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
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DH
DH = heat transferred between the system and its
surroundings during a constant-pressure process
DE = q + w
DE = D in internal E of a system
q = heat absorbed by the system from surroundings
w = work done on the system by the surroundings
(see Chapter 5 for details)
Chemical
Thermodynamics
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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
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Spontaneous Processes
Processes that are
spontaneous in one
direction are
nonspontaneous in
the reverse
direction.
Chemical
Thermodynamics
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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
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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,
DS = 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
DS =
T
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.2 (p. 807)
The element, Hg, is a silvery liquid at room
temperature. The normal freezing point of
mercury is -38.9oC, and its molar enthalpy of
fusion is DHfusion = 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?
Hint: Calculate q, as you have done before,
then reverse the sign and divide by T in K.
(DSsys = -2.44 J/K)
Chemical
Thermodynamics
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Practice Exercise 19.2
The normal boiling point of ethanol, C2H5OH,
is 78.3oC, 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?
(-163 J/K)
Chemical
Thermodynamics
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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:
DSuniv = DSsystem + DSsurroundings = 0
For irreversible processes:
DSuniv = DSsystem + DSsurroundings > 0
Chemical
Thermodynamics
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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
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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
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Entropy on the Molecular Scale
• The change in entropy for a process,
then, is
DS = k lnWfinal  k lnWinitial
lnWfinal
DS = k ln
lnWinitial
• 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
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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
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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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Sample Exercise 19.3 (p. 814)
Predict whether DS is positive or
negative for each of the following
processes, assuming each occurs at
constant temperature:
a) H2O(l)  H2O(g)
b) Ag+(aq) + Cl-(aq)  AgCl(s)
c) 4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
d) N2(g) + O2(g)  2 NO(g)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.3
Indicate whether each of the following
reactions produces an increase or
decrease in the entropy of the system:
a) CO2(s)  CO2(g)
b) CaO(s) + CO2(g)  CaCO3(s)
c) HCl(g) + NH3(g)  NH4Cl(s)
d) 2 SO2(g) + O2(g)  2 SO3(g)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.4 (p. 815)
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 25oC
b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25oC
c) 1 mol of HCl(g) or 1 mol or 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 100oC
and 0.5 atm
b) 1 mol of H2O(s) at 0oC or 1 mol of H2O(l) at 25oC
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
Chemical
Thermodynamics
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Where we’ve been:
Where we’re going:
DH, including Hess’s
Law – the heat
component
DH and DS  Gibbs
Free Energy – will a
reaction occur?
DS, including
#microstates – the
entropy component
Application to
equilibrium
Application to
electrochemistry
(after Spring Break)
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
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Standard Entropies
Larger and more complex molecules have
greater entropies.
Chemical
Thermodynamics
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Entropy Changes
Entropy changes for a reaction can be
estimated in a manner analogous to that by
which DH is estimated:
DS = nDS(products) — mDS(reactants)
where n and m are the coefficients in the
balanced chemical equation.
Chemical
Thermodynamics
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Entropy Changes in Surroundings
• Heat that flows into or out of the
system changes the entropy of the
surroundings.
• For an isothermal process:
DSsurr =
qsys
T
• At constant pressure, qsys is simply
DH 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,
DSuniverse = DSsystem + DSsurroundings
• For spontaneous processes
DSuniverse > 0
Chemical
Thermodynamics
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Entropy Change in the Universe
• Since DSsurroundings =
qsystem
and qsystem = DHsystem
T
This becomes:
DHsystem
DSuniverse = DSsystem +
T
Multiplying both sides by T, we get
TDSuniverse = DHsystem  TDSsystem
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Gibbs Free Energy
• TDSuniverse is defined as the Gibbs free
energy, DG.
• When DSuniverse is positive, DG is
negative.
• Therefore, when DG is negative, a
process is spontaneous.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Gibbs Free Energy
1. If DG is negative, the
forward reaction is
spontaneous.
2. If DG is 0, the system
is at equilibrium.
3. If DG is positive, the
reaction is
spontaneous in the
reverse direction.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.6 (p. 821)
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 DHo = 180.7 kJ and DSo = 24.7 J/K.
Is the reaction spontaneous under these
circumstances?
(no, DG = 173.3 kJ)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.6
A particular reaction has DHo = 24.6 kJ
and DSo = 132 J/K at 298 K. Calculate
DGo. Is the reaction spontaneous under
these conditions?
(Yes, DGo = -14.7 kJ)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Standard Free Energy Changes
Analogous to standard enthalpies of
formation are standard free energies of
formation, DG.
f
DG = nDGf (products)  mDG f(reactants)
where n and m are the stoichiometric
coefficients.
Chemical
Thermodynamics
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Standard Conditions
•
•
•
•
•
Solid
Liquid
Gas
Solution
Elements
pure solid
pure liquid
1 atm pressure
1 M concentration
standard free energy of
formation of an element in
its free state = zero
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.7 (p. 823)
Use data from Appendix C to calculate the
standard free-energy change for the following
reaction at 298 K:
a) P4(g) + 6 Cl2(g)  4 PCl3(g)
(-1102.8 kJ)
b) What is DGo for the reverse of the above
reaction?
