlined in Table 16.2.
∆H ◦
<0
<0
>0
>0
∆S ◦
>0
<0
>0
<0
reaction is spontaneous
always
at low temperatures
at high temperatures
never
example
2 H2 (g) + O2 (g) −→ 2 H2 O(g)
3 H2 + N2 −→ 2 NH3 (g)
N2 O4 (g) −→ 2 NO2 (g)
N2 + O2 −→ 2 NO2 (g)
∆H ◦
−50.5
−22.1
13.9
16.2
298 K × ∆S ◦
9.4
−14.1
12
−8.6
∆G◦
−59.9
−8.0
1.3
29.8
Notice that the reactions which are spontaneous only above or below a certain temperature have
identical signs for ∆H ◦ and ∆S ◦ . In these cases there will be a unique temperature at which ∆H ◦ =
−T ∆S ◦ and thus ∆G◦ = 0, corresponding to chemical equilibrium. This temperature is given by
T =
∆H ◦
∆S ◦
(17)
and corresponds to the point at which the lines representing ∆H ◦ and −T ∆S ◦ cross in a plot of these
two quantities as a function of the temperature.
This important relation must immediately be qualified in two ways. First, it applies only to heterogeneous reactions such a phase changes and other reactions in which products and reactants remain in
separate phases so they cannot mix (the reason for this will be explained further on). Secondly, the values
of both ∆H ◦ and ∆S ◦ are themselves temperature dependent, both increasing with the temperature.
This means that substitution of 298 ◦ K -values of these quantities into Eq 17 will give only an estimate
of the equilibrium temperature if this differs greatly from 298◦ K.
Chem1 General Chemistry
13
Chemical Energetics
• How free energies determine melting and boiling points
100
200
300
liquid
Gû
solid
400 ûK
vapor
Dashed parts of plot show
extrapolation of free energies
beyond temperatures of stable
existence of the phase.
The free energy of any phase always decreases with temperature at a rate that is proporational fo the molar
volume of the phase. The melting and boiling temperatures are those at which the free energy of one phase
becomes the smallest, making that phase the only stable one.
Figure 13: Free energy of solid, liquid and gaseous water as a function of temperature.
Nevertheless, the general conclusions regarding the temperature dependence of reactions such as
those shown in Table 16.2 are always correct, even if the equilibrium temperature cannot be predicted
accurately.
Problem Example 4
Estimate the temperature at which liquid water will be in equilibrium with its vapor at 1 atm pressure.
The temperature at which ∆G◦ = 0 is given by
T =
∆H ◦
44030 J mol−1
◦ =
∆S
118.91 J K−1 mol−1
and in this case is (44030 J mol−1 )/118.91 J K−1 mol−1 = 370 K.
16.3
How free energies determine melting and boiling points
As the temperature rises, T ∆S ◦ increases faster than ∆S ◦ , so the free energy always falls with temperature. Furthermore, it falls off fasster for gases than for liquids, and for liquids faster than for solids. In
Fig. 13 the free energies of each of the three phases of water are plotted as a function of temperature. At
any given temperature, one of these phases will have a lower free energy than any of the others; this will
of course be the stable phase at that temperature. The cross-over points where two phases have identical
free energies are the melting and boiling temperatures. At any other temperature, there is only a single
stable phase.
17
Free energy and concentration
The free energy of a pure liquid or solid at 1 atm pressure is just its molar free energy of formation G◦ ,
multiplied by the number of moles present. For gases and substances in solution, we have to take into
account the concentration (which, in the case of gases, is normally expressed in terms of the pressure).
We know that the lower the concentration, the greater the entropy, and thus the smaller the free energy.
Chem1 General Chemistry
14
Chemical Energetics
• The free energy of a gas: Standard states
17.1
The free energy of a gas: Standard states
The free energy of a gas depends on its pressure; the higher the pressure, the higher the free energy. Thus
the free expansion of a gas, a spontaneous process, is accompanied by a fall in the free energy. Such an
expansion is an example of an entropy-driven process, since the enthalpy of expansion of an ideal gas is
zero. Substituting pressures for volumes in Eq 13, we can express the change in free energy when a gas
undergoes a change in pressure from P1 to P2 as
µ ¶
P1
∆G = ∆H − T ∆S = 0 − RT ln
P2
How can we evaluate the free energy of a specific sample of a gas at some arbitrary pressure? First,
recall that the standard molar free energy G◦ that you would look up in a table refers to a pressure of
1 atm. The free energy per mole of our sample is just the sum of this value and any change in free energy
that would occur if the pressure were changed from 1 atm to the pressure of interest:
¶
µ
P2
◦
G = G + RT ln
1 atm
which we normally write in abbreviated form
G = G◦ + RT ln P
17.2
(18)
The free energy of a solute
All substances, given the opportunity to form a homogeneous mixture with other substances, will tend
to become more dilute. As discussed previously, this is simply a consequence of statistics; there are more
equally probable ways of arranging one hundred black marbles and one hundred white marbles, than
two hundred marbles of a single color. The entropy increase that occurs when a the concentration of a
substance decreases through dilution or mixing is given by Eq 14.
