Download Energetics and Equilibria

UNIVERSITY OF CAMBRIDGE
DEPARTMENT OF CHEMISTRY
IA CHEMISTRY 2002/3
Energetics and Equilibria
Lecturer: Dr James Keeler
EMAIL: [email protected]
1.
Introduction ............................................................................................ 4
2.
The First Law of Thermodynamics........................................................... 5
2.1
2.2
2.3
2.4
2.4.1
2.4.2
2.5
3.
What is heat?..............................................................................................................................5
What is work?.............................................................................................................................5
What is internal energy?........................................................................................................6
State functions and path functions....................................................................................6
State functions.................................................................................................................6
Path functions..................................................................................................................7
Sign conventions........................................................................................................................7
Gas expansions ........................................................................................ 7
3.1
3.2
3.3
3.4
3.4.1
3.4.2
3.5
4.
Gas laws.........................................................................................................................................7
How to think about a gas expansion.................................................................................7
Expansion against constant external pressure..............................................................7
Expansion doing maximum work.......................................................................................7
Reversible and irreversible processes.................................................................7
Reversible isothermal expansion of an ideal gas..............................................7
Heat changes in gas expansions.........................................................................................7
Entropy................................................................................................... 7
4.1
4.1.1
4.1.2
4.1.3
5.
Entropy changes – considering the system and the surroundings separately7
Entropy change of the system....................................................................................7
Entropy change of the surroundings.......................................................................7
Entropy change of the Universe...............................................................................7
Internal energy, enthalpy and heat capacity ............................................. 7
5.1
5.2
5.2.1
5.3
5.3.1
5.3.2
5.4
6.
Differential forms.......................................................................................................................7
Constant volume processes...................................................................................................7
Heat capacities...............................................................................................................7
Constant pressure processes – enthalpy..........................................................................7
Heat capacity at constant pressure........................................................................7
Variation of enthalpy with temperature................................................................7
Measurement and tabulation of heat capacities.........................................................7
Measuring entropy .................................................................................. 7
6.1
6.1.1
7.
7.1
8.
8.1
8.2
8.3
8.4
8.5
8.6
Absolute entropies.....................................................................................................................7
Converting entropies from one temperature to another................................7
Gibbs energy........................................................................................... 7
Gibbs energy and the Universal entropy.........................................................................7
Chemical changes.................................................................................... 7
The ∆ r symbol.............................................................................................................................7
Standard states............................................................................................................................7
Enthalpies of formation...........................................................................................................7
Entropy and Gibbs energy changes for reactions........................................................7
Variation of ∆rH with temperature....................................................................................7
Variation of ∆rS with temperature.....................................................................................7
1
9.
The Master Equations..............................................................................7
9.1
9.2
9.3
Variation of G with p (and V), at constant T ...............................................................7
Variation of G with T, at constant p.................................................................................7
Variation of S with V (and p), at constant T ................................................................7
10.
Chemical potential – dealing with mixtures...........................................7
10.1
10.2
10.3
10.4
10.5
The mixing of ideal gases.....................................................................................................7
Reacting mixtures.....................................................................................................................7
Definition of chemical potential........................................................................................7
Variation of chemical potential with pressure.............................................................7
Chemical potentials of solids and solutions.................................................................7
11.
Equilibrium constants..........................................................................7
12.
Chemical Equilibrium .........................................................................7
12.1
12.2
12.2.1
12.2.2
12.3
12.3.1
12.4
13.
13.1
14.
14.1
14.1.1
14.1.2
14.2
14.3
15.
15.1
15.2
15.3
15.3.1
15.3.2
16.
16.1
16.2
16.2.1
16.3
16.3.1
16.3.2
16.3.3
16.4
Condition for chemical equilibrium..................................................................................7
Relation between K and ∆ rG 0..............................................................................................7
What is ∆r G0?...................................................................................................................7
Equilibria involving solids.........................................................................................7
Variation of Kp with temperature.......................................................................................7
Measuring ∆r H0 indirectly.........................................................................................7
Variation of equilibrium with pressure............................................................................7
Applications in biology .........................................................................7
ATP..................................................................................................................................................7
Microscopic basis of entropy ................................................................7
Distributions.................................................................................................................................7
Ways of arranging things............................................................................................7
The Boltzmann distribution........................................................................................7
Entropy and distributions........................................................................................................7
Entropy at absolute zero.........................................................................................................7
Phase equilibria...................................................................................7
Phase diagrams...........................................................................................................................7
Criterion for phase stability and equilibria....................................................................7
Equation of a phase boundary..............................................................................................7
Interpretation of the Clapeyron equation............................................................7
Integration of the Clapeyron equation..................................................................7
Electrochemistry..................................................................................7
Introduction...................................................................................................................................7
Cell conventions........................................................................................................................7
Examples of using the cell conventions.................................................................7
Thermodynamic parameters from cell potentials.......................................................7
Concept of maximum work..........................................................................................7
Relation of the cell potential to ∆r Gcell...................................................................7
Relation of the cell potential to ∆r Scell and ∆r Hcell..............................................7
The Nernst equation.................................................................................................................7
2
16.4.1
16.4.2
16.4.3
16.4.4
16.4.5
16.5
16.6
16.6.1
16.6.2
16.6.3
16.6.4
16.7
16.8
16.9
16.9.1
16.9.2
16.9.3
16.9.4
17.
17.1
17.2
18.
18.1
18.2
Chemical potentials and activities.........................................................................7
Derivation of the Nernst equation...........................................................................7
Nernst equation for half cells....................................................................................7
Standard half-cell potentials.....................................................................................7
Tabulation of standard half-cell potentials.........................................................7
The spontaneous cell reaction.............................................................................................7
Types of half cells.....................................................................................................................7
Metal/metal ion...............................................................................................................7
Gas/ion................................................................................................................................7
Redox ...................................................................................................................................7
Metal/insoluble salt/anion.........................................................................................7
Liquid junctions..........................................................................................................................7
Standard functions for ions....................................................................................................7
Applications.................................................................................................................................7
Redox equilibria..............................................................................................................7
Solubility product of AgI..............................................................................................7
Thermodynamic parameters of ions.......................................................................7
Measurement of concentration................................................................................7
Appendix: Formal thermodynamics using partial derivatives................ 7
The Master Equations..............................................................................................................7
Maxwell's relations...................................................................................................................7
Appendix: Formal proofs ..................................................................... 7
Variation of ∆rH with temperature....................................................................................7
Variation of ∆rS with temperature.....................................................................................7
Books:
You will find that any general Physical Chemistry text book has a section on
Thermodynamics and Equilibrium which are the subject of these lectures.
The problem is that most books tend to cover far more than we need at the
moment, and also there is a tendency to couch the discussion in very
mathematical language, which can sometimes obscure the underlying
principles. So, when you consult a book, be prepared to have to be selective.
Probably the best book for this course is G J Price Thermodynamics of
Chemical Processes (Oxford Chemistry Primers no. 56, Oxford University
Press, 1998). Another useful book is P W Atkins The Elements of Physical
Chemistry (OUP, any edition). This is a simpler version of Physical
Chemistry (OUP, any edition) also by P W Atkins, which is an excellent
book although it is rather comprehensive .
Atkins has also written a "coffee table book" called The Second Law
(Scientific American Library, 1994) which is a well illustrated and very
readable informal account of thermodynamics and its applications.
3
1. Introduction
In the Michaelmas Term course Introduction to Energetics and Kinetics we
saw that some processes could be classed as natural or spontaneous. Such
processes take place on their own without any intervention from us. The
reverse of such processes do not happen spontaneously, but we can force
them to happen by intervening.
We saw that it is the Second Law of Thermodynamics which governs
whether or not a process will take place.
Second Law:
The entropy of the Universe increases in a spontaneous process.
Entropy, S, is a property of matter which can be associated with randomness
or disorder. So, for example, the entropy of a gas is greater than that of a
solid. We also saw that the entropy of a substance increases as it is heated.
Formally, entropy is defined as
δqrev
T
where dS is the change in entropy when an amount of heat, δq, is absorbed
reversibly at temperature T. One of our first tasks in these lectures will be to
define exactly what we mean by a reversible heat change, and so make it
possible to use this relationship to work out the entropy change of physical
processes and to find the absolute entropies of substances.
We saw that it was convenient to separate the Universe into two parts: the
system (the thing we are interested in) and the surroundings (the rest of the
Universe). The entropy change of the Universe can then be calculated from
dS =
universe
∆Suniv = ∆Ssys + ∆Ssurr .
=
system + surroundings
We have already seen that it is quite easy to work out the entropy change of
the surroundings from the heat change; here we will discuss more about how
the entropy change of the system can be determined.
The Second Law is easier to apply if we express it in terms of the Gibbs
energy, G, and we saw that the Gibbs energy decreases in a spontaneous
process. We will have more to say here about the Gibbs energy and how it
can be used to understand chemical equilibrium. Ultimately our aim will be
to prove the very useful relationship between the standard Gibbs energy
change for a reaction, ∆r G0 , and the equilibrium constant, K:
∆r G0 = –RT ln K
where
∆r G0 = ∆r H0 – T ∆r S0 .
These relationships are probably the most useful relationship in the whole of
thermodynamics.
Finally, we will look at two further applications of these basic ideas: the
first is to phase diagrams, which help us to understand the equilibria
between gases, liquids and solids, the second is to electrochemistry, which
enables us to find out about the thermodynamic properties of ions in
4
solution.
How do we know that the Second Law is true? Like any physical law
(such a Newton's Laws) it cannot be "proved". The law really is a statement
about experience ; the "proof", such that it is, is that nobody has ever
observed an event which contradicts the Second Law.
All laws have their limitations – for example Newton's Laws do not apply
to individual particles, like electrons and nuclei (here we need quantum
mechanics). The Second Law of Thermodynamics only applies to bulk
matter, it does not apply to individual atoms and molecules.
Before we look at entropy in more detail we need to sharpen up our
understanding of energy in its different forms: this is the domain of the First
Law of Thermodynamics.
2. The First Law of Thermodynamics
The First Law of Thermodynamics is a statement about the conservation of
different forms of energy. Like the Second Law, it is based on experience.
In words, the First Law says that energy cannot be created or destroyed
but is just transformed from one form into another.
The "forms" of energy are heat, work and internal energy. Heat and
(mechanical) work are perhaps familiar concepts, whose precise definitions
we will consider in the following sections. Internal energy is something that
objects posses – it is a property of matter. The internal energy of a system
can be changed by supplying heat to it or by the system doing work.
The First Law can be expressed mathematically in the following way. If
we take a system from state A to state B, then there is a definite change in the
internal energy, ∆U: ∆U = UB – UA where UA and UB are the internal
energies of the system in state A and state B. Such a change can be brought
about by the system absorbing a certain amount of heat, q, and a certain
amount of work, w, being done on the system. The First Law says that the
size of these three different kinds of energy are related:
•
∆U = q + w
q is the heat absorbed by the
system: it will be positive if heat
flows into the system (an
endothermic
process)
and
negative if heat flows out of the
system
(an
exothermic
change). w is the work done
on the system: it will be
negative if the system does
work on its surroundings.
{First Law}
2.1 What is heat?
Heat is not really a thing – it is not a fluid or a substance. Rather it is the
means by which energy is transferred from a hotter body to a cooler one in
order to equalize their temperatures.
It is very hard to talk about heat without using words which imply that it
"flows" between objects. Inevitably we will fall into this error, but we must
remember that heat is something that is "done" to an object, not something
which "flows" into an object.
Heat is a form of energy and so in SI it has units of Joules.
q
"doing heat"
x
F
2.2 What is work?
Work is done when a force moves. A force F moving a distance x does
work Fx. An example would be the work done pushing an object along a
rough surface, where the force is due to friction. Like heat, work is
5
"doing work"
something that is done to objects – for example when they are compressed
by the action of an external force.
Like heat, work is a form of energy and in SI it has units of Joules.
interaction
energy
bond
energy
kinetic
energy
Internal energy is the sum of
the kinetic energy of the
particles, together with their
energies
of
interaction,
including bond energies.
2.3 What is internal energy?
Internal energy is quite distinct from heat and work. It is a property that an
object possesses; different substances have different internal energies, and
the amount of internal energy also depends on temperature, pressure and so
on.
The internal energy is locked up inside a substance in the kinetic energy of
the particles and the interactions between the particles, particularly chemical
bonds. Internal energy is closely analogous to potential energy in mechanical
systems – it is stored up energy.
Internal energy is also closely related to temperature. For an ideal gas (in
which there are no interactions between particles) all of the internal energy is
present as the kinetic energy of the particles. If the temperature is raised, the
particles move more quickly and the internal energy is therefore raised.
Similarly, for solids and liquids an increase in temperature implies an
increase in internal energy.
2.4 State functions and path functions
There is another important distinction between U on the one hand and q and
w on the other. U is a state function, whereas q and w are path functions.
State functions play a particularly important role in thermodynamics and
several such functions will be introduced in this course.
2.4.1 State functions
A state function is one whose value depends only on the state of the
substance under consideration; it has the same value for a given state, no
matter how that state was arrived at.
By state we mean specified temperature, pressure, chemical composition
and so on – in fact all the necessary variables to define the state of the
system. For example, the state of an ideal gas is specified completely by the
number of moles, the temperature and pressure. For a mixture of ideal
gases, we would also need to know the amount of each substance present.
For chemical reactions the idea of a state function is used in Hess's Law.
For example, consider the cycle
1
C + O2
CO2
2
3
CO +
1
/2 O 2
The change in (internal) energy on going from C + O2 to CO2 is the same
whether it is done directly, step (1), or via step (2) and then step (3).
We have not proved that U is a state function. We will just appeal to
6
experience and say that all we know about heat, work and internal energy in
chemical and physical processes confirms that U is indeed a state function.
2.4.2 Path functions
The value which a path function takes depends on the path which the system
takes going from A to B.
For example, consider the reaction between hydrogen and oxygen to give
water: H2 + 12 O2 → H2 O. We could just let the hydrogen burn, releasing heat
and doing no work. Alternatively, we could react the gases in a fuel cell,
thereby creating electrical energy which could be used to drive a motor and
hence do mechanical work, such as lifting a weight. The amount of heat and
work done depends on how the overall conversion H2 + 12 O2 → H2 O is
carried out, that is on the path taken. Hence the description of heat and work
as path functions.
Having now seen what U, q and w are and the distinction between state
and path functions, we can interpret the First Law in the following way:
Α
→ B
∆U = U B − U A
=q+w
State A has a certain amount of internal energy, UA , and similarly state B has
internal energy UB . In going from A to B there is a change in internal energy
∆U, = UB – UA , which is the same regardless of the path taken between A
and B. This energy can appear as heat or work, and these can be in any
proportions provided that their sum, q+w, is equal to ∆U. The way in which
the change in internal energy is partitioned between heat and work will
depend on the path taken.
What the First Law really does is to establish the idea of internal energy,
U, as a state function.
2.5 Sign conventions
Heat and work are signed quantities, and we need to have a convention about
what the signs mean. The convention we use is that positive changes are
those done to the system.
For heat, q is the heat absorbed by the system. It is therefore positive for a
change in which heat is absorbed by the system, that is an endothermic
process, and negative for a change in which heat is given out by the system,
an exothermic process.
For work, w is the work done on the system. This means that when a gas
is compressed by an external force the work is positive, whereas when a gas
expands by pushing against an external force the work is negative.
Sometimes it is convenient to think about the work done by the system.
This is given the symbol w' and is simply the reverse of the work done on
the system: w' = –w.
In going from state A to B there is a definite change in the value of a state
function, such as U, as it has a particular value for A and for B; it is thus
acceptable to write this change as ∆U, meaning (UB – UA ). In contrast, for a
path function such as heat, was cannot talk about its value for state A or state
7
w1
UB
∆U
q1
UA
B
w2
w3
q2
A
q3
A change from A to B is
accompanied by an increase in
internal energy ∆U. This can
appear as heat or work, in any
combination, provided that ∆U
= q + w. In the diagram, upward
pointing
arrows
represent
positive
quantities,
and
downward
pointing
arrows
represent negative quantities.
B and hence a change in q "∆q". All we can specify is the amount of heat, q,
involved in going from A to B. The ∆ symbol is therefore only appropriate
for state functions.
3. Gas expansions
To explore the idea of heat and work as path functions, and to illustrate the
important concept of reversibility, we will look at the work done when a gas
expands. The results that we will obtain turn out to be practically useful and
lead to ideas which are applicable in a much wider context. However, before
starting on this we will remind ourselves about gas laws.
3.1 Gas laws
For n moles of a gas, the pressure, p, volume, V, and absolute temperature,
T, are related by what is called an equation of state or gas law. An ideal gas
is one which has the following equation of state
Exercise 1
For non-ideal gases the
equation of state is much
more complex and different
for each gas.
gas
cylinder
pext
pint
•
pV = nRT
{Ideal gas law}
–1
–1
where R is the gas constant (8.3145 J K mol ). At a molecular level such
a law is expected for a gas which is composed of molecules (or atoms)
which do not interact significantly with one another and are negligibly small
compared to the space between the molecules.
This gas law embodies Boyle's Law, which states that, at constant
temperature, the volume and pressure are inversely proportional to one
another and Charles' Law, which states that for a fixed volume the pressure
is proportional to the temperature.
If the molecules start to interact with one another the gas ceases to be ideal
and no longer obeys the ideal gas law. For simplicity, we will assume that
all of the gases we are dealing with are ideal – an approximation which is
reasonable provided we stick to modest pressures and temperatures.
3.2 How to think about a gas expansion
We start out by thinking of a gas confined inside a cylinder by a piston. The
gas, cylinder and piston form our system, and, so that we can just focus on
the work done by the gas, we will assume that the piston has no mass and
that it moves without friction.
The pressure of the gas inside the cylinder is pint, and the pressure of gas
outside is pext. Both gas pressures exert a force on the piston; the higher the
pressure the higher the force.
Suppose that the pressure outside is lower than the pressure inside. The
forces on the piston will be such that it moves out. What is the work done?
The key point to understand here is that the force which the system moves
against is that due to the external pressure. Of course, the fact that the
internal pressure is higher is the reason that the piston moves out, but
nevertheless the force which is being pushed back is that due to the external
pressure.
8
Suppose that the piston has area A. Then, as pressure is force per unit
area, the force on the piston due to the external pressure is pextA. If the piston
moves out by a small distance dx (we are using the language of calculus)
then the small amount of work done by the gas, dw', is
pressure × area
= (force/area) × area
= force
dw' = force × distance
= pext A dx
The quantity Adx is the volume by which the gas inside the piston has
increased (recall that the volume of a cylinder is the area of the base times the
height – here the base area is A and the height is dx). Writing this volume as
dV, the work done is
•
dw' = pext dV
area, A
dx
[1]
We can now use this relationship to find the work done in some special
cases.
If we want to think in terms of the work done on the gas, w, then as
w' = –w (Section 2.5), the relationship becomes
The volume swept out by the
piston of area A when it moves
through a distance dx is A dx.
dw = − pext dV
If the external pressure is zero then clearly no work is done i.e. no work is
done expanding against a vacuum.
3.3 Expansion against constant external pressure
If the external pressure is constant, Eq. [1] is easy to integrate . If the gas
expands between volume Vi and Vf, then the work done is found by
integrating between these two limits:
Vf
∫ dw' = ∫ p
ext
a
Vf
dV
pext
Vi
b
pext
Vi
Vf
= pext ∫ dV
Vi
= pext [V ]Vfi = pext (Vf − Vi )
V
On the second line we have assumed that pext is constant, and so it can be
moved outside the integral. The left hand side we will just call w', the work
done. So, the final result is
w' = pext (Vf − Vi )
{const. external press.}
This expression assumes that throughout the expansion the internal pressure
is always greater than the external pressure; remember that as the gas
expands the internal pressure will fall.
If the external pressure exceeds the internal pressure, the gas will be
compressed; the final volume will be less than the initial volume and so w' is
negative. This is as expected, as in this situation work is done on the gas.
3.4 Expansion doing maximum work
For a given initial gas pressure in the cylinder, how can we arrange things so
that the maximum amount of work is done? As it is the external pressure
9
Visualization
of
expansion
against a constant external
pressure. To start with the
internal pressure is greater
than the external pressure, so
the pisto n wants to move out
but is restrained from doing so
by the peg a; the volume is V i.
When peg a is removed the
piston moves out until it hits
peg b; the volume is now V f .
Example 1
which provides the force against which the gas pushes, it follows that to
maximize the work the external pressure needs to be as high as possible.
However, this pressure cannot exceed the internal pressure, otherwise the gas
would not expand but would instead be compressed.
So, to obtain the maximum work the external pressure needs to be
infinitesimally less that the internal pressure.
If we start with this situation, the gas will expand and as a result the
internal pressure will fall. Eventually the internal pressure will fall to the
value of the external pressure, and the expansion will stop. To continue the
expansion the external pressure needs to be lowered, again by an amount that
makes it infinitesimally less that the internal pressure. Again, the gas
expands until the pressures are equalized, and then once more we have to
lower the external pressure.
In summary, the condition for doing the maximum work is that the
external pressure should always be kept infinitesimally less than the internal
pressure.
reversible processes:
• infinitely slow
• at equilibrium
• do maximum work
irreversible processes
(spontaneous):
• go at finite rate
• not at equilibrium
• do less than
the maximum work
3.4.1 Reversible and irreversible processes
An expansion in which the external pressure is always just less than the
internal pressure does the maximum work but takes place infinitely slowly.
At any point the expansion can be turned into a compression simply by
making the external pressure infinitesimally greater that the internal
pressure. Such a process, whose direction can be changed by an
infinitesimal change in some variable (in this case the pressure), is said to be
reversible.
In contrast, an expansion against a fixed external pressure proceeds at a
finite rate until the internal and external pressures are equalized. Such a
process cannot be reversed by an infinitesimal change in the external
pressure – rather it would need to be increased by a significant amount until
it exceeds the internal pressure. Such a process is said to be irreversible.
Spontaneous processes are inherently irreversible.
We have seen in Section 3.4 that a reversible expansion of a gas does the
maximum work. Although we only demonstrated it for the case of a gas
expansion, it turns out to be generally true that any reversible process does
the maximum work:
a reversible process is one in which the work is a maximum
or the work done in a reversible process is a maximum
The converse is that an irreversible process always does less work than the
corresponding reversible processes.
Reversible processes are also called equilibrium processes. The idea is
that in a reversible process, such as a gas expansion, the system is always in
balance (the external and internal pressures are only infinitesimally different)
and so essentially in equilibrium.
10
3.4.2 Reversible isothermal expansion of an ideal gas
If we assume that we have an ideal gas and that it is expanding reversibly we
can integrate Eq. [1] to find out the work done. To make the process
reversible the external pressure is essentially equal to the internal pressure so
Vf
∫ dw' = ∫ p
int
dV
[2]
Vi
For an ideal gas
pint V = nRT
where V is the volume, n is the number of moles, R is the gas constant and T
is the temperature. Rearranging this gives us an expression for pint in terms
of V
nRT
V
This can then be substituted into Eq. [2] and integrated
pint =
Vf
nRT
dV
V
Vi
∫ dw' = ∫
Vf
1
dV
V
Vi
= nRT ∫
= nRT [ln V ]Vfi = nRT ln
V
Vf
Vi
On the second line we have imposed the condition of constant temperature,
and so have taken the T outside the integral.
So, the work done in the reversible, isothermal expansion of an ideal gas
is
w' = nRT ln
Vf
Vi
{isothermal, reversible, ideal gas}
3.5 Heat changes in gas expansions
Having calculated the work done in a gas expansion, we can use the First
Law to say something about the heat changes.
11
Isothermal me ans at constant
temperature. For example, we
might
achieve
this
by
surrounding the system with a
large
water
bath
whose
temperature
is
carefully
regulated. Heat flows into or
out of the water bath so as to
keep the objects of interest at a
constant temperature.
B
∆U
qirrev
qrev
For any gas expansion ∆U will be a certain fixed value, regardless of
whether or not the expansion is reversible or irreversible (recall that U is a
state function). Hence from the First Law
∆U = q + w or ∆U = q − w'
A
wirrev
wrev
Going from state A to state B
has the same ∆U regardless of
the path. In the reversible case
the work has its greatest
absolute value; in the case
shown above wrev is more
negative that wir rev. It follows
from the First Law that the heat
absorbed in the reversible
process is a maximum; in this
case q rev is more positive than
q irrev .
Exercises 2 & 3,
Example 2
it follows that if w' is a maximum in a reversible change, so too must q be a
maximum (the signs get a bit confusing here – it is best to think of the
absolute values of q and w' both being a maximum).
So, in a reversible process the heat absorbed is a maximum.
For the special case of an ideal gas, it turns out that U only depends on
temperature. At a molecular level, this is because the internal energy of an
ideal gas is present entirely as kinetic energy of the molecules (Section 2.3).
If the gas is heated, that is given energy, the only place that the energy can be
stored is in the motion of the molecules. As there are no interactions
between the molecules, altering the pressure or volume at constant
temperature does not affect the internal energy.
So, for an isothermal expansion of an ideal gas ∆U = 0, and hence from
the First Law 0 = q – w' or w' = q. In Section 3.4.2 we have calculated the
value of w' so we immediately know the value of q.
What is happening in this isothermal expansion is that the work done in
the expansion is exactly compensated for by the heat flowing in. If no heat
flowed in, some of the internal energy would have to be converted to work,
and as a result the temperature of the gas would drop. However, we
specified that the process had to be isothermal, so heat must be supplied in
order to maintain the temperature (internal energy) of the gas.
4. Entropy
In Section 1 we defined the entropy change, for an infinitesimal process, as
δqrev
T
For a finite change in which the system is taken from state A to state B the
entropy change is
dS =
qrev
T
We can now see why in this definition it is necessary to specify the
conditions under which the heat change is taken: different paths have
different heat changes. Although we will not demonstrate this, the above
definition of entropy makes it a state function.
∆S =
4.1 Entropy changes – considering the system and the
surroundings separately
The Second Law states that the entropy of the Universe must increase in a
spontaneous process. In Section 1 we introduced the idea of thinking about
the Universe as composed of the thing we are interested in, the system, and
the rest, the surroundings.
12
The entropy change of the universe can be therefore be separated into the
entropy change of the system, ∆Ssys , and the entropy change of the
surroundings, ∆Ssurr
∆Suniv = ∆Ssys + ∆Ssurr
In the next sections we will look at how we can determine each of these
quantities.
4.1.1 Entropy change of the system
Using the definition of entropy from Section 1, the entropy change of the
system is
∆Ssys =
qrev,sys
Tsys
The reversible subscript is important. Even if the process is not actually
reversible we must calculate the entropy change using the heat change that
would be observed if the process takes place reversibly.
For example, consider the gas expansions described in Section 3. The
work done in an irreversible expansion of an ideal gas from Vi to Vf is
pext(Vf – Vi ) so, as ∆U = 0 for an isothermal expansion of an ideal gas, the
heat is pext(Vf – Vi ). If the same expansion is brought about reversibly the
work done and the heat absorbed is nRT ln Vf Vi . The entropy change of the
gas in both cases is computed using the reversible heat
∆Ssys =
=
qrev,sys
Tsys
nRTsys ln(Vf Vi )
Tsys
= nR ln(Vf Vi )
{isothermal}
In subsequent sections, we will see some other ways of calculating the
entropy change. All these methods involve determining the reversible heat,
although sometimes by an indirect method.
4.1.2 Entropy change of the surroundings
Using the definition of entropy from Section 1, the entropy change of the
surroundings is
∆Ssurr =
qrev,surr
Tsurr
where qrev,surr is the heat absorbed by the surroundings under reversible
conditions, and Tsurr is their temperature.
However, remember that the surroundings are very large – the rest of the
Universe – so any heat transferred to or from them really has no significant
effect. If the surroundings absorbs 100 kJ it is unaffected and its
temperature will not rise, and so it can equally well give up the 100 kJ and
still remain unaffected.
In effect, therefore, all heat exchanged with the surroundings is reversible
13
surroundings
from the point of view of the surroundings. The qualification is important.
From the point of view of the system 100 kJ is a lot of heat, and absorbing
this is likely to be an irreversible process. It is only because of the size of the
surroundings that the heat exchanged with it is reversible.
Now, the heat that the surroundings absorbs comes from the system so:
qsys = − qsurr
system
Heat
flowing
from
the
surroundings to the system is
an endothermic process from
the points of view of the system
and an exothermic process
from the point of view of the
surroundings. q sys = –q surr .
In words, a process which is endothermic for the system (qsys is positive and
heat is absorbed) is exothermic from the point of view of the surroundings
as the heat absorbed by the system must have been given out by the
surroundings (qsurr is negative). Heat flows the other way from the point of
view of the surroundings.
Putting this all together, we can write ∆Ssurr as
∆Ssurr =
Exercises 4 & 5
− qsys
Tsurr
[3]
qsys is the actual heat change between the system and the surroundings
which, since they are very large, receives it reversibly.
Equation [3] is very convenient, as it enables us to calculate the entropy
change of the surroundings (a very big and complex object!) just by
knowing the heat absorbed by the system, which is easy enough to measure,
and the temperature of the surroundings.
Exchanging heat with the surroundings is a way of altering its entropy –
in particular, exothermic processes give heat to the surroundings and hence
increase its entropy.
4.1.3 Entropy change of the Universe
Using Eq. [3] we can write the entropy change of the Universe as
∆Suniv = ∆Ssys + ∆Ssurr
= ∆Ssys −
heat transfer to
surroundings,
increase in entropy
of surroundings
increase in order of system
decrease in entropy of system
The creation of order in the
system, which is accompanied
by a decrease in its entro py,
will only be spontaneous if the
process is exothermic enough
that the entropy of the
surroundings
increases
sufficiently to compensate for
the reduction in entropy of the
system.
qsys
Tsurr
This result enables us to understand a lot about what makes some processes
spontaneous and some not.
Endothermic processes decrease the entropy of the surroundings by
taking in heat; ∆Ssurr = –qsys /T is negative. They can be spontaneous
(∆Suniv > 0) provided that ∆Ssys is sufficiently positive to overcome the
negative ∆Ssurr term. In practice this means that spontaneous endothermic
processes must involve an increase in entropy (disorder) of the system.
For example, ice melting is a spontaneous endothermic process and, as
expected, the system's entropy increases (solid → liquid, increase in
disorder). Likewise dissolving solid NH4 NO3 in water is endothermic, and it
is also accompanied by an increase in the entropy of the system (solid +
liquid → solution, increase in disorder). In both cases the increase in entropy
of the system more than compensates for the decrease in entropy of the
surroundings.
Processes which create order in the system (∆Ssys < 0) can be
14
spontaneous provided they are sufficiently exothermic that the entropy
change of the surroundings, the –qsys /T term (which is positive for an
exothermic process), overcomes the negative ∆Ssys .
For example, ice freezing involves a decrease in the entropy of the
system, but the process is exothermic so that the entropy of the surroundings
increases. Provided that the temperature of the surroundings are low enough
the increase in the entropy of the surroundings will be large enough to
compensate for the decrease in entropy of the water as it freezes and so the
process will be spontaneous.
Put loosely, if we want to create order, we have to "pay" by increasing the
entropy of the surroundings, and we do this by giving out heat. The heat is
a kind of "tax" we have to pay to create order.
In the next few sections we will introduce some new state functions
which will enable us to use the ideas of this section in a more practical way.
We will also see how we can measure the entropy of a substance.
Example 3
5. Internal energy, enthalpy and heat capacity
5.1 Differential forms
So far we have written the First Law as ∆U = q + w, where the ∆ implies a
finite change in U. We could just as well have written it for an infinitesimal
change in U, using the language of calculus:
dU = dq + dw
To be strict, we ought not to write dq and dw as q and w are not state
functions so to talk of infinitesimal changes in their values is not appropriate;
what dq and dw represent are infinitesimal amounts of heat and work (more
properly represented as δq and δw), whereas dU represents an infinitesimal
change in internal energy. However, for convenience we will ignore this
distinction and write dq and dw.
In Section 3 we found that the work of expanding a gas was
dw = – p dV, so, if this is the only kind of work done, the First Law can be
written
dU = dq – p dV
{pV work only} [4]
Unless we say otherwise, we will assume that the only kind of work done is
that due to gas expansions, and so use the form of the First Law given in Eq.
[4].
The definition of entropy can be written in differential form as
dqrev
T
where, for the reasons explained above, we have written dq rather than δq.
dS =
15
Work can take many forms,
such as that due to gas
expansions, due to moving
charges (a current) in an
electrical field or due to
expanding a surface. Work due
to
gas
expansions
is
sometimes called " pV work".
q
If heat is supplied to a gas in a
sealed container
(constant
volume) no work is done as the
gas cannot expand. All of the
heat therefore ends up as
internal energy of the gas. As
the internal energy rises, so
does the temperature of the
gas.
5.2 Constant volume processes
If we imagine a process taking place in a sealed container, that is at constant
volume, then no work can be done as the gas cannot expand. In terms of
Eq. [4] the pdV term is zero (dV = 0 as the volume does not change) and so
dU = dqconst. vol
{constant volume}
This says that the heat absorbed under conditions of constant volume is equal
to the change in internal energy; in other words, all the heat goes into internal
energy. If we want to measure ∆U, therefore, all we have to do is measure
the heat change under constant volume conditions.
It is important to realize that, for a given change, ∆U has a certain value
because U is a state function. In the special case of a constant volume
process ∆U is equal to the heat absorbed, but under other conditions
although ∆U is no longer equal to the heat it still has a well defined value.
5.2.1 Heat capacities
When heat is supplied to an object its temperature usually increases; the
more heat that is supplied the greater the temperature rise. The relationship
between the heat supplied, q, and the temperature rise, ∆T, is
q = c ∆T
where c is the heat capacity of the object. For infinitesimal quantities this
relationship becomes dq = c dT .
The larger the object the greater the heat capacity. If we are talking about
chemical substances it is convenient to use the molar heat capacity, C, which
is the amount of heat needed to raise one mole of the substance through one
degree. C has units of J K–1 mol–1. Using this molar quantity the above
equation becomes
q = n C ∆T
where n is the number of moles.
For a process taking place at constant volume, we have seen in Section
5.2 that the heat is equal to the change in internal energy: ∆U = qconst. vol , so
we can write
∆U = n CV ∆T
where CV is the molar heat capacity at constant volume.
Writing this in the differential form we have dU = dqconst. vol , so
dU = CV dT
{constant volume}[5]
where we are now assuming that dU is the change in internal energy per
mole (i.e. n = 1). Equation [5] can be rearranged to
CV =
Typical values (in J K mol–1 and
at 1 bar and 298 K) for CV for
gases: noble gases 12.5, H2
20.4, O2 21.0, CO 2 28.5
 dU 
 dT  constant volume
[6]
We have written "constant volume" to remind ourselves that this only
applies to constant volume processes. Mathematically, Eq. [6] can be
expressed as a partial derivative:
16
 ∂U 
CV = 

