An Introduction to Thermodynamics phần 8 pdf

(
)
(
)
( )
torr
0.119 atm 10 x 56.1
0125.0
ln840,10
4-
2/1
2/12
2
2
2
2
==
==
−==∆
=
O
O
o
OAg
OAg
P
PK
KRTG
a
aa
K

(
)
(
)
10-
2
10
x
78
.
1
ln55900
2
=
−==∆
=
−++
K
KRTG
a
aa
K
o
CaF
FCa
7/10/07 4- 92
This is a quadratic equation, with the solution x = 0.0428 atm. The equilibrium condition may
now be more explicitly interpreted for this example. If 2 atm of ammonia comes to equilibrium
with its decomposition products, the equilibrium state will be 1.9572 atm of NH , together with
3

0.0214 atm of N and 0.0642 atm of H . Slight further decomposition, at these concentrations,
2 2
would cause no change in free energy; there is no tendency for further decomposition because the
system is now at equilibrium. An equilibrium constant for a reaction involving gases, in which
pressures appear as an approximation for activities, is often written K (and, because pressures
P
are substituted for dimensionless activities, K may have apparent dimensions of powers of
P
pressure).
Consider now the problem of finding the equilibrium pressure of oxygen above Ag O at
2
25 C.
o
Ag O 6 2 Ag + ½ O
2 2
Thus silver oxide, placed in an evacuated system, should decompose until the oxygen pressure
reaches 0.119 torr. In an atmosphere containing 20% oxygen the reaction would be driven in the
other direction, so that silver should react with oxygen to form silver oxide. The slow rate of this
reaction (in either direction) is attested by the durability of silver goods. An equilibrium constant
for dissociation of a compound is often written K or, when the dissociation is into ions, K .
diss ionization
For the solubility of CaF in water at 25 C,
2
o
CaF 6 Ca + 2 F
2
++ -
The equilibrium constant for dissolution of a slightly soluble salt in water is called the solubility
product constant, or K .
sp

The activity of solid CaF is 1. Assume that activities of the ions are numerically equal to
2
concentrations. From the dissociation equation we find that if the concentration of Ca is s, the
++
concentration of F will be 2s, so long as there is no other source of these ions than CaF .
-
2
Inserting these symbols into the equation above gives
Ms
s
10 x 6.7
1.0
10 x 78.1
4
4-
-10
3
=
=
M
s
10 x 5.3
0.5 x 1.0
10 x 78.1
9-
-10
=
=
7/10/07 4- 93
1.78 x 10 = s(2s)

-10 2
s = 3.5 x 10 M
-4
This is the solubility of CaF , because each mole of CaF going into solution gives one mole of
2 2
Ca ions. Introduction of activity coefficients would probably change the calculated value by
++
something on the order of 20%, which is often negligible in determining solubilities. (Reported
experimental values differ by much more than this for some compounds, but such accuracies are
entirely adequate for some applications.)
To find the solubility of CaF in 0.5 M NaCl the deviation of activity coefficients from 1
2
should not be neglected. The equations may be written
K = 1.78 x 10 = (a ++)(a -) = (γ ++)(c ++)(γ -) (c -)
-10 2 2 2
Ca F Ca Ca F F
=
Q
(γ)
Q
(c)
Experimental measurements indicate that
Q
(γ), the correction factor, is on the order of 0.1 for
such a solution. Solving for the concentrations as before gives
The increase in solubility when the other ions (Na and Cl ) are added is called the salt effect. The
+ -
added salt lowers the activity coefficients of all ions present.
The solubility of CaF in 0.5 M NaF will be smaller because of the common-ion effect. This
2

is apparent when we substitute numbers into the equations obtained above.
K = 1.78 x 10 =
Q
(γ)(c ++)(c -)
-10 2
Ca F
Now the concentration of fluoride is no longer equal to twice the concentration of calcium ion,
but is 0.5 + 2s when the Ca has the concentration s. The correction term may be assumed the
++
same as for NaCl. Then
1.78 x 10 = 0.1s(0.5 + 2s)
-10 2
Assume first that 2s is negligible with respect to 0.5. Then
This answer confirms the assumption that taking s negligible with respect to 0.5 gives a self-
consistent, and therefore mathematically valid, solution to the problem.
An orderly procedure in working problems of chemical equilibrium will greatly decrease the
2
PH
32
3
2
a
aa
K
HP
eq
=
2
PH
3

