4/19/2016 1 Copyright 2011 Pearson Education, Inc. Chapter 17 Free Energy and Thermodynamics Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Copyright 2011 Pearson Education, Inc. First Law of Thermodynamics • You can’t win! • First Law of Thermodynamics: Energy cannot be created or destroyed the total energy of the universe cannot change though you can transfer it from one place to another DE universe = 0 = DE system + DE surroundings 2 Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc. First Law of Thermodynamics • Conservation of Energy • For an exothermic reaction, “lost” heat from the system goes into the surroundings • Two ways energy is “lost” from a system converted to heat, q used to do work, w • Energy conservation requires that the energy change in the system equal the heat released + work done DE = q + w DE = DH + PDV • DE is a state function internal energy change independent of how done 3 Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc. The Energy Tax • You can’t break even! • To recharge a battery with 100 kJ of useful energy will require more than 100 kJ because of the Second Law of Thermodynamics • Every energy transition results in a “loss” of energy an “Energy Tax” demanded by nature and conversion of energy to heat which is “lost” by heating up the surroundings 4 Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc. Heat Tax fewer steps generally results in a lower total heat tax 5 Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc. Thermodynamics and Spontaneity • Thermodynamics predicts whether a process will occur under the given conditions processes that will occur are called spontaneous nonspontaneous processes require energy input to go • Spontaneity is determined by comparing the chemical potential energy of the system before the reaction with the free energy of the system after the reaction if the system after reaction has less potential energy than before the reaction, the reaction is thermodynamically favorable. • Spontaneity ≠ fast or slow 6 Tro: Chemistry: A Molecular Approach, 2/e
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4/19/2016
1
Copyright 2011 Pearson Education, Inc.
Chapter 17
Free Energy
and
Thermodynamics
Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MACopyright 2011 Pearson Education, Inc.
First Law of Thermodynamics
• You can’t win!
• First Law of Thermodynamics: Energy
cannot be created or destroyed
the total energy of the universe cannot change
though you can transfer it from one place to another
DEuniverse = 0 = DEsystem + DEsurroundings
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First Law of Thermodynamics• Conservation of Energy
• For an exothermic reaction, “lost” heat from the system goes into the surroundings
• Two ways energy is “lost” from a system converted to heat, q
used to do work, w
• Energy conservation requires that the energy change in the system equal the heat released + work done DE = q + w
DE = DH + PDV
• DE is a state function internal energy change independent of how done
3Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
The Energy Tax
• You can’t break even!
• To recharge a battery with 100 kJ of
useful energy will require more than
100 kJ
because of the Second Law of
Thermodynamics
• Every energy transition results in a
“loss” of energy
an “Energy Tax” demanded by nature
and conversion of energy to heat which
is “lost” by heating up the surroundings
4Tro: Chemistry: A Molecular Approach, 2/e
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Heat Tax
fewer steps
generally results
in a lower total
heat tax
5Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Thermodynamics and Spontaneity
• Thermodynamics predicts whether a process will occur under the given conditions
processes that will occur are called spontaneous
nonspontaneous processes require energy input to go
• Spontaneity is determined by comparing the chemical potential energy of the system before the reaction with the free energy of the system after the reaction
if the system after reaction has less potential energy than before the reaction, the reaction is thermodynamically favorable.
• Spontaneity ≠ fast or slow
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Comparing Potential Energy
The direction of
spontaneity can
be determined
by comparing
the potential
energy of the
system at the
start and the
end
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Reversibility of Process• Any spontaneous process is irreversible because
there is a net release of energy when it proceeds in that direction
it will proceed in only one direction
• A reversible process will proceed back and forth between the two end conditions
any reversible process is at equilibrium
results in no change in free energy
• If a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction
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Thermodynamics vs. Kinetics
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Diamond → Graphite
Graphite is more stable than diamond, so the conversion
of diamond into graphite is spontaneous – but don’t worry,
it’s so slow that your ring won’t turn into pencil lead in
your lifetime (or through many of your generations)
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Spontaneous Processes
• Spontaneous processes occur because they
release energy from the system
• Most spontaneous processes proceed from a
system of higher potential energy to a system at
lower potential energy
exothermic
• But there are some spontaneous processes that
proceed from a system of lower potential energy to
a system at higher potential energy
endothermic
• How can something absorb potential energy, yet
have a net release of energy?
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Melting Ice
12
Melting is an
Endothermic process,
yet ice will
spontaneously melt
above 0 °C.
When a solid melts, the
particles have more
freedom of movement.
