without changing their effect, and therefore decreases the pressure on the walls, requiring the
“supplement” to P to fit the ideal gas equation form; see Phys. Teach. 34, 248-249 (April, 1996).
The reader with some knowledge of special relativity may recognize that the total
16
energy of any system is measured by its mass, multiplied by the square of the speed of light.
However, it would be necessary to measure masses about a million times more accurately than is
now possible to be able to determine energies to the accuracy required in thermochemistry.
7/10/07 1- 27
It should also be possible to relate the volume of any liquid or solid to its temperature and
pressure, or to express such other properties as refractive index, heat capacity at constant volume
or pressure, thermal conductivity, heats of vaporization or fusion, or vapor pressures of solids or
liquids, in terms of the temperature and pressure. Some of these equations will be encountered in
later chapters.
Thermochemistry

The application to chemical reactions of the principles developed thus far is called
thermochemistry. In particular, the heats of reaction are measured and tabulated and from these
and from measured heat capacities the enthalpy changes are calculated for other reactions or for
other experimental conditions.
HESS’S LAW. The enthalpy change for a chemical reaction, such as the oxidation of sulfur dioxide
to sulfur trioxide —
2 SO (g) + O &6 2 SO (liq)
2 2 3
— can be expressed as the difference between the enthalpies of the initial and final states.
∆H = H - H
reaction final initial
= H(2 SO ) - H(2 SO ) - H(O )
3 2 2
There is no way within thermodynamics of measuring an absolute energy, or an absolute
16
enthalpy. Only energy, and enthalpy, changes can be determined. However, knowing that these

3

(SO - S - O ) + (½ O - ½ O ) &6 (SO - S - 3/2 O )
2 2 2 2 3 2
and therefore
∆H = ∆H (SO ) - ∆H (SO ) - ∆H (½ O )
reaction form form form
3 2 2
= -437.94 kJ/mol -296.90 kJ/mol - 0
= -141.04 kJ/mol(SO liq)
3
(Notice that the heat of formation of any element, in its standard state, is necessarily zero.)
An entirely equivalent way of obtaining the same numbers is to consider the enthalpy of each
compound on a scale taken with reference to the elements. Such enthalpy values are called
standard enthalpies of the compounds; they are identical with the standard enthalpies of
formation.
Hess’s law can often be applied to find heats of reaction that could not be directly measured
experimentally. For example, the reaction of two molecules of ethylene, C H , to form
2 4
cyclobutane, C H , would not readily occur quantitatively under conditions conducive to
4 8
measurement of the heat of reaction. But both ethylene and cyclobutane can be burned in
oxygen, and subtraction of these reactions gives the reaction equation desired.
2 C H + 8 O &6 4 CO + 4 H O
2 4 2 2 2
C H + 8 O &6 4 CO + 4 H O
4 8 2 2 2
Subtraction of the second from the first gives

2 C H &6 C H

2
1

) reactants(products12
T
T
PP
dTCCHH
7/10/07 1- 29
found by calculating enthalpy changes along an arbitrary path. The total enthalpy change is
independent of this choice of path. The method is known as Kirchhoff’s law.
Assume that ∆H is known for a reaction at a temperature T and the ∆H at another temperature,
1
T , is to be found. Starting with the hot reactants at T (Figure 4), the reaction could be carried
2 2
out isothermally to obtain products at the same temperature. An alternative path would be to
cool the reactants to the temperature T , carry out the reaction isothermally at T , and warm the
1 1
products to T . The heat of reaction at T is already known and if the heat capacities at constant
2 1
pressure are known, the enthalpy changes can be calculated for the processes of cooling reactants
and warming products. This path must give the same ∆H as the isothermal reaction at T .
2

or,
because interchanging limits of an integral will change the sign,

If the difference in heat capacities is independent of temperature, this may be rewritten in the form
∆H = ∆H + [C - C ](T - T ) (18)
2 1 P 2 1

368
= - 296.1 J/mol
Sometimes there will be a phase transition during the warming or cooling process. Sulfur has
a phase change at 95 C, at which point rhombic sulfur goes to monoclinic sulfur; the monoclinic
o
sulfur melts at 119 C. The enthalpy changes are 11.78 and 39.24 kJ/mol. The heat of reaction
o
for liquid sulfur burning in oxygen to form SO at 119 C (392 K) can be calculated as follows (see
2
o
Figure 5).

