15.9

a.


b.

K = 7.2 ´
10^{2}



15.10

[SO_{2}] = 0.0124 M
[O_{2}] = 0.031 M
[SO_{3}] = ?
Solving for [SO_{3}]
gives
[SO_{3}]^{2}
= K [SO_{2}]^{2}[O_{2}]
= (2.8 ´
10^{2})( 0.0124)^{2}(0.031) = 1.33 ´ 10^{3}
[SO_{3}] = = 0.037 M


15.11

The problem states that the system is at equilibrium, so we simply
substitute the equilibrium concentrations into the equilibrium expression
to calculate K_{c}.
Step 1: Calculate the concentrations of the components in units of
mol/L. The molarities can be
calculated by simply dividing the number of moles by the volume of the
flask.
Step 2: Once the molarities are known, K_{c} can be found by
substituting the molarities into the equilibrium expression.
Think About It:

If you forget to convert moles
to moles/liter, will you get a different answer? Under what circumstances will the two
answers be the same?



15.12

Using the mole
amounts and volume given in the problem, we calculate the concentrations of
H_{2} and CH_{3}OH.
[H_{2}]
= 0.0021 M
[CH_{3}OH]
= 0.00062 M
The equilibrium expression is
Solving for
[CO] gives
[CO] = = 8.8 ´ 10^{3} M


15.21

a.

With K_{c}
= 10, products are favored at equilibrium. Because the coefficients for both A and
B are one, we expect the concentration of B to be 10 times that of A at
equilibrium. Diagram (a) is the best choice with 10 B
molecules and 1 A molecule.

b.

With K_{c} = 0.10, reactants are favored at
equilibrium. Because the coefficients
for both A and B are one, we expect the concentration of A to be 10 times
that of B at equilibrium. Diagram (d) is the best choice with 10 A
molecules and 1 B molecule.
You can calculate K_{c} in each case without
knowing the volume of the container because the mole ratio between A and
B is the same. Volume will cancel from the K_{c} expression. Only moles of each component are needed
to calculate K_{c}.



15.22

Note that we are comparing similar reactions at equilibrium – two
reactants producing one product, all with coefficients of one in the
balanced equation.
a.

The reaction, A + C AC has the largest equilibrium constant. Of the three diagrams, there is the
most product present at equilibrium.

b.

The reaction, A
+ D AD has the smallest equilibrium constant. Of the three diagrams, there is the
least amount of product present at equilibrium.



15.23

When the equation for a
reversible reaction is written in the opposite direction, the equilibrium
constant becomes the reciprocal of the original equilibrium constant.


15.24

Using
Equation 15.4 of the text: K_{P} = K_{c}(0.0821 T)^{Dn}
^{ }
where, Dn = 2  3 = 1
and T = (1273
+ 273) K = 1546 K
K_{P} = (2.24 ´ 10^{22})(0.0821
´
1546)^{1} = 1.76 ´
10^{20}


15.25

Strategy:

The relationship between K_{c} and K_{P} is given by Equation
15.4 of the text. What is the
change in the number of moles of gases from reactant to product? Recall that
Dn
= moles of gaseous products

moles of gaseous reactants
What unit of temperature should we use?

Solution:

The relationship between K_{c} and K_{P}
is given by Equation 15.4 of the text.
K_{P} = K_{c}(0.0821
T)^{Dn}
Rearrange the equation relating K_{P} and K_{c},
solving for K_{c}.
Because T = 623 K and Dn =
3 
2 = 1, we have:



15.26

We can write the equilibrium expression from the balanced equation and
substitute in the pressures.
Think About It:

Do we need to know the
temperature?



15.27

The equilibrium expressions are:
a.

Substituting the given
equilibrium concentration gives:
or 8.2
´
10^{2}

b.

Substituting the given
equilibrium concentration gives:
Think About It:

Is there a relationship
between the K_{c}
values from parts (a) and (b)?




15.28

The equilibrium expression for
the two forms of the equation are:
The
relationship between the two equilibrium constants is
K_{P} can be found as shown
below.
K_{P} = K_{c}'(0.0821 T)^{Dn} = (2.6 ´ 10^{4})(0.0821
´
1000)^{1} = 3.2 ´
10^{2}


15.29

Because
pure solids do not enter into an equilibrium expression, we can calculate K_{P} directly from the
pressure that is due solely to CO_{2}(g).
Now,
we can convert K_{P} to K_{c} using the following
equation.
K_{P} = K_{c}(0.0821 T)^{Dn}


15.30

We substitute the given
pressures into the reaction quotient expression.
The calculated value of Q_{P}
is less than K_{P} for
this system. The system will change
in a way to increase Q_{P}
until it is equal to K_{P}. To achieve this, the pressures of PCl_{3}
and Cl_{2} must increase,
and the pressure of PCl_{5} must decrease.
Think About It:

Could you actually
determine the final pressure of each gas?