H
3
2
a
a
7/10/07 4- 94
opportunity for mistakes. The recommended steps are as follows:
1. Write the chemical reaction, even though it is a familiar one.
2. Write out the equilibrium constant expression,
Q
(a), in terms of activities and set this
equal to K .
eq
3. Determine the equilibrium constant for the reaction as written, either from the constant
given or from standard-state free energies. Note that reversing a reaction inverts the equilibrium
constant, and that multiplication of the equation by an integer raises the equilibrium constant to
that power.
4. Make the appropriate substitutions of symbols (that is: concentrations, pressures, mole
fractions, or unity) for each of the activities. Include activity coefficient if necessary (either
numerically or symbolically).
5. Determine, from information given and the stoichiometry of the chemical reaction, what is
known about concentrations or pressures.
6. Solve the algebraic equation.
7. Examine the solution to be sure that the answer found is the answer to the question asked
and is a reasonable value in terms of the physical problem as stated and the approximations made
in the solution of the problem.
Consider, for example, the decomposition of the gas phosphine, PH (∆G = 18.24 kJ/mol at
3
o
25 C), to give white phosphorus (solid) and hydrogen gas. If 5 atm of PH in a closed vessel

o
3
comes to equilibrium with its decomposition products, what will be the final pressure of PH ?
3
The specific steps of the solution, as described above, are:
1. 2 PH W 2 P + 3 H
3 2
2.
3. ∆G = - 36.5 kJ = - 8.314 x 298 ln K
o
K = 2.51 x 10
6
4. a = 1; 2.51 x 10 =
P
6
5. 2 PH 6 2 P + 3 H
3 2
5 - 2 x 3 x x = pressure, atm
2.51 x 10 = (3x) /(5 - 2x)
6 3 2
6. Find the solution to the equation 2.51 x 10 (5 - 2x) = 27x .
6 2 3
Because K is large, the reaction must be shifted well to the right, so x will be nearly equal to 5/2,
eq
and 5 - 2x will be very sensitive to x. Therefore, as a first approximation, let x = 5/2 on the right-
hand side and find (5 - 2x):
(
)
(
)

22
1

1

ln
RT
GST
RT
G
RT
S
dT
Td
R
G
dT
Gd
RTRT
G
dT
d
dT
Kd
oooo
ooo
∆+∆
=
∆
+

∆
=
∆
−
∆−
=








∆
−=
−
(18)
ln
2
RT
H
dT
Kd
o
∆
=
(19) ln
211
2

TRT
TH
K
K
o
∆∆
=
7/10/07 4- 95
(5 - 2x) = 27(5/2) /2.51 x 10 = 1.68 x 10
2 3 6 -4
x = 2.49
This calculated value is in excellent agreement with the value assumed, so the calculation is self-
consistent.
7. The final pressure of PH is 5 - 2x = 0.013 atm and of H , 3x = 7.48 atm. This is a
3 2
reasonable answer for the problem given, because ∆G (PH ) = 18.24 J/mol indicates that PH is
o
3 3
quite unstable.
Temperature Dependence of Equilibrium Constants
The equilibrium constant does vary with temperature. From equation 17, ln K = - ∆G /RT,
o
so we can find the temperature dependence. Take the derivative with respect to temperature
(holding the pressure constant at its standard value), employing equation 30 of Chapter 2.
and therefore
The last equation can be integrated, assuming ∆H is constant with temperature, to give
o
In these equations K is the equilibrium constant at the temperature T , K the equilibrium
2 2 1
constant at T , ∆H is the enthalpy change for the reaction with all reactants and products in their

