Chapter 2 The First Law of Thermodynamics

1
Chapter 2 The First Law of Thermodynamics
Chapter 1 was mainly concerned with macroscopic
properties such as pressure, volume, and
temperature. We have seen how some of the
relationships between these properties of
ideal and real gases can be interpreted in terms
of the behavior of molecules, that is, of the
microscopic properties. This kinetic-molecular
theory is of great value, but it is possible to
interpret many of the relationships between
macroscopic properties without any reference to
the behavior of molecules, or even without
assuming that molecules exist. This is what is
done in the most formal treatments of the
science of thermodynamics, which is concerned
with the general relationships between the
various forms of energy, including heat. At
first it might appear that the formal study of
thermodynamics, with no regard to molecular
behavior, would not lead us very far. The
contrary, however, is true. It has proved
possible to develop some very far-reaching
conclusions on the basis of purely
thermodynamic arguments, and these conclusions
are all the more convincing because they do
not depend on the truth or falsity of any
theories of atomic and molecular behavior.
Pure thermodynamics starts with a small number
of assumptions that are based on very
well-established experimental results and makes
logical deductions from them, finally giving a
set of relationships that are bound to be true
provided that the original premises are true.
2
2.1 Origin of the First Law
The Nature of Heat heat is a form of energy
There are three laws of thermodynamics (aside
from the zeroth law, which was mentioned in
Section 1.4). The first law is essentially the
principle of conservation of energy.
seventeenth century heat is a manifestation of
the motions of the particles of which matter is
composed, but the evidence was not then
compelling.
nineteenth century work performed by humans is
derived from the food they eat. precise
experiments on the interconversion of work and
heat, under a variety of conditions, were
carried out .
eighteenth century heat is a substance, and even
listed it as one of the chemical
elements. expenditure of work causes the
production of heat. a numerical value for the
amount of heat generated by a given amount of
work was obtained.
The energy of the universe remains constant,
which is a compact statement of the first law of
thermodynamics. Both work and heat are
quantities that describe the transfer of
energy from one system to another.
If two systems are at different temperatures,
heat can pass from one to the other directly, and
there can also be transfer of matter from one to
the other. Energy can also be transferred from
one place to another in the form of work. No
matter how these transfers occur, the total
energy of the universe remains the same.
3
2.2 States and State Functions
The distinction between a system and its
surroundings is particularly important in
thermodynamics, since we are constantly concerned
with transfer of heat between a system and its
surroundings. We are also concerned with the work
done by the system on its surroundings or by the
surroundings on the system. In all cases the
system must be carefully defined.
Certain of the macroscopic properties have fixed
values for a particular state of the system,
whereas others do not. Suppose, for example, that
we maintain 1 g of water in a vessel at 25 ?C and
a pressure of 105 Pa (1 bar) it will have a
volume of close to 1 cm3. These quantities, 1 g
of H2O, 25 ?C , 105 Pa , and 1 cm3, all specify
the state of the system. Whenever we satisfy
these four conditions, we have the water in
precisely the same state, and this means that the
total amount of energy in the molecules is the
same. As long as the system is in this state, it
will have these particular specifications.
These macroscopic properties that we have
mentioned (mass, pressure, temperature, and
volume) are known as state functions or state
variables.
4
Characteristic of a State Function
Once we have specified the state of a system by
giving the values of some of the state
functions, the values of all other state
functions are fixed. Thus, in the example just
given, once we have specified the mass,
temperature, and pressure of the water, the
volume is fixed. So, too, is the total energy in
the molecules that make up the system, and
energy is therefore another state function. The
pressure and temperature, in fact, depend on
the molecular motion in the system. When the
state of a system is changed, the change in any
state function depends only on the initial and
final states of the system, and not on the path
followed in making the change. For example, if
we heat the water from 25 ?C to 26 ?C, the change
in temperature is equal to the difference
between the initial and final temperatures
The way in
which the temperature change is brought about has
no effect on this result. This example may seem
trivial, but it is to be emphasized that not all
functions have this characteristic. For
example, raising the temperature of water from 25
?C to 26 ?C can be done in various ways the
simplest is to add heat. Alternatively, we could
stir the water vigorously with a paddle until
the desired temperature rise had been
achieved this means that we are doing work on
the system. We could also add some heat and do
some work in addition. This shows that heat and
work are not state functions.
5
Suppose that there is a point A on the earth's
surface that is 1000 m above sea level and
another point B that is 4000 m above sea level.
The difference, 3000 m, is the height of B with
respect to A. In other words, the difference in
height can be expressed as
where hA and hB are the heights
of A and B above sea level. Height above sea
level is thus a state function, the difference ?h
being in no way dependent on the path chosen.
However, the distance we have to travel in order
to go from A to B is dependent on the path we
can go by the shortest route or take a longer
route. Distance traveled is therefore not a
state function.
6
2.3 Equilibrium States and Reversibility
Thermodynamics is directly concerned only with
equilibrium states, in which the state functions
have constant values throughout the system. It
provides us with information about the
circumstances under which nonequilibrium states
will move toward equilibrium, but by itself it
tells us nothing about the nonequilibrium states.
Figure 2.1
Suppose, for example, that we have a gas
confined in a cylinder having a frictionless
movable piston (Figure 2.1). If the piston is
motionless, the state of the gas can be
specified by giving the values of pressure,
volume, and temperature. However, if the gas
is compressed very rapidly, it passes through
states for which pressure and temperature cannot
be specified, there being a variation of these
properties throughout the gas the gas near to
the piston is at first more compressed and
heated than the gas at the far end of the
cylinder. The gas then would be said to be in a
nonequilibrium state.
Pure thermodynamics could not deal with such a
state, although it could tell us what kind of a
change would spontaneously occur in order for
equilibrium to be attained.
The criteria for equilibrium are very important.
The mechanical properties, the chemical
properties, and the temperature must be uniform
throughout the system and constant in time. The
force acting on the system must be exactly
balanced by the force exerted by the system, as
otherwise the volume would be changing.
7
If we consider the system illustrated in Figure
2.1, we see that for the system to be at
equilibrium the force F exerted on the
piston must exactly balance the pressure P of
the gas if A is the area of the piston,
PA F
If we increase the force, for example, by adding
mass to the piston, the gas will be
compressed if we decrease it, by removing
mass, the gas will expand. Suppose that we
increase the force F by an infinitesimal amount
dF. The pressure that we are exerting on the gas
will now be infinitesimally greater than the
pressure of the gas (i.e., it will be PdP).