We can similarly define the Gibbs free energy of dilution or mixing by substituting this equation into
the definition of ∆G:
µ ¶
C1
∆Gdil = ∆H dil − RT ln
C2
If the substance in question forms an ideal solution with the other components, then ∆H dil is by definition
zero, and we can write 3
µ ¶
C2
∆Gdil = RT ln
(19)
C1
This tells us that the dilution of a substance from an initial concentration C1 to a more dilute concentration C2 is accompanied by a decrease in the free energy, and thus will occur spontaneously.
By the same token, if the initial concentration is smaller, the process will not take place spontaneously.
The “un-dilution” of a solution can be forced to occur by supplying an amount of energy (in the form of
work) given by ∆Gmix . An important practical example of this is the metabolic work performed by the
kidney in concentrating substances from the blood for excretion in the urine.
To find the free energy of a solute at some arbitrary concentration, we proceed in very much the
same way as we did for a gas: we take the sum of the standard free energy, and any change in the free
3 A more rigorous treatment of the thermodynamics of mixing requires that we take into account the dilution of the
solvent as well as that of the solute, and yields a slightly more complicated formula containing mole fractions rather than
concentrations.
Chem1 General Chemistry
15
Chemical Energetics
• 18 Free energy and the equilibrium constant
energy that would accompany a change in concentration from the standard state to the actual state of
the solution. Using Eq 19 it is easy to derive an expression analogous to Eq 18
G = G◦ + RT ln C
(20)
which gives the free energy of a solute at some arbitrary concentration C in terms of its value G◦ in its
standard state.
Although this expression has the same simple form as Eq 18, its practical application is fraught
with difficulties, the major one being that it doesn’t usually give values of G that are consistent with
experiment, especially for solutes that are ionic or are slightly soluble. In such solutions, intermolecular
interactions between solute molecules and between solute and solvent bring back the enthalpy term that
we left out in deriving Eq 18 (and thus Eq 20). In addition, the structural organization of the solution
becomes concentration dependent, so that the entropy depends on concentraiton in a more complicated
way than is implied by Eq 14.
Instead of complicating Eq 20 by trying to correct for all of these effects, chemists have chosen to
retain its simple form by making a single small change:
G = G◦ + RT ln a
(21)
This equation is guaranteed to work, because a, the activity of the solute, is its thermodynamically
effective concentration. The relation between the activity and the concentration is given by
a = γC
(22)
where γ is the activity coefficient. As the solution becomes more dilute, the activity coefficient approaches
unity:
lim γ = 1
C→0
The price we pay for the simplicity of Eq 21 is that the relation between the concentration and the
activity at higher concentrations can be quite complicated, and must be determined experimentally for
every different solution.
The question of what standard state we choose for the solute (that is, at what concentration does G◦
apply, and in what units is it expressed?) is one that you will wish you had never asked. We might
be tempted to use a concentration of 1 molar, but a solution this concentrated would be subject
to all kinds of intermolecular interaction effects, and would not make a very practical standard
state. These effects could be eliminated by going to the opposite extreme of an “infinitely dilute”
solution, but by Eq 20 this would imply a free energy of minus infinity for the solute, which would
be awkward. Chemists have therefore agreed to define the standard state of a solute as one in which
the concentration is 1 molar, but all solute-solute interactions are magically switched off, so that γ
is effectively unity. Since this is impossible, no solution corresponding to this standard state can
actually exist, but this turns out to be only a small drawback, and seems to be the best compromise
between convenience, utility, and reality.
