Chemistry part 30, Julia Burdge,2e (2009) ppsx

734 CHAPTER
18
Entropy, Free Energy, and
Equilibrium
The
reaction
generally
is
the
system.
Therefore,
t.S
~n
is
t.S
~ys
.
•
Recall
that
here
per mole
means
per
mole
of
reaction
as
written
[
H~

Section
5.6
].
•
•
•
•
Think
About
It
The
results
are
consistent
with
the
fact
that
production
of
gases
causes
an
increase
in entropy.
In
part
(a),
the
increa

se
in moles
of
gas gives a
large positive
value
for
LlS
:'xn
' In
part
(b),
the
decrease
in
mole
s
of
gas
gives a large negative value
for
LlS
:'x
n-
In
part
(c),
where
there is
no

change
in
the
number
of
moles
of
gas
in
the
reaction,
LlS
~xn
is
po
sitive
but
is
also fairly small.
In
general,
in cases
where
there
is
no
net
change
in
the

number
of
moles
of
gas in a
reaction
,
we
cannot
predict
whether
LlS
:'x
n will
be
positive
or
negative-but
we
can
predict
that
it
will
be
a relatively small number.
This
make
s s
ense

given
that
gases
invariably have
greater
entropy
than
liquids
and
solids.
For
reactions
involving
only
liquids
a
nd
sol
id
s,
predicting
the
sign
of t.
S"
is
more
difficult,
but in
many

such
cas
es
an
increase
in
the
total
number
of
molecules
and/or
ions
is
accompan
ied
by
an
increase
in
entropy.
The
standard entropy values
of
a large number
of
substances have been measured in
11K·
mol.

To
·
caiCi.iiate
· tiie' s't'lliidarci"entrcipy '
di.ange
·
for
'
li
' reaction
(ilS~n)'
we look up the standard entro-
pies
of
the products and reactants and use Equation 18.7. Sample Problem 18.2 demonstrates this
approach.
• • •
••
•
•
•
•
•
•
·
•
•
·
. .

SampleProble
l11
.
J8.2
'.'

From
the
s
tandard
entr
o
py
values in
Appendix
2,
calculate
the
st
andard
entropy
changes
for
the
following
reaction
s
at
25°C:
(a)

CaC0
3
(s) •
CaO(s)
+ CO
2
(g)
(b) N
2
(g) + 3H
2
(g) •
2NH
3
(g)
(c) H
2
(g) + CI
2
(g) •
2HCl(g)
:
Strategy
Look
up
standard
entropy
values a
nd
u

se
Equation
18.7 to
calculate
LlS
~xn
'
Just
as
we
did
:
when
we
calcula
ted
sta
ndard enthalpies
of
r
eac
tion,
we
consider
sto
ichiometric
coefficients to
be
· . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ' .' . . . .
dimensionless-giving

LlS
:'x
n
unit
s
of
J/K .
mol.
Setup
From
Appendix
2, SO
[CaC0
3
(s)) =
92.9
J
/K
.
mol,
SO[CaO(s)] = 39.8 JIK .
mol
,
SO[CO
zeg)
] =
213.6
JIK .
mol,
SO[N

2
(g)] = 191.5 J/K .
mol,
SO[H
2(g)
] = 131.0 J/K . mol,
SO
[NH
3(g)] =
193.0
JIK .
mol,
SO[CI
2
(g)] =
223.0
JIK .
mol
,
and
SO
[HCl
(g)] = 187.0 JIK . mol.
Solution
(a) M:'xn = [SO(
CaO
) + SO
(C
0
2

)]
- [SO(
CaC0
3
) ]
= [(39.8 JIK .
mol
) +
(213.6
J/K . mol)] - (92.9 J/K .
mol
)
=
160
.5 JIK .
mol
(b) M
:'x
n = [2S0(
NH
3
) ] -
[SO
(N
2
) +
3S
0(H2)]
= (2)(193.0 JIK .
mol

) - [(191.5 J/K .
mol
) + (3)(131.0 J/K . mol)]
= - 198.5
JIK·
mol
(c)
M:'xn = [2S0(
HCl
)] - [SO
(H
2
)
+ SO(CI
2
)]
= (2)(187.0 J/K .
mol
) - [(131.0 J/K .
mol
) + (223.0 JIK . mol)]
=
20.0
JIK . mol
Practice Problem
Calculate
the
sta
ndard
entropy

change
for
the
following
reaction
s
at
25°C.
Predict
first
whether
ea
ch
one
will
be
po
sitive, negative,
or
too close to call.
(a)
2C0zCg)

2CO(g) + 0
2(g)
(b)
30
2
(g) •
20

3
(g)
•
Entropy Changes
in
the
Surroundings
Next we discuss how
ilS
~
urr
is calculated.
When
an exothermic process takes
place
in the sys-
tem, the heat transferred to the surroundings increases the motion
of
the molecules in the
surroundings. Consequently, there is an increase in the
number
of
microstates and the entropy
of
the surroundings increases. Conversely, an endothermic process in the system absorbs
heat
from the surroundings and so decreases the entropy
of
the surroundings by slowing molecular
motion.

Remember
that for constant-pressure processes, the heat released
or
absorbed,
q,
is
equal to the
enthalpy change
of
the system,
ilH
sys
[
~~
Sectio
n 5.3] .
The
change in entropy for
the surroundings,
ilS
s
um
is directly proportional to
ilH
sy
s:
The minus sign indicates that a negative enthalpy change in the system (an exothermic process)
corresponds to a positive entropy change in the surroundings. For an
endothermic process, the
enthalpy change in the system is a positive number and corresponds to a negative entropy change

in the surroundings.
SECTION 18.3 The Second and Third Laws
of
Thermodynamics 735
In addition to being directly proportional
to
b.H
sys
,
b.S
s
urr
is
inversely propOltional to
temperature:
Combining the two expressions gives
Equation 18.9
With Equations 18.8 and 18.9, we can calculate the entropy change in both the system and sur-
roundings for a chemical reaction, and we can determine whether the reaction is spontaneous.
Consider the synthesis
of
ammonia at
25
°
C:
b.H~xn
=
-92
.6
kJ/mol

From Sample Problem 18.2(
b)
, we have
b.S
~
y
s
(- 92.6 kJ/mol) into Equation 18.9, we get
-198.5
J/K . mol, and substituting
b.H
~
y
s
A _
-(-92.6
X 1000) J/
mol
_ .
uS
surr
- 298 K - 311 J/K mol
The entropy change for the universe is
=
-199
J/K·
mol + 311
J/K·
mol
= 112 J/K . mol

Because
b.S
~niv
is
positive, the reaction will be spontaneous at 25°
C.
Keep in mind, though, that
just because a reaction is spontaneous does not mean that it will occur at an observable rate. The
synthesis
of
ammonia is, in fact, extremely slow at room temperature. Thermodynamics can tell
us whether or not a reaction will occur spontaneously under specific conditions, but it does not
tell us how fast it will occur.
Third Law
of
Thermodynamics
Finally, we consider the third law
of
thermodynamics briefly in connection with the determina-
tion
of
standard entropy. So far we have related entropy to microsta
te
s the greater the number
of
microstates a system possesses, the larger is the entropy
of
the syste
m.
Consider a perfect crystal-

line substance at absolute zero
(0 K). Under these conditions, there is essentially
no
molecular
motion and the number
of
microstates
(W)
is 1 (there is only one way
to
arrange the atoms or
molecules to form a perfect crystal). From Equation 18.2, we write
S
=
kin
W
=klnl=O
According to the third law
of
thermodynamics, the entropy
of
a perfect crystalline substance is
zero at absolute zero. As temperature increases, molecular motion increases, causing an increase
in the number
of
microstates. Thus, the entropy
of
any substance at any temperature above 0 K
is greater than zero.
If

the crystalline s
ub
stance is impure or imperfect in any way, then its
entropy is greater than zero even at
0 K because without perfect order there is more than one
microstate.
The important point about the third law
of
thermodynamics is that it enables us to determine
the
absolute entropies
of
substances. Starting with the knowledge that the entropy
of
a pure crys-
. . .

.

. . . .

.,

. .
talline substance is zero at 0 K, we can measure the increase in entropy
of
the substance when it
is heated. The change in entropy
of
a s

ub
stance,
b.S,
is the difference between the final and initial
entropy values:
b.S
=
Sf
-
Si
where
Si
is zero
if
the substance starts at 0 K. Therefore, the measured change in entropy is equal
to the absolute entropy at the new temperature.
f
Altho
ugh
the
complete
details
of
these
measurements
are
beyond
the
scope
of this

book, entropy
changes
are
determined
in
part
by
measur
i
ng
the heat
capacity
of a
substance
[
~~
Section
5.4]
as
a function of
absolute
temperature.
736
CHAPTER
18 Entropy, Free Energy,
and
Equilibrium
Figure 18.4 Entropy increas
es
in

a substance as temperature increases
from absolute zero.
•
Solid
Liquid
Melting
(
~Sfu
s)
Boiling
(~Sv
a
p)
T
em
perature (K)
Gas
The
entropy values arrived at in this way are
ca
lled absolute entropies because they are true
values-unlike
standard enthalpies
of
formation, which are derived using an arbitrary reference.
Becau
se
the tabulated values are determined at I atm, we usually refer to absolute entropies as
standard entropies, So. Figure 18.4 shows the increase in entropy
of

a substance as temperature
increases from absolute zero.
At
0 K, it
ha
s a zero entropy value (assuming that it is a perfect
crystalline subs
tance
). As it is heated, its entropy increases gradually at first because
of
greater
molecular motion within the crystal.
At
the melting point, there is a large increase
in
entropy as
the solid is transformed into the liquid. Further heating increases the entropy
of
the liquid again
due to increa
se
d molecular motion.
At
the boiling point, there is a large increase in entropy as a
result
of
the liquid-to-vapor transition.
Beyond
that temperature, the entropy
of

the gas continues
to increase with increasing temperature.
Checkpoint 18.3 The Second and Third Laws
of
Thermodynamics
18.3.1
U s
ing
data from Appendix
2,
calculate
t:.s
o (in 1
1K
· mol) for
the following reaction:
18.3.2 Using data from Appendix
2,
calculate
t:.s
o (
in
11
K·
mol
) for
the
fo
ll
ow

in
g reactio
n:
2NO(g) + 0
2(g)
+
.
2NOig)
CH
4
(g) +
20ig)
+.
COig)
+
2H
2
0(I)
a) 145.3 11
K·
mol
b)
-145.3IIK·mol
c) 59.7 J/K .
mol
d)
-59.7I
I
K·
mol

e)
-421.2IIK·
mol
a)
107.7 11
K·
mol
b)
-107.7I
I
K·
mol
c)
2.6 11
K·
mol
d) 242.8 11
K·
mol
e)
-2
42.8 11
K·
mol
Gibbs
Free
Energy
According to the se
cond
law

of
thermodynamics,
D.S
univ
> 0 for a spontaneous process. We are
usually concerned with and usually
measure, however, the properties
of
the
system
rather than
tho
se
of
the surroundings
or
those
of
the universe overall.
Ther
efore,
it
is convenient to have a
thermodynamic function that enables us to determine whether
or
not a process is spontaneous by
considering the sys
tem
alone.
We begin with Equation 18.5. For a spontaneous process,

D.S
u
niv
=
D.S
sys
+
D.S
surr
> 0
SECTION
18.4
Gibbs Free Energy 737
Substituting -
fUi
s
y/T
for
~SSUIT>
we
write
~H
s
ys
~Suniv
=
~S
s
y
s

+ - T > 0
Multiplying both sides
of
the equation
by
T gives
T~Suni
v
=
T~S
sys
-
~H
s
ys
> 0
Now
we
have an equation that expresses the second law
of
thermodynamics (and predicts whether
or
not a
proce
ss is spontaneous) in terms
of
only the system. We no
longer
need to consider the
surroundings.