The gas will therefore be compressed. We can make
dP as small as we like, and at all stages
during the infinitely slow compression we are
therefore maintaining the gas in a state of
equilibrium. We refer to a process of this kind
as a reversible process. If we reduce the
pressure to P-dP, the gas will expand
infinitely slowly, that is, reversibly.
Figure 2.1
Reversible processes play very important roles in
thermodynamic arguments. All processes that
actually occur are, however, irreversible since
they do not occur infinitely slowly, there is
necessarily some departure from equilibrium.
8
2.4 Energy, Heat, and Work
According to which the total amount of energy in
the universe is conserved, suppose that we add
heat q to a system such as a gas confined in a
cylinder (Figure 2.1). If nothing else is done
to the system, the internal energy U increases by
an amount that is exactly equal to the heat
supplied
This increase in internal energy is the increase
in the energy of the molecules which comprise
the system. Suppose instead that
no heat is transferred to the system but that by
the addition of mass to the piston an amount
of work w is performed on it. The system then
gains internal energy by an amount equal to
the work done
In general, if heat q is supplied to the
system, and an amount of work w is also performed
on the system, the increase in internal energy
is given by
A statement of the first law of thermodynamics
9
We can understand the law by noting that a
collection of molecules, on absorbing some heat,
can store some of it internally, and can do
some work on the surroundings. According to the
IUPAC convention the work done by the system is
-w, so that
It is, of course, necessary to employ
the same units for U, q, and w. The SI unit of
energy is the joule (J kg m2 s-2) it is the
energy corresponding to a force of one newton (N
kg m s-2) operating over a distance of one
metre.
1 cal4.184 J
qgt0, heat is added to the system qlt0, heat flow
out of the system wgt0, work is done on the
system wlt0, work is done by the system
Heat and work sign convention
10
We should note that the above equation leaves
the absolute value of the internal energy U
indefinite, in that we are dealing with only the
energy change ?U, and for most practical
purposes this is adequate. Absolute values can in
principle be calculated, although they must
always be referred to some arbitrary zero of
energy. Thermodynamics is concerned almost
entirely with energy changes. The internal
energy U is a state function of the system that
is, it depends only on the state of the system
and not on how the system achieved its particular
state. Earlier we saw that a change from one
state to another, such as from 25 ?C to 26
?C, can be achieved by adding heat, by doing
work, or by a combination of the two. It is
found experimentally that however we bring about
the temperature rise, the sum qw is always
the same. In other words, for a particular change
in state, the quantity ?U, equal to qw, is
independent of the way in which the change is
brought about. This behavior is characteristic of
a state function. This example demonstrates
that heat q and work w are not state functions
since the change can be brought about by
various divisions of the energy between heat and
work only the sum qw is fixed.
11
The definite integral of a state function such
as U, is an exact differential because it has
a value of U2-U1 ?U, which is independent of
the path by which the process occurs. A state
function such as the internal energy U has a
significance in relation to a particular state.
heat and work are the integrals of inexact
differentials. These integrals are not fixed
but depend on the process by which the change
from state 1 to state 2 occurs. It would
therefore be wrong to write q2-q1 ?q or w2-w1
?w the quantities ?q and ?w have no meaning.
Heat and work make themselves evident only
during a change from one state to another and
have no significance when a system remains in
a particular state they are properties of the
path and not of the state.
slashes indicating that the integrals are inexact
12
If U were not a state function, we could have
violations of the principle of conservation
of energy, something that has never been
observed. To see how a violation could occur,
consider two states A and B, and suppose that
there are two alternative paths between them.
Suppose that for one of these paths U is 10 J
and for the other, 30 J
We could go from A to B by the first
path and expend 10 J of heat. If we then
returned from B to A by the second path, we would
gain 30 J. We would then have the system in
its original state, and would have a net gain of
20 J. Energy would therefore have been created
from nothing. The process could be continued
indefinitely, with a gain of energy at each
completion of the cycle. Many attempts have been
made to create energy in this way, by the
construction of perpetual motion machines of the
first kind, but all have ended in
failure-patent offices are constantly rejecting
devices that would only work if the first law
of thermodynamics were violated! The inability to
make perpetual motion machines provides
convincing evidence that energy cannot be
created or destroyed.
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Nature of Internal Energy
In purely thermodynamic studies it is not
necessary to consider what internal energy really
consists of however, most of us like to have
some answer to this question, in terms of
molecular energies. There are contributions to
the internal energy of a substance from 1. the
kinetic energy of motion of the individual
molecules, 2. the potential energy that arises
from interactions between molecules, 3. the
kinetic and potential energy of the nuclei and
electrons within the individual
molecules. The precise treatment of these factors
is somewhat complicated and it is a great
strength of thermodynamics that we can make use
of the concept of internal energy without having
to deal with it on a detailed molecular basis.
14
Nature of Work
There are various ways in which a system may do
work, or by which work may be done on a system.
For example, if we pass a current through a
solution and electrolyze it, we are performing
one form of work electrical work. Chemical
work is usually, but not quite always, involved
when larger molecules are synthesized from
smaller ones, as in living organisms. Osmotic
work is the work required to transport and
concentrate chemical substances. It is
involved, for example, when seawater is purified
by reverse osmosis and in the formation of the
gastric juice, where the acid concentration is
much higher than that of the surroundings.
Mechanical work is performed, for example, when
a weight is lifted.
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Figure 2.2
One simple way in which work is done is when an
external force brings about a compression of a
system. Suppose, for example, that we have an
arrangement in which a gas or liquid is
maintained at constant pressure P, which it
exerts against a movable piston (Figure 2.2).
Suppose that the force is increased by an
infinitesimal amount dF, so that the piston
moves infinitely slowly, the process being
reversible. If the piston moves to the left a
distance l, the reversible work wrev on the
system is wrev Fl
PAl However, Al is the volume swept out by the
movement of the piston, that is, the decrease
in volume of the gas, which is -?V. The work
done on the system is thus
wrev -P?V
In order for the system to be at equilibrium we
must apply a force P to the piston, the force
being related to the pressure by the
relationship F PA where A is the area of the
position.