∆H = - ∆H - C (m) (119 - 95) - ∆H - C (r) (95 - 25) - C (O ) (119 - 25)
392 fusion P tr P P 2
+ ∆H + C (SO ) (119 - 25)
298 P 2
∆H = - 39,240 - 25.9 x 24 - 11,780 - 23.7 x 70 - 29.4 x 94 - 296,900 + 41.9 x 94 J/mol
392
= - 349.0 kJ/mol

Note that temperature differences can be found without conversion to the Kelvin scale.
Both Hess’s law and Kirchhoff’s law are simply applications of the principle that changes in
a state function, such as the enthalpy, are completely determined by the initial and final states.
17
This principle is combined with the equation arising from the first law that shows that i f the
pressure is constant, the enthalpy change will be equal to the heat absorbed by the system. Thus
the “heat of reaction,” by which we mean ∆H (at a particular temperature, pressure, and
reaction
concentrations of reactants and products), is only equal to the heat absorbed i f the reaction
proceeds at constant pressure (and at the specified temperature and concentrations). It is

a. Find the specific heat of Pb (molar mass = 207.19).
b. Find the specific heat of Cu (molar mass = 63.546).
c. Find ∆E for 25 g of Cu when it is warmed from 20 C to 35 C.
o o
2. Estimate the final temperature when 10 cm of iron (density 7.86 g/cm ) at 80 C is added to 30
3 3 o
ml of water at 20 C.
o
3. Find the work done when 3 mol of N gas at 4 atm pressure expands slowly at 25 C against a
2
o
constant pressure of 1 atm.
4. Calculate the work done when 0.25 mol of SO at 27 C and 1 atm expands reversibly and
2
o
isothermally to a final pressure of 0.20 atm. Find Q for the gas during this process.
5. Calculate the final temperature if an ice cube (25 g) at - 5 C is added to a cup of coffee (150
o
ml, or “2/3 cup”) at 90 C, neglecting heat loss to the surroundings.
o
6. The heat of vaporization of benzene, C H , is 30.72 kJ/mol and the normal boiling point is
6 6
80.1 C.
o
a. What is ∆E if the vapor is an ideal gas?
vap
b. Part of the thermal energy absorbed by a liquid as it vaporizes to a gas is required to
perform work on the atmosphere as the substance expands to a vapor. What fraction of the total
Q goes into work against the atmosphere when benzene (density about 0.88 g/cm ) vaporizes at
3

torr is named after Evangelista Torricelli, who invented the barometer in the 17 century.
th
7/10/07 2- 33
at 150 C. The standard heats of formation, at 25 C, of ethylene and ethane are 52.4 and - 84.0
o o
kJ/mol, respectively. Heat capacities (C ) of ethylene, hydrogen, and ethane within this
P
temperature interval are approximately 42.9, 28.8, and 52.5 J/mol·K. All three compounds are
gases at room temperature and above.
13. Benzene, C H , is reduced commercially with hydrogen to give cyclohexane, C H .
6 6 6 12
C H + 3 H 6 C H
6 6 2 6 12
At room temperature (25 C) the benzene and cyclohexane are liquids and the heat of reaction is
o
- 205.5 kJ/mol. Benzene boils at 80.1 C with a heat of vaporization of 47.8 kJ/mol; cyclohexane
o
boils at 80.7 C with a heat of vaporization of 33.0 kJ/mol. The average heat capacity of liquid
o
benzene is 136.0 J/mol·K, of liquid cyclohexane 154.9 J/mol·K, of benzene vapor 82.4 J/mol·K,
of cyclohexane vapor 150 J/mol·K, and of hydrogen 28.8 J/mol·K. Find ∆H for the reduction of
benzene with hydrogen at 150 C.
o
[0.44 cal/g·K; 0.41; 26.7 35.8 cal/mol·K; 6.94 cal/mol·K]Check numbers
14. Air is approximately 20% O and 80% N , by volume. Find the effective molar mass of air.
2 2
Calculate the density of air at 25 C and 1 atm in kg/m , assuming it acts as an ideal gas.
o 3
15. A gaseous compound, containing only sulfur and fluorine, is 62.7% (by weight) sulfur. At
27 C and 750 torr the density of the gas is 4.09 g/L. What is the molecular formula of the

The first law prohibits certain perpetual-motion machines, called “perpetual-motion
1
machines of the first kind.” The more interesting attempts, called “perpetual-motion machines of
the second kind,” do satisfy the first law.
7/10/07 2- 34
2 The Second Law of Thermodynamics
During the 19 century several men, of quite different backgrounds and interests, struggled
th
with the basic problems of thermodynamics. Brilliant flashes of understanding were followed by
years of doubting, testing, and interpreting. Among the fundamental difficulties was a confusion
between two concepts. One of these was the concept of energy. The other, which is related to
the general concept of equilibrium and the direction of changes with respect to equilibrium, was
stated first, but the language was such that it was long misunderstood and therefore rejected. It is
now accepted as the second law of thermodynamics.
Very few people today who have any acquaintance with modern science would doubt the
following generalizations:
1. Perpetual-motion machines don’t work.
2. Bodies in equilibrium have the same temperature. When two bodies in contact have
different temperatures, energy flows, as heat (Q), from the warmer to the cooler body.
3. Bodies in equilibrium have the same pressure. When two bodies in contact, at the same
temperature, have different pressures, the body at the higher pressure tends to expand and
compress the body at the lower pressure.
None of these statements is required by the first law, which requires energy balance in any
1
process but does not say whether a process will actually occur. For example, both exothermic
and endothermic reactions are known that do proceed without external forcing. There is, despite
the variety in these generalizations, a similarity of pattern and intent — saying, for example,
whether a specific process will or will not occur — that suggests a common basis. The basis is
found in the postulate known as the second law of thermodynamics.
The Spread Function