For
convenience,
we
can
rearrange the preceding equation, multiply through
by
-1,
and replace
the>
sign with a < sign:
-
T~Suni
v
=
~H
sys
-
T~S
s
y
s
< 0
According to this equation, a process carried out at constant pressure and temperature is spontaneous
if
the changes
in
enthalpy and entropy
of
the sys
tem

are such that
fUi
sys
-
T~Ssy
s
is less than zero.
To express the spontaneity
of
a,process
more
directly,
we
introduce another thermodynamic
function called the
Gibbs!
free
energy
(G),
or
simply
free
energy.
G =
H-
TS
Equation 18.10
Each
of
the

term
s
in
Equation 18.10 pertains to the system. G has units
of
energy
ju
st as
Hand
TS do. Furthermore, like enthalpy and entropy, free energy is a state function.
The
change
in free
energy,
~G,
of
a
system
for a process that occurs at constant temperature is
~G= ~H-
T~S
Equation 18.11
Equation 18.11 enables us to predict the spontaneity
of
a process using the
change
in enthalpy, the
change
in
entropy, and the absolute temperature.

At
constant temperature and pressure, for pro-
cesses that are spontaneous as written (in the forward direction),
~G
is negative.
For
processes that
are not spontaneous as written
but
that are spontaneous in the reverse direction,
~G
is
po
sitive.
For

,

,
systems
at
equilibrium,
~G
is zero.
•
~G<
0
•
~G>
0

• ~G = 0
The
reaction is spontaneous in the forward direction (and nonspontaneous in
the reverse direction).
The
reaction is nonspontaneous in the forward direction (and spontaneous in
the reverse direction).
The
system is
at
equilibrium.
Often
we
can predict the sign
of
~G
for a process
if
we
know the signs
of
~H
and
~S.
Table 18.3
shows
how
we
can use Equation 18.11 to
make

such predictions.
Based
on
the
information
in
Table 18.3, you may
wonder
what constitutes a "low" or a
"high"
temperature.
For
the
example
given in the table, O°C is the temperature that divides high
from
low. Water freezes spontaneously
at
temperatures below O°C, and ice melts spontaneously
at
temperatures above
O°C.
At
O°C, a system
of
ice and water is at equilibrium.
The
temperature that
In
this

context,
free
energy
is
the
energy
available
to
do
work
.
Thus,
if a
particular
pr
oces
s
is
accompanied
by
a
release
of
usable
energy
(
i.e.,
if
LiG
is

negative),
this
fact
alone
guarantees
that
it
is
spontaneous,
and
there
is
no
need
to
consider
what
happens
to
the
rest
of
the
universe
.
When
aH
Is
And
as

Is
aG
Will
Be
And
the
Process
Is
Example
Negative
Positive
Negative
Positive
Positive
Negative
Negative
Positive
Negative
Positive
Negative
when
T~S
<
fUi
Po
sitive when
T~S
>
~H
Negative when

T~S
>
fUi
Po
sitive
when
T~S
<
fUi
Always spo
nt
aneous
Always nonspontaneous
Spontaneous at low
T
Nonspontaneous
at
high
T
Spontaneous at
high
T
Nonspontaneous
at
low T
2H
2
0
2
(aq)

+.
2H
2
0(l)
+
°z(g)
30
z
(s) •
20
3
(g)
•
Hz0(l)
• HzO(s) (freezing
of
water)
2HgO(s)
+.
2Hg(l)
+ 0 z(g)
1.
Jo
siah Willard Gibbs (1839-1903). American physicist. One
of
the founders of thermodynamics. Gibbs
was
a modest
and private individual who spent almost a
ll

hi
s professional life at
Yale
Universi
ty.
Because he published most
of
his work
in
obscure journals, Gibbs never
ga
ined the eminence that
hi
s contemporary and admirer Jam
es
Maxwell did. Even today,
ve
ry
few
people o
ut
side
of
chemistry and physics have ever heard of Gibbs.
738 CHAPTER 18 Entropy,
Free
Energy, and Equilibrium
•
Think
About

It
Spontaneity is
favored by a release
of
energy (tlH
b
ei
ng negative) and by an increase
in entropy
(tlS being positive).
When both quantities are positive,
as in this cas
e,
only the entropy
change favors spontaneit
y.
For
an endothermic process such as
thi
s,
which requires the input
of
he
at, it should make sense that
adding more heat by increas
in
g
the temperature will shift the
equilibrium
to

the right, thus
making it
"more spontaneous."
divides "high" from "low" depends, though, on the individual reaction. To determine that tempera-
ture, we must set
t1G
equal to 0 in Equation 18.11 (i.e., the equilibrium condition):
0=
t1H
-
Tt1S
Rearranging to solve for T yields
T=
t1H
t1S
The
temperature that divides high from low for a particular reaction can now
be
calculated
if
the
values
of
t1H
and
t1S
are known. Sample
Pr
obl
em

18
.3 demonstrates the use
of
this approach .
Sample Problem 18.3
According to Table IS.3, a reaction will be spontaneous only at
hi
gh temperatures
if
both
tlH
and
tlS are positive. For a reaction in w
hi
ch
tlH
= 199.5 kJ/mol and tlS = 476 J
/K
. mol, determine the
temperature (in
0
c)
above which the reaction
is
spontaneous.
Strategy
The
temperature that divides high from low
is
the temperature at which

tlH
=
TtlS
(
tlG
= 0). Therefore, we use Equation IS.11, substituting 0 for
tlG
and solving for T to determine
temperature in kelvins; we
th
en convert to degrees Celsius.
Setup
Solution
tlS = ( 476 J ) ( 1 kJ ) = 0.476
kJ/K
. mol
mol,
K 1000 J
T =
D.H
= 199.5
kl
l
mol
= 419 K
tlS 0.476
kllK
. mol
= (419 - 273) = 146°C
Practice Problem A reaction will

be
spontaneous only at low temperatures
if
both
tlH
and
tlS
are
negative. For a reaction
in
which
tlH
= - 3S0. 1 kJ/mol and tlS =
-95.001IK·
mol, determine the
temperature (in
0
c)
below which the reaction is spontaneous.
Standard Free-Energy Changes
The
introduction
of
the term
~G
O
enabl
es
us
to

The
"
sia
'
ndardfree~eneiijY
'
of'riiaction
'
(t1G~X
ll
)
is the free energy change for a reaction when
it
write Equation 18.11
as
occurs under standard-state
conditions-that
is, when reactants in their standard states are con-
~G
O
= M ? - T/:,S verted to products in their standard states.
The
conventions used by chemists to define the standard
states
of
pure substances and solutions are
•
Ga
ses

• Liquids
• Solids
• Elements
• Solutions
1 atm pressure
Pure liquid
Pure solid
The
mo
st stable allotropic form at 1
atm
and
25
°C
1 molar concentration
To
calculate
t1G
~xn
'
we start with the general equation
aA
+ bB
-_.
cC + dD
The standard free-energy change for this reaction is given by
Equation 18.12
t1G
~xn
= [ct1G

r
(C) +
dt1G
r
(D)) -
[at1G
r
(A) + bt1G
r
(B))
Equation 18.12 can be generalized as follows:
Equation 18.13
t1G
~xn
=
lnt1G
r
(
pr
oducts) -
lmt1G
r
(reactants)
•
•
SECTION
18.4 Gibbs Free Energy 739
where m and n are stoichiometric coefficients. The term
!1G
J is the standard free energy

of
for-
mation
of
a compound that is, the free-energy change that occurs when 1 mole
of
the compound
is synthesized from its constituent elements, each in its standard state.
For
the combustion
of
graphite,
the standard free-energy change (from Equation 18.13) is
!1G~xn
=
[!1G
H
C0
2
)]
-
[!1G
~
(C,
graphite) +
!1G
~
(02)]
As with standard enthalpy
of

formation, the standard free energy
of
formation
of
any element (in
its most stable allotropic form at 1 atm) is defined as zero. Thus,
!1GJ?(C,
graphite) = 0 and
Therefore, the standard free-energy change for the reaction in this case is equal to the standard free
energy
of
formation
of
CO
2
:
!1G
~xn
=
!1Gf(C0
2
)
Appendix 2 lists the values
of
!1G
J?
at 25°C for a number
of
compounds.
Sample

Problem 18.4 demonstrates the calculation
of
standard free-energy changes.
Sample Problem 18.4
Calculate
the standard free-energy
changes
for
the following
re
actions at 25°C:
(a)
CH
4
(g) +
20z(g)
• COz(g) + 2H
2
0 (I)
(b)
2MgO(s)
•
2Mg(s)
+ Oz(g)
Strategy
Look
up the
!:J.G
'f
values for the reactants

and
products in each equation, and
use
Equation
18.13 to solve
for
!:J.G
~x
n'
Setup
From
Appendix
2,
we
have the following values: !:J.Gf[CH
4
(g)] =
-5
0.8
kllmol
,
!:J.Gf[COz(g)] =
-3
94.4
kllmol,
!:J.Gf[H
2
0(l)]
=
-237.2

kJ/mol, and !:J.Gf[MgO(s)] = - 569.6
kl
lmo!.
All
the
other
s
ubstance
s are
elements
in
their standard states and have, by definition,
!:J.G'f
=
O.
Solution
(a)
!:J.G
~n
= (!:J.Gf[C0
2
(g)] +
2!:J.Gf[H
2
0(
I)]) - (!:J.Gf[CHig)] + 2!:J.Gf[02(g)])
=
[(
-39
4.4

kllmol)
+ (2)(
-237.2
kJ/mol)] - [(
-50.8
kl
lmol) + (2)(0
kl
lmo
l)]
= -
818.0
kl
l
mol
(b)
!:J.G
~xn
= (2!:J.G'f[Mg(s)] + !:J.Gf[02(g)]) - (2
!:J.G
f[
MgO
(s)])
= [(2)(0
kllmol
) + (0
kllmol)]
-
[(2)(-569.6
kJ/mol)]