18
Free energy and the equilibrium constant
Consider a homogeneous chemical reaction of the form
A + B −→ C + D
in which all component are gases at the temperature of interest. If the standard free energies of the
products are less than those of the reactants, ∆G◦ for the reaction will be negative and the reaction
Chem1 General Chemistry
16
Chemical Energetics
• 18 Free energy and the equilibrium constant
reactants
DG=0 at
equilibrium
composition
DG
DGû
DG
standard free energy
change
products
free energy due to
mixing of products
with reactants
system composition
Figure 14: Relation between G, ∆G, and ∆G◦
will proceed to the right. But how far? If the reactants are completely transformed into products, the
equilibrium constant would be infinity. The equilibrium constants we actually observe all have finite
values, implying that even if the products have a lower free energy than the reactants, some of the latter
will always remain when the process comes to equilibrium.
In order to understand how equilibrium constants relate to ∆G◦ values, assume that all of the reactants
are gases, so that the free energy of gas A, for example, is given at all times by
GA = G◦A + RT ln PA
(23)
The free energy change for the reaction is sum of the free energies of the products, minus that of the
reactants:
(24)
∆G = GC + GD − GA − GB
Expanding each term on the right as in Eq 23,
∆G = G◦C + RT ln PC + G◦D + RT ln PD − G◦B + RT ln
1 atm
1 atm
− G◦A + RT ln
PB
PA
(25)
We can now express the G◦ terms collectively as ∆G◦ , and combine the logarithmic terms into a single
fraction
PC PD
(26)
∆G = ∆G◦ + RT ln
PA PB
which is more conveniently expressed in terms of the reaction quotient QP :
∆G = ∆G◦ + RT ln QP
(27)
Let’s pause at this point to make sure you understand the significance of this equation. First, recall that
the free energy G is a quantity that becomes more negative during the course of any natural process.
Thus as a chemical reaction takes place, G only falls and will never become more positive. Eventually a
point is reached where any further transformation of reactants into products would cause G to increase.
At this point G is at a minimum (see the plot of G as a function of the extent of reaction in Fig. 14),
and no further change can take place; the reaction is at equilibrium.
The quantity ∆G that is given by Eq 27 tells us how far the free energy can fall before its minimum
value is reached. At the beginning of the reaction, ∆G will have a negative value, indicating that change
is possible. When the reaction reaches equilibrium, ∆G = 0, and no further change will take place.
Chem1 General Chemistry
17
Chemical Energetics
• The equilibrium constant
According to Eq 27, the capacity for change as expressed by ∆G is determined by two quantities.
One, ∆G◦ , is a constant for the particular reaction at a fixed temperature. ∆G◦ expresses the difference
between the free energies of the pure products and the pure reactants.
The other part, the logarithmic term containing Q, is a variable whose value will increase as the
composition of the system changes– as reactants are transformed into products. The system composition
affects the free energy in two ways. First, since the free energy is an extensive property, it is clear that
the free energies of the products will make very little contribution to the total free energy of the system
near the beginning of the reaction when relatively few product molecules are present. Similarly, the free
energies of the reactants will not be very important when the reaction mixture consists mostly of product
molecules.
The other effect of system composition is to alter the concentrations of the various components. As
discussed previously, dilution of one substance by another is accompanied by an increase in the entropy,
and thus a decrease in the free energy. For a homogeneous reaction of the form A + B −→ C + D,
the entropy of mixing is greatest, and the free energy of mixing is most negative, at the point where half
the reactants have been transformed into products.
We are now in a position to answer the question posed earlier: if ∆G◦ for a reaction is negative,
meaning that the free energies of the products are more negative than those of the reactants, why will
some of the latter remain after equilibrium is reached? The answer is that no matter how low the free
energy of the products, the free energy of the system can be reduced even more by allowing some of
the products to be contaminated (i.e., diluted) by some reactants. Owing to the entropy associated with
mixing of reactants and products, no homogeneous reaction will be 100% complete.
18.1
The equilibrium constant
Now let us return to Eq 27, which we reproduce here:
∆G = ∆G◦ + RT ln QP
As the reaction approaches equilibrium, ∆G becomes less negative and finally reaches zero. At equilibrium, ∆G = 0 and Q = K, so we can write
∆G◦ = −RT ln KP
(28)
in which KP , the equilibrium constant expressed in pressure units, is the special value of Q that corresponds to the equlibrium composition.
This relation is one of the most important in chemistry, because it relates the equilibrium composition
of a chemical reaction system to measurable physical properties of the reactants and products. If you
know the entropies and the enthalpies of formation of a set of substances, you can predict the equilibrium constant of any reaction involving these substances without the need to know anything about the
mechanism of the reaction.