 ∂T  V
•
{definition of CV } [7]
The "curly d", ∂, indicates that only the variation of U with T is being
considered, and the subscript V indicates that volume is to be held constant.
Equation [7] is taken as the definition of the constant volume heat capacity.
5.3 Constant pressure processes – enthalpy
Processes taking place under constant pressure conditions are more
common, especially in chemistry, where open apparatus "on the bench" can
be considered to be at constant pressure.
If we heat a gas at constant external pressure it will expand and by doing
so it will do work. Thus, some of the heat is converted into work and some
is converted to internal energy. For a given amount of heat, the increase in
the internal energy will be less in a constant pressure process than in a
constant volume one on account of part of the heat being converted to work.
It will be convenient to have a state function whose value is equal to this
heat supplied at constant pressure; this function will be the constant pressure
analogue of U.
The function we need is called the enthalpy, H and it is defined as
•
H = U + pV
{definition of H} [8]
From this it follows that H is a state function as it is defined in terms of other
state functions (U) and variables (p and V) which define the state of the
system.
It will be useful to alter the form of Eq. [8] so that it is expressed in terms
of the change in H, dH. The procedure is as follows: suppose that p changes
by a small amount dp giving a new value (p + dp), likewise V changes by a
small amount dV giving a new value (V + dV), and likewise U changes by a
small amount dU giving a new value (U + dU). As a result of the changes
in p, V and U, it is clear that H will also change by a small amount dH giving
a new value (H + dH). So, from Eq. [8] we have
( H + d H ) = (U + d U ) + ( p + d p)(V + d V )
.
Multiplying this out gives
H + d H = U + pV + d U + pd V + Vd p + d p d V .
We ignore the last term, dp dV, because it is the square on an infinitesimal
quantity and also note that as H = U + pV, the term H on the left and
(U + pV) on the right will cancel to give
dH = dU + pdV + Vdp
[10]
Equation [10] is called the complete differential of Eq. [8].
Another way of thinking about how to take the complete differential is
just to differentiate both sides of Eq. [8]
d H = d U + d ( pV )
and recognise that pV is a product of two functions and so needs to be
differentiated using the usual rule
17
q
w'
Supplying heat to a gas held at
constant pressure (here held in
a cylinder by a piston) results in
the gas expanding and so doing
work against the external
pressure. Some of the heat
supplied increases the internal
energy and some appears as
work of expansion.
d H = d U + pdV + Vdp
[10]
Once we have seen how this works we can go straight from relationships
like Eq. [8] to their complete differentials without doing the intermediate
steps.
Now we substitute in the First Law, in the form of the expression for dU
from Eq. [4], into Eq. [10]
dH = dq − pdV + pdV + Vdp
= dq + Vdp
Finally, we impose the condition of constant external pressure, which means
that the term Vdp is zero (the change in p, dp, is zero), to give
dH = dqconst. press.
What we have shown is that the enthalpy change is equal to the heat change
measured under conditions of constant pressure. With hindsight we can see
that this is precisely why H is defined as it is (Eq. [8]) so that it will have this
useful property.
Just as with ∆U, we must not forget that ∆H has a defined value for a
particular change, regardless of whether or not it is at constant pressure. It is
only if the pressure is constant that q is equal to ∆H.
Chemists tend to talk of "heat of reaction" by which they normally mean
the enthalpy of reaction. Enthalpies are the quantities which are most
commonly tabulated.
5.3.1 Heat capacity at constant pressure
For a process taking place at constant pressure, the heat absorbed is equal to
the change in H: dH = dqconst. press. (Section 5.3). Following a similar line of
argument as in Section 5.2.1, we have
•
Since in a constant pressure
process some of the heat
supplied ends up as work of
expansion,
more heat
is
needed
to
raise
the
temperature of a gas at
constant pressure than is
needed at constant volume i.e.
Cp > CV . For an ideal gas
Cp – C V = R.
dH = Cp dT
{constant pressure}
where Cp is the constant pressure molar heat capacity. Note that in this
equation as the heat capacity is expressed per the mole, the enthalpy is also
per mole. As with CV , the usual definition of Cp is as a partial derivative
 ∂H 
Cp =  
 ∂T  p
{definition of Cp } [11]
5.3.2 Variation of enthalpy with temperature
Suppose we know the (molar) enthalpy, H(T1 ), of a substance at a particular
temperature, T1 , but we want to know its (molar) enthalpy, H(T2 ), at another
temperature T2 . This is quite a common situation as data are often tabulated
at only a few temperatures, and these may not correspond to the temperature
of interest.
Heat capacities are the key to finding how H varies with T. We start with
Eq. [11]
 ∂H 
Cp =  
 ∂T  p
and rearrange this to give
18
[11]
dH = Cp dT
{constant pressure} [12]
We have dropped the partial derivative symbol and moved to working with
normal derivatives, but have made ourselves a mental note that all of what
follows is only valid at constant pressure.
Both sides of Eq. [12] can be integrated. First the left-hand side
T2
∫ dH = [ H ]
T2
T1
T1
= H (T2 ) − H (T1 )
This needs a little explanation. The integral of dx is just x, so in the same
way the integral of dH is H. However, the limits of integration are T1 and T2 ,
which means we need to evaluate H at these two temperatures. These values
are written H(T1 ), meaning "the value of H at temperature T1 ", and likewise
H(T2 ).
The right-hand side of Eq. [12] is also integrated, and we make the
additional assumption that Cp does not vary with temperature, so it can be
taken outside the integral
T2
T2
∫ C dT = C ∫ dT
p
p
T1
T1
= Cp [T ]T12
T
= Cp [T2 − T1 ]
Putting the left- and right-hand sides together we have
H (T2 ) − H (T1 ) = Cp [T2 − T1 ]
or
H (T2 ) = H (T1 ) + Cp [T2 − T1 ]
[13]
This is a practical relationship: if we know H at one temperature and the heat
capacity, we can work out H at any other temperature (assuming that it is
valid to consider Cp as constant in this range). Note that as Cp is a molar
quantity, Eq. [13] gives the change in molar enthalpy. To compute the
change for n moles, we simply multiply the term in Cp by n so that the terms
becomes nCp [T2 – T1 ].
It turns out that we can only measure changes in enthalpy rather than the
absolute values of the enthalpies of substances. However, we will see in
Section 8.5 how a modified form of Eq. [13] can be used to convert ∆H
values from one temperature to another
19
Exercises 6 & 7
heat
temperature
The heat
capacity
at a
particular temperature is the
slope at that temperature of a
plot of heat supplied against
temperature.
5.4 Measurement and tabulation of heat capacities
Heat capacities turn out to be surprisingly useful when it comes to
manipulating various thermodynamic quantities. They are easy to measure,
as all we need to do is measure the temperature rise for a know heat input.
The heat capacities of a great many substances have been measured and are
available in tabulated form.
It is found that the heat capacity does vary with temperature, although
generally not too strongly. To get around this, heat capacities are often
expressed in the parametrized form
c
T2
We have written the heat capacity as C(T), meaning that "C is a function of
T". Armed with the values of the parameters a, b and c, the heat capacity can
be evaluated at any temperature.
C(T ) = a + bT +
6. Measuring entropy
6.1 Absolute entropies
Entropies can be evaluated from measurements of heat capacities in the
following way.
The definition of entropy is, in differential form
dqrev
T
If we think about a process at constant pressure the heat change is equal to
the enthalpy change: dqconst. press. = dH (Section 5.3), so
dS =
dH
[14]
T
We do not need to specify "reversible" as H is a state function and takes the
same value for any path. Next we use the definition of the constant pressure
heat capacity to express dH in terms of Cp :
dS =
 ∂H 
Cp =  
 ∂T  p
and hence dH = Cp dT
[11]
Substituting this into Eq. [14] gives
dS =
In principle there are other
ways of finding entropies, but
this method is the one most
commonly used
as
heat
capacities are relatively simple
to measure.
Cp d T
T
dS Cp
=
dT
T
or
{const. pressure}
To find the entropy at temperature T* both sides are integrated from T = 0 up
to T*
T*
∫ dS =
0
S(T *) − S(0) =
20
C p (T )
∫0 T dT
T*
C p (T )
dT
T
0
T*
∫
[15]
To evaluate the left-hand side a similar line of argument has been used as in
the Section 5.3.2. We have written the heat capacity as Cp (T) to remind
ourselves that Cp depends on T. Over a small temperature range it may be
acceptable to assume that it is constant, but not over a wide range.
The right-hand side of Eq. [15] has to be evaluated graphically. We make
measurements of Cp as a function of temperature and then plot Cp (T)/T
against T all the way from 0 to T*. The area under the curve is S(T*)–S(0).
By convention, the entropy at absolute zero is set to zero. In fact this is
not an arbitrary choice, but is founded on an understanding of entropy as a
statistical concept. We will look at this further in Section 14.
Measuring heat capacities right down to absolute zero (or, in practice, as
close as we can get) is not a particularly easy task but it has been done for
many substances. The entropy values determined in this way are called
absolute entropies.
There may well be phase changes (solid to liquid, for example) over the
range of temperature for which we want to plot the graph. At the precise
temperature of the phase change the heat capacity goes to infinity (supplying
heat does not change the temperature) and it is also commonly observed that
there is a jump in the plot. We will see in Section 15.3.1 that the entropy
associated with a phase change, ∆Spc, is related to the enthalpy of the phase
change, ∆Hpc, by
∆Spc =
∆Hpc
Tpc
These extra entropy changes need to be included in the calculation of absolute
entropies.
S(T *) =
∆Hpc
C p (T )
dT + ∑
T
phase changes Tpc
0
T*
∫
where the summation symbol, Σ, means to sum over all phase changes.
Usually, entropies are tabulated as molar quantities, in units J K–1 mol–1.
6.1.1 Converting entropies from one temperature to another
Absolute entropies vary with temperature but tend to be tabulated at just one
or two temperatures. It will therefore be convenient to have a way of
converting the value of the entropy from one temperature to another, just as
we did for enthalpies in Section 5.3.2.
The starting point is Eq. [14] from Section 6.1. As before we integrate
this but this time just between T1 and T2 . Provided that these two
temperatures are not too different, we can assume that Cp will be constant
and so take it out of the integrand:
21
Cp
T
0
T* T
Evaluating entropy: the shaded
area is S (T*) – S(0).
Cp
T
0 T
Tpc T*
A phase change at temperature
Tpc results in a discontinuity in
the Cp/ T against T plot. The
entropy change due to the
phase change needs to be
added to the shaded area in
order to find S (T*).
T2
T2
Cp
dT
T
T1
∫ dS = ∫
T1
T2
1
dT
T
T1
S(T2 ) − S(T1 ) = Cp ∫
S(T2 ) − S(T1 ) = Cp [ln T ]T12
T
S(T2 ) − S(T1 ) = Cp ln
Example 4
Exercise 8
T2
T1
or S(T2 ) = S(T1 ) + Cp ln
T2
T1
[16]
Equation [16] is a practical relationship for converting entropies from one
temperature to another. Note that as Cp is a molar quantity Eq. [16] gives us
the change in the molar entropy. To compute the change for n moles we
simply multiply the term in Cp by n: nCp ln(T2 /T1 ).
7. Gibbs energy
The Second Law gives us a criterion for which processes are spontaneous,
and which are at equilibrium (reversible) – that is ∆Suniv > 0 for a
spontaneous process and ∆Suniv = 0 for a process which has come to
equilibrium. However, it is not convenient to have to calculate the entropy
change of the Universe each time we want to think about a chemical or
physical process.
We will show in this section that the Gibbs energy, G, is a convenient
way of using the Second Law. G is defined as
•
G = H – TS
{definition of G} [17]
and we will see that in a spontaneous process G of the system decreases and
reaches a minimum at equilibrium. The convenience is that by employing the
Gibbs energy we only have to think about the system, and not the Universe.
Sometimes the Gibbs energy is called "the Gibbs free energy", "the Gibbs
function "or simply "the free energy".
7.1 Gibbs energy and the Universal entropy
Starting from the definition of G, Eq. [17], we can form the complete
differential as we did in Section 5.3.
dG = dH – TdS – SdT
Now, we consider a process at constant temperature, so that the SdT term is
zero. Then, we divide each side by –T to give
−
dGsys
Tsys
=−
dHsys
Tsys
+ dSsys
{const. T} [18]
To remind ourselves that all these quantities refer to the system the subscript
"sys" has been added.
Now, consider the expression for the Universal entropy found in Section
4.1.3
22
∆Suniv = ∆Ssurr + ∆Ssys = −
qsys
+ ∆Ssys
Tsurr
Writing this in differential form we have
dSuniv = −
dqsys
+ dSsys
Tsurr
We impose the further two conditions that (1) the process is taking place at
constant pressure, so dqsys = dH and (2) the system and the surroundings are
at the same temperature; then
dSuniv = −
dHsys
+ dSsys
Tsys
[19]
Comparing Eqs. [18] and [19] we see that –dGsys /T is exactly the same
thing as dSuniv.
It therefore follows that as Suniv increases in a spontaneous process, dSuniv
must be positive and so must –dGsys /T (T is always positive). Therefore
dGsys must be negative, i.e. G decreases in a spontaneous process. For
reversible (equilibrium) processes dSuniv is 0, and so is dGsys . All of this is
true under conditions of constant pressure and temperature, as these are
restrictions we have introduced earlier in this argument.
G falls in a spontaneous process i.e. dG < 0;
at equilibrium G reaches a minimum, i.e. dG = 0
(at constant temperature and pressure).
The usual units for G are kJ mol–1, i.e. it is an energy, just as U and H.
We can write Eq. [18] for finite changes as
−
∆Gsys
∆Hsys
=−
+ ∆Ssys
Tsys
Tsys
[20]
–∆Gsys /Tsys is the entropy change of the Universe and –∆Hsys /Tsys is the
entropy change of the surroundings. For a process to be spontaneous, ∆Suniv
must be positive so ∆Gsys must be negative.
Equation [20] can be written another way by multiplying both sides by –T
•
∆Gsys = ∆Hsys − T∆Ssys
[21]
We commented in Section 4.1.3 that a process which is endothermic can be
spontaneous provided ∆Ssys is sufficiently positive to compensate for the
reduction in entropy of the surroundings. In terms of Eq. [21] we would say
that an endothermic process can be spontaneous provided the ∆Ssys term is
large enough that the –T∆Ssys term overcomes the positive ∆H sys and so
makes ∆Gsys negative. Likewise, a process which has a negative ∆Ssys can be
spontaneous provided it is sufficiently exothermic that the negative ∆H sys
term overcomes the positive –T∆Ssys term. Thinking about ∆G sys is really
equivalent to thinking about ∆Suniv.
8. Chemical changes
We now need to extend what we have developed so far to cover chemical
23
Exercise 9
reactions and then chemical equilibrium. The idea of talking about the
"∆H of a reaction" is a familiar one, but to proceed we need to define more
carefully what we mean by this. In this section we will introduce the symbol
∆r to distinguish changes which accompany chemical reactions.
8.1 The ∆r symbol
Consider the balanced chemical equation:
CH4 + H2 O → CΗ3OH + H2
What this "means" is that methane and water combine in the molar ratio 1:1
to give methanol and hydrogen in the same ratio.
The enthalpy change for this reaction, ∆r H, is defined as
∆r H = H(CΗ3OH) + H(Η2) – H(CΗ4) – H(Η2O)
where H(CH3 OH) is the enthalpy of one mole of CH3 OH and so on for the
other species. This relationship takes the usual "products – reactants" form.
∆r H is the heat that would be absorbed (at constant pressure) if one mole
of CH4 reacted with one mole of H2 O and was converted completely to one
mole of CH3 OH and one mole of H2 .
∆r H is in many ways a hypothetical quantity. If we mixed CH4 with H2 O
and let them react the heat change would not be ∆r H as either they might not
react at all, or even if they did they might not go completely to products.
Only if a reaction goes 100% to products will the observed heat change be
equal to ∆r H.
A general chemical reaction can be represented by the balanced equation
νA A + νB B + ... → νPP + νQQ + ...
The "∆H" values you have
probably used up to now for
reactions should properly be
called ∆r H values.
where A, B ... are the reactants with stoichiometric coefficients νA , νB ..., and
likewise for the products P, Q ...
∆r H for this reaction is defined as
∆ r H = ν P HP + ν Q HQ + K − ν A HA − ν B HB − K
[22]
where HA is the molar enthalpy of A and so on.
A short hand way of writing Eq. [22] is to use the summation symbol to
indicate the sum over the products and the reactants
∆r H =
∑ν H
i
products, i
i
–
∑ν H
j
reactants, j
j
[22]
here νi are the stoichiometric coefficients for the products and the sum over
"products, i" means adding a term νiHi for each of the products; likewise for
the term summed over reactants.
∆r S and ∆r G are defined in the same way:
∆rS =
∆ rG =
∑ν S – ∑ν S
i i
products, i
j
reactants, j
∑ν G – ∑ν G
i
products, i
i
j
reactants, j
[23]
j
j
[24]
Again, these quantities refer to the change in S and G which would be
observed if the reactants, in their correct molar ratios, were converted
24
completely to products.
The definition of G is G = H – TS (Eq. [17], Section 7), so each term in
Eq. [24] can be written as Hi – TSi to give
∆ rG =
∑ ν {H
i
products, i
i
− TSi } –
∑ ν {H
j
reactants, j
j
− TS j
}
[24]
The terms on the right can be regrouped to give