= 1139 kJ/mol
Practice Problem
Calculate
the standard free-energy changes for the following reactions at 25°C:
(a) H
2
(g) + Br2(l)
+.
2HBr(g)
(b)
2C
2
H
6
(g) +
70
2
(g) • 4COzeg) +
6H
2
0(I)
Using
L1G
and
L1G
O
to
Solve Problems
It
is the sign

of
!1G,
the free-energy change, not the sign
of
!1G
o,
the standard free-energy change,
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
that indicates whether
or
not a process will occur spontaneously under a given set
of
conditions.
What the
sign
of
!1G
o tells us is the same thing that the magnitude
of
the equilibrium constant
(K)
tells us
[
~~
Section
15.2]. A negative
!1G
o value corresponds to a large K value (products favored
at equilibrium), whereas a positive
!1G

o value corresponds to a small K value (reactants favored at
equilibrium).
Like equilibrium constants,
!1G
o values change with temperature.
One
of
the uses
of
Equa-
tion 18.11 is to determine the temperature at
which
a particular equilibrium will begin to favor
a desired product.
For
example, calcium oxide (CaO), also called quicklime, is an extremely
valuable inorganic substance with a variety
of
industrial uses, including water treatment and
Think
About
It
Note
that,
like st
andard
enthalpies
of
formation (
DJI

f),
standard free
energies
of
formation
(!:J.G7)
depend
on
the state
of
matter.
Using
water
as an example,
!:J.G
f[H
2
0(l)]
=
-237.2
kl
l
mol
and
!:J.G
f[H
2
0(g)]
=
-228.6

kJ/mo!.
Always
double-check
to
make
sure
you
have
se
lected the right value
from the table.
,
The
sign
of
!:J.G
O
does
indicate
whether or not
a
process
is
spontaneous
when all
reactants
a
na
products
are

in
their
standard
states,
but
this
is
very
seldom
the
case.
740 CHAPTER
18
Entropy, Free Energy, and
Equilibrium
•
pollution control.
It
is prepared by heating
lime
stone
(CaC0
3
),
which
decompose
s at a high
temperature:
The
reaction is reversible, and under the right conditions, CaO and

CO
2
readily recombine to form
CaC0
3
again. To
pr
event this from happening in the industrial preparation, the system is never
maintained at equilibrium; rather,
CO
2
is constantly removed as
it
forms, shifting the equilibrium
from
left to right, thus promoting the formation
of
calcium oxide.
An important piece
of
information for the chemist responsible for maximizing
CaO
produc-
tion is the temperature at which the decomposition equilibrium
of
CaC0
3
begins to favor products .
We can
mak

e a reliable estimate
of
that temperature as follows. First
we
calculate
b H
o and
b S
o for
the reaction at
2S
o
C,
using the data in Appendix
2.
To
determine
b H
o, we apply Equation S
.1
9:
b W
=
[b H
t'
(CaO) + b Ht'(C0
2
)]
- [b Ht'(CaC0
3

)]
= [(
-63S.6
kJ/mol) + (- 393.S kJ/mol)] -
(-1206.9
kJ/mol)
= 177.8 kJ/mol
Next we apply Equation 18.8 to find
b S
o:
b S
o = [SO(CaO) + SOCO?)] - SO
(CaC0
3
)
= [(39.8 JIK . mol) + (213.6 J/K . mol)] - (92.9 J/K . mol)
= 160.S J/K . mol
From
Equation 18.11,
we
can write
and we obtain
. .

. . . . . . . . . . .

. . . . . . .

.

. .
Be
carefu
l with
units
in
problems
of
this
type.
b G
o = (177.8 kJ/mol) - (298 K)(0.160S kJ/K . mol)
5"
val
ue
s
are
·tabulated
using
joules,
whe
r
eas
Ll.H
\'
value
s
are
tabulated using
kilojoules.

= 130.0 kJ/mol
Becau
se
b G
o is a large positive numbe
r,
the reaction does not favor product formation at
2S
oC
(298 K). And, becau
se
b H
o and
b S
o are both positive, we know that
b G
o will
be
negative (product
formation will
be
favored) at high temperatures. We can determine what constitutes a high tem-
perature for this reaction by calculating the temperature at which
b G
o
is
zero.
or
(177.8 kJ/mol)(1000
JlkJ)

0.160S kJ/K . mol
= 1108 K (83S°C)
•
At
temperatures higher than 83S o
C,
b G
o becomes negative, indicating that the reaction would
then favor the formation
of
CaO and CO?
At
840°C (1113 K), for example,
b G
o =
b H
o -
Tb S
o
= 177.8 kJ/mol - (1113 K)(0.160S kJ/K . mol)
(
l~okg
J
=
-0,8
kJ/mol
At
still higher temperatures,
b G
o becomes increasingly negative, thus favoring product formation

even more. Note that in this example we used the
b H
o and
b S
o values at
2S
oC to calculate changes
to
b G
o at much higher temperatures. Because both
b H
o and
b S
o actually change with tempera-
ture, this approach does not give us a truly accurate value for
b G
o, but it does give us a reasonably
good estimate.
Equation 18.11 can also be used to calculate the change in entropy that accompanies a phase
change.
At
the temperature at which a phase change occurs (i.e., the melting point or boiling point
of
a substance), the system is at equilibrium
(b G
= 0). Therefore, Equation 18.11 becomes
SECTION 18.4 G
ibbs
Free Energy 741
or

0=
tlH
-
TtlS
tlS
=
tlH
T
Consider the ice-water equilibrium. For the ice-to-water transition,
tlH
is the molar heat
of
fusion
(see Table 12.8) and
T is the melting point.
The
entropy change is therefore
6010 I
lmo
l
tlS
ic
e
+.
wat
er
= 273 K = 22.0 I
lK
· mol
Thus, when 1 mole

of
ice melts at OD
C,
there is an increase in entropy
of
22.0 Il
K·
mol. The
increase in entropy is consistent with the increase in microstates from solid to liquid. Conversel
y,
for the water-to-ice transition, the decrease in entropy is given by
•
tlS
. =
-6010
Il
mol
=
-220
I
lK
· I
water • Ice 273 K . mo
The same approach can be applied to the water-to-steam transition.
In
this case,
tlH
is the heat
of
vaporization and T is the boiling point

of
water. Sample Problem 18.5 examines the phase transi-
tions in benzene.
Sample
Problem 18.S
The molar heats
of
fusion and vaporization
of
benzene are 10.9 and 31.0
kJ
/mol, respectively.
Calculate the entropy changes for the solid-to-liquid and liquid-to-vapor transitions for benzene. At
1
atm
pr
essure, benzene melts at 5.5°C and boils at 80.1 DC.
Strategy
The
solid-liquid transition at the melting point and the liquid-vapor transition at the boiling
point are
equilibrium processes. Therefore, because
D.G
is zero at equilibrium, in each case we
can use Equation 18.11, substituting
0 for
D.G
and solving for
D.S,
to determine the entropy change

associated with the process.
Setup
The
melting point
of
benzene is 5.5 + 273.15 = 278.7 K and the boiling point is 80
.1
+
273.15 = 353.3 K.
Solution
D.S
fu
s
=
D.H
ru
s
T melting
10.9
kJ
/mol
278.7 K
= 0.0391
kJ
/K .
molar
39
.1
J/K . mol
D.H

v
ap
D.
S y
ap
= -
'-
T boil
in
g
31.0 kJ/mol
353.3 K
= 0.0877
kJ
/K . mol or 87.7 J
IK
. mol
Practice Problem
The
molar heats
of
fu
sion and vap
or
ization
of
argon are 1.3 and 6.3
kJ
/mol,
respectivel

y,
and argon's melting point and boiling point are
-190
°C a
nd
-186
°C, respectivel
y.
Calculate the entropy changes for the fusion and vaporization
of
argon.
Checkpoint
18.4
Gibbs Free Energy
18.4.1 A reaction for which
D.H
and
D.S
are
both negative is
a) nonspontaneous at all temperatures
b) spontaneous at
a
ll
temperatures
c) spontaneous at
hi
gh temperatures
d) spontaneous at low temperatures
e) at equilibrium

18.4.2
At
what t
empe
rature
(i
n 0
c)
does a
re
act
ion
go
from being nonspontaneous
to spontaneous
if
it has
D.H
=
171
kl
lmol and
D.S
= 161 J/K . mol?
a)
270°C
b) 670°C
c) 1100°C
d) 790°C
e) 28°C

Think
About
It
For the same
substance,
D.S
v
ap
is always
significantly larger than
D.S
fus
'
The
change in number
of
microstates
is always bigger in a liquid-to-gas
transition than in a solid-to-liquid
transition.
•
742 CHAPTER
18
Entropy, Free Energy, and
Equilibrium
•
Eve
n for a
react
ion that s

tarts
with
all
reactan
ts
and
products
in
their
standard
states,
as
soon
as
the
reaction
begin
s, the
conc
entra
tions
of
all
species
change
and
standard-
st
ate
conditions

no longer
exist.
The
Q
used
in
Equation
18
.1
4
can
be
eith
er Q,
(for
reactions
that take
plac
e
in
so
lution) or
Qp
(for
reactions
that
take
place
in
the

gas
ph
as
e
).
• • • • •
18.4
.3 Using data from Appendix 2, calculate
LlG
o (
in
kJ/mol) at 25°C for the
reaction:
CH
4
(g) +
20
2
(g) • CO
2
(g)
+ 2H
2
0(l
)
a)
-580
.8
kJ
/mol

b) 580.8 kJ/mol
c)
-572.0
kJ/mol
d) 572.0 kJ/mol
e)
-818.0
kJ
/mol
18.4.4 Calculate
LlS
vap
(in J/K . mol) for the
vaporization
of
bromine:
Br
2
(l)
-_.
Br
2(g)
LlH
vap
=
31
kJ
/mol, and the boiling
point
of

bromine is 59°C.
a)
9311K'mol
b) 0.53 11K· mol
c)
11
11
K·
mol
d)
l.0
X 10
4
J
/K
. mol
e)
2.
1 X
10
2
11K . mol
Free Energy
and
Chemical Equilibrium
Reactants and products in a chemical reaction are almost always in something other than their
standard state
s-
that is, solutions usually have concentrations other than 1 M and gases usually
have

pre
ssures other than 1 atm. To determine whether or not a reaction is spontaneous, there-
fore, we
mu
st take into account the actual concentrations and/or pressures
of
the species involved.
And
although we can determine
t:: G
o from tabulated values, we need to know
t:: G
to determine
spontaneity.
Relationship Between
LlG
and
LlGO
The
relationship between
t:: G
and
t:: G
o,
which is derived from thermodynamics, is
Equation 18.14
t:: G
=
t:: G
o +