Note ?Vlt0, the gas compresses ? wrevgt0 ?Vgt0, the
gas expands ? wrevlt0
16
Example 2.1
Suppose that a chemical reaction is caused to
occur in a bulb to which is attached a capillary
tube having a cross-sectional area of 2.50 mm2.
The tube is open to the atmosphere (pressure
101.325 kPa), and during the course of the
reaction the rise in the capillary is 2.40 cm.
Calculate the work done by the reaction system.
Solution
The volume increase is 2.50
?10-6 m2?2.40?10-2 m 6.00?10-8 m3 The work done
by the system, which following the IUPAC
convention must be written as -w, is P?V
-w P?V 1.01325?l05 Pa?6.00?10-8 m3
6.08?10-3 J PaN m-2N mJ
17
Figure 2.3
If the pressure P varies during a volume change,
we must obtain the work done by a process of
integration. The work done on the system while an
external pressure P moves the piston so that the
volume of the gas changes by an infinitesimal
volume dV is If, as
illustrated in Figure 2.3, the volume changes
from a value V1 to a value V2, the reversible
work done on the system is
P is constant
P is not constant
expressing pressure as a function of V before
performing the integration
Note V1gtV2 (i.e., we have compressed the gas)
and the reversible work is positive.
18
Figure 2.4
We have already noted that work done is not a
state function, and this may be further stressed
with reference to the mechanical work of
expansion. The previous derivation has shown
that the work is related to the process
carried out rather than to the initial and
final states. We can consider the reversible
expansion of a gas from volume V1 to volume
V2, and can also consider an irreversible
process, as illustrated in Figure 2.4.
Figure 2.4a shows the expansion of a gas, in
which the pressure is falling as the volume
increases. The reversible work done by the
system is given by the integral
Figure 2.4b presents that the process is
performed irreversibly by instantaneously
dropping the external pressure to the final
pressure P2. The work done by the system is now
against this pressure P2 throughout the whole
expansion, and is given by

• The irreversible work is less than the reversible
  work.
• In both processes the state of the system has
  changed from A to B, the work done is different.
• The work done by the system in a reversible
  expansion from A to B represents the maximum
• work that the system can perform in changing
  from A to B.

19
Example 2.2
Suppose that water at its boiling point is
maintained in a cylinder that has a frictionless
piston. For equilibrium to be established, the
pressure that must be applied to the piston is 1
atm (101.325 kPa). Suppose that we now reduce the
external pressure by an infinitesimal amount in
order to have a reversible expansion. If the
piston sweeps out a volume of 2.00 dm3, what is
the work done by the system?
Solution
The external pressure remains constant at 101.325
kPa, and, therefore, the reversible work done by
the system is -wrevP?V 101 325 Pa?2.00
dm3202.65 Pa m3 Since Pa kg m-1 s-2 (see
Appendix A), the units are kg m2 s-2 J thus
the work done by the system is 202.65 J.
20
Problem 2.8
A balloon is 0.50 m in diameter and contains air
at 25 ?C and 1 bar pressure. It is then filled
with air isothermally and reversibly until the
pressure reaches 5 bar. Assume that the pressure
is proportional to the diameter of the balloon
and calculate (a) the final diameter of the
balloon and (b) the work done in the process.
Solution
(a)
(b)
21
For many purposes it is convenient to express
the first law of thermodynamics with respect to
an infinitesimal change. We have
where again the symbol
denotes an inexact differential. However, if
only PV work is involved and P is a constant,
may be written as -P dV, where dV is the
infinitesimal increase in volume thus,
Processes at Constant Volume
If an infinitesimal process occurs at constant
volume, and only PV work is involved, where
the subscript V indicates that the heat is
supplied at constant volume. (Note that under
these circumstances dqv is an exact differential,
so that the d has lost its slash.) This
equation integrates to The increase of internal
energy of a system in a rigid container (i.e., at
constant volume) is thus equal to the heat qv
that is supplied to it.
22
Processes at Constant Pressure Enthalpy
In most chemical systems we are concerned with
processes occurring in open vessels, which
means that they occur at constant pressure rather
than at constant volume. For an
infinitesimal process at constant pressure the
heat absorbed dqP is given by If the process
involves a change from state 1 to state 2, this
equation integrates as follows
P is constant
Enthalpy is defined as HUPV
This equation is valid only if the work is all PV
work.
Under these circumstances the increase in
enthalpy ?H of a system is equal to the heat
qP that is supplied to it at constant pressure.
Since U, P, and V are all state functions, that
enthalpy is also a state function. ?Hgt0?a
positive amount of heat is absorbed by the
system, endothermic processes. ?Hlt0?a positive
amount of heat is evolved by the system,
exothermic processes.
23
Problem 2.2
The densities of ice and water at 0 ?C are 0.9168
and 0.9998 g cm-3, respectively. If ?H for the
fusion process at atmospheric pressure is 6.025
kJ mol-1, what is ?U? How much work is done on
the system?
Solution
1 mol of ice has a volume of 18.01 g/0.9168 g
cm-319.64 cm3 1 mol of water has a volume of
18.01 g/0.9998 g cm-318.01 cm3
24
Heat Capacity
The amount of heat required to raise the
temperature of any substance by 1 K (which of
course is the same as 1 ?C) is known as its heat
capacity, and is given the symbol C its SI
unit is J K-1. The word specific before the
name of any extensive physical quantity refers to
the quantity per unit mass. The term specific
heat capacity is thus the amount of heat
required to raise the temperature of unit mass of
a material by 1 K if the unit mass is 1 kg,
the unit is J K-1 kg-1, which is the SI unit for
specific heat capacity. The word molar before
the name of a quantity refers to the quantity
divided by the amount of substance. The SI
unit for the molar heat capacity is J K-1
mol-1. Since heat is not a state function,
neither is the heat capacity. It is therefore
always necessary, when stating a heat capacity,
to specify the process by which the
temperature is raised by 1 K.
25
Isochoric Process
The heat capacity related to a process occurring
at constant volume (an isochoric process) this
is denoted by CV and is defined by
qV is the heat supplied at constant volume.
Note CV,m represents the molar heat capacity at
constant volume.
Isobaric Process
The heat capacity related to a process occurring
at constant pressure (an isobaric process) this
is denoted by CP and is defined by
qP is the heat supplied at constant volume.