∆ r G =  ∑ ν i Hi – ∑ ν j H j  − T  ∑ ν i Si – ∑ ν j S j 
reactants, j
reactants, j
products, i

products, i

The term in the first brace is ∆r H (compare Eq. [22]), and the term in the
second brace is ∆r S (compare Eq. [23]). So
•
∆r G = ∆r H – T∆r S
[25]
This relationship is used to calculate ∆r G.
The key point is that ∆r H, ∆r S and ∆r G all refer to a chemical process
specified by a stoichiometric equation.
8.2 Standard states
A key idea in thermodynamics is that of a standard state.
The standard state of a substance is the pure form
at a pressure of one bar and at the specified temperature
1 bar corresponds to a pressure of 105 N m–2 which is very close to, but not
quite the same as, a pressure of 1 atmosphere.
Here are some examples of standard states –
hydrogen gas: the gas at a pressure of 1 bar
solid NaCl: the pure solid
liquid water: the pure liquid
gaseous water: the gas at a pressure of 1 bar
The standard state is usually denoted by adding a superscript 0 or the
"underground" symbol e.g. S0 (H2 ) or S (H2 ) denote the standard entropy of
H2 .
It is often thought, erroneously, that the standard state implies a certain
temperature; this is not the case. The value that the standard entropy, Gibbs
energy or enthalpy takes depends on the temperature, which must be stated.
Standard changes in enthalpy etc. of reactions are defined in the same way
as in Section 8.1. For example:
∆r H0 =
∑ν H
i
products, i
0
i
–
∑ν H
j
reactants, j
0
j
The standard enthalpy of reaction, ∆r H0 , is the heat that would be absorbed,
under conditions of constant pressure, if the reactants, in their correct molar
ratios and in their standard states, were converted completely to products,
also in their standard states.
8.3 Enthalpies of formation
The standard enthalpy of formation of a compound, ∆fH0 , is the standard
25
enthalpy change for a reaction in which one mole of the compound is formed
from its constituent elements, each in their reference states.
The reference state is the most stable state of that element at a pressure of
1 bar. For example, at 298 K the reference state of nitrogen is N2 gas and the
reference state of carbon is graphite.
So the standard enthalpy of formation of liquid methanol is the standard
enthalpy change for the reaction
C(graphite) + 2H2 (g, 1 bar) + 12 O2 (g, 1 bar) → CΗ3OH(l)
With this definition the standard enthalpy of formation of elements in
their reference states is zero.
The standard enthalpy change for any reaction can be computed from
standard enthalpies of formation in a way which is best visualized as a cycle:
νA A
∆rH0
νBB
0
–νA∆fH (A)
0
–νB∆ fH (B)
νPP
νQ Q
νP∆ fH0(P)
νQ∆fH0(Q)
ELEMENTS in their reference states
The arrows on the left show the conversion of the reactants to their elements
in their reference states. The enthalpies are therefore minus the standard
enthalpies of formation; note that these have to be multiplied by the
stoichiometric coefficients νi. The arrows on the right show the formation of
the products from the elements in their reference states.
From the cycle it follows that
∆ r H 0 = ν P ∆ f HP0 + ν Q ∆ f HQ0 − ν A ∆ f HA0 − ν B ∆ f HB0
or, in general,
•
∆r H0 =
∑ν ∆ H
i
products, i
f
0
i
–
∑ν ∆ H
j
reactants, j
f
0
j
which in words says "standard enthalpy of formation of products minus
reactants". All of these relationships are consequences of the fact that
enthalpy is a state function.
8.4 Entropy and Gibbs energy changes for reactions
We have already seen in Section 8.1 that the entropy change for a reaction
can be calculated as
∆rS =
∑ν S – ∑ν S
i i
products, i
j
reactants, j
j
Values of the standard (absolute) entropies of compounds are tabulated and
so this relationship can be used directly to find ∆r S0 .
26
Having found ∆r H0 and ∆r S0 , ∆r G0 is computed from Eq. [25]
∆r G0 = ∆r H0 – T∆r S0
Alternatively, we can compute ∆r G0 directly from the values of standard
Gibbs energy of formation. The standard Gibbs energy of formation of a
compound, ∆fG0 , is the Gibbs energy change for a reaction in which one
mole of the compound is formed in its standard state from its constituent
elements, each in their reference states. The standard Gibbs energy of
formation of elements in their reference states is therefore taken as zero.
As with ∆r H0 , the standard Gibbs energy change for any reaction can then
be computed from standard Gibbs energies of formation
∆ rG0 =
∑ν ∆ G
i
products, i
f
0
i
–
∑ν ∆ G
j
reactants, j
f
0
j
Using these relationships and tables of thermodynamic data we can
compute ∆r H0 , ∆r S0 and ∆r G0 for any chemical reaction.
8.5 Variation of ∆rH with temperature
In Section 5.3.2 we saw that enthalpy varies with temperature in a way
which depends on the heat capacity. If we assume that the heat capacity does
not vary with temperature in the range T1 to T2 then we found
H (T2 ) = H (T1 ) + Cp [T2 − T1 ]
[13]
Suppose we know the value of ∆r H at one temperature, T1 , but that what we
want to know is the value at another temperature, T2 . We can use the
following cycle to make this conversion
T2 :
νAA
νBB
∆rH(T2)
ν ACp(A)[T1 – T2] νBCp(B)[T1 – T2]
T1 :
ν AA
νB B
νPP
ν P Cp(P)[T 2 – T1 ]
∆ rH(T1)
νP P
ν QQ
νQCp (Q)[T 2 – T1]
νQQ
First we convert the reactants from T2 to T1 ; the enthalpy changes are given
by Eq. [13] and the expressions are shown by the arrows on the left. Cp( A ) is
the constant pressure heat capacity of species A and so on; note that the
change in enthalpy has to take into account the stoichiometric coefficients, νi.
Then the reaction takes place at T1 . Finally, the products are converted from
T1 to T2 – the arrows on the right. From the diagram it follows that
∆ r H (T2 ) = ν A Cp( A ) [T1 − T2 ] + ν BCp( B) [T1 − T2 ] + ∆ r H (T1 )
+ ν P Cp( P ) [T2 − T1 ] + ν Q Cp( Q ) [T2 − T1 ]
Equation [27] can be rearranged to
27
[27]
Example 5
[
]
∆ r H (T2 ) = ∆ r H (T1 ) + ν P Cp( P ) + ν Q Cp( Q ) − ν A Cp( A ) − ν BCp( B) [T2 − T1 ] [28]
The quantity in the first set of square brackets is defined, in analogy to the
definition of ∆r H, as ∆r Cp
∆ r Cp = ν P Cp( P ) + ν Q Cp( Q ) − ν A Cp( A ) − ν BCp( B)
•
{defn. of ∆r Cp }
So Eq. [27] becomes
∆ r H (T2 ) = ∆ r H (T1 ) + ∆ r Cp [T2 − T1 ]
[29]
This is a very useful expression as it enables us to convert enthalpy changes
for reactions from one temperature to another.
There is clearly a strong analogy between Eq. [29] and the corresponding
relationship, Eq. [13], for H. In fact, because of the way that ∆r H, ∆r S, and
∆r G are defined, any relationship for H, S and G will hold equally well for
∆r H, ∆r S, and ∆r G provided that all the quantities are adjusted to the "∆r "
form.
For example
 ∂H 
Cp =  
 ∂T  p
 ∂∆ H 
goes over to  r  = ∆ r Cp
 ∂T  p
The formal steps in establishing this connection are given in the Appendix,
Section 18.1.
8.6 Variation of ∆rS with temperature
The argument here is just the same as it was in the previous section. We use
Eq. [16] (Section 6.1.1) to covert entropies from one temperature to another
S(T2 ) = S(T1 ) + Cp ln
T2
T1
[16]
The cycle is
ν AA
T2:
(A)
νACp
T1:
ν BB
∆ rS(T2)
(B)
ln(T1/ T2) νBCp ln(T1/ T2)
νAA
ν BB
ν PP
(P)
ν PC p ln(T2/ T1)
∆rS(T1)
ν PP
νQQ
(Q)
νQCp
ln(T2/ T1)
νQQ
so
∆ r S(T2 ) = ν A Cp( A ) ln
T1
T
T
T
+ ν BCp( B) ln 1 + ∆ r S(T1 ) + ν P Cp( P ) ln 2 + ν Q Cp( Q ) ln 2
T2
T2
T1
T1
which simplifies to
28
∆ r S(T2 ) = ∆ r S(T1 ) + ∆ r Cp ln
T2
T1
This is a useful relationship as it enables us to convert ∆r S values from one
temperature to another. Note that it assumes that ∆r Cp does not vary with
temperature.
In this case the basis of the calculation is that
Cp
 ∂S 
  =
 ∂T  p
T
∆C
 ∂∆ S 
goes over to  r  = r p
 ∂T  p
T
The formal steps in establishing this are given in the Appendix, Section 18.2
9. The Master Equations
We are almost in a position to derive the key relationship between the
standard Gibbs energy change and the equilibrium constant:
∆ r G 0 = − RT ln K
[30]
∆r G0 = ∆r H0 – T∆r S0
[31]
where
To do this we first need to introduce the Master Equations and then the
concept of chemical potential. These tools are needed in order to be able to
describe the thermodynamics of mixtures, such as reactions at equilibrium.
In Section 5.1 we wrote the First Law in differential form as dU = dq +
dw, and then went on to see that if the only work done is due to gas
expansions the First Law could be written
dU = dq – pdV
{pV work only} [4]
Imagine, for the moment, a reversible process: by definition, dS = dqrev /T
(Section 4), so it follows that dqrev = TdS. Substituting this into Eq. [4] we
have
•
dU = TdS – pdV
[32]
Now, U is a state function, so dU is the same for a given change, regardless
whether or not is takes place reversibly, so Eq. [32] is true for all changes.
This seems a bit paradoxical, as Eq. [32] appears to have been derived
under the assumption that we are thinking about a reversible process. The
point is that if the process is reversible, TdS is dq and –pdV is dw; if the
process is irreversible then the two terms cannot be identified directly with
dq and dw, but their sum is still equal to dU.
Equation [32] is some times called the First and Second Laws combined,
and it is the first of the thermodynamic Master Equations. We will find that
these Master Equations are very useful in relating one thermodynamic
quantity to another.
The second Master Equation is developed by starting with the definition
of enthalpy, H (Eq. [8], Section 5.3) and then taking its complete differential
(Section 5.1)
H = U + pV
29
[8]
Example 6
Exercises 10 – 13
hence
dH = dU + pdV + Vdp
[10]
Equation [32], the first Master Equation, is used to substitute for dU in Eq.
[10] to give
dH = TdS − pdV + pdV + Vdp
hence dH = TdS + Vdp
[33]
Equation [33] is the second of the Master Equations.
The next Master Equation is found by taking the definition of G, Eq. [17]
(Section 7), forming the complete differential and them substituting in for
dH from Eq. [33]
G = H − TS
dG = dH − TdS − SdT
dG = TdS + Vdp − TdS − SdT
hence dG = Vdp − SdT
[34]
Equation [34] is the most useful of the Master Equations we will derive.
We will use it immediately to find out how G varies with pressure and
temperature.
9.1 Variation of G with p (and V), at constant T
We start with Eq. [34], one of the Master Equations
dG = Vdp − SdT
[34]
and impose the condition of constant temperature, so that dT = 0
dG = Vdp
{const. T} [35]
which can be written, using the notation of partial derivatives, as
 ∂G 
  =V
 ∂p  T
To integrate Eq. [35], and hence find a useful equation for how G varies
with p, we need to recognise that V varies with p according to the ideal gas
equation
pV = nRT
so
V=
nRT
p
Substituting this expression for V into the right-hand side of Eq. [35] and
integrating gives
30
dG =
p2
nRT
dp
p
p2
nRT
dp
p
p1
∫ dG = ∫
p1
p2
1
p
p1
[G] = nRT ∫ dp
p2
p1
G( p2 ) − G( p1 ) = nRT [ln p] p12
p
G( p2 ) − G( p1 ) = nRT ln
p2
p1
where we have written G(p) to indicate that G depends on pressure, and on
the third line we have used the fact that T is constant, so it can be taken
outside the integral.
The usual way of writing the final expression is to take p1 as the standard
pressure, p0 , of 1 bar and to write G(p0 ) as G0 , the Gibbs energy at standard
pressure
G( p) = G 0 + nRT ln
p
p0
{const. T} [36]
For one mole, n = 1, the quantities all become molar quantities, indicated by
a subscript m:
Gm ( p) = Gm0 + RT ln
p
p0
{const. T} [37]
The pressures can be in any units we like, as the only important thing is the
ratio between them. If we choose to write them in bar, p0 = 1 and it is
tempting to write
Gm ( p) = Gm0 + RT ln p
{assumes p in bar}
Apart from the objection that we appear to be taking the logarithm of a
dimensioned quantity, this form can also lead us into difficulties so it is best
to use Eq. [37].
Since, for an ideal gas, pressure is inversely proportional to volume (at
constant temperature) , it follows that p2 p1 = V1 V2 and so
G(V2 ) − G(V1 ) = nRT ln
V1
V2
or G(V2 ) = G(V1 ) + nRT ln
V1
V2
9.2 Variation of G with T, at constant p
Again, we start with the Master Equation, Eq. [34]
dG = Vdp − SdT
[34]
and this time impose the condition of constant pressure, so that Vdp = 0
dG = − SdT
{const. p} [38]
which can be written, using the notation of partial derivatives.
31
These relationships apply at
"constant temperature". This
does not mean that they apply
at one unique temperature;
rather it means that they apply
at any temperature, provided
that it remains constant during
the change.
At different
temperatures G and G0 will be
different, but they are still
related by Eq. [37].
 ∂G 
  = −S
 ∂T  p
[39]
We have already seen in Section 6.1 that S itself varies with temperature, so
integrating Eq. [38] is not straightforward.
In fact there is a special relationship, called the Gibbs-Helmholtz equation,
which we can derive from Eq. [34] and which we will need later on. The
derivation starts by considering the derivative of the function G/T with
respect to T. As G depends on T, to find the derivative we have to recognize
that we are differentiating a product of two functions
d
dG
[40]
(GT –1 ) = T –1 − T –2G
dT
dT
At constant pressure, which we will assume from now on, Eq. [39] tells us
that dG/dT = –S; we substitute this on the right of Eq. [40]. We also know
that, by definition, G = H – TS; we substitute this for G on the right of Eq.
[40]. The result is
d
GT –1 ) = T –1 ( – S ) − T –2 ( H – TS )
(
dT
= − T –1S − T –2 H + T –1S
= − T –2 H
Tidying this up gives the Gibbs-Helmholtz equation
d  G
H
=− 2
dT  T 
T
{const. pressure} [41]
9.3 Variation of S with V (and p), at constant T
In Section 9.1 we showed that, for an ideal gas at constant temperature
G( p2 ) = G( p1 ) + nRT ln
p2
p1
[42]
and in Section 9.2 that
 ∂G 
  = −S
 ∂T  p
[39]
So, differentiating both sides of Eq. [42] with respect to T at constant
pressure we have
d
d
d 
p2 
G( p2 ) =
G( p1 ) +
 nRT ln 
dT
dT
dT 
p1 
– S( p2 ) = − S( p1 ) + nR ln
S( p2 ) = S( p1 ) + nR ln
Exercise 14
p2
p1
p1
p2
or
S(V2 ) = S(V1 ) + nR ln
V2
V1
Where we have used Eq. [39] on the second line and have written S(p) to
indicate that S depends on p. On the last line we have once again used the
fact that p is inversely proportional to V.
The final equation says that the entropy of an ideal gas increases as the gas
32
expands – just as we would expect.
10. Chemical potential – dealing with mixtures
We now want to use thermodynamics to understand chemical equilibrium,
which means that we must deal with the thermodynamic properties of
mixtures of substances (reactants and products). However, none of the tools
we have discussed so far are directly applicable to mixtures, so we need to
develop a few more. The thermodynamic function known as chemical
potential turns out to be the key to dealing with mixtures; in this section we
will explore the ideas that lead to a definition of chemical potential. The
mixing of ideal gases will be used as an illustration.
10.1 The mixing of ideal gases
We know that ideal gases mix spontaneously, so the mixing must be
accompanied by a reduction in the Gibbs energy. Our first task will be to
understand how this reduction arises and then calculate the Gibbs energy
change of mixing.
Consider two gases, A and B, separated by a partition but both at the
same temperature and pressure, p; let there be nA moles of A and nB moles
of B. When the partition is removed the gases mix completely. The final
pressure is still p and the temperature does not change.
The crucial thing that has changed is the partial pressure of the gases.
The partial pressure of gas i in a mixture is the pressure that it would exert if
it occupied the whole volume on its own. For a mixture of ideal gases, the
partial pressure of i is given
pi = xi ptot
where ptot is the total pressure and xi is the mole fraction of i.
The mole fraction of i, xi, is given by
xi =
ni
ntot
ntot = ∑ ni
i
where ni is the number of moles of i and ntot is the total number of moles.
Before mixing, when gas A is in its separate compartment, its partial
pressure is p. After mixing, the mole fraction of A, xA , is nA /(nA + nB ), so
the partial pressure of A, pA , is
 nA 
pA = 
p
 nA + nB 
and similarly for B
 nB 
pB = 
p
 nA + nB 
As a result of its definition, the mole fraction of any substance in a mixture is
always less than one, so the partial pressures of the gases in the mixture are
always less than their pressures before mixing. We will show that it is this
33
nA
nB
A
B
p
p
p
reduction in pressure which leads to the change in Gibbs energy.
In section 9.1 we saw that the molar Gibbs energy of a pure gas varies
with pressure according to Eq. [37]
Gm ( p) = Gm0 + RT ln
p
p0
[37]
If we are considering ideal gases, then there is no difference between a pure
gas at pressure p and the same gas at partial pressure p in a mixture; this is
because ideal gases do not interact. Thus, we can say that the molar Gibbs
energy of i in a mixture depends on its partial pressure, pi, in exactly the
same way as Eq. [37]
0
Gm, i ( pi ) = Gm,
i + RT ln
pi
p0
[37A]
From this relationship we see that the Gibbs energy falls as the partial
pressure falls. So, the Gibbs energy of A in the mixture is less than its
Gibbs energy before mixing; the same is true of B. So, in conclusion, the
lowering of the partial pressures of the gases which takes place on mixing
results in a reduction of the Gibbs energy of each gas and hence of the
mixture as a whole; this is why gases mix.
We can now compute the change in Gibbs energy, ∆Gmix , on mixing the
two gases. The method is to compute the Gibbs energy of the mixed gases,
Gmixed , and subtract the Gibbs energy of the separated gases, Gseparated
∆Gmix = Gmixed – Gseparated
Recalling that there are nA moles of A and nB of B, we can compute G separated
using the molar Gibbs energies of A and B at pressure p, Gm,A (p) and
Gm,B (p); these in turn can be computed using Eq. [37]
Gseparated = nA Gm,A ( p) + nBGm,B ( p)
 0
 0
p
p
= nA  Gm,A
+ RT ln 0  + nB  Gm,B
+ RT ln 0 


p 
p 
In the same way, Gmixed can be computed from the molar Gibbs energies of
A at partial pressure pA and similarly for B; this time we need Eq. [37A]
Gmixed = nA Gm,A ( pA ) + nBGm,B ( pB )
 0
 0
p 
p 
= nA  Gm,A
+ RT ln A0  + nB  Gm,B
+ RT ln B0 


p 
p 
 0
 0
p
p
= nA  Gm,A
+ RT ln xA 0  + nB  Gm,B
+ RT ln x B 0 


p 
p 
On the third line we have used pA = xA ptot (here ptot = p). From these
expressions we can compute ∆Gmix
34
∆Gmix = Gmixed − Gseparated
xA
 0
 0
p
p
= nA  Gm,A
+ RT ln xA 0  + nB  Gm,B
+ RT ln x B 0 


p 
p 
 0
 0
p
p
− nA  Gm,A
+ RT ln 0  − nB  Gm,B
+ RT ln 0 


p 
p 
0.2
0.4
0.6
0.8
1.0
∆Gmix
0.0
∆Gmix of two ideal gases
= nA RT ln xA + nB RT ln xB
In a mixture, xA and xB are always less than one; the ln terms will therefore
be negative and so ∆Gmix will always be negative. As we set out to show,
the mixing of ideal gases (at constant temperature) is always accompanied by
a reduction in the Gibbs energy. A plot of ∆Gmix against xA is shown
opposite.
Exercises 15 & 16
10.2 Reacting mixtures
So far, we have established that two ideal gases will always mix. The next
question is what would happen if the two species A and B could interconvert
chemically, according to the equilibrium
B
A
Such a situation would occur if A and B were two isomers of the same
compound (such as propene and cyclopropane).
Qualitatively we know that A and B would interconvert until the
equilibrium mixture is reached, at which point there would be no further
change. This approach to equilibrium is a spontaneous process and so must
be accompanied by a reduction in the Gibbs energy. At the equilibrium
point, the Gibbs energy is a minimum.
In the previous section we found an expression for the Gibbs energy of
the mixture:
 0
 0
p
p
Gmixed = nA  Gm,A
+ RT ln xA 0  + nB  Gm,B
+ RT ln x B 0 


p 
p 
It is easier to understand this if we divide both sides by (nA + nB )
 nA   0
Gmixed
p   nB   0
p
=
  Gm,A + RT ln xA 0  + 
  Gm,B + RT ln x B 0 
nA + nB  nA + nB  
p   nA + nB  
p 
 0
 0
p
p
= xA  Gm,A
+ RT ln xA 0  + xB  Gm,B
+ RT ln x B 0 


p 
p 
where on the second line we have used the definition of xi. Remember that
as A and B interconvert, both (nA + nB ) and p remain constant. A plot of
Gmixed /(nA + nB ) as a function of xA is shown opposite; to make this plot we
have used the fact that (xA + xB ) = 1.
The plot shows a minimum in G at around xA = 0.4. What this means is
that if A and B are mixed at an initial composition of xA = 0.2 the conversion
of B into A will be spontaneous as it is accompanied by a decrease in Gibbs
energy; the result in an increase in xA . This process goes on until the
minimum is reached; from this point no further increase in xA is possible as
35
0.0
0.2
0.4
xA
0.6
0.8
1.0
Gmixed/(n A + n B ) plotted as a
function of x A . It has been
assumed,
arbitrarily,
that
G0m,A > G0m,B .
this would result in an increase in G.
If the initial composition is xA = 0.6 the conversion of A into B is
spontaneous, and so xA decreases until the minimum is reached. Either way,
the system ends up at the composition of the minimum. The point to notice
is that even though the Gibbs energy of pure B is less than that of pure A, a
mixture of A and B can have even lower Gibbs energy than pure B.
The shape of this curve is a little difficult to guess just from looking at the
equation, but it turns out always to be of the form given above. All that
changes is the position of the minimum. This is illustrated below for
0
0
different ratios of Gm,A
to Gm,B
.
G0m,A = G0m,B
0.0
0.2
0.4
xA
0.6
G0m,A = 2G0m,B
0.8
1.0
0.0
0.2
0.4
xA
0.6
0.8
G0m,A = 4G0m,B
1.0
0.0
0.2
0.4
0.6
xA
0.8
1.0
Gmixed/(n A + n B ) plotted as a function of x A for different ratios of G0m,A to G0m,B .
0
0
We see that the larger Gm,A
becomes relative to Gm,B
the further the
0
equilibrium position moves towards B (i.e. smaller xA ). If Gm,A
were to
0
become very much larger than larger Gm,B the equilibrium position would be
practically at xA = 0 i.e. pure B.
In the section 12 we will determine where this minimum is and discover,
0
0
remarkably, that its position is simply determined by (Gm,B
− Gm,A
) . Before
this, though, we need to tidy up some loose ends concerning the Gibbs
energies of mixtures.
10.3 Definition of chemical potential
If we move away from mixtures of ideal gases, Eq. [37A] is not adequate as
when there are interactions between species the Gibbs energy of gas A will
not just depend on its partial pressure but also on the identity of gas B. To
handle this situation we need to introduce chemical potentials.
The chemical potential of substance i, µi, is defined as
•
 ∂G 
µi =  
 ∂ni  p, T , n j
What this means in words is that µi gives the change in the Gibbs energy of
a mixture when the number of moles of substance i, ni, is changed by a
small amount dni under the conditions that pressure, temperature and the
number of moles of all other substances present (denoted nj) are held
constant.
A particular number of moles of each substance is described as a certain
composition; changing the amount of any one of the substances results in a
change in the composition. The above definition of the chemical potential
36
can therefore be described as the change in Gibbs energy when the amount
of i is varied at constant composition; the change in i must be infinitesimal in
order that the composition remains constant.
Physically we can imagine taking a large amount of the mixture of
substances we are interested in, measuring its Gibbs energy, then adding a
tiny amount δA of substance A and then measuring the Gibbs energy again.
The chemical potential of substance A is (change in Gibbs energy)/δA .
Chemical potentials can be used to find how changing the composition of
a mixture affects the Gibbs energy: if the number of moles of each substance
change by dni, then the resulting change in Gibbs energy of the mixture (at
constant pressure and temperature) is
•
dG = µ A dnA + µ BdnB + µC dnC + K
[44]
where µi is the chemical potential of substance i. The chemical potentials
themselves depend on temperature, pressure and composition.
Equation [44] can be integrated (at constant composition) to enable us to
calculate the Gibbs energy of a mixture:
•
G = nA µ A + nB µ B + nC µC + K
[43]
where ni is the number of moles of substance i present in the mixture.
Using chemical potentials, the Master Equation, Eq. [34], can be
modified to allow for changes in pressure, temperature and composition:
dG = Vdp − SdT + µ A dnA + µ BdnB + µC dnC + K
10.4 Variation of chemical potential with pressure
For n moles of a pure ideal gas we saw in Section 9.1 that the Gibbs energy
varies with pressure, at constant temperature, according to
G( p) = G 0 + nRT ln
p
p0
[36]
In this simple case, µ is just dG/dn at constant temperature and pressure
µ ( p) =
dG( p)
dn
dG 0 d 
p
=
+  nRT ln 0 
dn dn 
p 
µ ( p) = Gm0 + RT ln
{const. p and T}
p
p0
dG0 /dn is just the change in G0 with number of moles which, for a pure
substance, is simply the molar Gibbs energy, Gm0 i.e. the Gibbs energy per
mole.
Comparing this last result with Eq. [37]
Gm ( p) = Gm0 + RT ln
p
p0
[37]
we see that, for an ideal gas, the chemical potential is the same thing as the
molar Gibbs energy. We can also identify Gm0 with the standard chemical
37
G1
nA, nB ...
δA
G2
nA, nB ...
µA =
G2 – G1
δA
Visualization of the meaning of
chemical potential. The Gibbs
energy of the mixture is
measured with a " G meter". The
chemical
potential
of
substance A is the change in G
on adding a very small amount
of A, δA , to the mixture such
that its composition does not
change
potential, µ0 , that is the chemical potential of the gas at standard pressure.
Hence, we can write
µ ( p) = µ 0 + RT ln
p
p0
[45A]
As we argued before, for a mixture of ideal gases each component behaves
just as if it was present on its own at its partial pressure. So, the chemical
potential of species i, µi, varies with its partial pressure, pi, in an analogous
way to Eq. [45A]
•
µi ( pi ) = µi0 + RT ln
pi
p0
{const. T} [45]
µi0 is the standard chemical potential of substance i; it is the chemical
potential of i present at standard pressure (1 bar) and in its pure state.
For ideal gases, Eqs. [37A] are [45] are trivially different, but we will use
[45] as it is the one which can be adapted to deal with non-ideal situations
which we may encounter in the future. µi0 varies with temperature, so in
Eq. [45] the appropriate value of µi0 for temperature T must be used.
10.5 Chemical potentials of solids and solutions
A solid is always present in the standard state, so its chemical potential is
simply µi0 .
The chemical potential of a solute (a species in solution) depends on its
concentration, ci, in a way analogous to that in which the chemical potential
of a gas depends on its partial pressure.
ci
{const. T} [46]
c0
where c0 in the standard concentration, and µi0 is the chemical potential when
the solute is present at the standard concentration. Usually concentrations are
expressed in mol dm–3 and the standard state is taken as 1 mol dm–3.
In fact, Eq. [46] only applies to ideal solutions, which, like ideal gases, are
ones in which there are no interactions between the species present. In
practice, solutions, especially if the solutes are ions, are rarely ideal so Eq.
[46] needs to be modified to include the concept of activities. The topic of
the thermodynamics of non-ideal solutions is outside the scope of this
course.
µi (ci ) = µi0 + RT ln
11. Equilibrium constants
In the previous section we found that for the simple A
B equilibrium
the Gibbs energy is minimised at a particular ratio of A to B. This ratio is, of
course, related to the equilibrium constant. In the next section we will find
the position of equilibrium, and hence the equilibrium constant, for a general
chemical reaction. First, though, we will recap on how equilibrium constants
are defined.
If we take a balanced chemical equation
38
νA A + νB B + ... → νPP + νQQ + ...
then the equilibrium constant, K, is given by "products over reactants"
K=
[P]ν [Q]ν K
[A]ν [B]ν K
P
Q
A
B
[47]
where [A] means the concentration of A and so on. The point about the
equilibrium constant is that it has a fixed value for a given reaction at a given
temperature.
A reaction mixture at equilibrium will always have
concentrations such that the right hand side of Eq. [47] is equal to the
particular value of K for that reaction, regardless of the initial concentration of
the reagents.
There are different ways in which concentration can be expressed, and
hence different equilibrium constants. Also, we will find that we need to
define K in way which makes it dimensionless.
One way to write the equilibrium constant is in terms of partial pressures;
for gases, these are measures of concentrations. Each partial pressure is
divided by the standard pressure, p0 , so as to make the whole expression
dimensionless. The resulting equilibrium constant is called Kp .
Kp =
ν
ν
ν
ν
P
Q
 pP   pQ 
 0  0 K
p  p 
A
B
 pA   pB 
 0  0 K
p  p 
[48]
Another commonly encountered form is Kc, the equilibrium constant in
terms of concentrations; again, division by c0 , the standard concentration,
make the equilibrium constant dimensionless; typically, c0 is 1 mol dm–3.
ν
νQ
P
 cP   cQ 
  K
 c0   c0 
Kc =
νA
νB
 cA   cB 
K
 c0   c0 
[49]
The convention is that solids in the equilibrium do not contribute a term to
the equilibrium constant. Why this is so, and indeed why there is such a
thing as the equilibrium constant , is the subject of the next section.
12. Chemical Equilibrium
12.1 Condition for chemical equilibrium
In section 10.2 we calculated explicitly how the Gibbs energy of the simple
A
B equilibrium varies with the ratio of A to B (i.e. the
composition). We found that this variation was a result of the changing
partial pressures of A and B, and that there was a minimum in the Gibbs
energy which corresponds to equilibrium. We also found that at this
minimum the Gibbs energy is lower than that of either pure A or pure B.
39
Example 7
Exercise 17
To generalise these ideas for any reaction it is useful to introduce the
concept of extent of reaction, z, which is the number of moles of reactants
which have passed to products. For example, in the reaction
G
νA A + νB B + ... → νPP + νQQ + ...
a
b
z
Illustration of how G changes
as a function of the extent of
reaction, z. At point a dG/dz is
negative and so there is a
thermodynamic driving force
causing the reaction to proceed
further. At point b, dG/dz is
zero; there is thus no driving
force and the system has come
to equilibrium.
if the extent of reaction is z moles, it means that νA z moles of A have reacted
with νB z moles of B to give νPz moles of P and νQz moles of Q. The plot
opposite shows how G of the reaction mixture varies as a function of z.
The condition for equilibrium, that is the minimum in G as a function of
extent of reaction, can be expressed as
 dG 
=0
 dz 
{const. p and T}
All we need to do, therefore, is to find an expression for G in terms of the
concentrations and then minimize it with respect to z.
Suppose that the reaction proceeds by a certain amount, and let the change
in number of moles of A be dnA and likewise for the other species. The
change in G is given by Eq. [44] in Section 10.4.
dG = µ P dnP + µQ dnQ + K µ A dnA + µ BdnB + K
[50]
The dnA etc. can all be expressed in terms of a change in the extent of
reaction, dz, and the stoichiometric coefficients
dnP = +ν P dz
dnQ = +ν Q dz
dnA = –ν A dz
dnB = −ν Bdz
The terms for P and Q are positive, as these are products which increase as
the reaction proceeds. The terms for A and B are negative, as these are
reactants which decrease as the reaction proceeds.
Using these, Eq. [50] can be expressed in terms of z
dG = ν P µ P dz + ν Q µ Q dz + K − ν A µ A dz − ν B µ B dz – K
Rearranging gives
dG
= ν P µ P + ν Q µQ + K − ν A µ A − ν B µ B – K
dz
but we have already seen that dG/dz is zero at equilibrium, so
•
at equilibrium : ν P µ P + ν Q µQ + K − ν A µ A − ν B µ B – K = 0
[51]
This is the general condition for equilibrium and we will use this to find an
expression for the equilibrium constant. It should be recalled that the
chemical potentials depend on composition.
12.2 Relation between K and ∆rG0
We will assume for the moment that all the species are gases. Each chemical
potential in Eq. [51] depends on the partial pressure of the gas, in the way
found in Section 10.4 Eq. [45]
µi ( pi ) = µi0 + RT ln
pi
p0
{const. T} [45]
Using this expression we substitute for µP etc. in Eq. [51] to give
40
p 
p 


