Entropy and the number of
microstates
Boltzmann proposed that the entropy was
proportional to lnW. That is
S = k ln W
where the proportionality constant k is called the
Boltzmann constant.
1321
A system with few microstates has low entropy.
A system with a large number of microstates has a
high entropy.
1322
The notion of order and disorder is related to
probability. A probable event is one that can
happen in many ways. An improbable event is one
that can happen in only one or a very few ways.
1323
The notion of order and disorder is related to
probability. A probable event is one that can
happen in many ways. An improbable event is one
that can happen in only one or a very few ways.
Consider the gas expansion experiment. Suppose
that initially we have only one gaseous atom or
molecule in the left flask.
1324
The notion of order and disorder is related to
probability. A probable event is one that can
happen in many ways. An improbable event is one
that can happen in only one or a very few ways.
Consider the gas expansion experiment. Suppose
that initially we have only one gaseous atom or
molecule in the left flask. Since the two flasks have
equal volume, the probability of finding the atom or
molecule in either flask after “expansion” is ½.
1325
If the number of molecules is increased to two, the
probability of finding both molecules in the same
flask is ½ x ½ = ¼.
1326
If the number of molecules is increased to two, the
probability of finding both molecules in the same
flask is ½ x ½ = ¼.
If there were N molecules, the probability of finding
all N molecules in the same flask is
1327
If the number of molecules is increased to two, the
probability of finding both molecules in the same
flask is ½ x ½ = ¼.
If there were N molecules, the probability of finding
all N molecules in the same flask is
1 x 1 x 1 x 1 x.....  1
2 2 2 2
2
N factors








N







1328
If the number of molecules is increased to two, the
probability of finding both molecules in the same
flask is ½ x ½ = ¼.
If there were N molecules, the probability of finding
all N molecules in the same flask is
1 x 1 x 1 x 1 x.....  1
2 2 2 2
2
N factors








N







When N = 6.02 x 1023 (i.e. one mole) the probability
that all the molecules would be in the same flask
1
after expansion  10181200000000000000000000  0
1329
1330
Because entropy is a direct measure of the order of
a system, the change in entropy, ΔS, as a result of
the expansion, is given by
ΔS  Sfinal  Sinitial
1331
Because entropy is a direct measure of the order of
a system, the change in entropy, ΔS, as a result of
the expansion, is given by
ΔS  Sfinal  Sinitial
Since the system is more disordered after
expansion, Sfinal is greater than Sinitial, so that ΔS  0.
1332
Because entropy is a direct measure of the order of
a system, the change in entropy, ΔS, as a result of
the expansion, is given by
ΔS  Sfinal  Sinitial
Since the system is more disordered after
expansion, Sfinal is greater than Sinitial, so that ΔS  0.
We can define a spontaneous process as one that
in itself, would lead to an increase of entropy of the
universe, and could therefore occur without an
input of energy from any other system.
1333
Second Law of Thermodynamics
1334
Second Law of Thermodynamics
Second Law of Thermodynamics: The entropy of
the universe increases in a spontaneous process.
1335
Second Law of Thermodynamics
Second Law of Thermodynamics: The entropy of
the universe increases in a spontaneous process.
Important Point: The term spontaneous as used in
the context of the second law of thermodynamics
refers only to the possibility that a change can
occur without being driven by a continuous input
of energy from an external source; it says nothing
about whether that process will actually occur at a
perceptible rate.
1336
Reversible Changes
1337
Reversible Changes
Everyday usage: A process that can occur in either
direction. For example, the heating and cooling of
an object.
1338
Reversible Changes
Everyday usage: A process that can occur in either
direction. For example, the heating and cooling of
an object.
In thermodynamics, this definition is refined.
Thermodynamic definition
1339
Reversible Changes
Everyday usage: A process that can occur in either
direction. For example, the heating and cooling of
an object.
In thermodynamics, this definition is refined.
Thermodynamic definition
Reversible Process: A process that can be reversed
by an infinitesimal change in a variable.
1340
Example: Irreversible change
Assume the piston can slide in and out without friction. The sudden
addition of a large mass to the piston will cause the piston to move
inward quickly. There will be considerable jostling of the gas molecules.
There will be a major disturbance of the equilibrium of the system.1341
Example: Reversible change
Microbe lands on piston
(enlarged so you can see it)
The piston moves inward an
infinitesimal distance.
