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Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Chapter 19
Chemical
Thermodynamics
John D. Bookstaver
St. Charles Community College
Cottleville, MO
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
First Law of Thermodynamics
• You will recall from Chapter 5 that
energy cannot be created nor
destroyed.
• Therefore, the total energy of the
universe is a constant.
• Energy can, however, be converted
from one form to another or transferred
from a system to the surroundings or
vice versa.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Spontaneous Processes
• Spontaneous processes
are those that can
proceed without any
outside intervention.
• The gas in vessel B will
spontaneously effuse into
vessel A, but once the
gas is in both vessels, it
will not spontaneously
return to vessel B.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Spontaneous Processes
Processes that are
spontaneous in one
direction are
nonspontaneous in
the reverse
direction.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Spontaneous Processes
• Processes that are spontaneous at one
temperature may be nonspontaneous at other
temperatures.
• Above 0 C it is spontaneous for ice to melt.
• Below 0 C the reverse process is spontaneous.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.1
Predict whether the following processes are
spontaneous as described, spontaneous in the
reverse direction, or in equilibrium: (a) When a
piece of metal heated to 150 °C is added to
water at 40 °C, the water gets hotter. (b) Water
at room temperature decomposes into H2(g)
and O2(g), (c) Benzene vapor, C6H6(g), at a
pressure of 1 atm condenses to liquid benzene
at the normal boiling point of benzene, 80.1
Chemical
°C.
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.1 Solution
Solve: (a) This process is spontaneous. Whenever two objects at different
temperatures are brought into contact, heat is transferred from the hotter
object to the colder one. Thus, heat is transferred from the hot metal to the
cooler water. The final temperature, after the metal and water achieve the
same temperature (thermal equilibrium), will be somewhere between the
initial temperatures of the metal and the water.
(b) Experience tells us that this process is not spontaneous—we certainly
have never seen hydrogen and oxygen gases spontaneously bubbling up
out of water! Rather, the reverse process—the reaction of H2 and O2 to
form H2O—is spontaneous.
(c) By definition, the normal boiling point is the temperature at which a
vapor at 1 atm is in equilibrium with its liquid. Thus, this is an equilibrium
situation. If the temperature were below 80.1 °C, condensation would be
spontaneous.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.1
• Under 1 atm pressure CO2(s) sublimes
at –78 °C. Is the transformation of
CO2(s) to CO2(g) a spontaneous
process at –100 °C and 1 atm
pressure?
• Answer: No, the reverse process is
spontaneous at this temperature.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Reversible Processes
In a reversible
process the system
changes in such a
way that the system
and surroundings
can be put back in
their original states
by exactly reversing
the process.
Chemical
Thermodynamics
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Irreversible Processes
• Irreversible processes cannot be undone by
exactly reversing the change to the system.
• Spontaneous processes are irreversible.
Chemical
Thermodynamics
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Entropy
• Entropy (S) is a term coined by Rudolph
Clausius in the 19th century.
• Clausius was convinced of the
significance of the ratio of heat
delivered and the temperature at which
it is delivered, q .
T
Chemical
Thermodynamics
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Entropy
• Entropy can be thought of as a measure
of the randomness of a system.
• It is related to the various modes of
motion in molecules.
Chemical
Thermodynamics
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Entropy
• Like total energy, E, and enthalpy, H,
entropy is a state function.
• Therefore,
S = Sfinal  Sinitial
Chemical
Thermodynamics
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Entropy
For a process occurring at constant
temperature (an isothermal process), the
change in entropy is equal to the heat that
would be transferred if the process were
reversible divided by the temperature:
qrev
S =
T
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.2
• The element mercury, Hg, is a silvery
liquid at room temperature. The normal
freezing point of mercury is
–38.9 °C, and its molar enthalpy of
fusion is ΔHfusion = 2.29 kJ/mol. What is
the entropy change of the system when
50.0 g of Hg(l) freezes at the normal
freezing point?.
Chemical
Thermodynamics
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Practice Exercise 19.2
• The normal boiling point of ethanol,
C2H5OH, is 78.3 °C, and its molar
enthalpy of vaporization is
38.56 kJ/mol. What is the change in
entropy in the system when 68.3 g of
C2H5OH(g) at 1 atm condenses to liquid
at the normal boiling point?
• Answer: –163 J/K
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Second Law of Thermodynamics
The second law of thermodynamics
states that the entropy of the universe
increases for spontaneous processes,
and the entropy of the universe does
not change for reversible processes.
Chemical
Thermodynamics
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Second Law of Thermodynamics
In other words:
For reversible processes:
Suniv = Ssystem + Ssurroundings = 0
For irreversible processes:
Suniv = Ssystem + Ssurroundings > 0
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Second Law of Thermodynamics
These last truths mean that as a result
of all spontaneous processes the
entropy of the universe increases.
Chemical
Thermodynamics
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Entropy on the Molecular Scale
• Ludwig Boltzmann described the concept of
entropy on the molecular level.
• Temperature is a measure of the average
kinetic energy of the molecules in a sample.