RTln
Q
where
R is the gas constant (8.314 11K· mol or 8.314 X
10
-
3
kllK
. mol), T is the absolute tem-

,
pe
'
ratiiie
'
at
w
hI
ch'
the
'
reac"ii"on
'
takes
'
pi
ace
:
and
'Q is the reaction quotient

[
~~
Secti
on
15.2]
. Thus,
t:: G
depends on two terms:
t:: G
o and RT In
Q.
For
a given reaction at temperature
T,
the value
of
t:: G
o is fixed
but
that
of
RT In Q can vary because Q varies according to the composition
of
the reaction mixture.
Con
sider the following equilibrium:
Using Equation 18.13 and information from Appendix 2, we find that
t:: G
o for this reaction at
25°C is 2.60

kl
lmol.
The
value
of
t:: G,
however, depends on the pressures
of
all three gaseous spe-
cies.
If
we
start with a mixture
of
gases in which P
H2
= 2.0 atm,
PIz
=
2.0
atm, and
PHI
= 3.0 atm,
the reaction quotient,
Qp,
is
(P
HI
)2
Qp

= (
PHz)(PIz
)
= 2.25
Using this value in Equation
18
.14 gives
(3.0)2
9.0
(2.0)(2.0) 4.0
t:: G
= 2.60
kl
+
mol
8.314
X
10
-
3
kl
(298 K)(ln 2.25)
K'mol
= 4.3
kllmol
Because
t:: G
is positive, we conclude that, starting with the
se
concentrations, the forward reaction

will not occur spontaneously as written. Instead, the
reverse reaction will occur spontaneously
and the sys
tem
will reach equilibrium
by
consuming part
of
the
HI
initially present and producing
more
H 2 and I
2
.
If,
on the other hand, we start with a mixture
of
gases in which
PH
= 2.0 atm, PI = 2.0 atm,
2 2
and
PHI
= 1.0 atm, the reaction quotient,
Qp,
is
(P
HI
)2

Qp
=
(PHz)(PIz)
= 0.25
(1.0)2
(2.0)(2.0)
1
- -
4
SECTION 18.5
Free
Energy and Chemical Equilibrium 743
Using this value in Equation 18.14 gives
AG
= 2.6
kJ
+ 8.314 X
10-
3
kJ
(298 K)(ln 0.25)
mol
K .
mol
= - 0.8 kJ/mol
With a negative value for
AG,
the reaction will be spontaneous as written in the forward direc-
tion.
In

this case, the
system
will achieve equilibrium
by
consuming so
me
of
the H2 and
12
to
produce
more
HI.
Sample
Problem
18.6 uses AGo and the reaction quotient to determine in which direction a
reaction is spontaneous .
•
Sample Problem 18.6
The
equilibrium constant, K
p
,
for
th
e reaction
is
0.1l3
at 298 K, which corresponds to a standard free-energy change
of

5.4 kllmo!. In a certain
experiment, the initial pressures are
P
N
0 = 0.453 atm and P
NO
= 0.122 atm. Calculate 6.G for the
2 4 2
reaction at these pressures, and predict the direction in which the reaction will proceed spontaneously
to establish equilibrium.
Strategy
Use the partial pressures
of
N
2
0
4
and
N0
2
to calculate the reaction quotient
Qp,
and then
use Equation 18.14 to calculate
6.G.
Setup
The
reaction quotient expression is
Solution
(P

NO
/
(0. 122l
Qp = = = 0.0329
P
N
0 0.453
2 4
6.G =
6.Go
+
RT
In
Qp
= 5.4
kl
+ (8.3
14
X
10-
3
kJ (298 K)(In 0.0329)
mol
K·
mol
= 5.4
kl
lmol - 8.46
kl
lmol

=
-3
.1
kJ/mol
Because
6.G
is negative, the reaction proceeds spontaneously fr
om
left to right to
rea
ch equilibrium.
Practice Problem A
6.G
o for the reaction
HzCg)
+ 1
2
(g)
:;:::,
~.
2HI(g)
is
2.60
kllmol
at 25°C. Calculate 6.G, and predict the direction in which the reaction is s
pont
aneous
if
the starting concentrations are P
H2

= 3.5 atm, P
I2
= 1.5 a
tm
, and
PHI
= 1.75 atm.
Practice Problem B What is the minimum partial pressure of 12 required for the preceding reaction
to be spontaneous in the forward direction at
25°C
if
the partial pressures
of
H2 and HI are 3.5 and
1.75 atm, respectively?
Relationship Between
~Go
and K
By
definition, AG = 0 and Q = K
at
equilibrium, where K is the equilibrium constant. Thus,
AG
= AGO +
RTln
Q (Equation 18.14)
become
s
Think
About

It
Remember, a
reaction with a positive
6.G
o value
can be spontaneous
if
the starting
concentrations
of
reactants and
pr
oducts are such that Q <
K.
•
-
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
0 =
AG
O +
RTlnK
or
AG
O = -
RTln
K
Equation
18.15
In
this

equation
and
the
one that f
ollows
, K
IS
_
for
reaction
s that
take
place
in
solution
an
d
Ko
f
or
reactions
that
take
place
in
the
gas
p
hase
.

744 CHAPTER 18 Entropy,
Free
Energ
y,
and Equilibrium
•
,
Multimedia
Chemic
al
Equilibrium
equilibrium.
.,. "
Multimedia
Ch
emical
Kinet
i
cs
- interact
iv
e
react
i
on
coordinate
diag
ram.
•
The

sign
of
6
GO
tells
us
the
same
thing
that the
magnitude
of
K
tells
us.
The
sign
of
6G
tells
us
the
same
thing
as
the
comparison
of
Q
and

K
values
[
~~
Section 15.4] .
Figure 18.5 (a)
!:lG
o <
O.
At
equilibrium, there is a significant
conversion
of
reactants to products.
(b)
!:lG
o >
O.
At equilibrium,
reactants are favored over product
s.
In
both cases, the net reaction toward
equilibrium is from left to right
(reactants to products)
if
Q < K and
right to left (products to reactant
s)
if

Q >
K.
At equilibrium, Q =
K.
According to Equation 18.15, then, the larger K is, the more negative
!1G
o is. For chemists, Equation
18.15 is one
of
the most important equations in thermodynamics because it enables us to find the
equilibrium constant
of
a reaction
if
we know the change
in
standard free energy, and vice versa.
It is significant that Equation 18.15 relates the equilibrium constant to the standard free-
energy change,
!1G
o, rather than to the actual free-energy change,
!1G.
The actual free-energy
change
of
the system varies as the reaction progresses and becomes zero at equilibrium. On the
other hand,
!1G
o, like
K,

is a constant for a particular reaction at a given temperature. Figure 18.5
shows plots
of
the free energy
of
a reacting system versus the extent
of
the reaction for two reac-
tions. Table 18.4 summarizes the relationship between the
magnitude
of
an equilibrium constant
and the
sign
of
the corresponding
!1G
o. Remember this important distinction:
It
is the sign
of
!1G
and not that
of
!1G
o that determines the direction
of
reaction spontaneity. The sign
of
!1G

o only
tells
us
the relative amounts
of
products and reactants when equilibrium is reached,
not
the direc-
. . .


. . .


. . . .
tion the reaction must go in order to reach equilibrium.
G
O(
reactants)
-,.
-~
- - - - - - - - - - -

- - - -

- - - -

- - - - - - - -

t:.G

O
is
negative


+-
G°(products)
Equilibrium
po
sition
(products favored)
I
I
I
I
I
I
I
I
I
~~

v
~ t~ v~ _
/
Q<K
Q=K
Q> K
t:.G
< 0

t:.G
= 0
t:.G
> 0
f Extent
of reaction
+ 1
(a)

t:.G
O
is
positive
G
O(
reactants)
,.
~~-

- - - - - - - - - - - - - - - - - - - - - - - - - - - -

I
I
I
I
I
I
I
I
I

v
t
Q< K
Q=K
t:.G
< 0
t:.G
= 0
Equilibrium position
(reactants favored)
/
v
Q>K
t:.G
> 0
f
Extent of reaction
1
(b)
GO
(products)
SECTION 18.5 Free Energy and Chemical
Equilibrium
745
K
>1
= 1
<1
In
K

Positive
o
Negative
Negative
o
Po
sitive
Result
at
Equilibrium
Products are favored.
Neither products
nor
reactants are favored.
Reactants are favored.
For
reactions with very large
or
very small equilibrium
co
nstants, it
can
be very
difficult-
sometimes impossible to determine K values by
mea
suring the concentrations
of
the reactants
and products. Consider, for example, the formation

of
nitric oxide from molecular nitrogen and
molecular oxygen:
•
At
25°C, the equilibrium constant, K
p
,
is
The
very small value
of
Kp
means that the concentration
of
NO
at equilibrium will be exceedingly
low and, for all intents and purposes, impossible to measure directly.
In
such a case, the equilib-
rium constant is
more
conveniently determined using
6.G
o,
which
can
be
calcu
lated either from

tabulated
6.G
t'
values
or
from 6.Ho and 6.so.
Sample
Problems 18.7 and 18.8 show how to use
6.G
o to calculate K and how to use K to
calculate
6.G
o, respectively.
Sample Problem 18.7
Using data from Appendix
2,
calculate the equilibri
um
constant, K
p
,
for the following reaction at
25
°C:
Strategy
Use data from Appendix 2 and Equation 18.13 to calculate
t:: G
o for the reaction.
Then
use

Equation
18
.15
to solve for Kp.
Setup
Solution
t:: G
o =
(2
t:: G
HH
2
(g)] +
t:: GH0
2(
g)]) -
(2
t:: G
f!
[H
2
0 (I)])
= [(2)(0
kJ
/mol) + (2)(0 kJ/mol)] - [(2)
(-
237.2
kJ
/mol)]
= 474.4

kJ
/mol
t:: G
o = - RT
In
Kp
474.4 kJ = _ 8.314 X
10
-
3
kJ (298
K)
In
Kp
mol
K·
mol
-1
9
1.5
=
In
Kp
= 7 X
10
-
84
Practice Problem Using data from Appendix 2, calculate the equilibrium constant, K
p
,

for the
following reaction at
25
°
C:
Think
About
It
This is an
extremely small equilibrium
constant, which is consistent with
the large, positive value
of
t:: G
o.
We know from everyday experience
that water does not de
co
mpose
spontaneously into its constituent
elements at
25
°
C.
746 CHAPTER 18 Entropy, Free Energy, and Equilibri
um
•
Think
About
It

The
relatively
large, positive
!J.G
o, like the very
small
K value, corresponds to a
process that lies very far to the left.
Note that the
K in Equation 18.15
can be any type
of
Kc
(K
a' K
b
,
K
sp
'
etc.) or
Kp.
Sample Problem 18.8
The equilibrium constant, K
sp
'
for the dissolution
of
silver chloride in water at 25°C,
AgCl(s)