There will be essentially no disturbance of the
equilibrium of the system.
1342
Example: Reversible change
Microbe flies off piston
The piston moves outward
an infinitesimal distance.
There will be essentially no disturbance of the
equilibrium of the system.
1343
Example of reversible change
The melting of ice:
1344
Example of reversible change
The melting of ice:
If T > 0 oC
H2O(s)
H2O(l)
1345
Example of reversible change
The melting of ice:
If T > 0 oC
H2O(s)
If T < 0 oC
H2O(s)
H2O(l)
H2O(l)
1346
Example of reversible change
The melting of ice:
If T > 0 oC
H2O(s)
If T < 0 oC
H2O(s)
If T = 0 oC
H2O(s)
H2O(l)
H2O(l)
H2O(l)
1347
Reversible change:
0 oC
Add very small
(infinitesimally small)
amount of heat
0 oC
Ice cube is
infinitesimally
smaller
1348
Reversible change:
0 oC
0 oC
Ice cube
returns to
original mass
Add very small
(infinitesimally small)
amount of heat
Remove very small
(infinitesimally small)
amount of heat
0 oC
Ice cube is
infinitesimally
smaller
0 oC
Equilibrium is maintained for these infinitesimal changes.
1349
Irreversible change:
25 oC
Place block of ice on table top at
room temperature.
Liquid water forms
on table top.
1350
Irreversible change:
25 oC
Place block of ice on table top at
room temperature.
Liquid water forms
on table top.
To go backward, we would have to first lower the
temperature of the water formed to 0 OC and then convert it
back to ice at 0 oC. So the path backwards is different
(because of the temperature change involved) than the path
forward.
1351
Irreversible change:
25 oC
Place block of ice on table top at
room temperature.
Liquid water forms
on table top.
To go backward, we would have to first lower the
temperature of the water formed to 0 OC and then convert it
back to ice at 0 oC. So the path backwards is different
(because of the temperature change involved) than the path
forward.
Equilibrium is not maintained for this process.
1352
Entropy change for a reversible process
1353
Entropy change for a reversible process
The entropy change for a reversible process is
defined by:
ΔS  qrev
T
1354
Entropy change for a reversible process
The entropy change for a reversible process is
defined by:
ΔS  qrev
T
where T is the temperature (assumed fix) of the
process, and qrev is the heat added under reversible
conditions.
1355
Entropy change for a reversible process
The entropy change for a reversible process is
defined by:
ΔS  qrev
T
where T is the temperature (assumed fix) of the
process, and qrev is the heat added under reversible
conditions. This result is particularly useful for
finding ΔS for phase transitions such as:
H2O(s)
H2O(l)
1356
Entropy change for a reversible process
The entropy change for a reversible process is
defined by:
ΔS  qrev
T
where T is the temperature (assumed fix) of the
process, and qrev is the heat added under reversible
conditions. This result is particularly useful for
finding ΔS for phase transitions such as:
H2O(s)
H2O(l)
qrev for the process is equal to ΔHmelting .
1357
The second Law for a reversible process
The second Law of thermodynamics for a reversible
process is:
ΔSuniv  ΔSsys  ΔSsurr  0
where univ = universe, sys = system, and surr =
surroundings.
1358
For an isolated system (no energy or matter
exchange with surroundings):
ΔSsys  0
reversible change
ΔSsys  0
irreversible change
1359
Absolute entropies and the Third Law of
Thermodynamics
1360
Absolute entropies and the Third Law of
Thermodynamics
The Third Law of Thermodynamics: The entropy of
a perfect crystalline substance is zero at the
absolute zero of temperature.
1361
Absolute entropies and the Third Law of
Thermodynamics
The Third Law of Thermodynamics: The entropy of
a perfect crystalline substance is zero at the
absolute zero of temperature.
Thus, if we assume that solid dinitrogen at 0 K has a
perfectly ordered structure, and that it is free of
impurities, then
S (N2 at 0 K) = 0 JK-1mol-1
1362
Note that it is not ΔS but S that has the value
0 JK-1mol-1. In thermodynamics, it is most common
to encounter variables such as ΔH , that is, changes
in some thermodynamic variable.
1363
Note that it is not ΔS but S that has the value
0 JK-1mol-1. In thermodynamics, it is most common
to encounter variables such as ΔH , that is, changes
in some thermodynamic variable. For the entropy,
we have an important exception. Values of S for
specific substances (at different temperatures) are
referred to as absolute entropies.