ν P  µ P0 + RT ln P0  + ν Q  µQ0 + RT ln Q0  + K


p 
p 
p 
p 


−ν A  µ A0 + RT ln A0  − ν B  µ B0 + RT ln B0  – K = 0


p 
p 
remember that these pi are the equilibrium pressures. Rearranging gives
[ν µ
P
0
P
]
+ ν Q µQ0 + K − ν A µ A0 − ν B µ B0 – K
pQ
pP
+
RT
ln
+K
ν
Q
p0
p0
p
p
−ν A RT ln A0 − ν B RT ln B0 – K = 0
p
p
+ν P RT ln
The quantity is square brackets is ∆r G0 for the reaction, that is the standard
Gibbs energy change (remember that µi0 is equivalent to the molar standard
Gibbs energy, Section 10.4).
•
∆ r G 0 = ν P µ P0 + ν Q µQ0 + K − ν A µ A0 − ν B µ B0 – K
The remaining terms can be manipulated to give
p 
∆ r G + RT ln P0 
p 
0
p 
− RT ln A0 
p 
νA
νP
p 
+ RT ln Q0 
p 
p 
− RT ln B0 
p 
∆ r G 0 + RT ln
νQ
+K
νB
–K= 0
ν
ν
ν
ν
P
Q
 pP   pQ 
 0  0 K
p  p 
A
B
 pA   pB 
 0  0 K
p  p 
=0
The quantity inside the ln is the equilibrium constant in terms of partial
pressures, Kp , as defined in Eq. [48], and so
•
∆ r G 0 = − RT ln K p
[52]
which is what we set out to prove.
There are a number of important points to note. (1) The ratio of pressures
inside the ln is only equal to the equilibrium constant if the pressures are
those at equilibrium; we assumed this was the case by starting with Eq. [51]
which is valid only at equilibrium. (2) ∆r G0 is a function of temperature, so
to use this equation to find the value of the equilibrium constant we need to
know ∆r G0 at the particular temperature of interest. (3) Equation [52]
essentially proves that the quantity we call the "equilibrium constant" is
indeed a constant at given temperature. We have not assumed this, but have
proved it by showing that K is related to ∆r G0 which takes a particular value.
If we considered a reaction in solution, then instead of Eq. [45] we would
use Eq. [46] to described how the chemical potentials vary with
concentration. The rest of the argument remains the same, and the final
result is
41
∆ r G 0 = − RT ln Kc
Example 8
As before, ∆r G0 is defined in terms of the standard chemical potentials,
which in this case refer to a standard state of 1 mol dm–3 rather than 1 bar
pressure.
12.2.1 What is ∆rG0 ?
∆r G0 is, like the other quantities with the ∆r symbol (Section 8.1) essentially
a hypothetical quantity. It is the change in Gibbs energy that would be
observed if the reactants, each in their standard states and in the molar ratios
indicated by the stoichiometric equation, were to be converted completely to
products, also in their standard states. It is not the "Gibbs energy change at
equilibrium".
We can regard ∆r G0 as the thermodynamic driving force of a reaction.
The size of sign of ∆r G0 determines how far the reaction will "go", and hence
the equilibrium constant. As discussed in Section 8 the value of ∆r G0 can be
found from tabulated data.
The diagram below shows how G of the reaction mixture varies for the
simple equilibrium
→B
A←
as a function of the mole fraction of B, xB ; we have assumed that we start out
with one mole of A and that the total pressure is 1 bar. As the reaction
involves no change in the number of moles, the pressure remains at 1 bar.
(a)
G
(b)
∆rG0 = 0
µ0A
0.0
You can reproduce the rough
form of these plots by holding
the ends of a piece of string in
either hand; it should hang
down giving a minimum whose
position depends
on
the
relative height of your hands!
(c)
µ0A
0.2
µ0B
∆rG0 < 0
µ0B
µ0B
0.4
xB
0.6
0.8
∆rG0 > 0
1.0
0.0
0.2
0.4
xB
0.6
0.8
µ0A
1.0
0.0
0.2
0.4
0.6
xB
0.8
1.0
If xB = 0 no B is present and the mixture is pure A, so the Gibbs energy is
just the standard chemical potential of A, µ A0 . If the reaction proceeds
completely to product, xB = 1 and the Gibbs energy is just the standard
chemical potential of B, µ B0 .
Plot (a) shows what happens to G for the case that ∆r G0 = 0. The Gibbs
energy of the reaction mixture falls as the amount of B increases, reaching a
minimum (indicted by the arrow) at xB = 0.5 . This minimum corresponds
to the equilibrium position; as expected, the composition at this point is 1:1 .
In plot (b) ∆r G0 is negative (the standard chemical potential of B is less
than that of A). As in plot (a), G falls as the amount of B increases but the
minimum in G occurs at a larger value of xB than before. As expected for a
reaction with a negative ∆r G0 , the equilibrium favours the products. In plot
(c) ∆r G0 is positive and, as expected, the equilibrium composition now
favours the reactants .
From these plots we can see how the value of ∆r G0 , which refers to the
42
pure reactants and products, affects the composition of the equilibrium
mixture; this is precisely what ∆r G0 = – RT ln K says.
12.2.2 Equilibria involving solids
Suppose we have an equilibrium involving solids as well as gases, for
example
CaCO3 (s)
CaO(s) + CO2 (g)
At equilibrium it follows from Eq. [51] in Section 12.1 that
µCaO + µCO 2 − µCaCO 3 = 0
The chemical potential of CO2 (g) varies with its partial pressure according to
Eq. [45]. The chemical potentials of the two solids are equal to their
standard chemical potentials, as they are present in the pure form (Section
10.5). Hence
0
0
µCaO
+ µCO
+ RT ln
2
pCO 2
p0
0
− µCaCO
=0
3
0
0
0
Recognising that ∆ r G 0 = µCaO
+ µCO
− µCaCO
we have
2
3
∆ r G 0 = − RT ln
pCO 2
p0
The quantity in the ln is the equilibrium constant; remember that we noted in
Section 11 that when writing equilibrium constants solids do not contribute a
term. What we have shown in this section is why this is so. Of course the
solid reactants and products do affect the value of the equilibrium constant
via their contribution to ∆r G0 .
12.3 Variation of Kp with temperature
As ∆r G0 varies with temperature, so does the value of Kp . One way of
approaching this is to compute ∆r G0 (via ∆r H0 and ∆r S0 ) at each new
temperature. However, another way is to use the van't Hoff isochore which,
as we will see, turns out to embody Le Chatalier's Principle in a
mathematical form.
We start from the Gibbs-Helmholtz equation, derived in Section 9.2
d  G
H
=− 2
dT  T 
T
{const. pressure} [41]
This is equally valid if we replace G by ∆r G0 and H by ∆r H0
d  ∆ rG0 
∆r H0
=
−