Instead of writing Eq 28 in terms of Kp , we can use any of the other forms of the equilibrium constant
such as Kc , Kx (mole fractions), KN (numbers of moles), etc. Remember, however, that for ionic solutions
especially, only the Ka , in which activities are used, will be strictly valid. Alternatively, we can solve
Eq 28 for the equilibrium constant, yielding
K = exp
¡ −∆G◦ ¢
RT
(29)
Notice that an equilibrium constant of unity implies a standard free energy change of zero, and that
positive values of ∆G◦ lead to values of K less than unity.
Chem1 General Chemistry
18
Chemical Energetics
• Equilibrium and temperature
5
10£
10ª
10
K
K
1
10ÑÁ
1
10Ѫ
0
10Ñ£
Ð5
5
-12.5
0
0
+12.5
Gû kJ /mol
Gû kJ /mol
Figure 15: Relation between ∆G◦ and K.
18.2
Equilibrium and temperature
We have already discussed how changing the temperature will increase or decrease the tendency for
a process to take place, depending on the sign of ∆S. This relation can be developed formally by
differentiating the relation
(30)
∆G◦ = ∆H ◦ − T ∆S ◦
with respect to the temperature:
d(∆G◦ )
= −∆S ◦
dT
(31)
More commonly, we want to know how a change in the temperature will affect the value of an
equilibrium constant whose value is known at some fixed temperature. By substituting Eq 30 into Eq 29
we obtain
¡ ∆S ◦ ¢
¡ −∆H ◦ ¢
K = exp
× exp
(32)
R
RT
Do you remember the Le Châtelier Principle? Here is its theoretical foundation with respect to the effect
of the temperature on equilibrium: if the reaction is exothermic (∆H ◦ < 0), then increasing T will make
the second exponential term smaller and K will decrease– that is, the equilibrium will “shift to the left”.
If dho > 0 then increasing T will make the exponent less negative and K will increase.
Suppose that the equilibrium constant has the value K1 at temperature T1 , and we wish to estimate
K2 at temperature T2 . Expanding Eq 28 in terms of ∆H ◦ and ∆S ◦ , we obtain
−RT1 ln K1 = ∆H ◦ − T1 ∆S ◦
and
−RT2 ln K2 = ∆H ◦ − T2 ∆S ◦
Dividing both sides of each expression by RT and subtracting, we obtain
ln K1 − ln K2 = −
∆H ◦
∆H ◦
+
RT1
RT2
which is most conveniently expressed as the ratio
ln
Chem1 General Chemistry
K1
∆H ◦
=−
K2
R
19
µ
1
1
−
T1
T2
¶
(33)
Chemical Energetics
• Equilibrium without mixing: coupled reactions
This is an extremely important relationship, but not just because of its use in calculating the temperature dependence of an equilibrium constant. Even more important is its application in the “reverse”
direction, to obtain a value of ∆H ◦ from two values of K measured at different temperatures. direct
calorimetric determinations of heats of reaction are not easy to make; relatively few chemists have the
equipment and experience required for this rather exacting task. Measurement of an equilibrium constant
is generally much easier, and often well within the capabilities of anyone who has had an introductory
Chemistry course.
Not only does Eq 33 provide a non-calorimetric means of measuring ∆H ◦ , but once its value is
determined, ∆S ◦ can be calculated from Eq 30, using a value of ∆G◦ calculated by substituting one of
the K values into Eq 28.
Problem Example 5
The vapor pressure of solid ammonium perchlorate was found to be 0.026 torr at 520 ◦ K and 2.32 torr
at 620 ◦ K. Addition of NH3 gas was observed to repress the vaporization, suggesting that the equilibrium under study is
NH4 ClO4 (s) −→ NH3 (g) + HClO4 (g)
Use this information to calculate ∆H ◦ and ∆S ◦ for this process.
Solution. In problems of this kind the first step is usually to express the equilibrium constant in
terms of the observed pressures. From the stoichiometry of this reaction it is apparent that Kp =
PN H3 PHClO4 , and that each of these partial pressures is just half the vapor pressure P , so that
KP = (P/(2 × 760))2 . This gives K1 = 2.92E − 10 and K2 = 2.33E − 6. Substituting these into
Eq 33 and solving for the enthalpy change gives ∆H ◦ = 241 kJ mol−1 . Next, we substitute the two
equilibrium constants into Eq 28 and obtain ∆G◦ 1 = 94.9 kJ mol−1 and ∆G◦ 2 = 56.1 kJ mol−1 . The
entropy change is found from Eq 31, using the non-calculus form ∆(∆G◦ )/∆T since the functional
relation between ∆G◦ and the temperature is not known. This gives ∆S ◦ = +388 J mol−1 K−1 .