More freedom of motion
increases the
randomness of the
system. When systems
become more random,
energy is released. We
call this energy,
entropyTro: Chemistry: A Molecular Approach, 2/e
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Factors Affecting Whether a
Reaction Is Spontaneous
• There are two factors that determine whether a
reaction is spontaneous. They are the
enthalpy change and the entropy change of
the system
• The enthalpy change, DH, is the difference in
the sum of the internal energy and PV work
energy of the reactants to the products
• The entropy change, DS, is the difference in
randomness of the reactants compared to the
products
13Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Enthalpy Change
DH generally measured in kJ/mol
• Stronger bonds = more stable molecules
• A reaction is generally exothermic if the bonds in the
products are stronger than the bonds in the
reactants
exothermic = energy released, DH is negative
• A reaction is generally endothermic if the bonds in
the products are weaker than the bonds in the
reactants
endothermic = energy absorbed, DH is positive
• The enthalpy change is favorable for exothermic
reactions and unfavorable for endothermic reactions
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Entropy
• Entropy is a thermodynamic function
that increases as the number of
energetically equivalent ways of
arranging the components increases, S
S generally J/mol
• S = k ln W
k = Boltzmann Constant = 1.38 x 10−23 J/K
W is the number of energetically equivalent
ways a system can exist
unitless
• Random systems require less energy
than ordered systems15Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
W
These are energetically
equivalent states for the
expansion of a gas.
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It doesn’t matter, in terms
of potential energy,
whether the molecules
are all in one flask, or
evenly distributed
But one of these states is
more probable than the
other two
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This macrostate can be achieved through
several different arrangements of the particles
Macrostates → Microstates
17
These
microstates all
have the same
macrostate
So there are six
different particle
arrangements
that result in the
same macrostate
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Macrostates and Probability
There is only one possible
arrangement that gives State
A and one that gives State B
There are six possible
arrangements that give State C
Therefore State C has
higher entropy than
either State A or State B
The macrostate with the
highest entropy also has the
greatest dispersal of energy
There is six times the
probability of having the
State C macrostate than
either State A or State B
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Changes in Entropy, DS
DS = Sfinal − Sinitial
• Entropy change is favorable when the result is a more random system
DS is positive
• Some changes that increase the entropy are
reactions whose products are in a more random state
solid more ordered than liquid more ordered than gas
reactions that have larger numbers of product molecules than reactant molecules
increase in temperature
solids dissociating into ions upon dissolving
19Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
combustion is largely exothermic, so the entropy of
the surroundings should increase significantly
DHsystem = −2044 kJ, T = 25 ºC = 298 K
DSsurroundings, J/K
Check:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DST, DH
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Practice – The reaction below has DHrxn = +66.4 kJ at
25 °C. (a) Determine the Dssurroundings, (b) the sign of
DSsystem, and (c) whether the process is spontaneous
2 O2(g) + N2(g) 2 NO2(g)
33Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – The reaction 2 O2(g) + N2(g) 2 NO2(g) has DHrxn
= +66.4 kJ at 25 °C. Calculate the entropy change of the
surroundings. Determine the sign of the entropy change in
the system and whether the reaction is spontaneous.
DHsys = +66.4 kJ, T = 25 ºC = 298 K
DSsurr, J/K, DSreact + or −, DSuniverse + or −
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DST, DH
the major difference is that there are
fewer product molecules than
reactant molecules, so the DSreaction
is unfavorable and (−)
because DSsurroundings is
(−) it is unfavorable
both DSsurroundings and DSreaction are (−),
DSuniverse is (−) and the process is
nonspontaneous
34Tro: Chemistry: A Molecular Approach, 2/e
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Gibbs Free Energy and Spontaneity
• It can be shown that −TDSuniv = DHsys−TDSsys
• The Gibbs Free Energy, G, is the maximum amount of work energy that can be released to the surroundings by a system for a constant temperature and pressure system
the Gibbs Free Energy is often called the Chemical Potential because it is analogous to the storing of energy in a mechanical system
DGsys = DHsys−TDSsys
• Because DSuniv determines if a process is spontaneous, DG also determines spontaneity
DSuniv is + when spontaneous, so DG is −
35Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Gibbs Free Energy, DG
• A process will be spontaneous when DG is
negative
DG will be negative when
DH is negative and DS is positive
exothermic and more random
DH is negative and large and DS is negative but small
DH is positive but small and DS is positive and large
or high temperature
• DG will be positive when DH is + and DS is −
never spontaneous at any temperature
• When DG = 0 the reaction is at equilibrium36Tro: Chemistry: A Molecular Approach, 2/e
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DG, DH, and DS
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Free Energy Change and Spontaneity
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Example 17.3a: The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has
DH = +95.7 kJ and DS = +142.2 J/K at 25 °C.