dT  T 
T2
Equation [52], ∆ r G 0 = − RT ln K p , can be rearranged to give
∆ rG0
= − R ln K p
T
and this can be substituted into the left-hand side of Eq. [53] to give
∆ H0
d
− R ln K p = − r 2
dT
T
(
)
43
[53]
which can be rearranged to give
d ln K p ∆ r H 0
=
{van't Hoff isochore} [54]
dT
RT 2
This equation says that if ∆r H0 is positive, i.e. an endothermic reaction,
d lnKp /dT is positive and so the equilibrium constant increases with
temperature. On the other hand, an exothermic reaction has a negative ∆r H0
and so d lnKp /dT is negative meaning that the equilibrium constant decreases
with increasing temperature. These are exactly the predictions made by Le
Chatalier's Principle – we have now put them on a firm footing.
We can go further and integrate Eq. [54] to find explicitly how the
equilibrium constant varies with temperature
•
d ln K p ∆ r H 0
=
RT 2
dT
∆ H0
d ln K p = r 2 dT
RT
T2
∆r H0
∫ d ln K p = R
T1
[ln K ]
T2
p T
1
T2
1
∫ T 2 dT
T1
∆ H 0  −1  2
= r
R  T  T1
T
∆r H0  1 1 
−
[55]
R  T2 T1 
On the third line we have assumed that ∆r H0 does not vary with temperature
and so can be taken outside the integral. This may not be a good
approximation, especially over a wide temperature range, as we know from
Section 8.5 that ∆r H0 does indeed vary with temperature. As usual, we have
written Kp (T) to remind ourselves that the equilibrium constant varies with
temperature.
Equation [55] is a practical recipe for converting the equilibrium constant
from one temperature to another. Alternatively, if we have measured values
of the equilibrium constant at two temperatures we can determine a value of
∆r H0 .
ln K p (T2 ) − ln K p (T1 ) = −
Example 9
12.3.1 Measuring ∆rH0 indirectly
This indirect way of measuring ∆r H0 , from the variation of Kp with
temperature, is in fact the commonest way of measuring ∆r H0 .
Recall that ∆r H0 is the heat absorbed, at constant pressure, when the
reactants are converted completely to products, under standard conditions.
There are very few reactions which actually go to completion under these
conditions. If a reaction does not go to completion, the measured heat will
not be equal to ∆r H0 .
Examples of reactions which go to completion, and so whose ∆r H0 can be
measured directly, are combustion of organic compounds and the formation
of some transition metal complexes:
44
C2 H5 OH + 3O2 → 3H2 O + 2CO2
Cu2+ (aq) + 4NH3 (aq) → Cu(NH3 )4
However, it is not possible to measure ∆r H0 directly for a reaction such as
the dimerization of NO2 in the gas phase
2NO2 (g) → N2 O4 (g)
for which the equilibrium constant is about 1; the reaction simply does not
go to completion.
If it is possible to measure a series of values of Kp at different
temperatures, ∆r H0 can be found using a graphical method. In the derivation
of Eq. [55] we integrated between two temperatures. If, instead, we found
the indefinite integral the result would be
ln K p =
∆r H0 1
dT
R ∫ T2
−∆ r H 0  1 
+ const.
R T
This implies that a plot of ln Kp against 1/T would be a straight line of slope
–∆r H0 /R. Note that again we have assumed that ∆r H0 does not vary with
temperature.
Having determined ∆r H0 we can go on to determine ∆r S0 at a particular
temperature in the following way. We know the value of Kp at each
temperature and so can find the corresponding value of ∆r G0 from
∆r G0 = –RT ln Kp ; then, using ∆r G0 = ∆r H0 – T∆r S0 , we can find a value for
∆r S0 at each temperature.
A slightly different approach is to start with ∆r G0 = –RT ln Kp and
∆r G0 = ∆r H0 – T∆r S0 to give
slope = –∆rH 0/R
ln Kp
∫ d ln K p =
1/T
Graphical
method
for
determining ∆r H0 from the
variation
of
Kp
with
temperature. The intercept at
1/T = 0 is ∆r S/R.
∆r H0 – T∆r S0 = –RT ln Kp
which rearranges to
−∆ r H 0 ∆ r S 0
+
RT
R
0
0
If we assume that neither ∆r H nor ∆r S vary with temperature, this
relationship implies that a plot of ln Kp against 1/T will be a straight line with
slope –∆r H0 /R and intercept ∆r S0 /R. The new feature here compared to the
previous derivation is the identification of the intercept as ∆r S0 /R.
ln K p =
12.4 Variation of equilibrium with pressure
Since Kp and ∆r G0 are related via ∆r G0 = –RT ln Kp , and ∆r G0 is defined at
standard conditions, which means standard pressure, it follows that Kp does
not vary with the actual pressure.
This does not mean that the composition of the equilibrium mixture does
not change when the pressure is altered – quite the contrary. For example in
a reaction such as
2NO2 (g) → N2 O4 (g)
45
Exercises 18 – 23
increasing the pressure moves the equilibrium to the right, as this is
associated with a reduction in the number of moles of gas (an application of
Le Chatalier's principle). One way of viewing what happens is to say that
when the pressure is changed, the composition changes in order to keep Kp
constant.
We can understand this quantitatively by thinking about a simple
equilibrium, such as the dissociation of a dimer, A2 :
A2 → 2A
The extent to which this reaction has gone to products can be characterised
by a parameter α, which is the degree of dissociation. If α = 1, the dimer is
completely dissociated; if α = 0, there is no dissociation. α can be described
as the fraction of A2 molecules which have dissociated.
Suppose that we start out with n0 moles of A2 in a reaction vessel whose
pressure can be controlled (for example by pushing in or pulling out a
piston). The system is allowed to come to equilibrium at a pressure ptot ; at
equilibrium the degree of dissociation is α.
It follows that the number of moles of A2 at equilibrium is (1 – α)n0 , and
the number of moles of A is 2αn0 ; the factor of two arises because each A2
molecule dissociates into two A molecules. The total number of moles is
thus (1 + α)n0 .
We now need to work out the partial pressures of the species, and this is
done using
pi = xi ptot
where pi is the partial pressure of species i and xi is the mole fraction of i
given by ni/ntot , where ntot is the total number of moles in the system.
Using this, the partial pressures of A2 and A, pA 2 and pA , are:
pA 2 =
n0 (1 − α )
(1 − α )
ptot =
p
n0 (1 + α )
(1 + α ) tot
pA =
2 n0α
2α
ptot =
p
n0 (1 + α )
(1 + α ) tot
Hence the equilibrium constant can be written
Kp
(p
= A
(p
A2
=
p0 )
p0
2
)
=
2
4α 2 (1 + α ) ptot
p0
2
(1 + α ) (1 − α ) ptot p02
4α 2 ptot
(1 − α 2 ) p0
When ptot is varied, we have already seen that Kp does not vary, so it must
be that α varies i.e. the degree of dissociation varies. We could solve this
equation to find how explicitly how α varies with pressure. To make this
solution easier, we will assume that α is small, α << 1, so (1 – α2 ) ≈ 1 and
hence
K p = 4α 2
hence α =
46
ptot
p0
K p p0
4 ptot
It is clear from this that as the pressure is increased, the degree of
dissociation falls i.e. increasing the pressure moves the equilibrium towards
the reactants, just as we would expect from Le Chatalier's principle.
13. Applications in biology
We are familiar with the idea that living organisms need "energy" in order to
carry out the chemical processes which are the basis of life. However, from
our understanding of the Second Law and the Gibbs energy we now see that
the key characteristic of these reactions is that they must be accompanied by
a reduction in Gibbs energy.
For many organisms the source of Gibbs energy is the oxidation of
glucose to water and carbon dioxide – a reaction which has ∆r G0 = –2880
kJ mol–1, a substantial quantity. It is this reduction in Gibbs energy which is
used to drive chemical reactions which would not otherwise "go".
The way Nature does this is to couple two reactions together. The
reaction we want to promote, which is accompanied by an increase in the
Gibbs energy, is coupled to a second reaction in which the Gibbs energy
decreases. Provided that this decrease is larger in size than the increase in
Gibbs energy of the first reaction, the two reactions taken together result in a
decrease in Gibbs energy and therefore "go". The trick is how do you couple
two reactions together? Nature has been working this out for a few
millennia!
In plants, photosynthesis produces glucose is formed from water and
carbon dioxide. The process has ∆r G0 = +2880 so how can it possible take
place? Again the clever trick which Nature uses here is the coupling of two
processes. In photosynthesis the energy of light is used to drive the
unfavourable formation of glucose; a very complex scheme is used to
achieve this coupling – we can only marvel at Nature's ingenuity at this
amazing feat.
13.1 ATP
In biological systems the molecule adenosine triphosphate (ATP) plays an
important role in supplying the Gibbs energy which is used to drive other
reactions, such a protein synthesis and active transport.
ATP releases Gibbs energy through a hydrolysis reaction involving
breaking one of the phosphate bonds to give ADP
NH2
N
O–
–
O
O–
O–
P
O P
O P
O
O
O
N
N
N
O–
N
–
O
O
OH
NH2
O
O–
P
O P
O
O
N
O
O
OH
OH
ADP
ATP
The reaction is
47
OH
N
N
ATP(aq) + H2 O(l) → ADP(aq) + HPO4 2–(aq) + H+(aq)
At physiological pH (about 7) and body temperature (37 °C) this reaction
has ∆r G0 = –30 kJ mol–1, ∆r H0 = –20 kJ mol–1, ∆r S0 = +34 J K–1 mol–1 and
–T∆r S0 = –10 kJ mol–1. The reaction is thus favoured, both energetically
through the ∆r H0 term and entropically through the ∆r S0 term.
Proteins are synthesised by coupling together amino acids, a reaction
which has an unfavourable ∆r G0 of about 17 kJ mol–1:
H2N
O
O
O
OH
H2N
H2N
OH
R'
R
amino acid 1, A1
R
R'
N
H
OH
H2O
O
peptide
amino acid 2, A2
The reaction can be made to "go" by coupling it with the hydrolysis of ATP:
(1)
ATP + H2 O → ADP + HPO4 2– + H+
(2)
A1 + A2 → peptide + H2 O
(1) + (2) ATP + H2 O + A1 + A2 →
ADP + HPO4 2– + H+ + peptide + H2 O
∆r G0 = –30 kJ mol–1
∆r G0 = 17 kJ mol–1
∆r G0 = –13 kJ mol–1
The result is that ∆r G0 is negative for the coupled reactions, which means that
the equilibrium lies well towards the products i.e. the reaction "goes". The
hard part is to make sure that the two reactions are physically coupled – it is
no use hydrolysing some ATP in one part of the cell and trying to make a
peptide bond in another part! Nature has crafted carefully designed enzymes
to ensure this coupling of the reactions.
Of course, if ATP is to be used as a source of Gibbs energy, there has to
be a way of converting the ADP back to ATP within the cell. This is done
by coupling the reverse of the hydrolysis of ATP to a reaction whose
∆r G0 is more negative than –30 kJ mol–1.
Many organisms utilize glucose as their primary energy source, and the
Gibbs energy is derived from the conversion of glucose into water and
carbon dioxide. The complete conversion of glucose to H2 O + CO2 yields a
∆r G0 of –2880 kJ mol–1. Nature has developed a set of reactions which
couple the formation of no less than 38 ATP molecules (from ADP) to the
complete oxidation of one glucose molecule.
14. Microscopic basis of entropy
We have several times used the idea that entropy is related to randomness or
disorder: the greater the disorder the greater the entropy. It is not necessary
to use this idea to develop thermodynamic relationships, such as we have
been doing throughout these lectures. However, our understanding of
thermodynamics is greatly enhanced by recognizing the molecular basis for
functions such as entropy, so we will look at this in this section. The topic –
called Statistical Thermodynamics – is a large and profound one, and we will
only touch on the main ideas at this point. Further details are covered in the
Part IB and Part II Chemistry.
48
14.1 Distributions
In the Shapes and Structures of Molecules course you came across the idea
that molecules have quantized energy levels. Each molecule has available to
it a set of energy levels associated with translation, rotation, vibration and
electronic structure. At any one moment, a given molecule is in a particular
energy level corresponding to a certain amount of translation, rotation etc.
In a macroscopic sample there are a very large number of molecules
(1020, say) and each of these has available to it very many energy levels. The
question is, how do the molecules distribute themselves amongst the
available energy levels?
To answer this question in detail is clearly an impossibly difficult task
simply because the numbers of molecules and levels is so high. However, it
turns out that such a large assembly of molecules behaves in a way which
can be described using statistics and the thermodynamic properties are
related to this statistical analysis; we do not need to know the details of what
each individual molecule is doing.
14.1.1 Ways of arranging things
We start by imagining a system which has a certain number of molecules,
occupying a fixed volume and with a certain amount of energy. Neither
molecules nor energy is allowed to flow into or out of the system.
To make the numbers tractable, we will suppose that there are just 16
molecules in our system, and that there are five energy levels with energy 0,
1, 2, 3, 4 units. We will also suppose that the system has 16 units of energy.
These numbers are very very much smaller than we would find in a real
system, but they are large enough to show the essential details.
One possible distribution of molecules amongst the energy levels is:
energy
0
1
2
3
4
population, ni
7
5
2
1
1
We can quickly confirm that all 16 molecules are accounted for
(7+5+2+1+1=16) and that the total energy, found by multiplying the
population of level i by its energy and summing over all levels, is 16 units
(7×0+5×1+2×2+1×3+1×4=16).
If we imagine that the 16 molecules are distinguishable (different colours,
different labels) then the number of ways, W, that a particular distribution can
be achieved is
W=
N!
n0 ! n1! n2 ! n3! n4 !
where n!= n × (n − 1) × (n − 2)K1† For the distribution above W = 1.7 × 107 .
Another distribution is
energy
0
†
1
2
3
4
You are not required to know how to derive this expression for W and will not be
expected to use it. Note that 0! is taken as 1.
49
4
3
2
1
0
The five energy levels available
in our hypothetical system.
12
0
0
0
4
energy
0
1
2
3
4
population, ni
11
0
0
4
1
population, ni
this has W = 1820.
Another is
this has W = 1.2 × 10 .
There are many other possible distributions.
At this point we introduce the hypothesis that the system has no
preference for one arrangement of molecules over any other. We can
imagine that, as the molecules jostle and collide with one another, the
molecules are being constantly moved from one level to another, thus
continuously changing their arrangement amongst the energy levels.
Each arrangement of molecules belongs to one of the possible
distributions. For example, in the third distribution given above, any one of
the 16 molecules could be in the level with energy 4. The number of
arrangements which correspond to the same distribution are the number of
ways, W, in which that distribution can be achieved.
The greater the number of ways (arrangements) that a particular
distribution can be achieved, the more likely that distribution will be
observed. In other words, as the molecules constantly rearrange themselves,
a distribution which can be achieved in a greater number of ways will occur
more often.
It turns out that one distribution has a larger value of W than all of the
others; in this case it is the first distribution given. This distribution is called
the most probable distribution.
As the number of particles becomes larger, this most probable
distribution becomes overwhelmingly more probable (i.e. has the largest
value of W) than any of the others and we can safely assume, to all intents
and purposes, that the particles are distributed in this way.
4
14.1.2 The Boltzmann distribution
This most probable distribution is the one observed at equilibrium; it is called
the Boltzmann distribution. In this, the number of molecules in level i, ni,
which has energy εi, is
kT2
kT1
Visualization of
how the
population of energy levels
depends on kT . The size of kT
is indicated in relation to the
energy levels by the arrow: T2 >
T1. Note how the molecules
move to higher levels as the
temperature increases, but the
population always tails off in
the higher energy levels.
ni = n0 exp
 −ε i 
 kT 
where it is assumed that the lowest energy level, with i = 0, has energy 0. k
is Boltzmann's constant, which in SI has the value 1.38 × 10–23 J K–1; k is
related to the gas constant, R: R = NA k, where NA is Avogadro's number.
It follows that kT has the dimensions of energy, making the argument of
the exponential dimensionless, as required.
In words, the Boltzmann distribution says that as the energy of a level
increases its population decreases. Levels whose energies are much greater
than kT have vanishingly small populations, whereas those whose energy is
50
less than or comparable to kT have significant populations. The ground state,
level 0, is always the most highly populated.
Thus, as the temperature is raised, molecules move to higher energy
levels. At very low temperatures, the molecules all cluster in the lowest
levels.
From a knowledge of the energy levels of individual molecules and the
way in which molecules are distributed amongst them we can calculate the
bulk properties of matter. The theory thus connects the microscopic world
of quantum mechanics with the macroscopic world of thermodynamics.
14.2 Entropy and distributions
Boltzmann postulated that the entropy was related to the number of ways that
a distribution could be achieved:
S = k ln W
[57]
Despite its simplicity, this is undoubtedly one of the most profound results
in physical science. This relationship puts into quantitative form the idea that
an increase in randomness represents an increase in entropy. Randomness is
measured as the number of ways, W, that a particular distribution can be
achieved; in practice, this distribution will be the most probable distribution.
We can see how Eq. [57] works in practice by considering some simple
cases. The first is simply heating a system: we know that this increases the
entropy. Heat supplied to a system is stored as internal energy, and this
means that the molecules themselves must acquire more energy which in
turn means that they must move to higher energy levels. As the molecules
spread themselves out more amongst the energy levels there are more ways
of achieving the resulting distribution, so from Eq. [57] the entropy goes up.
The second is expanding the system. Quantum mechanics tells us that as
the system is expanded the spacing of the energy levels decreases. Thus
there are more energy levels available to the molecules (that is, levels within
~kT of the ground state), so the molecules are distributed over more levels
and W is increased. Thus, as expected, increasing the volume increases the
entropy. This is the result we found in Section 9.3.
A physical transformation, such as solid going to liquid or gas, is also
associated with a large increase in the number of energy levels available to
the system. In a solid the molecules (or atoms) are only able to vibrate about
fixed positions in the lattice and so only vibrational energy levels are
available to them. In a liquid or gas the molecules are free to translate, and
this gives them access to the large number of translational energy levels; thus
W is larger and the entropy of the liquid or gas is much greater than that of
the solid.
14.3 Entropy at absolute zero
At absolute zero, kT goes to zero and all the molecules must be in the ground
state. There is thus only one way of arranging them so W = 1 and hence,
from Eq. [57], S = 0.
This is the reason for choosing S at absolute zero to be zero, as we did in
51
heat
W~2 × 104
W~107
Heating a system causes its
internal energy to rise and
hence the molecules to move to
higher levels. The arrangement
on the right can be achieved in
a larger number of ways that
that on the left, and is thus of
higher entropy.
expand
W~104
W~6 × 105
Expanding a system causes its
energy levels to move closer
together
and
hence
the
molecules
can
spread
themselves over more levels.
The arrangement on the right
can be achieved in a larger
number of ways that that on the
left, and is thus of higher
entropy. Note that energy is
conserved in this expansion.
Section 4. Having this zero point enables us to determine absolute entropies.
In contrast, there is no such natural zero for enthalpy or internal energy, so
we cannot determine these quantities absolutely.
15. Phase equilibria
A phase is a physically distinct state of a substance, such as the gaseous,
liquid and solid states. Which phase is the most thermodynamically stable
depends on the pressure and temperature. For example, at high temperatures
and low pressures the gaseous phase tends to be favoured, whereas at low
temperatures and high pressures the solid phase is favoured. At some
temperatures and pressures, two or more phases may exist in equilibrium
with one another.
For example, water exists as the gas phase for temperatures above 100°C
and pressures below 1 atmosphere. At temperatures below 0°C and for
modest pressures the solid phase (ice) is the most stable. At 1 atmosphere
and 100 °C the gas and liquid phases coexist, as they do for many other
particular combinations of pressures and temperature.
p
a
solid
liquid
b
c
gas
T
α
dni
β
15.1 Phase diagrams
The information about which phase or phases is present is often presented in
the form of a phase diagram, such as that shown in the margin. For a given
temperature and pressure the diagram shows which phase or phases are
present. The lines represent pressures and temperatures at which two phases
are in equilibrium.
At the temperature and pressure corresponding to point a the stable phase
is the liquid, at c the stable phase is the gas and at b the gas and liquid are in
equilibrium with one another. The point where all three lines intersect is the
unique temperature and pressure where all three phases are in equilibrium,
called the triple point.
15.2 Criterion for phase stability and equilibria
At given temperature and pressure, the phase with the lowest Gibbs energy
will be the one which is favoured. This is because the transformation from a
phase with a higher G to one with a lower G will be accompanied by a
decrease in G and hence be spontaneous (Section 7). Two (or more) phases
will be in equilibrium if there is no change in Gibbs energy when matter
flows between them.
To formulate these criteria more precisely we need to use chemical
potentials as these are the thermodynamic tools used to describe how Gibbs
energy varies with composition and the amount of substances present.
We will start by formulating a general principle about the stability of two
phases. Consider two phases, α and β which are held at a common
temperature and pressure. Each phase contains substance i which has
chemical potential µi(α ) when it is in phase α and µi( β ) when it is in phase β.
Suppose that dni moles of substance i are transferred between phases α
52
and β. We can use Eq. [44] from Section 10.3 to calculate the change in G
of the whole system.
dG = µ A dnA + µ BdnB + µC dnC + K