1
o
standard states, and ∆T = ∆T -∆T .
2 1
The similarity of equation 19 to the Clausius-Clapeyron equation (equation 6, Chapter 3) is
not accidental. Equations for solubility, for reaction rate constants, and for other quantities, have
equivalent forms. This form is to be expected because all of these physical properties depend on
the system surmounting an energy barrier (corresponding to the heat of reaction, the heat of
vaporization, the heat of solution, or some other such energy, or enthalpy, term) and therefore
they depend, in a fundamental way, on the Boltzmann distribution law, which is an exponential
expression relating numbers of molecules, energy states, and temperature, and which therefore
gives rise to a logarithmic temperature dependence.
( )
(22) ln a
n
RT
o
Q
F
−= EE
(23) ln K
n
RT
o
F
E =
7/10/07 4- 96
Electrochemistry
The best method of directly measuring the free-energy change of a chemical reaction is often
to carry out the reaction in an electrochemical cell. The free-energy change, at constant

temperature and pressure, is (eqn. 26, Chapter 2)
(T,P) dG = w + PdV = w ’
rev rev
or
∆G = W ’ (20)
rev
where W ’ is the electrical work. Electrical power, P, is I R = EI and electrical work is power
rev
2
multiplied by time, or potential multiplied by charge:
W ’ = Pt = EI t = Eq
rev
The potential, E, is in volts, the current I in amperes, the time t in seconds, the resistance, R, in
ohms, the charge q, in coulombs, and power, P, in watts, and the work w’ in joules. A convenient
unit of charge is a mole of electrons (6 x 10 electrons), called a faraday. One faraday, F, is
23
96,487 coulomb. If n mol of electrons (or n “equivalents”) are transferred at a potential
difference E the work done is W ’ = Eq = E(nF) and the free-energy change is
rev
∆G = - nFE (21)
Thus ∆G is obtained directly from E.
For example, we will find that the standard cell potential (all reactants and products present
at unit activity) for the reaction that transfers 2 electrons from copper to ferric iron (Fe ) is
+++
1.100 V. The standard free energy (per mol Cu 6 Cu ) is therefore
++
∆G = - nFE = - 2 x 96,487 x 1.100 = - 21,230 J/mol
o
Substitution of equation 21 into equation 7 gives
-nFE = - nFE + RT ln

Q
(a)
o
and therefore
This is now commonly called the Nernst equation. Similarly, by combining equation 21 with
equation 17 we obtain
( ) ( )
cQ
n
RT
Q
n
RT
o
lnln
FF
EE −−=
γ
( ) ( )
(24) ln 1 c
n
RT
o
i
Q
F
EE −==
γ
(25)
PP

T
n
T
G
S






∂
∂
=






∂
∆∂
−=∆
E
F
(26)
P
T
TnnH







∂
∂
+−=∆
E
FFE
7/10/07 4- 97
Writing
a = γ c
i i i
for each of the reactants and products, the function
Q
factors to give
Q
(a) =
Q
(γ)
Q
(c)
and thus
If each of the activity coefficients, γ , is unity, then
Q
(γ) = 1, ln
Q
(γ) = 0, and
i