:;::,
=
='
Ag
+(aq) +
Cl
- (aq)
is 1.6 X 10-
10.
Calculate
!J.G
o for the process.
Strategy
Use Equation 18.15 to calculate
!J.G
o.
Setup R = 8.314 X
10
-
3
kJlK . mol and T
= (25 + 273) = 298 K.
Solution
!J.G
o =
-RT
In
K
sp
= _ (8.314 X

10
-
3
kl
(298 K)
In
(1.6 X
10
-
10
)
K'mol
= 55.9
kllmol
Practice Problem Calculate
!J.G
o for the process:
BaF
2
(s)
:;::,
=
='
Ba
2+(
aq) + 2F-
(a
q)
The K sp
of

BaF
2 at 25°C is 1.7 X
10
-
6
Checkpoint
18.5
Free Energy and Chemical Equilibrium
18
.5.1 For the reaction
A(aq) + B(aq)
:;::,
=
='
C(aq)
6.GO =
-1.95
kl
lmol at 25°
C.
What
is
6.G
(in kJ/mol) at 25°C when the
concentrations are
[AJ
=
[BJ
=
0.315 M and [C] = 0.405 M?

a)
-1
.95
kllmol
b) 1.53 kJ/mol
c)
-5.43
kllmol
d) 8.
16
kl
lmol
e)
-12.1
kl
l
mol
18
.
5.2
Consider the reaction
X (g) + Y (g)
:+=
,
==
'
Z(g)
for which
6.G
o = - 1.2 kJ/

mol
at
25
°
C.
Without knowing the pressures
of
the
reactants and the product, we know that
this reaction
a) is spontaneous as written.
b) is nonspontaneous as written.
c) fa
vo
rs reactants at equilibrium.
d) favors products at equilibrium.
e) is at equilibrium.
18.5.3
The
!J.G
o for the reaction
is
-33.3
kJ/
mol
at
25
°
C.
What is the

val
ue
of
K
p?
a)
2
X 10-
6
b) 7
X 10
5
c)
3
X 10-
70
d)
1
e)
I X 10-
1
18.5.4
The
K sp for iron(III) hydroxide
[Fe(OH)3J is
1.1 X
10
-
36
at

25°
C.
For
the process
determine
6.G
o (in kJ/mol
')
at
25
°C.
a)
-2.73
X 10-
36
kJ/mol
b) - 17.2
kl
l
mol
c)
17.2kl
l
mo
l
d)
-205
kllmol
e) 205
kl

l
mo
l
Thermodynamics in Living Systems
Many biochemical reactions have a positive
I1G
o valu
e,
yet they are essential
to
the maintenance
of
life. In living systems, these reactions are coupled to an energetically favorable process, one
-
SECTION 18.6 Thermodynamics in Living Systems 747
that has a negative
!:::'Go
value. The principle
of
coupled reactions is based
on
a simple concept:
we
can use a thenllodynamically favorable reaction to drive an unfavorable one. Suppose, for
example, that we want to extract zinc from a zinc sulfide (ZnS). The following reaction will not
work because it has a large positive
!:::.G
o value:
ZnS(s) -

Zn(s) + S(s)
!:::.G
o = 198.3
kJ/mol
On
the other hand, the combustion
of
sulfur to form sulfur dioxide is favored because
of
its large
negative
!:::'G
o value:
S(s)
+
02(g)
-

S02(g)
!:::.G
o = - 300.1
kJ/mol
By coupling the two processes, we can bring about the separation
of
zinc from zinc sulfide. In
practice, this means heating ZnS in air so that the tendency
of
S to form
S02
will promote the

decomposition
of
ZnS:
ZnS(s) -

Zn(s) +
$01
$01+
02(g)
• S02(g)
ZnS(s) +
02(g)
• Zn(s) + SOz(g)
!:::.G
o = 198.3
kJ/mol
!:::.G
o = - 300.1
kJ/mol
!:::.G
o =
-101.8
kJ/mol
Coupled reactions
playa
crucial role in our survival. In biological systems, enzymes facili-
tate a wide variety
of
nonspontaneous reactions. In the human body, for example, food molecules,
represented by glucose (C

6
H
12
0
6
),
are converted to carbon dioxide and water during metabolism,
resulting in a substantial release
of
free energy:
!:::.G
o =
-2880
kJ/mol
In a living cell, this reaction does not take place in a single step; rather, the glucose molecule is
broken down with the aid
of
enzymes in a series
of
steps.
Much
of
the free energy released along
the way is used to synthesize adenosine triphosphate
(ATP) from adenosine diphosphate (ADP)
and phosphoric acid (Figure 18.6):
!:::.G
o =
31
kJ/mol

The
function
of
ATP is to store free energy until it is needed by cells. Under appropriate condi-
tions,
ATP undergoes hydrolysis to give ADP and phosphoric acid, with a release
of
31
kJ
/mo
l
of
free energy, which can
be
used to drive energetically unfavorable reactions, such as protein
synthesis.
Proteins are polymers made
of
amino acids. The stepwise synthesis
of
a protein molecule
involves the joining
of
individual amino acids. Consider the formation
of
the dipeptide (a unit
composed
of
two amino acids) alanylglycine from alanine and glycine. This reaction represents
the first step in the synthesis

of
a protein molecule:
alanine
+ glycine
-_.
alanylglycine
!:::'G
o = 29
kJ/mol
The
positive
!:::.G
o value means this reaction does not favor the formation
of
product, so only a little
of
the dipeptide would be formed at equilibrium. With the aid
of
an enzyme, however, the reaction
is coupled to the hydrolysis
of
ATP as follows:
ATP + H
2
0 + alanine + glycine
-_.
ADP + H
3
P0
4

+ alanylglycine
H
HO
Adenosine triphosphate
(ATP)
H
OH
H
HO
OH
Adenosine diphosphate
(ADP)
Figure 18.6
Structures
of
ATP
andADP .
•
748 CHAPTER
18
Entropy, Free Energy, and
Equ
i
librium
•
Glucose
ATP
Proteins
CO
2

+ H
2
0
ADP
Figure 18.7 Schematic representation
of
ATP synthesis and coupled reactions in living system
s.
The
conversion
of
glucose to carbon dioxide and water during metabolism releases free energy. The released
free energy is used to convert ADP to
ATP.
The
AT
P molecules are then used as an energy source
to
drive
un
favorable reactions such
as
protein synthesis from amino acid
s.
The overall free-energy change is given by
6 G
o =
-31
kJ/mol
+

29
kJ/mol
=
-2
kJ/mol, which
means that the coupled reaction now favors the formation
of
product and an appreciable amount
of
alanylglycine will be formed under these conditions. Figure 18.7 shows the ATP-ADP intercon-
versions that act as energy storage (from metabolism) and free-energy release (fromATP hydroly-
si
s)
to drive essential reactions.
APPLYING
WHAT
YOU'VE LEARNED 749
Applying
What
You've Learned
Ethylene (C
2
H
4
)
plays an important role in the ripening
of
certain fruits and vegetables.
It
is sometimes supplemented under controlled conditions at produce-processing facili-

ties, but it is also produced naturally by many fruit-bearing plants. Although the actual
process by which plants produce ethylene is a complex one, imagine that it is produced
by the simple breakdown
of
glucose according to the equation
Problems:
a)
Determine the sign
of
AS
o for the reaction.
[
~~
Sample
Problem
18.
1]
b) Calculate
ASo
for the reaction.
[
~~
Sample
Problem
18.2]
c) Using the tabulated
AG
~
values in Appendix 2, calculate
AG

O for the reaction.
[
~~
Sample
Problem
18.4]
d) Above what temperature (in
0
c)
is the reaction spontaneous? (You will first have to
calculate
AH
o using tabulated values. Assume that AHo and
AS
o do not depend on
temperature.)
[
~~
Sample
Problem
18.
3]
,
750 CHAPTER 18 Entropy,
Free
Energy, and Equilibrium
CHAPTER SUMMARY
Section 18.1
• A spontaneous process is
one

that occurs
under
a specified
se
t
of
conditions.
• A nonspontaneous process is
one
that
does not
occur
under
a
specified set
of
conditions.
• Spontaneous
proce
sses do
not
necessarily happen quickly.
Section 18.2
• Entropy is a thermodynamic state function
(of
ten described as a measure
of
disorder) that is related to the number
of
microstates in a system.

•
Whether
or
not
a process is spontaneous depends on the
change
in
enthalpy and the
change
in entropy
of
the system.
• Tabulated standard entropy va
lue
s are absolute val
ue
s.
Section 18.3
•
According
to the second law
of
thermodynamics, the entropy change
for the universe is
po
sitive for a spontaneous process and zero for an
equilibrium process.
•
According
to the third law

of
thermodynamics, the entropy
of
a
perfectly crystalline substance
at
0 K is zero.
KEyWORDS
Entropy (S),
726
Eq
uilibrium
proce
ss, 733
Free
energy, 737
Gibbs free energy
(G),
737
KEY EQUATIONS
Microstate,
728
Nonspontaneous
proce
ss,
726
Second
law
of
thermodynamics, 733

Section 18.4
•
The
Gibbs free energy
(G)
or simply the free energy
of
a sys
tem
is the
energy available to do work.
•
The
standard free energy
of
reaction
(D.G
;x,,
) for a reaction tells us
whether the equilibrium lies to the right (negative
D.G~xn)
or
to the left
(positive
D.G;'.n).
• Standardfree energies offormation (D.Gf) can
be
used
to calculate
standard free energies

of
reaction.
Section 18.5
•
The
free-energy
change
(
D.G
) is determined using the standard free-
energy change
(
D.G
O) and the reaction quotient (Q).
•
The
sign
of
D.G
tells
us
whether
the reaction is spontaneous under the
conditions
de
scribed.
•
D.G
o is related to the equilibrium constant,
K.