[44]
In this case dnA will be the number of moles leaving phase α, dni(α ) , and dnB
will be the number of moles arriving in phase β, dni( β ) . Hence the change in
G will be
dG = µi(α )dni(α ) + µi( β )dni( β )
As the number of moles leaving phase α is equal to the number arriving in
phase β we can write
dni( β ) = + dni
dni(α ) = − dni
where the first term is negative as i is leaving phase α and the second term is
positive as i is arriving in phase β. Thus, dG can be written
dG = − µi(α )dni + µi( β )dni
Rearranging this gives
dG
= µi( β ) − µi(α )
dni
In words this says that if µi(α ) > µi( β ) the derivative will be negative and so
substance i will be transferred from phase α to β as this results in a decrease
in G (and is therefore spontaneous).
In general, chemical potential depends on composition (Section 10.4) so
as substance i is transferred from one phase to another its chemical potential
will change in both phases. It may happen that at some point the chemical
potentials in the two phases are equal. At this point dG/dni = 0 which means
that the two phases have come to equilibrium.
The criterion, therefore, for substance i to be in equilibrium between
phases α and β is therefore
µi(α ) = µi( β )
[58]
Chemical potentials also vary with temperature and pressure, so which
phase has the lower chemical potential, and hence is the more stable, can be
affected by altering the temperature and pressure. This is exactly what a
phase diagram shows.
The graph in the margin shows how the chemical potentials of two phases
might vary with temperature (at fixed pressure). At temperatures below Te
phase α has the lower chemical potential, so it will be the only phase present
i.e. there will be no phase β there at all. Above Te , phase β has the lower
chemical potential, so it will be the only phase present. Only at Te are the
two chemical potentials equal and so only at this temperature are the two
phases in equilibrium.
If the pressure is varied, the graph of the chemical potentials against
temperature changes and so the two lines intersect at a different temperature.
There are thus a whole series of particular temperatures and pressures at
which the two phases are in equilibrium; this is the line on the phase
diagram.
53
µ
α
β
Te
T
Variation of the chemical
potentials of two phases with
temperature.
The chemical
potential of phas e α is shown
by the solid line and that of
phase β by the dashed line.
15.3 Equation of a phase boundary
We restrict ourselves now to phase equilibria involving just one substance.
In such cases, the chemical potential is the same as the molar Gibbs energy,
Gm , (Section 10.4), so the criterion for phases to be in equilibrium, Eq. [58],
becomes
Gm(α ) = Gm( β )
If this holds, the pressure and temperature must correspond to a point on a
line in the phase diagram. What we are going to do is to find the condition
for the change in temperature and pressure which moves us to another point
on the line (i.e. where the two phases are still in equilibrium). We will then
use this criterion to find the equation of the line.
If the pressure and temperature are changed by dp and dT, respectively,
the change in the Gibbs energy is given by the Master Equation (Section 9)
dG = Vdp − SdT
[34]
in this case
dGm(α ) = Vm(α )dp − Sm(α )dT and dGm( β ) = Vm( β )dp − Sm( β )dT
(α )
[59]
(α )
where Vm is the molar volume of the substance in phase α, and Sm is the
molar entropy.
If the two phases are still in equilibrium at the new temperature and
pressure, then it must be that Gm(α ) and Gm( β ) are equal at the new temperature
and pressure. So, the change in Gm(α ) must be equal to the change in Gm( β ) :
dGm(α ) = dGm( β )
Hence, from Eq. [59],
Vm(α )dp − Sm(α )dT = Vm( β )dp − Sm( β )dT
which can be rearranged to
(
)
(
dp Vm(α ) − Vm( β ) = dT Sm(α ) − Sm( β )
)
dp Sm( β ) − Sm(α )
=
dT Vm( β ) − Vm(α )
dp ∆Sm
=
dT ∆Vm
{Clapeyron equation} [60]
where ∆Sm = Sm( β ) − Sm(α ) , the change in molar entropy on going from phase α
to phase β and likewise ∆Vm = Vm( β ) − Vm(α ) is the change in molar volume.
15.3.1 Interpretation of the Clapeyron equation
The Clapeyron equation gives us the slope of the phase boundary line in the
phase diagram. For the solid-liquid boundary, ∆Vm is rather small (the
molar volumes of solids and liquids do not differ by very much), and so the
slope of the line is large. This is exactly what we see on the phase diagram.
54
In the case of water, the molar volume of ice is slightly greater than that of
liquid water (water expands on freezing), so ∆Vm for the solid → liquid
transition is negative. ∆Sm for this transition is positive (increase in
disorder), so the slope of the line on the phase diagram is negative. What
this means is that increasing the pressure actually decreases the temperature
at which ice melts to liquid water.
The solid → gas and liquid→ gas phase transitions (sublimation and
evaporation/boiling, respectively) are both accompanied by a large increase in
molar volume and an increase in molar entropy. So, in both cases the lines
have positive slopes. The slope of the solid → gas line is greater than that of
the liquid → gas line as the former transition has a larger increase in entropy
(a greater increase in disorder when going from a solid than when going
from a liquid).
When two phases are in equilibrium any heat absorbed or evolved in
going from one to the other must be under reversible conditions. It therefore
follows that the entropy of the phase change, ∆Spc, can be computed directly
from this reversible heat, qrev,pc
∆Spc =
qrev,pc
Tpc
=
∆Hpc
Tpc
where ∆Hpc is the enthalpy of the phase change which is equal to qrev,pc under
constant pressure conditions.
The Clapeyron equation, Eq. [60], can therefore be rewritten in terms of
the molar enthalpy change as
dp
∆Hm
=
dT T ∆Vm
{Clapeyron equation} [60]
15.3.2 Integration of the Clapeyron equation
Integrating Eq. [60] is not straightforward as ∆Vm depends on temperature
and pressure, and ∆Hm depends on temperature. To make progress, we will
need to consider some special cases and introduce extra assumptions.
Solid → gas and liquid → gas
In these cases we can assume that as the molar volume of a gas is so much
larger than that of a liquid or solid, ∆Vm ≈ Vm,gas. We will also assume that
the gas behaves ideally, so the relationship between its pressure and volume
is given by the ideal gas equation
pVm,gas = RT thus Vm,gas =
RT
p
Finally, we assume that, over the temperature range of interest, ∆Hm does
not vary with temperature. Using all these assumptions the integration can
proceed
55
In a glacier the high pressure
where the ice is in contact with
rock causes a lowering on the
freezing point of water. Under
the right conditions, therefore,
ice may melt at such pressure
points and the resulting liquid
water is thought to lubricate the
flow of the glacier
Exercise 24
dp
∆H m
=
dT TVm,gas
d p p ∆H m
=
dT
RT 2
1
∆H m
dp =
dT
p
RT 2
1
∫ p dp =
ln p =
∆H m 1
dT
R ∫ T2
−∆Hm 1
+ const.
R T
[61]
Equation [61] is the equation for the solid–gas or liquid–gas phase boundary
line on the phase diagram. It also implies that if we measured the pressure
of gas in equilibrium with a solid or liquid as a function of temperature and
plotted ln p against 1/T we would get a straight line with slope –∆Hm /R.
This is a method for determining the enthalpy (and entropy) of sublimation
or vaporization.
The integral can also be carried out between limits to give
p2 −∆Hm  1 1 
=
−
p1
R  T2 T1 
where p1 and p2 are the vapour pressures at temperatures T1 and T2 ,
respectively.
Solid → liquid
Here we assume that, over the temperature range of interest, neither ∆Hm nor
∆Vm vary with temperature and that ∆Vm does not vary with pressure. The
integration is then straightforward
ln
Example 10
dp =
∆Hm
dT
T∆ Vm
∆Hm
1
∫ dp = ∆V ∫ T dT
m
p=
Exercises 25 – 29
∆Hm
ln T + const.
∆ Vm
[62]
Equation [62] is the equation of the solid-liquid phase boundary. It implies
that a plot of p against ln T will be a straight line with slope ∆Hm /∆Vm .
Integrating between limits gives
p2 − p1 =
∆Hm T2
ln
∆ Vm T1
56
16. Electrochemistry
16.1 Introduction
Chemical reactions involving reduction and oxidation of ionic species can be
thought of as involving the transfer of electrons. For example, the reaction
Cu2+ (aq) + Zn(m) → Cu(m) + Zn2+ (aq)
can be thought of as taking place by two electrons being transferred from the
zinc metal to the copper cation. If we just mix the reactants there is no way
to "trap" these electrons, but if the reaction is set up in an electrochemical cell
we can arrange for the electron transfer to take place through an external
circuit. A suitable cell is shown opposite.
The key to capturing the electrons is to keep the Cu and Zn separate in
what are called two half cells. In the right-hand half cell Zn dissolves from
the electrode to form Zn2+ ions; the two electrons then move round the
external circuit where, on arrival at the Cu electrode, they pick up a Cu2+ ion
from the left-hand half cell. Overall the reaction is as above, but the electrons
travel round the external circuit rather than between the reacting ions.
This cell turns the energy available from the chemical reaction (in fact the
Gibbs energy) into an electrical current; this is of great practical importance
as this principle is the basis of electrical batteries. However, we will be more
interested not in the current that we can extract from the cell but the electric
potential (sometimes called the electromotive force, EMF, or loosely the
"voltage") that the cell develops between its electrodes under conditions of
zero current flow.
We will show that this potential is directly related to ∆r G for the cell
reaction. This is exceptionally useful, as it gives us a direct measurement of
∆r G, something which is rarely achieved by other methods. Using cells we
can therefore determine the thermodynamic parameters of ions in solution
and other related species.
We will also find that there is a relatively simple relationship (the Nernst
Equation) between the cell potential and the concentration of the species in
the solutions. Thus concentrations can be determined from measurements of
the cell potential. This has very important practical applications in the
construction of non-invasive sensors for measuring the concentration of ions
in solution.
16.2 Cell conventions
When we specify ∆r S for a chemical reaction we understand that this means
"entropy of products minus entropy of reactants". The fact we calculate it in
this way is simply a convention; we must adhere to this convention if we
want our values to agree with what every one else obtains. In the same way
there are certain conventions for dealing with cells which we must adhere to.
We usually think of a cell in terms of two half cells. For example, in the
cell described above the right-hand half cell consists of Zn(m) in contact with
Zn2+ (aq), and the left-hand cell consists of Cu(m) in contact with Cu2+ (aq).
57
e–
Cu
Cu2+
Zn
Zn2+
A
schematic
cell
which
"captures" the electrons from
the redox reaction between
Cu 2+ and Zn and results in the
flow of electrons round an
external circuit.
The two
compartments are separated
by a porous barrier.
Cu
Zn
V
Cu2+
Zn2+
The same cell, but this time
wired to measure the potential
(voltage) produced.
Modern
electronic voltmeters are able
to measure the voltage without
drawing significant current from
the cell.
The reactions that take place in these half cells include electrons in the
balanced chemical equations. So, in the case of the left-hand half cell, the
reaction is
Cu2 +(aq) + 2e– → Cu(m)
RHS: right-hand side; LHS: lefthand side.
This means that you connect
the positive (red) lead of the
voltmeter to the RHS electrode
Note that the charges as well as the chemical species balance. Such a
reaction is called a half-cell reaction.
The sign of the cell potential is important, so we need to have an agreed
convention about what a positive or negative sign means.
All of these concerns can be resolved by using the following set of
conventions:
(1) The cell is written on paper, thus identifying clearly the left- and righthand half cells.
(2) Each half cell reaction is written as a reduction i.e. with the electrons on
the left-hand side of the reaction.
(3) Having written the left- and right-hand half cell reactions as reductions
involving the same number of electrons, the conventional cell reaction
is found by taking (RHS half-cell reaction) – (LHS half-cell reaction).
(4) The cell potential is that of the RHS measured relative to the LHS.
In addition to these conventions, there is a short-hand way of writing cells
on paper. The electrodes and species involved are written out on a line, with
the left-hand electrode on the left and the right-hand on the right. So, for the
cell above the shorthand notation is
Cu(m) | Cu2+ (aq) || Zn2+ (aq) | Zn(m)
A vertical line, |, is used to separate different phases (here the solution from
the solid), a comma is used to separate species in the same phase and a
double vertical line, ||, indicates a junction between two solutions – often
called a liquid junction. An alternative symbol for this is |×| .
16.2.1 Examples of using the cell conventions
Example 1
For the cell
Zn(m) | Zn2+ (aq) || Cu2+ (aq) | Cu(m)
The RHS half-cell reaction is
Cu2+ (aq) + 2e– → Cu(m)
and the LHS half-cell reaction is
Zn2+ (aq) + 2e– → Zn(m)
Note that both are written as reductions and both involve the same number of
electrons. The conventional cell reaction is thus (RHS – LHS).
Remembering that chemical equations can be subtracted just in the same way
as algebraic equations, we find
58
Cu 2+ (aq ) + 2e − → Cu( m )
−
Zn 2+ (aq ) + 2e − → Zn( m )
≡ Cu 2+ (aq ) + 2e − − Zn 2+ (aq ) − 2e − → Cu( m ) − Zn( m )
≡ Cu 2+ (aq ) + Zn( m ) → Cu( m ) + Zn 2+ (aq )
where to get from the second to the third line we have moved negative terms
to the opposite side. The conventional cell reaction is thus
Cu 2+ (aq ) + Zn( m ) → Cu( m ) + Zn 2+ (aq )
Example 2
The cell
Pt(m) | H2 (g) | H+(aq) || Cd2+ (aq) | Cd(m)
has as hydrogen electrode on the left-hand side. In this electrode the species
involved are H+(aq) and H2 (g); the platinum just acts as in inert conductor to
"pick up" the electrons.
The RHS half cell reaction is
Cd2+ (aq) + 2e– → Cd(m)
and the LHS is
H+(aq) + e– →
1
2
H2 (g)
We note that there are not the same number of electrons involved, so either
the RHS equation will have to be halved, or the LHS doubled; we choose the
latter, although the former is just as valid a choice. The LHS then becomes
2H+(aq) + 2e– → H2 (g)
Then, taking (RHS – LHS) as before we find the conventional cell reaction is
Cd2+ (aq) + H2 (g) → Cd(m) + 2H+(aq)
We see from these examples that the conventional cell reaction is just a
result of how we write the cell down: if we swapped the two sides, the
conventional cell reaction would reverse. The conventional cell reaction may
not be the same as the reaction which takes place when a current is allowed
to flow – called the spontaneous cell reaction; we return to this point later on.
16.3 Thermodynamic parameters from cell potentials
We will now establish the important connection between the cell potential
produced by the cell and ∆r G for the conventional cell reaction.
16.3.1 Concept of maximum work
In all our discussions up to now we have purposefully restricted the work
term in the First Law only to that due to gas expansions ("pV work"). We
will now relax this and allow the possibility of other kinds of work, which
we will simply call "additional work". The work done by electrons moving
through a potential i.e. from one electrode to the other will be a source of this
additional work.
With this extra work term, the First Law can be written
59
Example 11(a)
dU = dq + dw pV + dwadd
where wpV is the pV work and wadd is other kinds of work. We proceed as
before by considering a reversible change, so that
dqrev = TdS
and
dw pV = − pdV
substituting these into the modified First Law gives
dU = TdS − pdV + dwadd,rev
[63]
We found before (Section 5.3) that
dH = dU + pdV + V dp
[64]
and that (Section 7.1)
dG = dH − T dS − S dT
so substituting for dH from Eq. [64] we find
dG = dU + pdV + V dp − T dS − S dT
Then, substituting for dU from Eq. [63] gives
dG = TdS − pdV + dwadd,rev + pdV + V dp − T dS − S dT
= dwadd,rev + V dp − S dT
Finally, we imagine a process taking place at constant pressure and
temperature (dp = 0, dT = 0), so that
dG = dwadd,rev
const. p and T
What this convoluted argument has shown is that the change in Gibbs
energy is the same as the reversible non-pV work. Recall from Section 3.4
that the reversible work is a maximum, so it follows that
The change in Gibbs energy is equal to the
maximum non-pV work available
16.3.2 Relation of the cell potential to ∆rGcell
We will now use the result of the previous section to establish the
relationship between the potential, E, generated by a cell and ∆r G for the
conventional cell reaction, ∆r Gcell. Imagine that we have a cell for which the
conventional cell reaction has been determined using the procedure given in
Section 16.2 and that this cell reaction involves n electrons.
We now let the cell reaction proceed by an amount dz, where z is the
extent of reaction, a concept introduced in Section 12.1. As n electrons are
involved in the cell reaction, it follows that ndz moles of electrons must
travel between the two electrodes of the cell (round an external circuit).
The charge carried by these electrons is –nFdz. The minus sign arises
because the electrons are negatively charged and F, the Faraday constant, is
the charge on one mole of fundamental charges; F = 96,485 C mol–1.
When a charge is moved through an electric potential work is done; this
work is given by charge × potential. In this case the potential is that
generated by the cell, E, so the work is –nFEdz. We have seen in the
previous section that we can identify this non-pV work with the change in the
Gibbs energy under conditions of reversibility, constant pressure and
temperature; therefore
60
dG = − nFEdz
const. p and T [65]
We now want to go on to relate the cell potential to ∆r Gcell. To do this we
use a similar approach to that used in Section 11 when discussing chemical
equilibrium.
Suppose that the conventional cell reaction can be written
νA A + νB B + ... → νPP + νQQ + ...
It was shown in Section 12.1 that if the reaction takes place by an amount dz,
the change in G, dG is
dG = ν P µ P dz + ν Q µ Q dz + K − ν A µ A dz − ν B µ B dz – K
which can be rearranged to
dG
[67]
= ν P µ P + ν Q µQ + K − ν A µ A − ν B µ B – K
dz
The quantity on the right of Eq. [67] can be identified as ∆r Gcell as it has the
form (products – reactants) and includes the stoichiometric coefficients.
∆ r Gcell = ν P µ P + ν Q µQ + K − ν A µ A − ν B µ B – K
[68]
This is entirely analogous to the way in which we defined ∆r G0 in Section
12.2. So, from Eqs. [67] and [68] we can write
dG
= ∆ r Gcell
dz
but Eq. [65] can be rearranged to
dG
= − nFE
dz
and so comparing these last two equations we find
•
∆ r Gcell = − nFE
[65]
[69]
This is the key result as it relates the measurable cell potential, E, to the
Gibbs energy change for the reaction.
16.3.3 Relation of the cell potential to ∆rScell and ∆rHcell
Once we have determined ∆r Gcel l it is easy to find ∆r Scell. Recall from
Section 9.2 that
 ∂G 
  = −S ;
 ∂T  p
this applies equally well to ∆r G and ∆r S. So, using Eq. [69], we have
 ∂∆ G 
∆ r Scell = − r cell 
 ∂T  p
 ∂ ( − nFE ) 
 ∂E 
= −
 = nF  
 ∂T  p
 ∂T  p
So the entropy change of the cell reaction can be found simply from the
temperature coefficient of the cell potential. In practice, all we would need to
do is to measure the cell potential over a range of temperatures, plot the
graph and then take the slope.
61
Once we have ∆r Gcell and ∆r Scell, ∆r Hcell is easily found from ∆r Gcell =
∆r Hcell – T∆r Scell. So, for reactions which can be set up to form an
electrochemical cell it is very straightforward to measure ∆r Gcell, ∆r Hcell and
∆r Scell.
16.4 The Nernst equation
The cell potential changes as the concentration of the species involved in the
cell reaction change. This should come as no surprise as ∆r Gcell, and hence
the cell potential, depends on the chemical potentials of the species involved
(Eq. [68]) and these in turn depend on the concentrations. The Nernst
equation predicts this dependence. Before deriving the Nernst equation we
need to consider how chemical potentials depend on concentration.
16.4.1 Chemical potentials and activities
We saw in Section 10.4 that the chemical potential, µi, of an ideal gas
depends on its partial pressure, pi:
p 
µi = µi0 + RT ln 0i 
p 
where µi0 is the standard chemical potential, which is the chemical potential
of pure i at the standard pressure, p0 (usually p0 = 1 bar = 105 N m–2). Solids
are essentially always in their pure forms so for such species µi = µi0 .
For solutions, the chemical potential is written as
µi = µi0 + RT ln ai
[70]
where ai is the activity of substance i. The activity is a dimensionless
quantity whose value is defined by Eq. [70]. At first, this seems like an odd
definition, but it turns out that chemical potentials are experimentally
measurable, for example from electrochemical cells.
As the concentration tends to zero, the activity can be approximated by the
concentration
c
ai →  0i  as ci → 0
c 
where c0 is the standard concentration; typically the concentrations are
measured in mol dm–3 so c0 = 1 mol dm–3. Particularly for ionic species, it is
found that even for very dilute solutions the concentration is still not a good
approximation to the activity. The reason for this is that there are long-range
interactions between the ions.
It is possible to develop a simple theory, the Debye Hückel Theory, which
predicts activities from concentrations ; this will be covered in the Part II
course Chemistry of Ions in Solution and at Interfaces. However, for the
present course we will simply assume that in calculations activities can be
replaced by concentrations. We will, however, continue to write activities in
our expressions.
16.4.2 Derivation of the Nernst equation
Consider once again an electrochemical cell whose conventional cell reaction
62
is
νA A + νB B + ... → νPP + νQQ + ...
As we saw in Section 16.3.2 if we define ∆r Gcell as
∆ r Gcell = ν P µ P + ν Q µQ + K − ν A µ A − ν B µ B – K
[68]
the cell potential can then be calculated from
∆ r Gcell = − nFE
[69]
All we need to do therefore is to substitute Eq. [70] for the µi into Eq. [68]
and then we will have an expression for the concentration dependence of the
cell potential.
∆ r Gcell = ν P µ P0 + ν P RT ln aP + ν Q µQ0 + ν Q RT ln aQ + K
− ν A µ A0 − ν A RT ln aA − ν B µ B0 − ν B RT ln aB – K
= ν P µ P0 + ν Q µQ0 − ν A µ A0 − ν B µ B0
( )
+ RT ln( aP ) P + RT ln aQ
ν
νQ
+ K − RT ln( aA )
νA
– RT ln( aB )
νB
–K
0
We define ∆ r Gcell
as
0
= ν P µ P0 + ν Q µQ0 − ν A µ A0 − ν B µ B0
∆ r Gcell
Using this, and combining the log terms in Eq. [70] gives
∆ r Gcell = ∆ r G
0
cell
( )
 ( a )ν P a ν Q K
P
Q
+ RT ln
νA
 ( aA ) ( aB )ν B K