The potential depends on the actual activities, or concentrations, of the reactants and products.
Not only ∆G and equilibrium constants can be determined electrochemically, but also ∆H and
∆S. From equation 21 and equation 30 of Chapter 2,
and therefore, from ∆H = ∆G + T∆S, we obtain
HALF-REACTION POTENTIALS. An electrical potential, or electrochemical potential, exists for any
chemical reaction in which electrons are transferred from one substance to another. Such
reactions are called oxidation-reduction reactions, or simply “redox” reactions. Removal of
electrons, as in Fe 6 Fe + 3e, or 2 I 6 I + 2e, is called oxidation; addition of electrons is
+3 -
2
called reduction.
Electrons enter or leave a device through a conductor called an electrode. The cathode
(from the Greek “down way”) is the electrode at which electrons enter any device; the anode
(from the Greek “up way”) is the electrode at which electrons leave any device. To push
electrons through a vacuum tube or electroplating cell, the cathode must be made negative. On
the other hand, the electrode at which an electrochemical cell releases electrons is called negative
(Figure 1) and this is, by definition, the anode.
It is not possible to measure half a reaction. Only complete reactions (cathode reaction +
anode reaction) can be measured. Nevertheless, it is convenient to consider the electrochemical
7/10/07 4- 98
potential of a reaction as a sum of the potentials of
two half reactions, each defined from equation 21
in terms of the free-electron changes. Let ∆G be
A
the free-energy change for the anode reaction
(electrons leaving, hence oxidation), ∆G be the
C
free-energy change for the cathode reaction
(electrons entering, hence reduction), and ∆G be
the free-energy change for the total reaction.

Free-energy changes are additive, so
∆G = - nFE = ∆G + ∆G = - n FE - n FE
A C A A C C

and because n = n = n for any balanced reaction,
A C
E = E + E (27)
A C

Note that the potentials are independent of n and
thus independent of the amount of reaction! That
is, the potential of a half-reaction such as
½ M + e
6
½ M
++
is identically the same as the potential of the half-
reaction
M + 2 e 6 M
++
In nearly all electrochemical cells there is a
small additional term that should be included on
the right-hand side of equation 27, arising from the
contact between dissimilar solutions. Like E and
A
E , these “junction potentials” cannot be
C
measured, nor can they be calculated by
thermodynamics. They can, however, be
Because of the inherent asymmetry of an

electrochemical cell, oxidation, occurring at the
anode, with release of electrons, and reduction,
occurring at the cathode, with removal of electrons
from the circuit, produces a potential difference
between anode and cathode that drives charges
around the circuit. Charge is conserved. There is
no “source” of charge, or “source” of current.
estimated, based on theories involving diffusion rates, and are usually sufficiently small in practice
7/10/07 4- 99
that they can be ignored for present purposes.
Although it is not possible to measure E or E separately, it is possible to measure total cell
A C
potentials. Then, by choosing an arbitrary value for some half-cell potential as a reference level,
other half-reaction potentials can be determined relative to this arbitrary reference. The half-
reaction chosen as the reference for this purpose is
2 e + 2 H 6 H
+
2
for which E (the potential with all reactants and products at unit activity) is arbitrarily assigned
o
the value zero. Then, for example, because the standard potential of the cell reaction
Zn + 2 H 6 H + Zn
+ ++
2
Table 1 STANDARD ELECTRODE POTENTIALS, 25 C (STANDARD REDUCTION POTENTIALS)
O
Reaction E (volt) Reaction E (volt)
o o
Li + e 6 Li -3.0401 Cu + e 6 Cu 0.153
+ ++ +

K + e 6 K -2.931 SO + 4 H + 2e 6
+ = +
4
Na + e 6 Na -2.71 H SO + H O 0.172
+
2 3 2
Mg + 2e 6 Mg -2.372 AgCl + e 6 Ag + Cl 0.22233
++ -
Al + 3e 6 Al -1.662 Cu + 2e 6 Cu 0.3419
+++ ++
Mn(OH) + 2e 6 Cu + e 6 Cu 0.521
2
+
Mn + 2 OH -1.5 I + 2e 6 2 I 0.5355
- -
2
Cr + 2e 6 Cr -0.913 MnO + e 6 MnO4 0.558
++ - =
4
Zn + 2e 6 Zn -0.7628 MnO + 2 H O + 3e
++ -
4 2
2 CO (g) + 2 H + 2 e 6 MnO + 4 OH 0.595
2 2
+ -
6 H C O (aq) -0.49 O + 2 H + 2e 6 H O 0.695
2 2 4 2 2 2
+
Fe + 2e 6 Fe -0.447 Fe + e 6 Fe 0.771
++ +++ ++