A negative
D.G
o
corresponds to a large
K;
a
po
sitive
D.G
o corresponds to a small
K.
Section 18.6
• In livi
ng
sys
tem
s, thermodynamically favorable reactions provide
the free energy needed to drive necessary
but
thermodynamically
unfa
vorable reactions.
Spontaneous process,
726
Standard
entropy,
729
Standard free energy
of
formation

(D.G)?),
739
Standard
free energy
of
reaction
(D.G~xn)'
738
Third
law
of
thermodynamics, 735
1
8.
1
number
of
micr
ostates = nX
18.2 S = k
In
W
18.3
18.4
18.5
18.6
18.7
18.8
18.9
18.10

18.11
18.12
18.13
18.14
18.15
D.S
=
Sf
-
Si
D.S
= k In
Wr
- k In Wi
For
a spontaneous proces
s:
D.S
univ
=
D.S
sys
+
D.S
s
urr
> 0
For
an equilibrium process:
D.S

u
niv
=
D.S
sys
+
D.S
surr
= 0
D.S;'.n
= [
cS
O(C) + dSO
(D)]
-
[aS
O(A) + bSO(
B)]
D.S
~
xn
=
LnS
O(product
s)
-
LmS
O(reactants)
-D.Hsys
D.S

s
urr
= T
G =
H-
TS
D.G
=
D.H
-
TD.S
D.G
;'.n
=
[cD.G
r(C) +
dD.G
)?(
D)] -
[aD.G
)?(
A)
+ MG)?(B)]
D.G~xn
=
LnD.G
)?(
product
s)
- LmD.G

)?(
r
eac
tants)
D.G
=
D.G
o +
RTln
Q
D.G
o = -
RTln
K
•
•
QUESTIONS
AND
PROBLEMS
QUESTIONS AND PROBLEMS
====================
=====-~~-
Section 18.1: Spontaneous Processes
Review
Questions
18.1
Explain what is meant by a
spontaneous process. Give two
examples each
of

spontaneous and nonspontaneous processes.
18.2 Which
of
the following processes are spontaneous and which
are nonspontaneous: (a)
di
ssolving table salt (NaCI) in hot soup,
(b) climbing Mt. Everest, (c) spreading fragrance in a room by
removing the cap from a perfume bottle, (d) separating helium
and neon from a mixture
of
the gases?
18.3 Which
of
the following processes are spontaneous and which are
nonspontaneous at a given temperature?
(a)
NaN0
3
(s) H
2
0.
NaN0
3
(aq)
saturated soln
(b)
NaN0
3
(s) H

2
0.
NaN0
3
(aq) unsaturated soln
(c)
NaN0
3
(s)
H2
0
•
NaN0
3
(aq)
supersaturated soln
Section 18.2: Entropy
Review
Questions
18.4 Define entropy. What are the units
of
entropy?
.
18.5 Define microstate. What is the relationship between microstates
and entropy?
18.6
How
does the entropy
of
a system change for each

of
the
following processes?
(a) A solid melts.
(b) A liquid freezes.
(c) A liquid boils.
(d) A vapor is converted to a solid.
(e) A vapor condenses to a liquid.
(f) A solid sublimes.
(g) A solid dissolves in water.
18.7
How
does the entropy
of
a system change for each
of
the
following processes?
(a) Bromine liquid vaporizes.
(b) Water freezes to form ice.
(c) Naphthalene, the key component
of
mothballs, sublimes.
(d)
Sugar crystals form from a supersaturated solution.
(e) A block
of
lead melts.
(f) Iodine vapor condenses to form solid iodine.
(g) Carbon tetrachloride dissolves in liquid benzene.

Problems
18.8 Referring to the setup in Figure 18.1, calculate the probability
of
all the molecules ending up in the same flask
if
the number is
(a)
6,
(b) 60, (c) 600.
18.9 Referring to the setup in Figure
18
.1, calculate the probability
of
all the molecules ending up in the
same
flask
if
the number
is
(a) 4, (b) 10, (c) 200.
18.10
Predict
whether the entropy change is positive or negative
for each
of
the following reactions. Give reasons for your
prediction
s.
(a)
2KCl0

4
(s)
-_I
2KCl0
3
(s)
+ 0 2(g)
(b) H
2
0 (g) • H
2
0
(l)
(c) 2Na(s) +
2H
2
0(l)
-_I
2NaOH(aq)
+ H
2
(g)
(d) N
2
(g) •
2N(g)
18.11 State whether the sign
of
the entropy change expected for each
of

the following processes will be positive or negative, and explain
your predictions.
(a)
PCI
3
(l) + CI
2
(g) • PCls(s)
(b) 2HgO(s) • 2Hg(l) +
02(g)
(c) H
2
(g) •
2H(g)
(d) U(s) +
3F
2
(g) • UF
6
(s)
Section 18.3: The Second and Third Laws
of
Thermodynamics
Review Questions
18.12 State the second law
of
thermodynamics in words, and expr

j-
mathematically.

18.13
State the third law
of
thermodynamics in words, and expl
ai
n
i-
usefulness in calculating entropy values.
Problems
18.14 For each pair
of
substances listed here, choose the one having
th~
larger standard entropy value at
25
°
C.
The same molar amoun
is used in the comparison. Explain the basis for your choice.
(a) Li(s) or
Li
(l),
(b) C
2
H
s
OH(l)
or
CH
3

0CH
3
(l)
(Hint:
Which
molecule can hydrogen-bond?), (c)
Ar(g)
or Xe(g), (d)
CO
(g)
or
CO
2
(g),
(e)
02(g
) or
03(g),
(f)
N0
2
(g)
or N
2
0ig).
18.15 Arrange the following substances
(1
mole each) in order
of
increasing entropy at

2s
o
e:
(a) Ne(g), (b)
S02(g),
(c) Na(s),
(d) NaCl(s), (e)
H
2
(g). Give the reasons for your arrangement.
18.16
Using the data in Appendix 2, calculate the standard entropy
changes for the following reactions at
2ye:
(a) S(rhombic) + 0 2(g) •
S02(g)
(b)
MgC0
3
(s) • MgO(s) + COzeg)
(c)
2C
2
H
6
(g) +
70
2
(g)
•

4C0
2
(g)
+
6H
2
0 (I)
18.17 U sing the data in Appendix 2, calculate the standard entropy
changes for the following reactions at
2YC:
(a) H
2
(g) + CuO(s)
-_I
Cu
(s) + H
2
0(g)
(b) 2Al(s) + 3ZnO(s) • AI
2
0
3
(s) +
3Zn(s)
(c)
CH
4
(g) + 20zeg) • COzeg) +
2H
2

0(l)
18.18 According to the second law
of
thermodynamics, the entropy
of
an irreversible process
in
an isolated system must always
increase.
On the other hand, it is well known that the entropy
of
li
ving systems remains small. (For example, the synthesis
of
highly complex protein molecules from individual amino acids is
a process that leads to a decrease in entropy.) Is the second law
invalid for
li
ving systems? Explain.
Section 18.4: Gibbs Free Energy
Review Questions
18.19
18.20
Definefr
ee
energy.
What
are its units?
Why
is it more convenient to predict the direction

of
a reaction
in terms
of
I1G
sys
instead
of
I1S
uni
v
? Under what conditions c
an
I1G
sys
be used to predict the spontaneity
of
a reaction?
752
CHAPTER
18 Entropy, Free Energy, and
Equilibrium
18.21
From
the following combinations
of
!:::Ji
and
!::,.S,
predict

if
a process will
be
spontaneous at a high or low temperature:
(a) both
!::,.H
and
!::,.S
are negative, (b)
!:::Ji
is negative and
!::,.S
is
positive, (c) both
!::"H
and
!::,.S
are positive, (d)
!::,.H
is positive and
!::,.S
is negative.
Problems
18.22 Calculate!::"Go for the following reactions at 25°C:
(a) Nz(g) +
02(g)
•
2NO(g)
(b) HzO(I) • HzO(g) •
(c) 2C

2
HzCg)
+
50
z
(g) •
4CO
z
(g)
+
2H
z
O(I)
(Hint:
Look
up the standard free energies
of
formation
of
the
reactants and products in Appendix 2.)
18.23
Calculate!::"Go for the following reactions at 25°
C:
(a)
2Mg(s)
+
02(g)

2MgO(s)

(b) 2S0zCg) +
02(g)
•
2S0
3
(g)
(c)
2C
2
H
6
(g) +
70
z
(g) •
4C0
2
(g)
+ 6H
z
O(
I)
(See Appendix 2 for thelIDodynamic data.)
18.24
From
the values
of
!::"H
and
!::,.S,

predict which
of
the following
reactions would
be
spontaneous at 25°
C:
reaction
A:
!::,.H
= 10.5
kllmol,
!::,.S
=
30
11K . mol; reaction
B:
!:::Ji
=
1.8
kl
l
mol
,
!::,.S
= - 113 11K · mol.
If
either
of
the reactions is

nonspontaneous at
25°C, at
wha
t temperature might it become
spontaneous?
18.25 Find the temperatures at which reactions with the following
!::"H
and
!::"S
values would
become
spontaneou
s:
(a)
!::,.H
=
-
126
kl
l
mol
,
!::,.S
= 84 11K . mol; (b)
!::,.H
=
-11.7
kllmol,
!::,.s
=

-105
11
K·
mol.
18.26
The
molar heats
affusion
and vaporization
of
ethanol are 7.61
and
26.0
kllmol,
respectivel
y.
Calculate the molar entropy
changes for the solid-liquid and liquid-vapor transitions for
ethanol. At 1 atm pressure, ethanol melts at -
11
7.3 °C and boils
at
78.3°C.
18.27
The
molar heats
of
fusion and vaporization
of
mercury are

18.28
18.29
18.30
23.4 and 59.0
kllmol
, respectively. Calculate the molar entropy
changes for the solid-liquid and liquid-vapor transitions for
mercury. At 1 atm pressure, mercury melts at
- 38.9°C and boils
at
357°C.
Consider the formation
of
a dimeric protein:
2P
, P
2
At 25°C, we have
!::,.H
o = 17
kJ/mol
and
!::"S
o = 65
11K
. mol. Is
the dimerization favored at this temperature?
Comment
on the
effect

of
lowering the temperature.
Doe
s your result explain why
so
me
enzymes lose their activities under cold conditions?
Use
the values listed in Appendix 2 to calculate!::"Go for the
following alcohol fermentation:
.
As an approximation,
we
can assume that proteins exist
either in the native (physiologically functioning) state or the
denatured state.
The
standard molar enthalpy and entropy
of
the denaturation
of
a certain
pr
otein are 512
kllmol
and
1.60
kl
lK
. mol, respectively.