Finally, we divide by –nF
( )
νQ
νP
0
∆ r Gcell ∆ r Gcell
RT  ( aP ) aQ K
=
+
ln

ν
ν
− nF
− nF
− nF  ( aA ) A ( aB ) B K
and use
∆ r Gcell = − nFE
0
∆ r Gcell
= − nFE0
to give the Nernst equation in its final form
•
( )
νQ
νP
RT  ( aP ) aQ K
E = E0 −
ln

ν
ν
nF  ( aA ) A ( aB ) B K
[71]
The Nernst equation relates the cell potential to the activities of the species
involved in the cell reaction. E0 is the standard cell potential which is related
to the standard Gibbs energy change of the reaction; E0 is the potential
developed by a cell in which all of the species involved are present at unit
activity.
For simplicity we have written the Nernst equation in terms of activities.
However, if a particular species is a gas its activity is replaced by (pi/p0 ) (as
µi = µi0 + RT ln( pi p 0 )). If a species is a solid, µi = µi0 and so there is no
contribution to the fraction on the RHS of Eq. [71].
The fraction on the RHS of the Nernst equation has the same form as an
equilibrium constant i.e. products over reactants. However, it is not an
equilibrium constant as the activities are not necessarily those at equilibrium.
63
16.4.3 Nernst equation for half cells
As derived, the Nernst equation (Eq. [71]) refers to the conventional cell
reaction which can be thought of as the difference between two half-cell
reactions. It is often convenient to think of the potential developed by each
half cell on its own, called the half-cell potential. Of course, we cannot
measure such a potential directly as any cell necessarily involves two halfcells.
A typical half-cell reaction (written as a reduction in line with the
convention described in Section 16.2) is
νL L + νM M + ... + ne– → νXX + νy Y + ...
The half-cell potential, E1 , is given by
2
E1 = E0, 1
•
2
2
νX
νY
RT  ( aX ) ( aY ) K
−
ln
ν
ν
nF  ( aL ) L ( aM ) M K
where E0, 1 is the standard half-cell potential which is defined as the potential
2
generated when all the species are present at unit activity. As before, the
fraction in the log has the familiar form of "products over reactants".
The cell potential is calculated by taking "right minus left", just as we did
when determining the conventional cell reaction in Section 16.2:
E = E1RHS − E1LHS
•
2
2
In the same way the standard cell potential can be found from the standard
half-cell potentials
E0 = E0RHS
− E0LHS
,1
,1
•
2
Example 11(b)
H2(g, p=1 bar)
2
Given that standard half-cell potentials are tabulated (see Section 16.4.5), this
is the way in which we can determine the standard potential of any cell.
16.4.4 Standard half-cell potentials
We noted in the previous section that it is not possible to measure a half-cell
potential on its own as the potential of any cell will always be the difference
of two half-cell potentials. We can draw up a table of standard half-cell
potentials by defining one of the half-cells to have zero potential and then
referencing all other half-cell potentials to this. Since all calculations involve
a difference in half-cell potentials, this arbitrary choice of a zero point will not
cause any problems.
The electrode whose standard potential is take as zero is the standard
hydrogen electrode (SHE). This consists of H2 gas at a pressure of 1 bar in
contact with an aqueous solution of H+ at unit activity; an inert Pt electrode is
used to make the electrical contact. In the usual notation the cell is
Pt(m) | H2 (g, p = 1 bar) | H+(aq, a = 1)
Pt
H+(aq, a = 1)
Schematic
diagram
of
a
standard hydrogen electrode.
and the half cell reaction is
H+(aq) + e– →
1
2
H2 (g) .
The standard potential of an electrode is defined as the potential developed
by a cell in which the left-hand electrode is the standard hydrogen electrode
64
and the right-hand electrode is the one under test; all species are present at
unit activity. The potential of such a cell is
E = E0RHS
− E0LHS
,1
,1
2
=E
RHS
0 , 12
2
−E
SHE
0 , 12
= E0RHS
,1
.
2
16.4.5 Tabulation of standard half-cell potentials
Extensive tabulations of standard half-cell potentials are available (see, for
example, the data sections in P W Atkins Physical Chemistry). As the cell
potential is essentially a value for ∆r G, we expect it to depend strongly on
temperature. Therefore the tabulated data are at a specified temperature, most
commonly 298 K.
16.5 The spontaneous cell reaction
In Section 16.2 we described the way in which the conventional cell reaction
can be determined. It is important to realise that this conventional reaction
depends only on how the cell is written down on paper. If we swap the RHS
and LHS electrodes, the conventional reaction will be the other way round.
For example we found in Section 16.2.1 that the cell
Zn(m) | Zn2+ (aq) || Cu2+ (aq) | Cu(m)
has the following half-cell reactions
RHS: Cu2+ (aq) + 2e– → Cu(m)
LHS: Zn2+ (aq) + 2e– → Zn(m)
and so the conventional cell reaction, given by (RHS – LHS), is
Cu 2+ (aq ) + Zn( m ) → Cu( m ) + Zn 2+ (aq ) .
On the other hand, if we wrote the cell as
Cu(m) | Cu2+ (aq) || Zn2+ (aq) | Zn(m)
The half-cell reactions are
RHS: Zn2+ (aq) + 2e– → Zn(m)
LHS: Cu2+ (aq) + 2e– → Cu(m)
and so the conventional cell reaction is
Cu( m ) + Zn 2+ (aq ) → Cu 2+ (aq ) + Zn( m ) .
The actual reaction that would take place if current were allowed to flow –
called the spontaneous cell reaction – is found by inspecting the sign of the
cell potential. If the potential is positive, it follows from ∆ r Gcell = − nFE that
∆r Gcell is negative and so the reaction is spontaneous in the direction indicated
by the conventional cell reaction. If the potential is negative, ∆r Gcell is
positive and so the reaction is spontaneous in the opposite direction to that
indicated by the conventional cell reaction.
For example, the cell
Cu(m) | Cu2+ (aq, a = 1) || Zn2+ (aq, a = 1) | Zn(m)
in which all species are present at unit activity develops a potential
65
E = E0RHS
− E0LHS
,1
,1
2
2
= E0 ( Zn, Zn
2+
) − E (Cu,Cu )
2+
0
= −0.76 − ( +0.34) = −1.10 V
The conventional cell reaction is
Conventional : Cu( m ) + Zn 2+ (aq ) → Cu 2+ (aq ) + Zn( m )
but as the cell potential is negative the spontaneous reaction is the reverse of
this
Spontaneous : Cu 2+ (aq ) + Zn( m ) → Cu( m ) + Zn 2+ (aq )
Example 11(c)
Note that cell potentials are affected by the concentrations of the species
involved, so under some circumstances the direction of the spontaneous cell
reaction can be reversed simply by altering the concentrations of the ions.
16.6 Types of half cells
There are many different types of half cells whose potentials are sensitive to
a wide variety of different species. For each half-cell electrode illustrated
below the half-cell reaction and the Nernst equation is also given.
16.6.1 Metal/metal ion
These consist of a metal in contact with a solution of its ions:
E = E0 ( Ag + /Ag) −
Ag + (aq ) + e – → Ag( m )
RT  1 
ln