1364
It is possible by experimental measurement to
determine the entropy that any substance has
above 0 K. In order to tabulate entropy data, the
most convenient reference temperature is 25 oC.
1365
It is possible by experimental measurement to
determine the entropy that any substance has
above 0 K. In order to tabulate entropy data, the
most convenient reference temperature is 25 oC.
The entropy of 1 mole of a substance at 1 bar is
called the standard entropy – symbol S0.
1366
It is possible by experimental measurement to
determine the entropy that any substance has
above 0 K. In order to tabulate entropy data, the
most convenient reference temperature is 25 oC.
The entropy of 1 mole of a substance at 1 bar is
called the standard entropy – symbol S0.
The term absolute standard entropy is also
employed.
1367
Entropy changes in chemical reactions
1368
Entropy changes in chemical reactions
The calculation of the entropy change in a chemical
reaction is similar to the calculation of the enthalpy
change for a reaction.
1369
Entropy changes in chemical reactions
The calculation of the entropy change in a chemical
reaction is similar to the calculation of the enthalpy
change for a reaction. Consider the reaction:
aA + bB
cC + dD
1370
Entropy changes in chemical reactions
The calculation of the entropy change in a chemical
reaction is similar to the calculation of the enthalpy
change for a reaction. Consider the reaction:
aA + bB
cC + dD
ΔS0  Σ n S0(products)  Σ n S0(reactants)
1371
Entropy changes in chemical reactions
The calculation of the entropy change in a chemical
reaction is similar to the calculation of the enthalpy
change for a reaction. Consider the reaction:
aA + bB
cC + dD
ΔS0  Σ n S0(products)  Σ n S0(reactants)
where Σ means “sum over” and “n” refers to the
moles of each species in the balanced chemical
equation.
1372
For the preceding reaction:
ΔS0  {c S0(C) + d S0(D)} – {a S0(A) + b S0(B)}
1373
For the preceding reaction:
ΔS0  {c S0(C) + d S0(D)} – {a S0(A) + b S0(B)}
Tables of S0 values are readily available (see the
Appendices in the text).
1374
For the preceding reaction:
ΔS0  {c S0(C) + d S0(D)} – {a S0(A) + b S0(B)}
Tables of S0 values are readily available (see the
Appendices in the text).
Sometimes it is only necessary to know if ΔS0  0
or ΔS0  0 for a reaction. (The next section will
clarify this).
1375
Consider the reaction:
NH4NO3(s)
N2(g) + 2 H2O(g) + ½ O2(g)
Is ΔS0 expected to be positive or negative? Why?
1376
Consider the reaction:
NH4NO3(s)
N2(g) + 2 H2O(g) + ½ O2(g)
Is ΔS0 expected to be positive or negative? Why?
ΔS0 = S0(N2) + 2 S0(H2O) + ½ S0(O2) – S0(NH4NO3)
1377
Consider the reaction:
NH4NO3(s)
N2(g) + 2 H2O(g) + ½ O2(g)
Is ΔS0 expected to be positive or negative? Why?
ΔS0 = S0(N2) + 2 S0(H2O) + ½ S0(O2) – S0(NH4NO3)
= “big” + 2 x “big” + ½ x “big” – “small”
>0
1378
Consider the reaction:
NH4NO3(s)
N2(g) + 2 H2O(g) + ½ O2(g)
Is ΔS0 expected to be positive or negative? Why?
ΔS0 = S0(N2) + 2 S0(H2O) + ½ S0(O2) – S0(NH4NO3)
= “big” + 2 x “big” + ½ x “big” – “small”
>0
Consider the reaction: 2 NO(g) + O2(g)
2 NO2(g)
Is ΔS0 expected to be positive or negative? Why?
1379
Consider the reaction:
NH4NO3(s)
N2(g) + 2 H2O(g) + ½ O2(g)
Is ΔS0 expected to be positive or negative? Why?
ΔS0 = S0(N2) + 2 S0(H2O) + ½ S0(O2) – S0(NH4NO3)
= “big” + 2 x “big” + ½ x “big” – “small”
>0
Consider the reaction: 2 NO(g) + O2(g)
2 NO2(g)
Is ΔS0 expected to be positive or negative? Why?
ΔS0 = 2 S0(NO2) – 2 S0(NO) – S0(O2)
1380