Chemical
Thermodynamics
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Entropy on the Molecular Scale
• Molecules exhibit several types of motion:
– Translational: Movement of the entire molecule from
one place to another.
– Vibrational: Periodic motion of atoms within a molecule.
– Rotational: Rotation of the molecule on about an axis or
rotation about  bonds.
Chemical
Thermodynamics
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Entropy on the Molecular Scale
• Boltzmann envisioned the motions of a sample of
molecules at a particular instant in time.
– This would be akin to taking a snapshot of all the
molecules.
• He referred to this sampling as a microstate of the
thermodynamic system.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Entropy on the Molecular Scale
• Each thermodynamic state has a specific number of
microstates, W, associated with it.
• Entropy is
S = k lnW
where k is the Boltzmann constant, 1.38  1023 J/K.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Entropy on the Molecular Scale
• The change in entropy for a process,
then, is
S = k lnWfinal  k lnWinitial
Wfinal
S = k ln
Winitial
• Entropy increases with the number of
Chemical
microstates in the system.
Thermodynamics
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Entropy on the Molecular Scale
• The number of microstates and,
therefore, the entropy tends to increase
with increases in
– Temperature.
– Volume.
– The number of independently moving
molecules.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Entropy and Physical States
• Entropy increases with
the freedom of motion
of molecules.
• Therefore,
S(g) > S(l) > S(s)
Chemical
Thermodynamics
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Solutions
Generally, when
a solid is
dissolved in a
solvent, entropy
increases.
Chemical
Thermodynamics
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Entropy Changes
• In general, entropy
increases when
– Gases are formed from
liquids and solids;
– Liquids or solutions are
formed from solids;
– The number of gas
molecules increases;
– The number of moles
increases.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.3
• Predict whether ΔS is positive or
negative for each of the following
processes, assuming each occurs at
constant temperature:
Chemical
Thermodynamics
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Practice Exercise 19.3
• Indicate whether each of the following
processes produces an increase or
decrease in the entropy of the system:
• Answer: (a) increase, (b) decrease, (c)
Chemical
decrease, (d) decrease
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.4
• Choose the sample of matter that has
greater entropy in each pair, and
explain your choice: (a) 1 mol of NaCl(s)
or 1 mol of HCl(g) at 25 °C, (b) 2 mol
of HCl(g) or 1 mol of HCl(g) at 25 °C,
(c) 1 mol of HCl(g) or 1 mol of Ar(g) at
298 K.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.4
Choose the substance with the greater entropy
in each case: (a) 1 mol of H2(g) at STP or
1 mol of H2(g) at 100°C and 0.5 atm, (b) 1 mol
of H2O(s) at 0 °C or 1 mol of H2O(l) at 25 °C,
(c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at
STP, (d) 1 mol of N2O4(g) at STP or 2 mol of
NO2(g) at STP.
Answers: (a) 1 mol of H2(g) at 100 °C and 0.5
atm, (b) 1 mol of H2O(l) at 25 °C, (c) 1 mol of
Chemical
SO2(g) at STP, (d) 2 mol of NO2(g) at STP Thermodynamics
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Third Law of Thermodynamics
The entropy of a pure crystalline
substance at absolute zero is 0.
Chemical
Thermodynamics
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Standard Entropies
• These are molar entropy
values of substances in
their standard states.
• Standard entropies tend
to increase with
increasing molar mass.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Standard Entropies
Larger and more complex molecules have
greater entropies.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Entropy Changes
Entropy changes for a reaction can be
estimated in a manner analogous to that by
which H is estimated:
S = nS(products) — mS(reactants)
where n and m are the coefficients in the
balanced chemical equation.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.5
• Calculate ΔSº for the synthesis of
ammonia from N2(g) and H2(g) at 298 K:
•
N2(g) + 3 H2(g) → 2 NH3(g)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.5
• Using the standard entropies in
Appendix C, calculate the standard
entropy change, ΔS°, for the following
reaction at 298 K:
• Al2O3(s) + 3 H2(g) → 2 Al(s) + 3 H2O(g)
• Answers: 180.39 J/K
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Entropy Changes in Surroundings
• Heat that flows into or out of the
system changes the entropy of the
surroundings.
• For an isothermal process:
Ssurr =
qsys
T
• At constant pressure, qsys is simply
H for the system.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Entropy Change in the Universe
• The universe is composed of the system
and the surroundings.
• Therefore,
Suniverse = Ssystem + Ssurroundings
• For spontaneous processes
Suniverse > 0
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Entropy Change in the Universe
• Since Ssurroundings =
qsystem
and qsystem = Hsystem
T
This becomes:
Hsystem
Suniverse = Ssystem +
T
Multiplying both sides by T, we get
TSuniverse = Hsystem  TSsystem
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Gibbs Free Energy
• TSuniverse is defined as the Gibbs free
energy, G.
• When Suniverse is positive, G is
negative.
• Therefore, when G is negative, a
process is spontaneous.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Gibbs Free Energy
1. If G is negative, the
forward reaction is
spontaneous.
2. If G is 0, the system
is at equilibrium.