F  aAg + 
16.6.2 Gas/ion
These consist of the gas in contact with a solution containing related ions; an
inert Pt electrode provides the electrical contact. The ions can be anions or
cations.
Cl 2 (g) + 2e → 2Cl (aq )
–
–
( )
 a – 2
RT
Cl
E = E0 (Cl 2 /Cl ) −
ln
2 F  pCl 2 p 0

–
(
)




The second example is similar to the hydrogen electrode, but instead of there
being H+ in the solution we have OH– . In order to balance the equation two
(solvent) water molecules are needed; we can always use solvent in this way
2 H 2 O(l) + 2e – → 2OH – (aq ) + H 2 (g)
{(
E = E0 (H 2 /OH – ) −
)(
)}
2
RT
ln aOH – pH 2 p 0
2F
As the water is the solvent we assume that it is essentially in its pure form
and so does not contribute a term to the fraction inside the logarithm.
16.6.3 Redox
The oxidized and reduced species are both present in solution; an inert Pt
electrode provides the electrical contact.
66
Fe 3 + (aq ) + e – → Fe 2 + (aq )
E = E0 ( Fe 3 + /Fe 2 + ) −
RT  aFe 2+ 
ln
F  aFe 3+ 
In the second example the solution is acid and solvent (water) is needed to
balance the equation
MnO 4 – (aq ) + 8H + (aq ) + 5e – → Mn 2 + (aq ) + 4H 2 O(l)
E = E0 (MnO 4 – /Mn 2 + ) −

aMn 2+
RT
ln
5F  a
 MnO 4 - aH +

8


( )
16.6.4 Metal/insoluble salt/anion
These consist of a metal coated with a layer of the insoluble salt formed by
the metal and the anion which is in the solution. The commonest example is
the combination silver, silver chloride, chloride anion:
RT
ln aCl –
F
Another commonly encountered electrode is that between liquid mercury,
solid mercurous chloride (mercury(I) chloride) and chloride anions. The
mercurous chloride is made up in a paste with liquid mercury and then
floated on top of mercury. This electrode is called the calomel electrode.
AgCl (s) + e – → Ag ( m ) + Cl − (aq )
E = E0 ( AgCl/Ag) −
Hg 2Cl 2 (s) + 2e – → 2 Hg (l) + 2Cl − (aq )
{( ) }
2
RT
E = E0 (Hg 2Cl 2 /Hg) −
ln aCl –
2F
As in earlier examples, some of the half-cell reactions involve the solvent.
For example the lead, lead oxide, hydroxide system
PbO(s) + H 2 O + 2e – → Pb( m ) + 2OH − (aq )
{(
)}
2
RT
ln aOH –
2F
The point about all of these electrodes is that their half-cell potentials depend
on the concentrations of anions. The AgCl/Ag electrode is a much more
convenient chloride electrode to use that the Cl2 /Cl– electrode.
E = E0 ( PbO/Pb) −
16.7 Liquid junctions
If it is the case that the solutions in the two half-cells are different, then a
problem arises in that for the cell to produce a potential the two solutions
must be in contact. However, they must not mix as then the reaction would
take place without the electrons moving round the external circuit. The "join"
between the two solutions is called a liquid junction.
Typically such a junction is handled in practice by having a porous barrier
between the two solutions; this allows contact but prevents rapid mixing.
The problem with this arrangement is that a potential, called a liquid junction
potential, may be set up across the barrier. It turns out that this potential
detracts from the cell potential and so the measured potential will not be
correct.
67
Ag(m)
layer of
insoluble
AgCl(s)
Cl–(aq)
Schematic
diagram
Ag,AgCl electrode.
of
a
Cu
Cu2+
sat. KCl
Zn
Zn2+
Schematic of a cell employing a
salt bridge to connect the two
half cells.
Example 12
Liquid junction potentials can be minimised by using a salt bridge. This
is a tube containing, typically, a concentrated solution of KCl or KNO3 ; the
ends of the tube dip in the solutions which form the cell and the KCl solution
is kept in the tube by glass sinters at each end. Such a salt bridge provides
electrical contact between the solutions and minimizes the liquid junction
potential.
16.8 Standard functions for ions
We saw in Section 8.3 that ∆fH0 and ∆fG0 of elements in their reference
states are defined as zero. For ions in solution, we introduce the additional
conventions that
∆fH0 (H+) = 0 and ∆fG0 (H+) = 0 .
In any experiment we will only be able to determine the difference of
enthalpies or Gibbs energies of ions, so in order to establish values for
individual ions we need to fix a zero point of the scale. Assigning the value
0 to H+ is arbitrary, but as only differences are measurable we are free to
choose the origin to be where we like.
16.9 Applications
As has already been commented on, the measurement of cell potentials is
essentially a measurement of ∆r G. All we have to do in order to measure
∆r G is to set up a cell with the appropriate conventional cell reaction and then
measure its potential. Further manipulations can be used to extend the
usefulness of such measurements. We will also see that often we can use
the standard half cell potentials directly to calculate further thermodynamic
parameters without actually setting up an experimental cell.
16.9.1 Redox equilibria
We saw in Section 16.5 that, under standard conditions, the following
reaction is spontaneous
Cu 2+ (aq ) + Zn( m ) → Cu( m ) + Zn 2+ (aq )
The standard half-cell potentials are: E0 (Zn,Zn2+ ) = –0.76 V and E0 (Cu,Cu2+ )
= +0.34 V at 298 K. We can see that the metal with the lowest potential out
of the pair will reduce the ions of the other. It is said that the (Zn,Zn2+ )
couple is more strongly reducing than (Cu,Cu2+ ).
For example, given that E0 (Mg,Mg2+ ) = –2.36 V it follows that Mg metal
will reduce Zn2+ ions i.e. the following reaction will be spontaneous
Zn 2+ (aq ) + Mg( m ) → Zn( m ) + Mg 2+ (aq ) .
In the same way, the half-cell potentials of any two couples can be compared
to deduce which is the stronger reducing agent. Note that if the species are
not present at unit activity it will be necessary to use the Nernst equation to
predict the cell potentials.
16.9.2 Solubility product of AgI
AgI is a sparingly soluble salt whose solubility is determined by the
68
solubility product, Ksp , which is defined as the equilibrium constant for the
dissolution reaction
→ Ag + (aq) + I − (aq)
AgI(s) ←
Ksp = aAg + aI –
We have written the equilibrium constant using activities rather than
concentrations. To find Ksp we will first find ∆r G0 for the above reaction and
then use ∆r G0 = –RTlnK.
From data tables we find the following two standard half-cell potentials
(at 298 K):
AgI(s) + e – → Ag ( m ) + I − (aq )
Ag + (aq ) + e – → Ag( m )
E0 ( AgI/Ag) = −0.15 V
E0 ( Ag + /Ag) = +0.80 V
Taking the second of these away from the first gives just the reaction we
want
AgI(s) → Ag + (aq ) + I − (aq )
The standard potential is given by
E0 = E0 ( AgI/Ag) − E0 ( Ag + /Ag) = −0.15 − ( +0.80) = −0.95 V
and this corresponds to a ∆r G0 value of −1 × F × ( −0.95) = 91.6 kJ mol –1
(
)
and so the equilibrium constant, Ksp , is exp −(91.6 × 10 3 ) RT = 8.8 × 10 –17 .
If we assume that the activities can be approximated by concentrations (in
mol dm–3), then Ksp = [Ag+][I– ]/ c02 , so the concentration of either Ag+ or I– in
the solution is 8.8 × 10 –17 = 9.3 × 10 –9 mol dm –3 (c0 = 1 mol dm–3). The
solubility is indeed very low.
Note that to find the solubility product we did not really use a cell as such,
but used the standard half-cell potentials as a source of ∆r G0 values.
16.9.3 Thermodynamic parameters of ions
Cell potentials can be used to determine the standard Gibbs energies of
formation of ions. Consider the two half-cell reactions
H + (aq ) + e – → 12 H 2 (g)
Ag + (aq ) + e – → Ag( m )
E0 (H + /H 2 ) = 0.00 V, by definition
E0 ( Ag + /Ag) = +0.80 V
Taking the second away from the first gives the reaction
H + (aq ) + Ag( m ) → 12 H 2 (g) + Ag + (aq )
for which E0 = E0 (H + /H 2 ) − E0 ( Ag + /Ag) = 0.00 − ( +0.80) = −0.80 V . The
corresponding ∆r G0 is = −1 × F × ( −0.80) = 77.2 kJ mol –1.
For the reaction, we can write ∆r G0 in terms of standard Gibbs energies of
formation ∆fG0 . Noting that ∆fG0 of elements in their reference states is
defined as zero and that ∆fG0 (H+) = 0, gives
∆ r G 0 = 12 ∆ f G 0 (H 2 (g)) + ∆ f G 0 ( Ag + (aq )) − ∆ f G 0 ( Ag( m )) − ∆ f G 0 (H + (aq ))
= ∆ f G 0 ( Ag + (aq ))
So, the standard Gibbs energy of formation of Ag+(aq) is 77.2 kJ mol–1.
69
Using this approach, the ∆fG0 values of other ions can be found.
As discussed in Section 16.3.3, the entropy (and hence the enthalpy)
change can be found from the temperature coefficient of the cell potential.
16.9.4 Measurement of concentration
The calomel electrode was discussed earlier; it has the half-cell reaction and
potential:
Hg 2Cl 2 (s) + 2e – → 2 Hg (l) + 2Cl − (aq )
Pt wire
Hg(l)
porous
plug
Hg2Cl2/Hg
paste
sat. KCl
glass
sinter
Schematic view of a calomel
electrode suitable for dipping
into other solutions.
{( ) }
2
RT
ln aCl –
2F
Frequently, this electrode is constructed in such a way that the Cl– is
provided by a saturated solution of KCl; the activity of the chloride is thus
fixed and so the electrode provides a fixed potential (at given temperature),
which we will denote Ecalomel . The KCl solution also conveniently forms a
salt bridge with a second solution. So, the calomel electrode can be
constructed in such a way that it can simply be "dipped" into other solutions
and provide the second electrode of a cell with a well defined half-cell
potential.
Consider the following cell consisting of a calomel electrode and a
Ag/AgI electrode dipping into the same solution which contains I– ions
E = E0 (Hg 2Cl 2 /Hg) −
Calomel || I– (aq) | AgI(s),Ag(m)
The RHS half cell reaction and Nernst equation are
AgI(s) + e – → Ag ( m ) + I − (aq )
RT
ln aI –
F
and so the cell potential is
E = E0 ( AgI/Ag) −
We will simply write the LHS potential as Ecalomel
RT
ln aI –
F
If E0 (AgI/Ag) and Ecalomel are known, we can use the value of the measured
cell EMF to determine the iodide activity ( ≈ concentration).
We can imagine this idea as being the basis of a simple device which
could be dipped into solutions in order to measure the concentration of iodide
ions. Typically, we would calibrate the potential produced by using solutions
of known concentration.
E = ERHS − ELHS = E0 ( AgI/Ag) − Ecalomel −
Exercises 30 – 37
70
17. Appendix: Formal thermodynamics using partial
derivatives†
We can obtain many thermodynamic relationships by using the properties of
state functions (exact functions) and partial derivatives. This section outlines
the way in which this more mathematical development is carried out. It is
important when taking this approach not to lose sight of the physical
meaning of the results – the manipulations are not merely mathematical, but
express underlying physical principles.
17.1 The Master Equations
The starting point is the First and Second Laws combined, Eq. [32]
(Section 9)
dU = TdS – p dV
[32]
The form of this relationship prompts us to consider U to be a function of S
and V as Eq. [32] says that a change in either of these variables will result in
a change in U.
If U is a function of S and V, it follows from the properties of partial
derivatives that
 ∂U 
 ∂U 
dU = 
 dS + 
 dV
 ∂S  V
 ∂V  S
[A1]
comparison of Eq. [A1] with [32] gives
 ∂U 
 ∂U 

 = T and 
 = −p
 ∂S  V
 ∂V  S
In words, the first relationship says that "the rate of change of internal energy
with entropy, at constant volume, is equal to the temperature". It is hard to
imagine a physical situation in which such a process could be realized.
However, other relations derived in the same way are more useful.
The definition of H is
H = U + pV
and so, forming the complete differential we find
dH = dU + pdV + Vdp
combining this with the Master equation for dU, Eq. [32] gives
dH = TdS + Vdp
[A2]
This Master equation prompts us to consider H as a function of S and p
 ∂H 
 ∂H 
dH =   dS +   dp
 ∂S  p
 ∂p  S
[A3]
comparison of Eq. [A2] with [A3] gives
†
You are not required to use the more mathematical approach outlined in this
Section, but you may do so if you wish.
Appendices 71
 ∂H 
 ∂H 
  = T and   = V
 ∂S  p
 ∂p  S
The thermodynamic function A, the Helmholtz (free) energy is defined as
A = U − TS
and so, forming the complete differential we find
dA = dU − TdS − SdT
combining this with Eq. [32] gives
dA = − pdV − SdT
[A4]
This is another of the Master Equations; it prompts us to consider A as a
function of V and T
 ∂A 
 ∂A 
dA =   dV +   dT
 ∂V  T
 ∂T  V
[A5]
comparison of Eq. [A4] with [A5] gives
 ∂A 
 ∂A 
  = – p and   = − S
 ∂V  T
 ∂T  V
These relationships seem more "physical" – it is possible to imagine
situations in which they might arise.
Finally, G is defined as
G = H − TS
and so, forming the complete differential we find
dG = dH − TdS − SdT
combining this with Eq. [A2] gives
dG = Vdp − SdT
[A6]
This Master Equation prompts us to consider G as a function of p and T
 ∂G 
 ∂G 
dG =   dp +   dT
 ∂T  p
 ∂p  T
[A7]
comparison of Eq. [A6] with [A7] gives
 ∂G 
 ∂G 
  = V and   = − S
 ∂T  p
 ∂p  T
We have already used both of these relationships, derived in a slightly more
informal manner.
All of the above apply to systems which consist of a single pure
substance. If we wish to consider mixtures we need to consider the effect
that composition has on the thermodynamic variables.
We imagine that G is a function of pressure, temperature and
composition; by composition we mean the number of moles, ni, of the
different substances present. Forming the complete differential we therefore
find
 ∂G 
 ∂G 
 ∂G 
dG( p, T , ni ) =   dp +   dT + ∑  
dni
 ∂T  p, ni
 ∂p  T , ni
i  ∂ni  p , T , n
j
Appendices 72
[A8]
where the summation is over all substances present. The subscript ni means
constant composition, and the subscript nj in the final derivative means the
amounts of all substances except i are held constant.
As before the first two derivatives can be identified as V and –S. The
definition of chemical potential, µi, is
 ∂G 
µi =  
 ∂ni  p, T , n j
so Eq. [A8] can be written
dG = Vdp − SdT + ∑ µ i dni
i
This is a complete expression for how G varies with pressure, temperature
and composition.
17.2 Maxwell's relations
If a function F(x,y) is an exact differential and a function of variables x and y
then
 ∂F 
 ∂F 
dF =   dx +   dy
 ∂x  y
 ∂y  x
= Mdx + Ndy
It follows that
 ∂M 
 ∂N 

 = 
 ∂y  x  ∂x  y
So, each master equation gives rise to a further relation, which as know as
Maxwell's relations.
dU = TdS – p dV
 ∂T 
 ∂p 
gives   = − 
 ∂V  S
 ∂S  V
[A9]
dH = TdS + V dp
 ∂T 
 ∂V 
gives   =  
 ∂p  S  ∂S  p
[A10]
dA = –pdV – S dT
 ∂p 
 ∂S 
gives   =  
 ∂T  V  ∂V  T
[A11]
dG = Vdp – S dT
 ∂S 
 ∂V 
gives   = − 
 ∂T  p
 ∂p  T
[A12]
These relationships are often useful in relating one quantity to another. For
example, suppose we want to know how S varies with p and constant T.
Equation [A12] tells us that the required variation in S is related to the
variation of V with T and constant p. For an ideal gas we know how these
Appendices 73
three quantities are related, V = nRT/p, and so
nR
 ∂V 
  =
 ∂T  p
p
so, using this in Eq. [A12] we have
− nR
 ∂S 
  =
 ∂p  T
p
which integrates to
S( p2 ) = S( p1 ) + nR ln
p1
p2
just as we found somewhat more indirectly in Section 9.3
18. Appendix: Formal proofs†
18.1 Variation of ∆rH with temperature
Consider the general reaction
νA A + νB B + ... → νPP + νQQ + ...
for which ∆r H is defined as (Section 8.1)
∆ r H = ν P HP + ν Q HQ + K − ν A HA − ν B HB − K
[A13]
We differentiate the left and right hand sides of Eq. [A13] and impose the
condition of constant pressure, to give
 ∂HQ 
 ∂∆ r H 
 ∂HP 
 ∂HA 
 ∂HB 

 = νP 
 + νQ 
 − νB
 −K
 + K − νA 
 ∂T  p
 ∂T  p
 ∂T  p
 ∂T  p
 ∂T  p
[A14]
The first derivative on the right we recognize as the constant pressure heat
capacity of P, Cp( P )
 ∂HP 

 = Cp( P )
 ∂T 
Likewise the derivatives on the right are the constant pressure heat capacities
of Q, A, B etc. So, the right hand side of Eq. [A14] simplifies to
 ∂∆ r H 

 = ν P Cp( P ) + ν Q Cp( Q ) + K − ν A Cp( A ) − ν BCp( B) − K
 ∂T  p
[A15]
In analogy to the definition of ∆r H, we now define the quantity ∆r Cp as
∆ r Cp = ν P Cp( P ) + ν Q Cp( Q ) + K − ν A Cp( A ) − ν BCp( B) − K
=
∑ ν C( ) – ∑ ν C( )
i
products, i
i
p
j
j
reactants, j
p
Applying this to the right-hand side of Eq. [A15] we find
†
You are not required to use the more formal approaches described in this Section,
but you may do so if you wish.
Appendices 74
 ∂∆ r H 

 = ∆ r Cp
 ∂T  p
[A16]
If it is assumed that ∆r Cp is independent of temperature in the range T1 to
T2 , Eq. [A16] can be integrated in exactly the same way as Eq. [11] was in
Section 5.3.2 to give
∆ r H (T2 ) = ∆ r H (T1 ) + ∆ r Cp [T2 − T1 ]
18.2 Variation of ∆rS with temperature
The argument here is just the same as it was in the previous section.
∆ r S = ν P SP + ν Q SQ + K − ν A SA − ν B SB − K
hence
 ∂SQ 
 ∂∆ r S 
 ∂SP 
 ∂SA 
 ∂SB 

 = νP 
 + νQ 
 − νB
 − K[A17]
 + K − νA 
 ∂T  p
 ∂T  p
 ∂T  p
 ∂T  p
 ∂T  p
From Section 4 we have
Cp
 ∂S 
  =
 ∂T  p T
Each term on the right of Eq. [A17] can be substituted by Cp /T to give
1
 ∂∆ r S 
ν C ( P ) + ν Q Cp( Q ) + K − ν A Cp( A ) − ν BCp( B) − K

 =
 ∂T  p T P p
{
}
∆ r Cp
T
If it is assumed that ∆r Cp is independent of temperature in the range T1 to T2 ,
the integration then proceeds exactly as in Section 6.1.1 to give
=
∆ r S(T2 ) = ∆ r S(T1 ) + ∆ r Cp ln
Appendices 75
T2
T1