(+1102.8 kJ)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.7
By using the data from Appendix C,
calculate DGo at 298 K for the
combustion of methane:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
(-800.7 kJ)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.8 (p. 823)
In Section 5.7 we used Hess’s law to calculate DHo
for the combustion of propane gas at 298 K:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DHo = -2220 kJ
Without using data from Appendix C, predict whether
DGo for this reaction is more negative or less
negative than DHo.
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?
(-2108 kJ)
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)
DHo = -2220 kJ
Would you expect DGo to be more
negative or less negative than DHo?
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Free Energy Changes
At temperatures other than 25°C,
DG° = DH  TDS
How does DG change with temperature?
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Free Energy and Temperature
• There are two parts to the free energy
equation:
 DH— the enthalpy term
– TDS — 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 (p. 826)
The Haber process for the production of ammonia involves the
following equilibrium:
N2(g) + 3 H2(g) D 2 NH3(g)
Assume that DHo and DSo for this reaction do not change with
temperature.
a) Predict the direction in which DGo for this reaction changes with
increasing temperature.
b) Calculate the values of DGo for the reaction at 25oC and 500oC.
(-33.3 kJ, 61 kJ)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.9
a) Using standard enthalpies of formation and
standard entropies in Appendix C, calculate
DHo and DSo at 298 K for the following
reaction:
2 SO2(g) + O2(g)  2 SO3(g)
(DHo = -196.6 kJ, DSo = -189.6 J/K)
b) Using the values obtained in part (a),
estimate DGo at 400 K.
(DGo = -120.8 kJ)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Free Energy and Equilibrium
Reminders:
DGo and Keq apply to standard conditions
(pure solid or liquid, gases at 1 atm and 1M solutions).
DG and Q (equilibrium quotient) apply to any
conditions.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Free Energy and Equilibrium
Under any conditions, standard or
nonstandard, the free energy change can be
found this way:
DG = DG + RT lnQ
(Under standard conditions, all concentrations are 1 M,
so Q = 1 and lnQ = 0; the last term drops out.)
R = 8.314 J/mol.K, T is in K
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.10 (p. 827)
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 DGo for the equilibrium in (a)?
c)
Use thermodynamic data in Appendix C and DGo =
DHo – TDSo to estimate the normal boiling point of
CCl4. (70oC)
Chemical
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). (330 K)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Free Energy and Equilibrium
• At equilibrium, Q = K, and DG = 0.
• The equation becomes
0 = DG + RT lnK
• Rearranging, this becomes
DG = RT lnK
or,
-DG
K = e RT
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Free Energy and Equilibrium
• If DGo < 0, then K >1
when ln K is +  more negative DGo  larger K  products are
favored
• If DGo = 0, then K = 1
• If DGo > 0, then K <1
when ln K is -  more positive DGo  smaller K  reactants
are favored
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.11 (p. 828)
We will continue to explore the Haber
process for the synthesis of ammonia:
N2(g) + 3 H2(g) D 2 NH3(g)
Calculate DG at 298 K for a reaction mixture
that consists of 1.0 atm N2, 3.0 atm H2, and
0.50 atm NH3.
(-44.9 kJ/mol)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.11
Calculate DG 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.
(-26.0 kJ/mol)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Exercise 19.12 (p. 829)
Use standard free energies of formation
to calculate the equilibrium constant Keq
at 25oC for the reaction involved in the
Haber process:
N2(g) + 3 H2(g) D 2 NH3(g)
(7 x 105)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Practice Exercise 19.12
Use data from Appendix C to calculate
the standard free-energy change, DGo,
and the equilibrium constant, K, at 298
K for the following reaction:
H2(g) + Br2(l) D 2 HBr(g)
(-106.4 kJ/mol; 4 x 1018)
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.
Sample Integrative Exercise 19 (p. 831)
a)
b)
c)
d)
e)
Consider the simple salts NaCl(s) and AgCl(s). We will examine the
equilibria in which these salts dissolve in water to form aqueous
solutions of ions:
NaCl(s) D Na+(aq) + Cl-(aq)
AgCl(s) D Ag+(aq) + Cl-(aq)
Calculate the value of DGo at 298 K for each of the preceding reactions.
The two values from part (a) are very different. Is this difference
primarily due to the enthalpy term or the entropy term of the standard
free-energy change?
Use the values of DGo to calculate Ksp values for the two salts at 298 K.
Sodium chloride is considered a soluble salt, whereas silver chloride is
considered insoluble. Are these descriptions consistent with the
answers to part (c)?
How will DGo for the solution process of these salts change with
increasing T? What effect should this change have on the solubility of
the salts?
Chemical
Thermodynamics
© 2009, Prentice-Hall, Inc.