Note CP,m represents the molar heat capacity at
constant pressure.
26
The heat required to raise the temperature of 1
mol of material from T1 to T2 at constant volume
is
CV,m is independent of temperature
The heat required to raise the temperature of 1
mol of material from T1 to T2 at constant
pressure is
CP,m is independent of temperature
For liquids and solids, ?Um and ?Hm are very
close to one another. Consequently, CV,m and
CP,m are essentially the same for solids and
liquids. For gases, however, the ?(PV) term is
appreciable, and there is a significant
difference between CV,m and CP,m.
For 1 mol of an ideal gas,
27
Problem 2.4
The latent heat of fusion of water at 0 ?C is
6.025 kJ mol-1 and the molar heat capacities
(CP,m) of water and ice are 75.3 and 37.7 J K-1
mol-1, respectively. The CP values can be taken
to be independent of temperature. Calculate ?H
for the freezing of 1 mol of supercooled water
at -10.0 ?C.
Solution
Heat the water from -10 ?C to 0 ?C
Freeze the water at 0 ?C
Cool the ice from 0 ?C to -10 ?C
28
2.5 Thermochemistry
The study of enthalpy changes in chemical
processes is known as thermochemistry.
Extent of Reaction
In dealing with enthalpy changes in chemical
processes it is very convenient to make use of a
quantity known as the extent of reaction it is
given the symbol ?. The extent of reaction must
be related to a specified stoichiometric equation
for a reaction. A chemical reaction can be
written in general as
It can also be written as
where ?A, ?B, ?Y, and ?Z are known as
stoichiometric coefficients. By definition
the stoichiometric coefficient is positive for a
product and negative for a reactant.
29
The extent of reaction is defined by
where ni,0 is the initial amount of the substance
i and ni is the amount at any time. What makes
the extent of reaction so useful is that it is
the same for every reactant and product. Thus
the extent of reaction is the amount of any
product formed divided by its stoichiometric
coefficient
It is also the change in the amount of any
reactant (a negative quantity), divided by its
stoichiometric coefficient (also a negative
quantity)
These quantities are all equal.
30
Example 2.3
When 10 mol of nitrogen and 20 mol of hydrogen
are passed through a catalytic converter, after a
certain time 5 mol of ammonia are produced.
Calculate the amounts of nitrogen and hydrogen
that remain unreacted. Calculate also the extent
of reaction a. on the basis of the
stoichiometric equation b. on the basis of the
stoichiometric equation
Solution
The amounts are
N2 H2 NH3
Initially 10 20
0 mol Finally 7.5
12.5 5 mol a. The extent of
reaction is the amount of ammonia formed, 5 mol,
divided by the stoichiometric coefficient
for NH3 The same answer is
obtained if we divide the amounts of N2 and H2
consumed, 2.5 mol and 7.5 mol, by the
respective stoichiometric coefficients b. The
extent of reaction is now doubled, since the
stoichiometric coefficients are halved
31
The SI unit for the extent of reaction is the
mole, and the mole referred to relates to the
stoichiometric equation. For example, if the
equation is specified to be the mole relates
to N2, 3H2, or 2NH3. If, therefore, the ?H
for this reaction is stated to be -46.0 kJ mol-1,
it is to be understood that this value
refers to the removal of 1 mol of N2 and 1 mol of
3H2, which is the same as 3 mol of H2. It
also refers to the formation of 1 mol of 2NH3,
which is the same as 2 mol of NH3. In other
words, the ?H value relates to the reaction
as written in the stoichiometric equation. This
recommended IUPAC procedure, which we shall use
throughout this book, avoids the necessity of
saying, for example, -46.0 kJ per mol of
nitrogen, or -23.0 kJ per mol of ammonia. It
cannot be emphasized too strongly that when this
IUPAC procedure is used, the stoichiometric
equation must be specified.
32
Standard States
Enthalpy is a state function, and the enthalpy
change that occurs in a chemical process depends
on the states of the reactants and products.
Consider, for example, the complete combustion
of ethanol, in which 1 mol is oxidized to carbon
dioxide and water
The enthalpy change in this
reaction depends on whether we start with liquid
ethanol or with ethanol in the vapor phase.
It also depends on whether liquid or gaseous
water is produced in the reaction. Another
factor is the pressure of the reactants and
products. Also, the enthalpy change in a
reaction varies with the temperature at which the
process occurs.
In giving a value for an enthalpy change it is
therefore necessary to specify (1) the state of
matter of the reactants and products
(gaseous, liquid, or solid if the last, the
allotropic form), (2) the pressure, and (3) the
temperature. If the reaction occurs in solution,
the concentrations must also be specified.
33
By general agreement the standard state of a
substance is the form in which it is most stable
at 25.00 ?C (298.15 K) and 1 bar (105 Pa)
pressure. For example, the standard state of
oxygen is the gas, and we specify this by writing
O2(g). Since mercury, water, and ethanol are
liquids at 25 ?C, their standard states are
Hg(l), H2O(l), and C2H5OH(l). The standard
state of carbon is graphite. These standard
states should be specified if there is any
ambiguity for example,
not involving standard states
If a reaction involves species in solution, their
standard state is 1 mol kg-1(1 m) for example,
34
Enthalpy changes depend somewhat on the
temperature at which the process occurs. Standard
thermodynamic data are commonly quoted for a
temperature of 25.00 ?C (298.15 K), and this can
be given as a subscript or in parentheses thus
The
superscript ? on the ?H? specifies that we are
dealing with standard states, so that a
pressure of 1 bar is assumed and need not be
stated. The subscript c on the ? is commonly
used to indicate complete combustion, and the
modern practice is to attach such subscripts to
the ? and not to the H. As emphasized in our
discussion of extent of reaction, the value
1357.7 kJ mol-1 relates to the combustion of 1
mol of ethanol, since that is what appears in the
equation. Standard thermodynamic values can be
given for a temperature other than 25 ?C for
example, we could give a value for ?H? (100 ?C),
and it would be understood that the pressure
was again 1 bar and that reactants and products
were in their standard states but at 100 ?C.