Comment
on the signs and
magnitudes
of
these quantities, and calculate the temperature
at which the denaturation becomes spontaneous.
18.31
Certain bacteria in the soil obtain the necessa
ry
energy for growth
by oxidizing nitrites to nitrates:
2N0
2
+ O
2

2NO
)
Given that the standard Gibbs free energies
of
formation
of
N0
2
and
NO
) are
-34.6
and - 110.5
kllmol,

respectively, calculate
the a
mount
of
Gibbs free energy released when 1
mole
of
NO
z
is
oxidized to 1 mole
of
NO
) .
Section 18.5: Free Energy and Chemical Equilibrium
Review
Questions
18.32 Explain the difference between
!::"G
and
!::"G
o.
18.33 Explain why Equation 18.15 is
of
great importance in chemistry.
18.34 Fill in the missing entries in the following table:
K
InK
!::"G
o

Result
at
equilibriu
ill
< 1
0
Product s are fa
are
v d
Problems
18.35 Calculate
Kp
for the following reaction at 25°
C:
!::"G
o = 2.60 kJ/
mol
18.36 For the autoionization
of
water at 25°C,
18.37
18.38
Kw is 1.0 X 10-
14
What
is
!::"G
o for the process?
Consider the following reaction at
25°C:

Fe(OH)z(s),
•
Fe
2+
(aq) +
20H
- (aq)
Calculate!::"Go for the reaction. Ksp for
Fe(OHh
is 1.6 X
10
-
14
Calculate!::"Go and Kp for the following equilibrium reaction at
25°
C:
18.39 (a) Calculate!::"Go and Kp for the following equilibrium reaction
at
25°C.
The
!::"G
f values are 0 for Clz(g),
-286
kl
l
mol
for
PCI
3
(g), and

-325
kllmol
for PCls(g).
18.40
18.41
18.42
PCls(g) ' • PCI
3
(g) +
CI
2
(g)
(b) Calculate!::"G for the reaction
if
the partial pressures
of
the
initial mixture are
PPC1
5
= 0.0029 atm,
PP
C1
, = 0.27 atm, and
P
C!
, = 0.40 atm.
The
equilibrium constant (Kp) for the reaction
H

2
(g) +
COzCg)
:;:.=='
HzO(g) +
CO
(g)
is 4.40 at 2000
K.
(a) Calculate!::"GO for the reaction. (b)
Calculate!::"G for the reaction when the partial pressures are
PH
, = 0.25 atm, P
eo
= 0.78 atm,
PH
0 = 0.66 atm, and
_ 2 2
P
eo
= 1.20 atm.
Consider the
de
com
po
sition
of
calcium carbonate:
CaC0
3

(s)
:;:.
=
='
CaO(s) +
CO
2
(g)
Calculate the pressure in atm
of
COz in an equilibrium process
(a) at
25°C and (b) at 800°C. Assume that
!::,.W
= 177.8
kllmol
and
!::"S
o = 160.5 11
K·
mol for the temperature range.
The
equilibrium constant Kp for the reaction
CO(g) +
Cl
z(g)
:;:,
=
='
COCI

2
(g)
is 5.62 X 10
3s
at 25°C. Calculate!::,.Gf for COCl
2
at
25°C
.
18.43
18.44
At
25°C,
!J.G
o
for the process
H
2
0(I)
+.
=::t:, H
2
0(g)
is 8.6 kJ/mol. Calculate the vapor pressure
of
water at this
temperature.
Calculate
!J.G
o

for the process
C(diamond)
+.
=::t:, C(graphite)
Is the formation
of
graphite from diamond favored at 25°C?
If
so,
why is it that diamonds do not
become
graphite on standing?
Section 18.6: Thermodynamics
in
Living Systems
Review
Questions
18.45
What
is a coupled reaction?
What
is its importance in biological
reactions?
18.46
What
is
the role
of
ATP in biological reactions?
Problems

18.47 Referring to the metabolic process involving glucose on
page
747, calculate the
maximum
number
of
moles
of
ATP that can
be
synthesized from
ADP
from the breakdown
of
1
mole
of
glucose.
18.48
In
the metabolism
of
glucose, the first step is the
con
ve
rsion
of
glucose to glucose 6-phosphate:
glucose
+ H

3
P0
4
+,
glucose 6-phosphate + H
2
0
!J.G
o
= 13.4 kJ/mol
Because
!J.G
o
is positive, this reaction does
not
favor the
formation
of
products. Show how this reaction can be
made
to proceed
by
coupling it with the hydrolysis
of
ATP.
Write an
equation for the coupled reaction, and estimate the equilibrium
constant for the coupled process.
Additional Problems
18.49 Explain the following nursery

rhyme
in
terms
of
the second law
of
thermodynamics.
18.50
Humpty
Dumpty sat on a wall;
Humpty
Dumpty
had
a great fall.
All the
King's
horses and all the
King's
men
Couldn't
put
Humpty
together again.
Calculate
!J.G
for the reaction .
at
25°C for the following conditions:
(a) [H+]
= 1.0 X 10-

7
M,
[OH-]
= 1.0 X
10-
7
M
(b) [H+] = 1.0 X 10-
3
M,
[OW]
= 1.0 X 10-
4
M
(c) [H +] = 1.0 X
10-
12
M,
[OH-]
= 2.0 X 10-
8
M
(d) [H+] = 3.5 M,
[OW]
= 4.8 X
10-
4
M
18.51
Which

of
the following thermodynamic functions are associated
only with the first law
of
thermodynamics:
S,
E,
G,
and H?
18.52 A student placed 1 g
of
each
of
three compounds A, B, and C
in
a container and found that after 1
week
no change
had
occun"ed.
Offer some possible explanations for the fact that no reactions
took place. Assume that
A,
B, and C are totally miscible liquids.
18.53
The
enthalpy change
in
the denaturation
of

a certain protein is
125
kJ/mo!.
If
the entropy change is 397 J/K " mol, calculate
the
minimum
temperature at which the protein would denature
spontaneously.
18.54
18.55
QUESTIONS
AND
PROBLEMS 753
Give a detailed example
of
each
of
the following, with an
explanation: (a) a thermodynamically spontaneous process, (b) a
process that would violate the first law
of
thermodynamics, (c) a
process that would violate the second law
of
thermodynamics,
(d) an irreversible process, (e) an equilibrium process.
Predict the signs
of
!J.H,

!J.S,
and
!J.G
of
the system for the
following processes at 1 atm: (a)
ammonia
melts at
-60
°C,
(b) ammonia melts at
-77.7
°C, (c)
ammonia
melts at -
100°e.
(T
he normal melting point
of
ammonia is
-77.7°e.)
18.56 Consider the following facts: Water freezes spontaneously at
-5
°C and 1 atm, and ice has a more ordered structure than liquid
water. Explain how a spontaneous process can lead to a decrease
in
entropy.
18.57
Ammonium
nitrate

(NH
4
N0
3
)
dissolves spontaneously and
endotherrnically in water.
What
can
you
deduce about the sign
of
!J.S
for the solution process?
18.58 Calculate the equilibrium pressure
of
CO
2
due to the
decomposition
of
barium carbonate
(BaC0
3
)
at 25°
e.
18.59 (a) Trouton's rule states that the ratio
of
the

molar
heat
of
vaporization
of
a liquid
(!J.H
vap)
to
it
s boiling point
in
kelvins is
approximately
90
J/K . mol. Use the following data to show that
this is the case and explain why Trouton's rule holds true:
T
bp
COc)
!J.H
vap(kJ/mol)
Benzene
80.1 31.0
Hexane 68.7
30.8
Mercury
357
59.0
Toluene

110.6 35.2
(b)
Use the values in Table 12.6 to calculate the
same
ratio for
ethanol and water. Explain why Trouton's rule does not apply to
these two substances as well as it does to other liquids.
18.60 Referring to Problem 18.59, explain why the ratio is considerably
smaller than
90 J/K .
mol
for liquid HF.
18.61 Carbon
mono
x
ide
(
CO
) and nitric oxide (NO) are polluting gases
contained
in
automobile exhaust.
Under
suitable conditions,
these gases can
be
made
to react to form nitrogen (N
2
)

and the
less harmful carbon dioxide
(C0
2
) .
(a) Write an equation for
this reaction. (b) Identify the oxidizing and reducing agents.
(c) Calculate the
Kp
for the reaction at
25°e.
(d)
Under
normal
atmospheric conditions, the partial pressures are
P
N2
= 0.80 atm,
P
eo
,
= 3.0 X
10-
4
atm
, P
eo
= 5.0 X
10-
5

atm, and P
NO
=
5.0 X
10-
7
atm. Calculate Q
p,
and predict the direction toward
which the reaction will proceed. (e) Will raising the temperature
favor the formation
of
N2 and CO
2
?
18.62
For
reactions can"ied out under standard-state conditions,
Equation 18.11 takes the form
!J.G
o
=
!J.H
o
-
T!J.S
o
.
(a) Assuming
!J.H

o
and
!J.S
o
are independent
of
temperature,
deri
ve
the equation
where
K I
and
K2 are the equilibrium constants at TI and T
z
,
respectivel
y.
(b) Giv
en
that at 25°C Kc is 4.63 X
10-
3
for the
reaction
!J.W = 58.0 kJ/mol
calculate the equilibrium constant at 65°C.
754
CHAPTER
18 Entropy, Free Energy, and

Equilibrium
18.63
The
K
,p
of
AgCl
is given in Table 17.4.
What
is
it
s value at 6Q°C?
[Hint:
You need the result
of
Problem
18.62(a) and the data in
Appendix 2 to calculate
LVi
o.J
18.64
Under
what
conditions does a s
ub
stance have a standard entropy
of
ze
ro?
Can

an
element
or
a
compound
ever have a negati
ve
standard entropy?
18.65
Water gas, a mixture
of
H2 and
CO
, is a fuel made by reacting
steam with red-hot coke (a by-product
of
coal distillation):
18.66
H
2
0 (g) +,C(s)
••
=='
CO(g) + H
lg)
Prom
the
data
in Appendix 2, estimate the temperature at which
the reaction begins to favor the formation

of
products.
Consider
the following
Br
iZl
nsted acid-base reaction
at
2So
C:
HF
(aq) +
Cnaq)
••
=='
HCI(aq) +
P-(aq)
(a) Predict
whether
K will
be
greater
or
smaller than
1.
(b)
Doe
s
!:1S
o or

!:1H
o
make
a greater contribution to
!:1GO?
(c) Is
!:1H
o likely
to
be
positive
or
negative?
18.67
The
pH
of
gastric
juice
is
about
1.00 and that
of
blood
pla
sma
is
7.40. Calculate the Gibbs free energy required to secrete a
mole
of

H+ ions
from
blood
pla
s
ma
to the sto
mach
at
37°
e.
18.68
Cry
stallization
of
sodium acetate from a su
per
saturated solution
occurs spontaneously (see
page
S09).
Ba
sed on this,
what
can
you deduce
about
the si
gn
s

of
!:1S
and
LVi
?
18.69
Consider
the thermal
decompo
sition
of
CaC0
3
:
The
equilibrium vapor pressures
of
CO
2
are
22
.6 rnmHg at
70
0°C
and 1829
mmHg
at
9S0°C. Calculate
the
standard enthalpy

of
the
reaction.
[Hint:
See
Problem
18.62(a).J
18.70 A certain reaction is spontaneous
at
72
°
e.
If
the enthalpy change
for the reaction is 19
kJ/mol,
what
is the
minimum
value
of
!:1S
(in
J/K . mol) for the reaction?
18.71
Predict
whether
the entropy
change
is positive

or
negative for
each
of
these reaction
s:
18.72
18.73
18.74
18.75
(a)
Zn(s) +
2HC
I(aq) • ' ZnCI
2
(aq) + H
2
(g)
(b) O(
g)
+ O(g) . ' 02(g)
(c)
NH
4
N0
3
(s) • ' N
2
0(g)
+