Notice that the sign of this entropy change could have been anticipated by inspection of the reaction
equation, since the volume increase will dominate other entropy differences.
18.3
Equilibrium without mixing: coupled reactions
You should now understand that for reactions which take place entirely in the gas phase or in solution,
the equilibrium composition will never by 100% products, no matter how much lower their free energy
relative to the reactants. As was explained on page 18 , this is due to dilution of the products by the
reactants.
If the reaction involves more than one phase of matter, this dilution, and the effects that flow from it,
may not be possible. As a simple example, consider an equilibrium mixture of ice and liquid water. The
concentration of H2 O in each phase is dependent only on the density of the phase; there is no way that
ice can be “diluted” with water, or vice versa. This means that all temperatures other than the freezing
point, the lowest free energy state will be that corresponding to pure ice or pure liquid (Fig. 16). Only
at the freezing point, where the free energies of water and ice are identical, can both phases coexist, and
they may do so in any proportion.
19
Coupled reactions
Two reactions are said to be coupled when the product of one of them is the reactant in the other:
A −→ B
Chem1 General Chemistry
B −→ C
20
Chemical Energetics
• Extraction of metals from their oxides
Ð1ûC
1ûC
DG
0
0ûC
0
(ice)
100
composition, % water
In a heterogeneous reaction such as the meltng of ice shown here, the “reactants” and “products” can coexist
only at a unique equilibrium temperature (horizontal line); at the melting point, the fraction of water in the form
of ice or liquid can have any value. At all other temperatures, the only stable composition is 100 percent solid or
liquid.
Figure 16: Free energy of H2 O as a function of solid-liquid fraction
If the standard free energy of the first reaction is positive but that of the second reaction is sufficiently
negative, then ∆G◦ for the overall process will be negative and we say that the first reaction is “driven”
by the second one.
This, of course, is just another way of describing an effect that you already know as the Le Châtelier
principle: the removal of substance B by the second reaction causes the equilibrium of the first to “shift
to the right”. Similarly, the equilibrium constant of the overall reaction is the product of the equilibrium
constants of the two steps. Nevertheless, the concept of coupled reactions is sufficiently important to
deserve special mention in at least two contexts.
19.1
Extraction of metals from their oxides
Since ancient times, the recovery of metals from their ores has been one of the most important applications
of chemistry to civilization and culture. The oldest, and still the most common smelting process for oxide
ores involves heating them in the presence of carbon. Originally, charcoal was used, but industrial-scale
smelting uses coke, a crude form of carbon prepared by pyrolysis (heating) of coal.
The basic reactions are
MO + C −→ M + CO
MO + 12 C −→ M + 12 CO2
MO + CO −→ M + CO2
(i)
(ii)
(iii)
Each of these can be regarded as a pair of coupled reactions in which the metal M and the carbon are
effectively competing for the oxygen atom. Using (i) as an example, we can write
MO −→ M + 12 O2
C + 12 O2 −→ CO
(iv)
(v)
At ordinary environmental temperatures, reaction (iv) is always spontaneous in the reverse direction
(that is why ores form in the first place!), so ∆G◦ (iv) will be positive. ∆G◦ (v) is always negative, but at
low temperatures it will not be sufficiently negative to drive (iv).
Chem1 General Chemistry
21
Chemical Energetics
• Extraction of metals from their oxides
Note: free energies for oxides are per
mole of O in the oxide.
Ð50
0
Al2O3
ë G ˜ kJ m olÑÁ
Ð40
0
TiO2
C + 1/2 O2 ® CO
Ð30
0
Ð20
0
C + O2 ® CO2
FeO
Ð10
0
PbO
0
CuO
+10
0
Ore can be reduced by carbon only if its
Gû is below that of one of the carbon
oxidation reactions. Practical refining
temperatures are generally limited to
about 1500ûK.