35
Measurement of Enthalpy Changes
The enthalpy changes occurring in chemical
processes may be measured by three main methods
Direct Calorimetry
Some reactions occur to completion and without
side reactions, and it is therefore possible to
measure their ?H? values by causing the reactions
to occur in a calorimeter(???). The
neutralization of an aqueous solution of a strong
acid by a solution of a strong base is an
example of such a process, the reaction that
occurs being
Combustion processes also frequently
occur to completion with simple stoichiometry.
When an organic compound is burnt in excess of
oxygen, the carbon is practically all converted
into CO2 and the hydrogen into H2O, while the
nitrogen is usually present as N2 in the final
products. Often such combustions of organic
compounds occur cleanly, and much thermochemical
information has been obtained by burning organic
compounds in calorimeters.
36
Indirect Calorimetry
Few reactions occur in a simple manner, following
a simple chemical equation, with the result that
the enthalpy changes corresponding to a simple
chemical equation often cannot be measured
directly. For many of these, the enthalpy changes
can be calculated from the values for other
reactions by making use of Hess's law. According
to this law, it is permissible to write
stoichiometric equations, together with the
enthalpy changes, and to treat them as
mathematical equations, thereby obtaining a
thermochemically valid result. For example,
suppose that a substance A reacts with B
according to the equation 1. AB?X ?H1 -10 kJ
mol-1 Suppose that X reacts with an
additional molecule of A to give another product
Y 2. AX?Y ?H2 -20 kJ mol-1 According to
Hess's law, it is permissible to add these two
equations and obtain 3. 2AB?Y ?H3 ?H1
?H2 -30 kJ mol-1 The law follows at once from
the principle of conservation of energy and from
the fact that enthalpy is a state function. Thus,
if reactions 1 and 2 occur, there is a net
evolution of 30 kJ when 1 mol of Y is produced.
In principle we could reconvert Y into 2A B by
the reverse of reaction 3. If the heat required
to do this were different from 30 kJ, we should
have obtained the starting materials with a net
gain or loss of heat, and this would violate the
principle of conservation of energy.
37
Example 2.4
The enthalpy changes in the complete combustion
of crystalline ?-D-glucose and maltose at 298 K,
with the formation of gaseous CO2 and liquid H2O,
are
?cH?/kJ mol-1
?-D-Glucose, C6H12O6(c)
-2809.1 Maltose, C12H22O11(c)
-5645.5 Calculate the enthalpy
change accompanying the conversion of 1 mol of
crystalline glucose into crystalline maltose.
Solution
38
Variation of Equilibrium Constant with Temperature
A third general method of measuring ?H? will be
mentioned here only very briefly, since it is
based on the second law of thermodynamics and is
considered in Section 4.8. This method is based
on the equation for the variation of the
equilibrium constant K with the temperature
If, therefore, we
measure K at a series of temperatures and plot ln
K against 1/T, the slope of the line at any
temperature will be -?H?/8.3145 J mol-1, and
hence ?H? can be calculated. Whenever an
equilibrium constant for a reaction can be
measured satisfactorily at various
temperatures, this method thus provides a very
useful way of obtaining ?H?. The method cannot
be used for reactions that go essentially to
completion, in which case a reliable K cannot
be obtained, or for reactions that are
complicated by side reactions.
39
Calorimetry
The heats evolved in combustion processes are
determined in bomb calorimeters, two types of
which are shown in Figure 2.5. A weighed sample
of the material to be burnt is placed in the cup
supported in the reaction vessel, or bomb, which
is designed to withstand high pressures. The
heavy- walled steel bomb of about 400 ml volume
is filled with oxygen at a pressure of perhaps
25 atm, this being more than enough to cause
complete combustion. The reaction is initiated
by passing an electric current through the
ignition wire. Heat is evolved rapidly in the
combustion process and is determined in two
different ways in the two types of calorimeter.
Figure 2.5
40
Figure 2.5a
In the type of calorimeter shown in Figure 2.5a,
the bomb is surrounded by a water jacket that
is insulated as much as possible from the
surroundings. The water in the jacket is
stirred, and the rise in temperature brought
about by the combustion is measured. From the
thermal characteristics of the apparatus the heat
evolved can be calculated. A correction is
made for the heat produced in the ignition
wire, and it is customary to calibrate the
apparatus by burning a sample having a known
heat of combustion.
41
Figure 2.5b
The type of calorimeter illustrated in Figure
2.5b is known as an adiabatic calorimeter. An
adiabatic process is thus one in which there
is no flow of heat. In the adiabatic
calorimeter this is achieved by surrounding
the inner water jacket by an outer water jacket
which by means of a heating coil is maintained
at the same temperature as the inner jacket.
When this is done the amount of heat supplied
to the outer jacket just cancels the heat loss
to the surroundings. This allows a simpler
determination of the temperature rise due to
the combustion the measured (Tfinal Tinitial)
is directly related to the amount of heat
evolved in the combustion.
42
By the use of calorimeters of these types, heats
of combustion can be measured with an accuracy
of better than 0.01. Such high precision is
necessary, since heats evolved in combustion
are large, and sometimes we are more interested
in the differences between the values for two
compounds than in the absolute values. Sometimes
the heat changes occurring in chemical reactions
are exceedingly small, and it is then
necessary to employ very sensitive calorimeters.
Such instruments are known as
microcalorimeters. Another type of
microcalorimeter is the continuous flow
calorimeter which permits two reactant
solutions to be thermally equilibrated during
passage through separate platinum tubes and
then brought together in a mixing chamber the
heat change in the reaction is measured.
43
Relationship between ?U and ?H
Bomb calorimeters and other calorimeters in which
the volume is constant give the internal energy
change ?U. Other calorimeters operate at constant
pressure and therefore give ?H values. Whether ?U
or ?H is determined, the other quantity is easily
calculated from the stoichiometric equation for
the reaction.
If all reactants and products are solids or
liquids, the change in volume if a reaction
occurs at constant pressure is quite small.
Usually 1 mol of a solid or liquid has a volume
of less than 1 dm3, and the volume change in a
reaction will always be less than 1 (i.e., less
than 0.01 dm3). At 1 bar pressure, with ?V0.01
dm3,
This is quite negligible compared with most heats
of reaction, which are of the order of
kilojoules, and is much less than the
experimental error of most determinations.
Note If gases are involved in the reaction,
however, either as reactants or products, ?U and
?H may differ significantly, as illustrated in
the following example.