2H
2
0(g)
(d)
2H
2
0
2
(l).
' 2HzO(l) + 0
2(g)
The
reaction
NH
3
(g) + HCI(g) ,
NH
4
Cl(s) proceeds
spontaneously
at
2S
oC even though there is a decrease in disor
der
in
the
system (gases are converted to a solid). Explain.
Use
the
following data to determine the normal boiling

point
, in
kelvins,
of
mercury.
What
assumptions
mu
st you
make
in
order
to
do the calculation?
Hg(l):
!:1H
'f
= 0 (
by
definition)
So =
77.4
JIK
.
mol
Hg(
g):
!:1H
'f
= 60.78

kJ
/
mol
So
= 174.7 J/K . mol
The
molar
heat
of
vaporization
of
ethanol is 39.3 kJ/mol, and
the
boiling
point
of
ethanol is 78.3°
e.
Calculate!:1S for the
vaporization
of
O.SO
mole
of
ethanol.
A certain reaction is known to have a
!:1G
o value
of
-122

kJ/moi.
Will
the
reaction necessarily
occur
if
the reactants are mixed
together?
18.76 In the
Mond
process for the purification
of
nickel, carbon
monoxide
is reacted with heated nickel to
produce
Ni(CO)4,
which is a gas and
can
therefore
be
separated
from
solid
impurities:
18.77
18.78
Ni(s)
+ 4CO(g) +=. :=::::!'
Ni(COMg)

Given that the standard free energies
of
formation
of
CO(g)
a
nd
Ni(CO)ig)
are
-137
.3 and
-S87.4
kJ/
mol,
respectively,
calculate the equilibrium constant
of
the reaction
at
80°e.
Ass
ume
that
!:1G
'f
is temperature independent.
Calculate!:1Go and
Kp
for
the following processes at

2S
o
C:
(a)
Hig)
+
Br
i l) ••
=='
2HBr(g)
(b)
!H2(g)
+
~Br2(l).
'
HBr(g
)
Account
for
the differences in
!:1G
o and Kp obtained for parts (a)
and (b).
Calculate the pressure
of
O
2
(in atm) over a
sample
of

NiO
at
2S
oC
if
!:1G
o =
212
kJ/
mol
for the reaction
NiO(s) ••
=='
Ni(s) +
~02(g)
18.79
Comment
on the statement:
"Just
talking about entropy increases
its val
ue
in the universe."
18.80
Por
a reaction with a negative
!:1G
o value,
which
of

the following
statements is fals
e?
(a)
The
equilibrium constant K is greater than
one
. (b)
The
reaction is spontaneous when all
the
reactants and
products are in their standard states. (c)
The
reaction is always
exothermic.
18.81
Consider
the reaction
18.82
18.83
18.84
N
2
(g) + 0
2(g)
:;::.
:=::::!' 2NO(g)
Given
that

!:1G
o for the reaction at
2S
oC
is 173.4 kJ/mol,
(a) calculate the standard free energy
of
formation
of
NO
and
(b) calculate
Kp
of
the reaction. (c)
One
of
the starting substances
in s
mog
formation is NO.
Assuming
that the temperature in a
running automobile
engine
is 1
100
°C, estimate
Kp
for

the given
reaction. (d) As farmers know, lightning helps to
produce
a better
crop.
Why
?
Heating copper(II)
oxide
at
400
°C
doe
s
not
produce
any
appreciable amount
of
Cu:
CuO(s)
••
=='
Cu(s) + W
2
(g)
!:1G
o = 127.2 kJ/mol
However,
if

this reaction is coupled to the conversion
of
graphite
to carbon monoxide,
it
becomes spontaneous.
Write
an equation
for the coupled process, and
ca
lculate the equilibrium constant
for the coupled reaction.
Consider
the
decomposition
of
magnesium carbonate:
MgC0
3
(s)
••
=='
MgO(s)
+
COig)
Calculate the temperature
at
which the decomposition begins to
favor products.
Assume

that both
!:1H
o and
!:1S
o are
independent
of
temperature.
(a)
Over
the years, there have been
numerou
s claims
about
"perpetual motion machines,"
machine
s
that
will
produce
useful
work
with no input
of
energy. Explain
why
the first law
of
thermodynamics prohibits the possibility
of

such a machine
existing. (b) Another kind
of
machine, sometimes called a
"perpetual motion
of
the second kind," operates as follows.
Suppose an ocean liner sails by scooping up water
from
the
ocean and then extracting
heat
from
the water, converting the
heat to electric power to run the ship, and
dumping
the water
back
into the ocean. This process does not violate the first law
of
thermodynamics, for no energy is
created-energy
from
the ocean is
just
converted to electric energy. Show that the
second law
of
thermodynamics prohibits the existence
of

such a
machine.
18.85
The
activity series in Table
4.6
shows that reaction (a) is
spontaneous whereas reaction (b) is nonspontaneous at
25°C:
(a) Fe(s) +
2H
+ •
Fe
2+
(aq) +
Hig)
(b) Cu(s) +
2H
+ • Cu
2
+(aq) +
Hig)
Use
the data in Appendix 2 to calculate the equilibrium constant
for these reactions and hence confirm that the activity series is
correct.
18.86
The
rate constant for the elementary reaction
18.87

18.88
18.89
18.90
18.91
is '(.1 X
10
9
1M
2
•
s at 25°C.
What
is the rate constant for the
reverse reaction at the same temperature?
The
following reaction was described as the cause
of
sulfur
deposits formed at volcanic sites.
It
may also be
used
to remove
50
2
from powerplant stack gases.
(a) Identify the type
of
redox reaction it is. (b) Calculate the
equilibrium constant

(Kp) at 25°C, and
comment
on
whether this
method
is
feasible for removing
50
2
,
(c) Would this procedure
become
more effective
or
less effective at a higher t
empe
rature?
Describe two ways that you could determine
I1G
o
of
a reaction.
The
following reaction represents the removal
of
ozone in the
stratosphere:
Calculate the equilibrium constant
(Kp) for this reaction. In view
of

the magnitude
of
the equilibrium constant, explain why this
reaction is
not
considered a major cause
of
ozone
depletion in the
absence
of
human-made pollutants such as the nitrogen oxides
and
CFCs.
Assume
the temperature
of
the
st
ratosphere is -
30
°C
and
I1G
? is temperature independent.
When
a native protein in solution is heated to a high enough
temperature, its polypeptide chain will unfold to become the
denatured protein.
The

temperature at which a large portion
of
the protein unfolds is called the melting temperature.
The
melting
temperature
of
a certain protein is found to
be
46
°C, and the
enthalpy
of
denaturation is
382
kl/mo!. Estimate the entropy
of
denaturation, assuming that the denaturation is a two-state
proce
ss; tnat is, native protein • denatured protein.
The
single polypeptide protein chain has 122 amino acids. Calculate
the entropy
of
denaturation
per
amino acid.
Comment
on your
result.

A 74.6-g ice cube floats in the Arctic Sea.
The
pressure and
temperature
of
the system and surroundings are at 1 atrn and O°C,
respectively. Calculate
I1S
sys
,
I1S
s
urn
and
I1S
univ
for the melting
of
the ice cube.
What
can you conclude about the nature
of
the
proce
ss from the value
of
I1S
univ
? (The
molar

heat
of
fusion
of
water is 6.01 kJ/mo!.)
18.92
Comment
on the feasibility
of
extracting copper from its ore
chalcocite
(CU2S)
by
heating:
CU
2S(S)
+. 2Cu(s) + S(s)
QUESTIONS
AND
PROBLEMS
Calculate the
I1G
o for the overall reaction
if
this process is
coupled to the conversion
of
sulfur to sulfur dioxide and
I1Gr(CU2S) =
-86.1

kl
/mo!.
18.93
Active transport is the
proce
ss
in
which a substance is
transferred from a region
of
lower concentration to one
of
high
er
concentration. This is a nonspontaneous process
and
must
be
coupled to a spontaneous process, such as the hydrolysis
of
ATP.
The
concentrations
of
K+ ions in the
blood
plasma and in ner 'e
ce
lls are 15
mM

and
400
mM
, respectively
(1
mM
= 1 X
10
-
3
M).
Use
Equation 18.14 to calculate I1G for the process at the
physiological temperature
of
37°C:
18.94
18.95
18.96
18.97
K+(15 mM) • K
+(
400
mM)
In this calculation, the
I1G
o term can
be
set to zero.
What

is the
ju
st
ification for this step?
Large quantities
of
hydrogen are needed for the synthesis
of
ammonia.
One
preparation
of
hydrogen involves the reaction
between carbon monoxide and
steam
at
300
°C in the presence
of
a copper-zinc catalyst:
Calculate the equilibrium constant
(Kp) for the reaction and the
temperature at which the reaction favors the formation
of
CO
and
H
2
0.
Will a larger Kp

be
attained at the
same
temperature
if
a
more
efficient catalyst is used?
Consider two carboxylic acids (ac
id
s that contain the
-COOH
group):
CH
3
COOH
(acetic acid,
Ka
= 1.8 X
10
-
5
)
and
CH
2
CICOOH
(chloroacetic acid,
Ka
= 1.4 X 10-

3
).
(a) Calculate
D.G
o for the ionization
of
these acids at 25°C. (b)
From
the
equation
I1Go =
I1H
o
-TI1S
o,
we
see that the contributions to the
D.G
o term are an enthalpy term
(D.
HO) and a temperature times
entropy term
(TD.S
O).
These
contributions are listed here for the
two acids:
CH
3
COOH

CH
2
CICOOH
l1Ho
(kJ/mol)
-0.57
-4.7
ns
o
(kJ/mol)
- 27.6
-
21.l
Which is the dominant term in determining the value
of
D.G
o (and
hence
Ka
of
the acid)? (c)
What
processes contribute to
I1H
O?
(Consider the ioni
za
tion
of
the acids as a Br0nsted acid-base

reaction.) (d) Explain why the
TI1S
o term is more negative for
CH
3
COOH.
Many
hydrocarbons exist as structural iso
mer
s,
which
are
compounds that have the same molecular formula
but
different
structures.
For
example,
both
butane and isobutane have the
sa
me
molecular formula
of
C
4
H
IO
(see Problem 12.20 on page
497). Calculate the

mole
percent
of
these molecules in an
equilibrium mixture at
25°C, given that the standard free energy
of
formation
of
butane is - 15.9 kJ/mol and that
of
isobutane is
- 18.0
kl
/mo!. Does
your
result s
upport
the notion that straight-
chain hydrocarbons (that is, hydrocarbons in which the C
atoms are joined along a line) are less stable than branch-chain
hydrocarbons?
•
One
of
the steps
in
the extraction
of
iron from its ore (FeO) is the

reduction
of
iron (II) oxide
by
carbon monoxide at 900°C:
FeO(s)
+ CO(g)
+.
=~.
Fe(s) +
CO
2
(g)
If
CO
is allowed to react with an excess
of
FeO, calculate
the mole fractions
of
CO
and
CO
2
at equilibrium. State any
assumptions.