6 10 60
one location is not much more than 1 in 10 , which is awfully close to zero for most purposes.
60 2
Now imagine a very small number of molecules (say 10 ) to which we add 10 molecules that
6
are “excited” — perhaps they are each vibrating, or rotating, or otherwise have some extra energy
that can be transferred to a neighboring molecule. After a very short time, the extra energy will
be spread throughout the 10 molecules, so there is no significant chance the energy will ever
6
again (in the lifetime of the universe) be collected in the original 10 molecules. With a more
typical number of molecules (such as 6 x 10 ) and many ways for each to store energy, it is not at
23
all unreasonable to say the probability of “unspreading” the energy goes quickly to zero.
If we have a small steel ball (which will typically have 10 or more atoms), one way of
23
storing energy in the ball is to start it rolling. Another very large group of ways of storing energy
in the ball is to let some of the atoms begin to vibrate. The mechanical energy of rolling may be
converted by rolling friction to thermal energy of internal motions. There are so many possible
internal motion states that we don’t expect the energy to ever spontaneously reappear as
mechanical energy of rolling.
Entropy
Fortunately, there is an easy way to keep track of S, the spread function, without even having
to count molecules. Any energy passing into an object by thermal energy transfer, Q, is “gone” in
terms of having been spread around. It shows up afterward only as an increase (perhaps slight) in
the temperature of the system. So the first step is to keep track of Q, and in particular, we want
Q for a reversible path, or one that is as close as possible to equilibrium.
Then we want to know whether this amount of energy is important to the system, or whether
there is already so much energy spread around that this addition is relatively unimportant. That
average energy we will find is nicely measured by the (absolute) temperature of the system. So
we measure the increase in average energy as Q and divide by the temperature, T.

(T) ∆E = Q + W = Q + W = Q + W (2)
rev rev max min
There can be only one value of Q and one value of W , so we can represent these, also, as
max min
values “independent of path”, which for the moment we will label as ∆A (= W ) and as ∆B (=
rev
Q ) and regard A and B as state functions.
rev
(T) ∆E = ∆A + ∆B
or
E = A + B (3)
In any isothermal process in which the energy of the system changes by ∆E, some part of this
energy change, equal to ∆A, must appear as work done on the system. The second energy term,
called ∆B for the moment, cannot be lost, because of the first law (conservation of energy);
however, it may be regarded as “spilled energy”, which is lost for useful purposes during the
transfer operation because it has become spread through the system.
From equation 3, two important functions are obtained. One, temporarily given the symbol
A, is called the Helmholtz free energy, or sometimes the work function, and is characterized, as
above, by the constant temperature equation
(5)
T
q
dS
rev
=
KJ/g 22.1
K 15.273
J/g 6.333
⋅===∆
T

The thermostat might be a very large block of metal, well insulated from its surroundings and
large enough so that the reasonable amount of thermal energy added or withdrawn will not
change its temperature. For the ice,
If the thermostat is at exactly the same temperature,
Taking the ice as the system and the thermostat as the surroundings, (∆S) = 0. Thus
system + surroundings
far it might appear as if entropy also is conserved (compare equation 1, Chapter 1). But we know
that unless there is at least a small temperature differential between the thermostat and the ice, the
ice will not melt. We cannot melt an ice cube by bringing it into contact with a large block of ice;
it must be touched to something warmer. If the temperature of the thermostat is higher than
273.15 K, ∆S will be smaller in magnitude than ∆S , and ∆S for the ice plus the
thermostat ice
thermostat, or system plus surroundings, will be positive. This is the essence of the second law of
thermodynamics, which states
(8) T
S
E
V
=






∂
∂
(8a)
1
TE

1 2 rev
the various amounts of thermal energy transferred and w , w , ··· w the various amounts of
1 2 rev
work done along these paths. Then, in general, q …q … ··· …q and w …w … ··· …w . But dE
1 2 rev 1 2 rev
is the same for each of these paths. That is, regardless of which path the system actually follows,

dE = q + w = q + w = ··· = q + w
1 1 2 2 rev rev
and, because q = T dS and w = - P dV (if the only work is work of expansion or compression),
rev rev
it follows that, for any path,
dE = T dS - P dV (7)
This equation is valid for any process occurring in a system of constant composition. (If there is a
change of chemical composition, as would be required if electrical work were produced, or a
change of concentrations, additional terms should be added to the first-law equation and to
equation 7.) At constant volume the second term of equation 7 becomes zero and we obtain
or
(10)
1
2
2
1
1
1
1
1
2
1
1





−= dE
TT
dS
dV
T
P
dE
T
dS +=
1
(12)
T
P
V
S
E
=






∂
∂
7/10/07 2- 39

1 2 1 2 1
that part one loses energy to part two. This proves that when two bodies in contact have
different temperatures, energy flows, as a thermal energy flow (Q), from the warmer to the cooler
body.
From equation 7 we can derive a relationship for the dependence of entropy on volume.
and therefore