44
Example 2.5
For the complete combustion of ethanol,
the amount of heat produced, as measured in a
bomb calorimeter, is 1364.47 kJ mo-1 at 25 ?C.
Calculate ?cH for the reaction.
Solution
Since the bomb calorimeter operates at constant
volume, ?cU -1364.47 kJ mol-1. The product of
reaction contains 2 mol of gas and the reactants,
3 mol the change ?n is therefore -1 mol.
Assuming the ideal gas law to apply, ?(PV) is
equal to ?nRT, and therefore to
The difference between ?U and ?H is now large
enough to be experimentally significant.
45
Problem 2.15
A sample of liquid benzene weighing 0.633 g is
burned in a bomb calorimeter at 25 ?C, and 26.54
kJ of heat are evolved. a. Calculate ?U per mole
of benzene. b. Calculate ?H per mole of benzene.
Solution
Molar mass of benzene 6?12.016?1.00878.11 g
mol-1
Heat evolved in the combustion of 1 mol 26.54
kJ?78.11 g mol-1/0.633 g3274.9 kJ
a.
b.
46
Temperature Dependence of Enthalpies of Reaction
Enthalpy changes are commonly tabulated at 25 ?C,
and it is frequently necessary to have the values
at other temperatures. These can be calculated
from the heat capacities of the reactants and
products. The enthalpy change in a reaction can
be written as
partial differentiation with respect to
temperature at constant pressure
Similarly,
For small changes in temperature the heat
capacities, and hence ?CP and ?CV, may be taken
as constant.
integrating between two temperatures T1 and T2
47
If there is a large difference between the
temperatures T1 and T2, it is necessary to take
into account the variation of CP with
temperature. This is often done by expressing the
molar value Cp,m as a power series Alternativ
ely, and somewhat more satisfactorily, we can use
an equation of the form
48
Example 2.6
Consider the gas-phase reaction
A bomb-calorimetric study of
this reaction at 25 ?C leads to ?H? -565.98 kJ
mol-1. Calculate ?H? for this reaction at 2000 K.
Solution
Note that when numerical values are given, it is
permissible to drop the subscript m from ?H?,
since the unit kJ mol-1 avoids ambiguity.
Remember that the mole referred to always
relates to the reaction as written.
49
Enthalpies of Formation
The total number of known chemical reactions is
enormous, and it would be very inconvenient if
one had to tabulate enthalpies of reaction for
all of them. We can avoid having to do this by
tabulating molar enthalpies of formation of
chemical compounds, which are the enthalpy
changes associated with the formation of 1 mol of
the substance from the elements in their
standard states. From these enthalpies of
formation it is possible to calculate enthalpy
changes in chemical reactions. We have seen
that the standard state of each element and
compound is taken to be the most stable form
in which it occurs at 1 bar pressure and at 25
?C. Suppose that we form methane, at 1 bar and
25 ?C, from C(graphite) and H2(g), which are the
standard states the stoichiometric equation
is It is found that ?H? for this reaction is
-74.81 kJ mol-1, and this quantity is known as
the standard molar enthalpy of formation ?fH?
of methane at 25 ?C (298.15 K). The term
standard enthalpy of formation refers to the
enthalpy change when the compound in its
standard state is formed from the elements in
their standard states.
The standard enthalpy of formation of any element
in its standard state is zero.
50
Enthalpies of formation of organic compounds are
commonly obtained from their enthalpies of
combustion, by application of Hess's law. When,
for example, 1 mol of methane is burned in an
excess of oxygen, 802.37 kJ of heat is
evolved, and we can therefore write
In addition, we have the following data
If we add reactions 2 and 3 and subtract reaction
1, the result is
Enthalpies of formation of many other compounds
can be deduced in a similar way. Appendix D gives
some enthalpies of formation. The values, of
course, depend on the state in which the
substance occurs, and this is indicated in the
table the value for liquid ethanol, for
example, is a little different from that for
ethanol in aqueous solution.
51
Included in Appendix D are enthalpies of
formation of individual ions. There is an
arbitrariness about these values, because
thermodynamic quantities can never be
determined experimentally for individual ions it
is always necessary to work with assemblies of
positive and negative ions having a net charge of
zero. For example, ?fH? for HCl in aqueous
solution is -167.15 kJ mol-1, but there is no way
that one can make experimental determinations
on the individual H and Cl-1 ions. The
convention is to take ?fH? for the aqueous ion H
in its standard state (1 mol kg-1) to be
zero. It then follows that, on this basis, the
value of ?fH? for the Cl-1 ion is -167.15 kJ
mo1-1. Then, since the ?fH? value for the NaCl
in aqueous solution is-407.27 kJ mo1-1, we
have In this way a whole set of values can be
built up. Such values are often known as
conventional standard enthalpies of formation
the word conventional refers to the value of
zero for the aqueous proton. In spite of the
arbitrariness of the procedure, correct values
are always obtained when one uses these
conventional values in making calculations for
reactions this follows from the fact that there
is always a balancing of charges in a chemical
reaction. Enthalpies of formation allow us to
calculate enthalpies of any reaction, provided
that we know the ?fH? values for all the
reactants and products. The ?H? for any reaction
is the difference between the sum of the ?fH?
values for all the products and the sum of the
?fH? values for all the reactants
52
Example 2.7
Calculate, from the data in Appendix D, ?H? for
the hydrolysis of urea to give carbon dioxide and
ammonia in aqueous solution
Solution
Subtraction of reactions 1 2 from reactions 3
4' then leads to the desired equation, and the
enthalpy change in the reaction is thus
53
Problem 2.17
A sample of liquid methanol weighing 5.27 g was
burned in a bomb calorimeter at 25 ?C, and 119.50
kJ of heat was evolved (after correction for
standard conditions). a. Calculate ?cHo for the
combustion of 1 mol of methanol. b. Use this
value and the data in Appendix D for H2O(1) and
CO2(g) to obtain a value for ?fHo (CH3OH,l),
and compare with the value given in the table. c.
If the enthalpy of vaporization of methanol is
35.27 kJ mol-1, calculate ?fHo for CH3OH(g).
Solution
Molar mass of CH3OH 12.014?1.00816.00 32.04
g mol-1 Amount of methanol 5.27 g/32.04 g mol-1
0.164 mol Heat evolved 119.50 kJ/0.164 mol
726.5 kJ mol-1
a.