3. If G is positive, the
reaction is
spontaneous in the
reverse direction.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.6
• Calculate the standard free energy
change for the formation of NO(g) from
N2(g) and O2(g) at 298 K:
•
N2(g) + O2(g) → 2 NO(g)
• given that ΔH° = 180.7 kJ and ΔS° =
24.7 J/K. Is the reaction spontaneous
under these circumstances?
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.6
• A particular reaction has ΔH° = 24.6 kJ
and ΔS° = 132 J/K at 298 K. Calculate
ΔG°.
• Is the reaction spontaneous under these
conditions?
• Answers: ΔG° = –14.7 kJ
• the reaction is spontaneous.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Standard Free Energy Changes
Analogous to standard enthalpies of
formation are standard free energies of
formation, G.
f
G = nGf (products)  mG f(reactants)
where n and m are the stoichiometric
coefficients.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.7
• (a) Use data from Appendix C to
calculate the standard free-energy
change for the following reaction at
298 K:
• P4(g) + 6 Cl2(g) → 4 PCl3(g)
• (b) What is ΔG°for the reverse of the
above reaction?
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.7
• By using data from Appendix C,
calculate ΔG° at 298 K for the
combustion of methane:
• CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g).
• Answer: –800.7 kJ
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.8
• In Section 5.7 we used Hess’s law to calculate ΔH°
for the combustion of propane gas at 298 K:
•
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
ΔH° = –2220 kJ
• (a) Without using data from Appendix C, predict
whether ΔG° for this reaction is more negative or
less negative than ΔH°.
• (b) Use data from Appendix C to calculate the
standard free-energy change for the reaction at 298
K. Is your prediction from part (a) correct?
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.8
• Consider the combustion of propane to
form CO2(g) and H2O(g) at 298 K:
• C3H8(g) + 5 O2(g) → 3 CO2(g) + 4
H2O(g). Would you expect ΔG° to be
more negative or less negative than
ΔH°?
• Answer: more negative
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Free Energy Changes
At temperatures other than 25°C,
G° = H  TS
How does G change with temperature?
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Free Energy and Temperature
• There are two parts to the free energy
equation:
 H— the enthalpy term
– TS — the entropy term
• The temperature dependence of free
energy, then comes from the entropy
term.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Free Energy and Temperature
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.9
• The Haber process for the production of ammonia
involves the equilibrium
• Assume that ΔH° and ΔS° for this reaction do not
change with temperature. (a) Predict the direction in
which ΔG° for this reaction changes with increasing
temperature. (b) Calculate the values ΔG° of for the
reaction at 25 °C and 500 °C.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.9
• (a) Using standard enthalpies of
formation and standard entropies in
Appendix C, calculate ΔH° and ΔS°
at 298 K for the following reaction: 2
SO2(g) + O2(g) → 2 SO3(g). (b) Using
the values obtained in part (a), estimate
ΔG° at 400 K.
• Answers: (a) ΔH° = –196.6 kJ, ΔS° Chemical
= –189.6 J/K; (b) ΔG° = –120.8 kJ Thermodynamics
© 2009, Prentice-Hall, Inc.
Free Energy and Equilibrium
Under any conditions, standard or
nonstandard, the free energy change
can be found this way:
G = G + RT lnQ
(Under standard conditions, all concentrations are 1 M,
so Q = 1 and lnQ = 0; the last term drops out.)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Free Energy and Equilibrium
• At equilibrium, Q = K, and G = 0.
• The equation becomes
0 = G + RT lnK
• Rearranging, this becomes
G = RT lnK
or,
-G
K = e RT
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.10
• As we saw in Section 11.5, the normal boiling
point is the temperature at which a pure liquid
is in equilibrium with its vapor at a pressure of
1 atm. (a) Write the chemical equation that
defines the normal boiling point of liquid
carbon tetrachloride, CCl4(l). (b) What is the
value of ΔG° for the equilibrium in part (a)?
(c) Use thermodynamic data in Appendix C
and Equation 19.12 to estimate the normal
Chemical
boiling point of CCl4.
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.10
• Use data in Appendix C to estimate the
normal boiling point, in K, for elemental
bromine, Br2(l). (The experimental value
is given in Table 11.3.)
• Answers: 330 K
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.11
• We will continue to explore the Haber
process for the synthesis of ammonia:
• Calculate ΔG at 298 K for a reaction
mixture that consists of 1.0 atm N2, 3.0
atm H2, and 0.50 atm NH3.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.11
• Calculate ΔG at 298 K for the reaction
of nitrogen and hydrogen to form
ammonia if the reaction mixture consists
of 0.50 atm N2, 0.75 atm H2, and 2.0
atm NH3.
• Answers: –26.0 kJ/mol
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.12
• Use standard free energies of formation to
calculate the equilibrium constant, K, at 25
°C for the reaction involved in the Haber
process:
• The standard free-energy change for this
reaction was calculated in Sample Exercise
19.9:
• ΔG° = –33.3 kJ/mol = –33,300 J/mol.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.12
• Use data from Appendix C to calculate
the standard free-energy change, ΔG°,
and the equilibrium constant, K, at 298
K for the reaction
• Answers: ΔG° = –106.4 kJ/mol, K = 4 × 1018
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.