Cr + 2e 6 Cr -0.407 Hg + 2e 6 2 Hg 0.7973
+++ ++ ++
2
Cd + 2e 6 Cd -0.4030 Ag + e 6 Ag 0.7996
++ +
AgI + e 6 Ag + I -0.15224 Hg + 2e 6 Hg 0.851
- ++ ++
2
Sn + 2e 6 Sn -0.1375 2 Hg + 2e 6 Hg 0.920
++ ++ ++
2
Pb++ + 2e 6 Pb -0.1262 NO + 4 H + 3e 6
3
- +
O + H O + 2e NO + 2 H O 0.957
2 2 2
6 HO + OH -0.076 Cr O + 14 H + 6e
2 2 7
- - = +
Cu(NH ) + 2e 6 2 Cr + 7 H O 1.350
3 4 2
++ +++
6 Cu + 4 NH3 -0.05 Cl + 2e 6 2 Cl 1.35827
2
-
Fe + 3e 6 Fe -0.037 MnO + 8 H + 5e 6
+++ - +
4
2 H + 2e 6 H 0.0000 Mn + 4 H O 1.507
+ ++

2 2
AgBr + e 6 Ag + Br- 0.07133 Cr + e 6 Ce 1.72
++++ +++
Hg Br + 2e 6 MnO + 4 H + 3e 6
2 2 4
= +
2 Hg + 2 Br- 0.13923 6 MnO + 2 H O 1.679
2 2
Sn + 2 e 6 Sn 0.151 H O + 2 H + 2e 6 2 H O 1.776
++++ ++ +
2 2 2
has been found to be 0.763 V, it follows that the standard potential for the half reaction
Zn 6 Zn + 2 e
++
An electric potential difference is the work required, per unit of charge, to move a charge
4
between two points, just as a gravitational potential difference is the work required, per unit of
mass, to move a mass between two points. The gravitational potential difference for a well is
positive if the top is measured with respect to the bottom, or negative if the bottom is measured
with respect to the top, and an electric potential that is positive if the cathode is measured with
respect to the anode will be negative if the reaction is reversed. (The “standard electrode
potential” includes a choice of sign in the definition, but “standard potential” simply means that
the reactants and products are in their standard states.)
For more half-reaction potentials see especially W. Latimer, Oxidation Potentials, 2
5 nd
ed., Prentice-Hall, Englewood Cliffs, N.J., 1952.
7/10/07 4- 100
must be 0.763 V. In turn, from the measured E for the reaction
o
Zn + Cu 6 Cu + Zn

++ ++
of 1.100 V, it follows that for
2 e + Cu 6 Cu
++
the standard potential is E = 0.337 V. We can proceed in this fashion to find the potential for
o
any half reaction on this arbitrary scale. Reversal of a cell reaction changes the sign of a
potential. The more positive a half-reaction potential, the greater the tendency of that half
4
reaction to proceed as written. (But it should be kept in mind that the reaction may be too slow
to observe or that some other spontaneous reaction may occur instead.)
Some standard potentials for reduction half-reactions are given in Table 1. According to the
Stockholm Convention, such standard reduction potentials are called standard electrode
potentials. If the reaction is written in reverse order, E will have the opposite sign and is called
o
the standard oxidation potential. Tabulations of half-reaction potentials are often given for
oxidation half-reactions.
5
CALCULATIONS OF CELL POTENTIALS. A few examples will illustrate the applications of standard
potentials of half reactions. Consider the cell reaction
2 Fe + Cu 6 Cu + 2 Fe
+++ ++ ++
This consists of the two half reactions,
2 e + 2 Fe 6 2 Fe
+++ ++
Cu 6 Cu + 2 e
++
The standard potential for the first is taken directly from Table 1, that for the second is obtained
from the table simply by changing the sign (because the reaction is the reverse of that given in the
V 130.1