Ag2O
0
2500
500
1000
1500
CO + 1/2 O2 ® CO2
2000
tem perature, ˜K
Figure 17: Ellingham diagram showing temperature dependence of oxidation reactions involved in metal
extraction
Chem1 General Chemistry
22
Chemical Energetics
• 20 Bioenergetics
The smelting process depends on the different ways in which the free energies of reactions like (iv)
and (v) depend on the temperature. This temperature dependence is almost entirely dominated by the
T ∆S term in the Gibbs function, and thus by the entropy change. The latter depends mainly on ∆ng ,
the change in the number of moles of gas in the reaction. Removal of oxygen from the ore is always
accompanied by a large increase in volume, so ∆S for this step is always positive and the reaction
becomes more spontaneous at higher temperatures. The temperature dependences of the reactions that
take up oxygen vary, however:
reaction
C + 12 O2 −→ CO
C + O2 −→ CO2
CO + 12 O2 −→ CO2
∆ng
1
2
0
− 12
d(∆G◦ )/dT
<0
0
>0
A plot of the temperature dependences of the free energies of these reactions, superimposed on similar
plots for the oxygen removal reactions (iv) is called an Ellingham diagram. In order for a given oxide
MO to be smeltable, the temperature must be high enough that ∆G◦ (iv) falls below that of at least one
of the oxygen-consuming reactions. The slopes of the lines on this diagram are given by
d(∆G◦ )
= −∆S ◦
dT
Examination of the Ellingham diagram in Fig. 17 illustrates why the metals known to the ancients
were mainly those such as copper and lead, which can be obtained by smelting at the relatively low
temperatures that were obtainable by the methods available at the time. Thus the bronze age preceded the
iron age; the latter had to await the development of technology capable of producing higher temperatures,
such as the blast furnace. Smelting of aluminum oxide by carbon requires too high temperatures to be
practical; commercial production of aluminum is accomplished by electrolysis of the molten ore.
20
Bioenergetics
Many of the reactions that take place in living organisms require a source of free energy to drive them.
The immediate source of this energy in heterotrophic organisms, which include animals, fungi, and most
bacteria, is the sugar glucose. Oxidation of glucose to carbon dioxide and water is accompanied by a
large negative free energy change.
C6 H12 O6 + O2 −→ 6 CO2 + 6 H2 O
∆G◦ = −2880 kJ mol−1
(34)
Of course it would not do to simply “burn” the glucose in the normal way; the energy change would be
wasted as heat, and rather too quickly for the well-being of the organism. Effective utilization of this
free energy requires a means of capturing it from the glucose, and releasing it in small amounts when
and where it is needed. This is accomplished by breaking down the glucose in a series of a dozen or more
steps in which the energy liberated in each step is captured by an “energy carrier” molecule, of which
the most important is adenosine diphosphate, known as ADP. At each step in the breakdown of glucose,
an ADP molecule reacts with inorganic phosphate (denoted by Pi ) and changes into ATP:
∆G◦ = +30 kJ mol−1
ADP + Pi −→ ATP
(35)
The 30 kJ mol−1 of free energy stored in each ATP molecule is released when the molecule travels to a site
where it is needed and loses one of its phosphate groups, yielding inorganic phosphate and ADP, which
eventually finds its way back the site of glucose metabolism for recycling back into ATP. The complete
Chem1 General Chemistry
23
Chemical Energetics
• 20 Bioenergetics
is
es )
h
t
in
yn gy
s
r
o
ot ene
h
p e
e
(fr
n t)
ti o ou
i ra gy
sp er
re en
ee
(fr
free energy
{CH2O} + O2
DG
AAAAAAAAAAAAA
CO2 + H2O
CO2 + H2O
Figure 18: Climbing the free energy hill: photosynthesis and respiration
breakdown of one molecule of glucose is coupled with the production of 38 molecules of ATP, according
to the overall process
C6 H12 O6 + 6 O2 + 38 Pi + 38 ADP −→ 38 ATP + 6 CO2 + 44 H2 O
(36)
For each mole of glucose metabolized, 38 × 30 = 1140 kJ of free energy is captured as ATP, representing
an energy efficiency of 1140/2880 = 0.40%. That is, 40% of the free energy available from the oxidation
of glucose is made available to drive other metabolic processes. The rest is liberated as heat.
Where does the glucose come from? Animals obtain their glucose from their food, especially cellulose
and starches that, like glucose, have the empirical formula {C H2 O}. Animals obtain this food by eating
plants or other animals. Ultimately, all food comes from plants, most of which are able to make their own
glucose from CO2 and H2 O through the process of photosynthesis. This is just the reverse of Eq 34 in
which the free energy is supplied by the quanta of light absorbed by chlorophyll and other photosynthetic
pigments.