54
b.
2?(2)(3)-(1) gives
c.
(4)(5) gives
55
Problem 2.28
The standard enthalpy of formation of the
fumarate ion is -777.4 kJ mol-1. if the standard
enthalpy change of the reaction fumarate2-(aq)
H2(g)?succinate2-(aq) is 131.4 kJ mol-1,
calculate the enthalpy of formation of the
succinate ion.
Solution
56
Bond Enthalpies
One important aspect of thermochemistry relates
to the enthalpies of different chemical bonds. As
a very simple example, consider the case of
methane, CH4. The standard enthalpy of formation
of methane is -74.81 kJ mol-1
We also know the following thermochemical values
enthalpy of sublimation of graphite
one-half the enthalpy of dissociation of hydrogen
Applying Hesss law
enthalpy of atomization of methane
Since each CH4 molecule has four C-H bonds, we
can divide 1663.5 by 4, obtaining 415.9 kJ mol-1,
and we call this the C-H bond enthalpy it is an
average quantity, and is commonly called the
bond strength.
57
A similar procedure with ethane, C2H6, leads to
an enthalpy of atomization of 2829.2 kJ mol-1.
This molecule contains one C-C bond and six C-H
bonds. If we subtract 6?415.9 2495.4 kJ
mol-1 as the contribution of the C-H bonds, we
are left with 333.8 kJ mol-1 as the C-C bond
enthalpy. However, if we calculate the
enthalpies of atomization of the higher paraffin
hydrocarbons using these values, we find that
the agreement with experiment is by no means
perfect. In other words, there is not a strict
additivity of bond enthalpies. The reason for
this is that chemical bonds in a given
molecule have different environments. On the
whole, enthalpies of atomization are more
satisfactorily predicted if we use the
following bond enthalpies rather than the ones
deduced from the data for CH4 and C2H6 C-H
413 kJ mol-1 C-C 348 kJ mol-1
By the use of similar procedures for molecules
containing different kinds of bonds, it is
possible to arrive at a set of bond enthalpies
that will allow us to make approximate estimates
of enthalpies of atomization and enthalpies of
formation. Such a set is shown in Table 2.2.
Values of this kind have proved very useful in
deducing approximate thermochemical information
when the experimental enthalpies of formation are
not available. These simple additive procedures
can be improved in various ways.
58
It is important to distinguish clearly between
bond enthalpies, the additive quantities we have
just considered, and bond dissociation
enthalpies. The distinction can be illustrated by
the case of methane. We can consider the
successive removal of hydrogen atoms from methane
in the gas phase
The first and last of these bond dissociation
enthalpies are known with some certainty the
values for the second and third reactions are
less reliable. What is certain is that the sum of
the four values must be 1663.5 kJ mol-1, the
enthalpy of atomization of CH4. Although all four
C-H bonds in methane are the same, the four
dissociation enthalpies are not all the same,
because there are adjustments in the electron
distributions as each successive hydrogen atom is
removed. The additive bond strength is the
average of these four dissociation enthalpies.
59
For the gaseous water molecule we have similarly
Note that it is easier to remove the second
hydrogen atom than the first. The bond enthalpy
is 926.9/2 463.5 kJ mol-1, which is the mean of
the two dissociation enthalpies.
Only in the case of diatomic molecules can bond
enthalpy be identified with the bond
dissociation enthalpy for example,
which is both the dissociation enthalpy and the
bond enthalpy.
60
Problem 2.34
From the bond strengths in Table 2.2, estimate
the enthalpy of formation of gaseous propane,
C3H8, using the following additional data

?Ho/kJ mol-1
C(graphite) ? C(g)
716.7 H2(g) ? 2H(g)
436.0
Solution
In propane there are 2 C-C bonds and 8 C-H
bonds heat of atomization is thus (2?348)(8?413)
4000 kJ mol-1
Then
Adding
61
Problem 2.31
Here is a problem with a chemical engineering
flavor Ethanol is oxidized to acetic acid in a
catalyst chamber at 25 ?C. Calculate the rate at
which heat will have to be removed (in J h-1)
from the chamber in order to maintain the
reaction chamber at 25 ?C, if the feed rate is
45.00 kg h-1 of ethanol and the conversion rate
is 42 mole of ethanol. Excess oxygen is
assumed to be available.
Solution
From Appendix D, ?fH? for ethanol, acetic acid
and H2O are -277.69, -484.5 and -285.83 kJ mol-1,
respectively.
Since ethanol is fed in at the rate of 45.00 kg
h-1, and only 42 mole of ethanol is converted,
the actual heat released in the reaction per hour
is (MW of ethanol 46.069 g mol-1)
Therefore, heat will have to be removed at the
rate of 197990 kJ h-1.
62
Problem 2.32
An ice cube at 0 ?C weighing 100 g is dropped
into 1 kg of water at 20 ?C. Does all of the ice
melt? If not, how much of it remains? What is the
final temperature? The latent heat of fusion of
ice at 0 ?C is 6.025 kJ mol-1, and the molar heat
capacity of water, Cp,m is 75.3 J K-1 mol-1.
Solution
Assume that all the ice melts. The process would
absorb 100 g ?6.025 kJ mol-1/18.02 g mol-1
33.44 kJ Suppose that the final temperature is t
?C then 33440 J 75.3 J K-1 mol-1 ? 100 g ?
(0-t) K/18.02 g mol-1 75.3 J K-1 mol-1 ? 1000 g
? (20-t) K/18.02 g mol-1 ? t 11
It is obvious that now not all of the ice will
melt. The final temperature of the water is now 0
?C, and if we suppose that x g of the ice melts,
the heat balance equation is x g ? 6025 J mol-1
75.3 J K-1 mol-1 ? 1000 g ? 20 K ? x 250 g
63
2.6 Ideal Gas Relationship
The various transformations that can be brought
about on ideal gases have played a very
important part in the development of
thermodynamics. There are good reasons for
devoting careful study to ideal gases. ?In
the first place, ideal gases are the simplest
systems to deal with, and they therefore
provide us with valuable and not too difficult
exercises for testing our understanding of
the subject. ?In addition, some of the simple
conclusions that we can draw for ideal
gases can readily be adapted to more complicated
systems, such as solutions. ?A direct
application of thermodynamics to solutions would
be difficult if we did not have the ideal
gas equations to guide our way.