1.0
01.0
ln
C/equiv 96,487 x equiv/mol 2
K 298K x J/mol 314.8
V 130.1
ln
=
⋅
−=
−=
++
++
Cu
Zn
Zn
Cu
o
aa
aa
n
RT
F
EE
( ) ( )
( ) ( )
(28) log
05915.0
ln C)(25
a log 2.306 x

96487 x
V 298.15 x 314.8
ln
o
aQ
n
aQ
n
RT
n
aQ
n
RT
=
=
F
F
Q
( )
( )
(28a)
log
05915.0
C25
oo
a
n
Q
−= EE
7/10/07 4- 101

table). Thus
E = 0.771 + (-0.337) = 0.434 V
o
The positive cell potential indicates that the original reaction will proceed as written, with
substances at unit activity.
The cell reaction shown in Figure 2 is

Zn 6 Zn++ + 2e and Cu++ + 2e 6 Cu
The potential, at 25 C, may be found as follows. From Table 1,
o
E = 0.763 + 0.337 = 1.100 V
o
assuming the solid metals, Zn and Cu, are at unit activity (pure
and not seriously strained). The activities of the ions have been
approximated by their concentrations.
The concentration correction factor, (RT/nF) ln
Q
(a),
appears sufficiently often that it is helpful to carry out a partial
evaluation. At room temperature, 25 C, we may write
o
Equation 22 may therefore be written
The Nernst equation may also be applied to half-reaction potentials. For example, the
potential of the reduction half-reaction
Zn + 2 e 6 Zn
++
would be
++
−=
Zn

Zn
a
a
n
RT
ln
o
F
EE
++
−−=
Zn
Zn
a
a
log
2
05915.0
763.0E
Zn
Zn
a
a
++
−= log
2
05915.0
763.0E
7/10/07 4- 102
or, at 25 C,

o
(The electrons should, in principle, also appear in this expression, but the “activity” of the
electrons is unknown, and when this reduction half-reaction is combined with an oxidation half-
reaction, the electron terms will drop out.) Similarly, for the oxidation half-reaction

Zn 6 Zn + 2 e
++

the potential, at 25 C, would be
o
Although space does not permit an adequate discussion of the problem of reversibility, it
should be pointed out that not all chemical reactions can be carried out in electrochemical cells,
and even many that can will not be reversed by a change of potential. The first requirement for a
well-behaved electrochemical cell must be that the reaction will proceed as desired and be capable
of reversal. To achieve quantitative agreement with the equations of thermodynamics, however,
the requirements are more stringent. The chemical reaction must be thermodynamically
reversible, so that an infinitesimal change of potential is sufficient to reverse the flow of current
and the direction of chemical reaction. Discussions of the problems encountered in irreversible
electrochemical systems may be found in more advanced works under the listings of
“overvoltage”, “polarization”, and “junction potentials.”
CELL NOTATION. A very convenient shorthand notation has been developed for writing
electrochemical cells. The cell shown in Figure 2 can be represented by

Zn/ZnSO (0.01 M)//CuSO (0.1 M)/Cu
4 4

A cell in which the reaction is the displacement of hydrogen by zinc can be written