This describes aerobic respiration, which evolved after the development of photosynthetic life on Earth
began to raise the concentration of atmospheric oxygen. Oxygen is a poison to most life processes at
the cellular level, and it is believed that aerobic respiration developed as a means to protect organisms
from this peril. Those that did not have literally “gone underground” and constitute the more primitive
anaerobic bacteria.
The function of oxygen in respiration is to serve as an acceptor of the electrons that glucose loses
when it undergoes oxidation. Other electron acceptors can fulfill the same function when oxygen is not
available, but none yields nearly as much free energy. For example, if oxygen cannot be supplied to
mammalian muscle cells as rapidly as it is needed, they switch over to an anaerobic process yielding
lactic acid instead of CO2 :
C6 H12 O6 + 2 ADP −→ 2 CH3 CH(OH)COOH
(37)
In this process, only 2 × 30 = 60 kJ of free energy is captured; the free energy of conversion of glucose
into lactic acid is −218 kJ mol−1 , so the efficiency is only 28% on the basis of this reaction, and it is
even lower in relation to Eq 34. In “aerobic” exercising, one tries to maintain sufficient lung capacity and
cardiac output to supply oxygen to muscle cells at a rate that promotes the aerobic pathway.
Chem1 General Chemistry
24
Chemical Energetics
• 20 Bioenergetics
The fall of the electron. Oxidation-reduction reactions proceed in a direction that allows the
electron to “fall” (in free energy) from a “source” to a “sink”. In the next chapter on electrochemistry
you will see how this free energy can manifest itself as an electrical voltage and be extracted from the
system as electrical work. In organisms, the free energy is captured in a series of coupled reactions as
described above, and eventually made available for the various free energy-consuming processes associated
with life processes.
Fig. 19 shows the various electron sources (foodstuffs) and sinks (oxidizing agents) on a vertical scale
that illustrates the relative free energy of an electron in each sink. Notice that carbohydrate is at the top
of this scale and dioxygen is at the bottom, indicating that O2 is the best possible electron acceptor in
terms of free energy yield.
Organisms that live in environments where oxygen is lacking, such as marshes, muddy soils, and the
intestinal tracts of animals, must utilize other electron acceptors to extract free energy from carbohydrate.
A wide variety of inorganic ions such as sulfate and nitrate, as well as other carbon compounds can serve
as electron acceptors, yielding the gaseous products like H2 S, NH3 and CH4 which are commonly noticed
in such locations. From the location of these acceptors on the scale, it is apparent that the amount of
energy any can extract from a given quantity of carbohydrate is much less. One reason that aerobic
organisms have dominated the Earth is believed to be the much greater energy-efficiency of oxygen as an
electron acceptor.
What did organisms use for food before there was a widespread supply of carbohydrate in the world?
Any of the electron sources near the upper left of the table can in theory serve this function, although
at reduced energy efficiency. As a matter of fact, there are still a number of these autotrophic bacteria
around whose “food” is CH4 , CH3 OH, FeCO3 , and even H2 !
c
°1996
by Stephen K. Lower; all rights reserved.
January 5, 1997
Please direct comments and inquiries to the author at lower@sfu.ca.
Chem1 General Chemistry
25
Chemical Energetics
• 20 Bioenergetics
electron
sources
electron
sinks
-8
300
CO2, H+
C6H12O6
H2
-6
CH3, H2O
H2S, H2O
CH3OH, H2O
FeCO3, H2O
CH4, H2O
-4
-2
N2, H+ CO , H+
2
NH4+
SO42Ð, H2
CH3O , H+
FeOOH, HCO3Ñ
CH3O, H+
NH4+, H2O
NO3Ñ, H+
100
0
pEû(w)
200
2
CH3OH, H+
CH4, H2O
4
0
6
NH3, H2O
NO3Ñ, H+
NO3Ñ, H2O
NO2Ñ, H+
8
MnO2, HCO3Ñ, H+
MnCO3, H2O
Ð100
10
12
NO3Ñ, H+
N2, H2O
O2, H+
H2O
14
Ð200
free energy released per mole of electrons transferred to hydrogen ion at unit activity
H+
Figure 19: Electron free energy levels in bioenergetics
Chem1 General Chemistry
26
Chemical Energetics