64
Reversible Compression at Constant Pressure
As a first example, we will consider reducing
the volume of an ideal gas by lowering its
temperature at constant pressure. We are going
to find how much work is done on the system
during this process, how much heat is lost,
and what the changes are in internal energy and
enthalpy.
Suppose that we have 1 mol of an ideal gas
confined in a cylinder with a piston, at a
pressure P1, a molar volume Vm,1, and an
absolute temperature T1. The isotherm (i.e., the
PV relationship) for this temperature is shown
in the upper curve in Figure 2.6a and the
initial state is represented by point A. We
now remove heat from the system reversibly, at
the constant pressure P1 until the volume has
fallen to Vm,2 (point B). This could be done
by lowering the temperature of the
surroundings by infinitesimal amounts, until the
temperature of the system is T2 the isotherm
for T2 is the lower curve in Figure 2.6a.
Figure 2.6a
65
As far as work and heat changes are concerned,
we could not obtain values unless we had
specified the path taken, since work and heat are
not state functions. In this example, we have
specified that the compression is reversible and
is occurring at constant pressure. The work
done on the system is
This is true whether the gas is ideal or not.
If the gas is ideal, use of the gas law PVm RT
(for 1 mol of gas) leads to the expression
Since V1gtV2, and T1gtT2, a positive amount of
work has been done on the system. This work is
represented by the shaded area in Figure 2.6a.
If the system had expanded isothermally and at
constant pressure, from state B to state A,
the shaded area would represent the work done by
the system.
66
The heat absorbed by the system during the
process A?B is given by
since the process occurs at constant pressure.
For an ideal gas, CP,m is independent of
temperature and this expression therefore
integrates to
Since T1gtT2, a negative amount of heat is
absorbed (i.e., heat is released by the system).
This amount of heat qP,m, being absorbed at
constant pressure, is the molar enthalpy change,
which is also negative
The molar internal energy change ?Um (also
negative for this process) is obtained by use of
the first law
It can easily be verified that these expressions
for ?Hm and ?Um are consistent with the
relationship
67
Problem 2.38
Two moles of oxygen gas, which can be regarded as
ideal with CP,m 29.4 J K-1 mol-1 (independent
of temperature), are maintained at 273 K in a
volume of 11.35 dm3. a. What is the pressure of
the gas? b. What is PV? c. What is CV,,m?
Solution
a. 1 mol in 22.7 dm3 at 273 K exerts a pressure
of 1 bar ? 2 mol in 11.35 dm3 at 273 K exerts
a pressure of 4 bar b. PV 4 bar ? 11.35 dm3
45.40 bar dm3 4.540 kJ c. CV,m CP,m-R
29.4-8.3145 21.1 J K-1 mol-1
68
Problem 2.40
Suppose that the gas in Problem 2.38 is heated
reversibly to 373 K at constant pressure. a. What
is the final volume? b. How much work is done on
the system c. How much heat is supplied to the
system? d. What is the increase in enthalpy? e.
What is the increase in internal energy?
Solution
a. V 11.35 ? 373/273 15.5 dm3 b. w P?V 4
bar ? (15.5-11.35) dm3 16.6 bar dm3 1.66
kJ c. q n ? CP,m ? ?T 2 ? 29.4 ? (373-273)
5880 J 5.88 kJ d. ?H 5880 J 5.88 kJ e. ?U
?H - P?V 5880 1660 4220 J 4.22 kJ 2 ?
21.1 ? 100
69
Reversible Pressure Change at Constant Volume
Figure 2.6b
Suppose, instead, that we take 1 mol of ideal
gas from the initial state P1, Vm,l, T1 to the
final state P2, Vm,2, T2 as shown in Figure
2.6b. The pressure P1 is taken to be higher
than P2, and to accomplish this at constant
volume we must remove heat until the
temperature is T2. Again, we bring about the
change reversibly.
The work done on the system is the area below the
line AC in Figure 2.6b and is zero. This is
confirmed by considering the integral
Since the process occurs at constant volume, the
heat absorbed is given by
which is negative since T1gtT2. This expression is
also ?Um
70
The value of ?Hm is obtained as follows
It is interesting to compare the changes that
occur in going from A to B (volume decrease at
constant pressure, Figure 2.6a) with those when
we go from A to C (pressure decrease at
constant volume, Figure 2.6b). The work and
heat values are different in the two cases.
However, the ?Um and ?Hm values are the same.
This means that the internal energy is the same
at point B on the T2 isotherm as it is at
point C the same is true of the enthalpy. This
result can be proved for any two points on an
isotherm. We thus reach the very important
conclusion that the internal energy and enthalpy
of an ideal gas depend only on the temperature
and remain constant under isothermal conditions.
71
Problem 2.39
Suppose that the gas in Problem 2.38 is heated
reversibly to 373 K at constant volume. a. How
much work is done on the system? b. What is the
increase in internal energy, ?U? c. How much heat
was added to the system? d. What is the final
pressure? e. What is the final value of PV? f.
What is the increase in enthalpy, ?H?
Solution
a. w P?V 0 b. ?U n ? CV,m ? ?T 2 ? 21.1
? (373-273) 4220 J 4.22 kJ c. q ?U w
4.22 kJ d. P 4 ? 373/273 5.47 bar 547
kPa e. P2V2 5.47 bar ? 11.35 dm3 62.08 bar
dm3 6.208 kJ f. ?H ?U P?V 4220
(6208-4540) 5888 J 5.89 kJ 2 ? 29.4 ?
(373-273)
72
Reversible Isothermal Compression
Figure 2.6c
Another process of very great importance is that
of compression of an ideal gas along an
isotherm (i.e., at constant temperature). Such
a process is illustrated in Figure 2.6c, the
temperature being written simply as T. The
initial conditions are P1, Vm,l and the final are
P2, Vm,2 with Vm,l gtVm,2 . We have just seen
that for an isothermal process in an ideal gas
The work done on the system in a reversible
compression is
Vm,l gtVm,2
The heat absorbed is found by use of the equation
for the first law
Heat is evolved during the compression. When we
compress a gas, we do work on it and supply
energy to it if the temperature is to remain
constant, heat must be evolved.
73
If 1 mol of gas has a volume Vm,i, the
concentration is
Similarly,