Calculate DG and determine if it is spontaneous.
39
Because DG is +, the reaction is not spontaneous
at this temperature. To make it spontaneous, we
need to increase the temperature.
DH = +95.7 kJ, DS = 142.2 J/K, T = 298 K
DG, kJ
Answer:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGT, DH, DS
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – The reaction Al(s) + Fe2O3(s) Fe(s) + Al2O3(s) has
DH = −847.6 kJ and DS = −41.3 J/K at 25 °C.
Calculate DG and determine if it is spontaneous.
40Tro: Chemistry: A Molecular Approach, 2/e
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Practice – The reaction Al(s) + Fe2O3(s) Fe(s) + Al2O3(s) has
DH = −847.6 kJ and DS = −41.3 J/K at 25 °C.
Calculate DG and determine if it is spontaneous.
Because DG is −, the reaction is spontaneous at
this temperature. To make it nonspontaneous, we
need to increase the temperature.
DH = −847.6 kJ and DS = −41.3 J/K, T = 298 K
DG, kJ
Answer:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGT, DH, DS
41Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Example 17.3b: The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has
DH = +95.7 kJ and DS = +142.2 J/K.
Calculate the minimum temperature it will be spontaneous.
42
the temperature must be higher than 673K for
the reaction to be spontaneous
DH = +95.7 kJ, DS = +142.2 J/K, DG < 0
T, K
Answer:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
TDG, DH, DS
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Practice – The reaction Al(s) + Fe2O3(s) Fe(s) + Al2O3(s) has
DH = −847.6 kJ and DS = −41.3 J/K.
Calculate the maximum temperature it will be spontaneous.
43Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – The reaction Al(s) + Fe2O3(s) Fe(s) + Al2O3(s) has
DH = −847.6 kJ and DS = −41.3 J/K at 25 °C.
Determine the maximum temperature it is spontaneous.
any temperature above 1775 C will make the
reaction nonspontaneous
DH = −847.6 kJ and DS = −41.3 J/K, DG > 0
T, C
Answer:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
TDG, DH, DS
2048 − 273 = 1775 C
44Tro: Chemistry: A Molecular Approach, 2/e
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Standard Conditions
• The standard state is the state of a
material at a defined set of conditions
• Gas = pure gas at exactly 1 atm pressure
• Solid or Liquid = pure solid or liquid in its
most stable form at exactly 1 atm pressure
and temperature of interestusually 25 °C
• Solution = substance in a solution with
concentration 1 M
45Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
The 3rd Law of Thermodynamics:
Absolute Entropy
• The absolute entropy of a substance is the amount of energy it has due to dispersion of energy through its particles
• The 3rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K
therefore, every substance that is not a perfect crystal at absolute zero has some energy from entropy
therefore, the absolute entropy of substances is always +
46Tro: Chemistry: A Molecular Approach, 2/e
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Standard Absolute Entropies
• S°
• Extensive
• Entropies for 1 mole of a substance at 298 K
for a particular state, a particular allotrope,
particular molecular complexity, a particular
molar mass, and a particular degree of
dissolution
47Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Standard Absolute Entropies
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Relative Standard Entropies:
States
• The gas state has a larger entropy than the
liquid state at a particular temperature
• The liquid state has a larger entropy than the
solid state at a particular temperature
SubstanceS°,
(J/mol∙K)
H2O (l) 70.0
H2O (g) 188.8
49Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Relative Standard Entropies:
Molar Mass
• The larger the molar
mass, the larger the
entropy
• Available energy states
more closely spaced,
allowing more dispersal
of energy through the
states
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Relative Standard Entropies:
Allotropes
• The less constrained
the structure of an
allotrope is, the
larger its entropy
• The fact that the
layers in graphite
are not bonded
together makes it
less constrained
51Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Relative Standard Entropies:
Molecular Complexity
• Larger, more complex
molecules generally
have larger entropy
• More available energy
states, allowing more
dispersal of energy
through the states
SubstanceMolar
Mass
S°,
(J/mol∙K)
Ar (g) 39.948 154.8
NO (g) 30.006 210.8
52
SubstanceMolar
Mass
S°,
(J/mol∙K)
CO (g) 28.01 197.7
C2H4 (g) 28.05 219.3
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Relative Standard Entropies
Dissolution
• Dissolved solids
generally have larger
entropy
• Distributing particles
throughout the mixture