Zn/Zn //H /H ,Pt
++ +

2

The platinum serves as the electrical contact to the solution, with hydrogen gas bubbled around
the platinum. Although there are minor differences among authors concerning separating
components with commas and shilling marks, generally one of these marks will separate
substances in direct contact. The // indicates a junction between two different solutions, which
are commonly connected by a “salt bridge”, such as a U tube containing concentrated KCl
There are very few conventions that need to be memorized, although these few can be
6
expressed in a bewildering variety of forms. One convention is that the anode (where the
electrons leave the cell) in on the left in these condensed cell descriptions. A mnemonic for
remembering this is a partial acrostic based on the word “always”: Anode Left W A Y
S. Another convention is that the term standard electrode potential means the same as
standard reduction potential. A third is that a positive potential for a cell corresponds to a
spontaneous reaction, as implied by equation 2.
7/10/07 4- 103
solution and dipping into the two solutions.
The direction of the cell reaction is determined by the convention that the electrode on the
left is the anode.
6
APPLICATIONS OF ELECTRODE POTENTIALS. Electrochemistry has important applications in
physical chemistry, analytical chemistry, descriptive inorganic chemistry, preparative inorganic
and organic chemistry, biochemistry, and metallurgy. Electrochemical measurements provide an
accurate means of obtaining the thermodynamic properties of a reaction, including ∆H, ∆S, and
∆G. These thermodynamic properties depend, in turn, on the exact nature of the reaction,
including the charges on the ions participating. Because the potentials depend on concentrations,
information is available on the concentrations of individual ions in solution for analytical purposes
or for determining solubilities.
Electrode potentials are often more convenient than free energies for prediction of which
reactions will be spontaneous. For example, oxidizing agents can be put in order of oxidizing

powers by reference to the standard potentials for the corresponding reactions. Often an
oxidizing agent can be found that is just strong enough to do the job desired and not so strong
that it will oxidize other substances present in the mixture. Electrode potentials are the basis of
studies of corrosion and electro-refining of metals.
Problems
1. The standard free energies of NO and N O at 25 C are 51.3 and 99.8 J/mol. Find the
2 2 4
o
equilibrium constant for the reaction
2 NO N O
2 2 4
2. For each of the following reactions, set up the equilibrium constant expression, in terms of
activities, and make the appropriate substitutions of symbols (concentrations, pressures, mole
fractions, or one) for the activities:
a. Cu + H S(aq) CuS + 2 H
++ +
2
b. 2 HgO 2 Hg + O
2
c. NaHCO + HCl CO (g) + H O + NaCl
3 2 2
H O
2
d. AcOEt + H O AcOH + EtOH
2
(ethyl acetate) (acetic acid) (ethanol)
7/10/07 4- 104
3. The standard free energies of benzene and cyclohexane are 124.5 and 28.5 J/mol at 25 C. Find
o
the equilibrium constant for the reduction of benzene with hydrogen at 25 C.

o
C H (liq) + 3 H C H (liq)
6 6 2 6 12
If equilibrium is reached with 5 atm H gas present, what is the mole fraction of benzene left in the
2
cyclohexane liquid produced?
4. What effect will the addition of an inert gas have on the equilibrium point of the gas-phase
reaction
N O 2 NO
2 4 2
if the volume is maintained constant?
5. What effect will the addition of an inert gas have on the equilibrium point of the gas-phase
reaction
N O 2 NO
2 4 2
if the total pressure is maintained constant?
6. For the reaction
CO + 2 H CH OH
2 3
∆G = -13.5 J at 700 K. Find the percentage decomposition of methanol, at this temperature and
o
constant volume, if the initial pressure of pure methanol is 5 atm.
7. Water and solid sulfur can react to form SO and H S according to the equation
2 2
H O (gas) + 3/2 S ½ SO + H S
2 2 2
The standard free energies are, for H O (gas), -228.6, for H S (gas) -33.4, and for SO (gas), -
2 2 2
300.1 J/mol, at 25 C.
o

a. Find the equilibrium constant for the reaction as written.
b. Calculate the pressures of SO and H S to be expected in equilibrium with solid sulfur and
2 2
moist air (vapor pressure of water = 23.8 torr) at 25 C, assuming gases to be ideal.
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8. The reaction
2 SO + O 2 SO
2 2 3
is of considerable industrial importance. Starting with one-fourth atm of SO vapor at 25 C, in a
3
o
fixed volume, find the equilibrium pressure of SO and hence the percentage dissociation.
3
9. The standard free energy of HI at 250 C is -10.8 J/mol.
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a. Find ∆G and the equilibrium constant for the reaction at 250 C,
o o
½ H + ½ I HI
2 2
b. Find ∆G and the equilibrium constant for the reaction, at 250 C,
o o
H + I 2 HI
2 2