SubstanceS°,
(J/mol∙K)
KClO3(s) 143.1
KClO3(aq) 265.7
53Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
The Standard Entropy Change, DS
• The standard entropy change is the difference
in absolute entropy between the reactants and
products under standard conditions
DSºreaction = (∑npSºproducts) − (∑nrSºreactants)
remember: though the standard enthalpy of
formation, DHf°, of an element is 0 kJ/mol, the
absolute entropy at 25 °C, S°, is always positive
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Example 17.4: Calculate DS for the
reaction
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
55
DS is +, as you would expect for a reaction with more gas
product molecules than reactant molecules
standard entropies from Appendix IIB
DS, J/K
Check:
Sol’n:
Conceptual
Plan:
Relationships:
Given:
Find:
DSSNH3, SO2, SNO, SH2O
Substance S, J/molK
NH3(g) 192.8
O2(g) 205.2
NO(g) 210.8
H2O(g) 188.8
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – Calculate the DS for the reaction
2 H2(g) + O2(g) 2 H2O(g)
Substance S, J/molK
H2(g) 130.7
O2(g) 205.2
H2O(g) 188.8
56Tro: Chemistry: A Molecular Approach, 2/e
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Example 17.4: Calculate DS for the
reaction
2 H2(g) + O2(g) 2 H2O(g)
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DS is −, as you would expect for a reaction with more
gas reactant molecules than product molecules
standard entropies from Appendix IIB
DS, J/K
Check:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DSSH2, SO2, SH2O
Substance S, J/molK
H2(g) 130.6
O2(g) 205.2
H2O(g) 188.8
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Calculating DG
• At 25 C
DGoreaction = SnDGo
f(products) - SnDGof(reactants)
• At temperatures other than 25 C
assuming the change in DHoreaction and DSo
reaction is negligible
DGreaction = DHreaction – TDSreaction
• or
DGtotal = DGreaction 1 + DGreaction 2 + ...
58Tro: Chemistry: A Molecular Approach, 2/e
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Example 17.6: The reaction SO2(g) + ½ O2(g) SO3(g) has
DH = −98.9 kJ and DS = −94.0 J/K at 25 °C.
Calculate DG at 125 C and determine if it is more or less
spontaneous than at 25 °C with DG° = −70.9 kJ/mol SO3.
59
because DG is −, the reaction is spontaneous at
this temperature, though less so than at 25 C
DH = −98.9 kJ, DS = −94.0 J/K, T = 398 K
DG, kJ
Answer:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGT, DH, DS
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – Determine the free energy change in the
following reaction at 298 K
2 H2O(g) + O2(g) 2 H2O2(g)
Substance DH, kJ/mol S, J/mol
H2O2(g) −136.3 232.7
O2(g) 0 205.2
H2O(g) −241.8 188.8
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Practice – Determine the free energy change in the
following reaction at 298 K
2 H2O(g) + O2(g) 2 H2O2(g)
61
standard energies from Appendix IIB, T = 298 K
DG, kJ
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGDHf° Sº of prod & react DH, DSº
DH = 211.0 kJ, DS = −117.4 J/K, T = 298 K
DG, kJ
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Standard Free Energies of Formation
• The free energy of formation (DGf°) is the change
in free energy when 1 mol of a compound forms
from its constituent elements in their standard
states
• The free energy
of formation of
pure elements in
their standard
states is zero
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Example 17.7: Calculate DG at 25 C
for the reactionCH4(g) + 8 O2(g) CO2(g) + 2 H2O(g) + 4 O3(g)
63
standard free energies of formation from Appendix IIB
DG, kJ
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGDGf° of prod & react
Substance DGf°, kJ/mol
CH4(g) −50.5
O2(g) 0.0
CO2(g) −394.4
H2O(g) −228.6
O3(g) +163.2
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – Determine the free energy change in the
following reaction at 298 K
2 H2O(g) + O2(g) 2 H2O2(g)
Substance DG, kJ/mol
H2O2(g) −105.6
O2(g) 0
H2O(g) −228.6
64Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Determine the free energy change in the
following reaction at 298 K
2 H2O(g) + O2(g) 2 H2O2(g)
65
standard free energies of formation from Appendix IIB
DG, kJ
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGDGf° of prod & react
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DG Relationships
• If a reaction can be expressed as a series of
reactions, the sum of the DG values of the
individual reaction is the DG of the total reaction
DG is a state function
• If a reaction is reversed, the sign of its DG value
reverses
• If the amount of materials is multiplied by a factor,
the value of the DG is multiplied by the same
factor
the value of DG of a reaction is extensive
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Example 17.8: Find DGºrxn for the following reaction