15.31

Strategy:

Because they are constant
quantities, the concentrations of solids and liquids do not appear in the
equilibrium expressions for heterogeneous systems. The total pressure at equilibrium that
is given is due to both NH_{3} and CO_{2}. Note that for every 1 atm of CO_{2}
produced, 2 atm of NH_{3} will be produced due to the
stoichiometry of the balanced equation.
Using this ratio, we can calculate the partial pressures of NH_{3}
and CO_{2} at equilibrium.

Solution:

The equilibrium expression for the
reaction is
The total pressure in the
flask (0.363 atm) is a sum of the partial pressures of NH_{3} and
CO_{2}.
Let the partial pressure of CO_{2} = x.
From the stoichiometry of the balanced equation, you should find
that Therefore, the partial pressure of NH_{3}
= 2x. Substituting into the equation for
total pressure gives:
3x
= 0.363 atm
Substitute the equilibrium
pressures into the equilibrium expression to solve for K_{P}.



15.32

If the CO pressure at
equilibrium is 0.497 atm, the balanced equation requires the chlorine
pressure to have the same value. The
initial pressure of phosgene gas can be found from the ideal gas equation.
Since
there is a 1:1 mole ratio between phosgene and CO, the partial pressure of
CO formed (0.497 atm) equals the partial pressure of phosgene reacted. The phosgene pressure at equilibrium is:
CO(g) + Cl_{2}(g) COCl_{2}(g)
Initial
(atm): 0 0 1.31
Change
(atm): +0.497 +0.497 0.497
Equilibrium
(atm): 0.497 0.497 0.81
The
value of K_{P} is then
found by substitution.


15.33

Let x be the initial pressure of
NOBr. Using the balanced equation,
we can write expressions for the partial pressures at equilibrium.
P_{NOBr} =
(1 
0.34)x =
0.66x
P_{NO} =
0.34x
The sum of these is
the total pressure.
0.66x + 0.34x + 0.17x =
1.17x =
0.25 atm
x
= 0.214 atm
The
equilibrium pressures are then
P_{NOBr} =
0.66(0.214) = 0.141 atm
P_{NO} =
0.34(0.214) = 0.073 atm
P=
0.17(0.214) = 0.036 atm
We
find K_{P} by
substitution.
K_{P} =9.6 ´ 10^{3}
The
relationship between K_{P}
and K_{c} is given by
K_{P} = K_{c}(RT)^{Dn}
We
find K_{c} (for this
system Dn = +1)
K_{c} =3.9 ´ 10^{4}


15.34

The target equation is the
sum of the first two.
H_{2}S ⇄ H^{+} +
HS
HS ⇄ H^{+} +
S^{2}^{ }
H_{2}S ⇄ 2H^{+} + S^{2}^{}
Since this is the case, the
equilibrium constant for the combined reaction is the product of the
constants for the component reactions (Section 15.3 of the text). The equilibrium constant is therefore:
K_{c} = K_{c}'K_{c}"
= 9.5 ´
10^{27}
Think About It:

What happens in the special case when the two
component reactions are the same?
Can you generalize this relationship to adding more than two
reactions? What happens if one
takes the difference between two reactions?



15.35

K =
(6.5 ´
10^{2})(6.1
´
10^{5})
K = 4.0
´ 10^{6}


15.36

Given:
For the overall reaction:


15.37

To obtain
2SO_{2} as a reactant in the final equation, we must reverse the
first equation and multiply by two.
For the equilibrium, 2SO_{2}(g) 2S(s) + 2O_{2}(g)
Now we can add the above
equation to the second equation to obtain the final equation. Since we add the two equations, the
equilibrium constant is the product of the equilibrium constants for the
two reactions.
2SO_{2}(g)
2S(s) + 2O_{2}(g)
2S(s) +
3O_{2}(g) 2SO_{3}(g)
2SO_{2}(g) + O_{2}(g)
2SO_{3}(g)


15.39

Strategy:

Use the law of
mass action to write the equilibrium expression. The equilibrium expression must yield K_{c} = 2 for the reaction
to be at equilibrium.

Setup:

The
equilibrium expression has the form of concentrations of products over
concentrations of reactants, each raised to the appropriate power.
Plug
in the concentrations of all three species to evaluate K_{c}.

Solution:

Diagram (a):
Diagram (b):
Diagram (c):
Diagram (d):
Since K_{c} = 2, diagrams (a), (b), and (c)
represent and equilibrium mixture of A_{2}, B_{2}, and
AB.



15.40

Strategy:

Use the law of
mass action to write the equilibrium expression. The mixtures at equilibrium will have
the same K_{c} value.

Setup:

The
equilibrium expression has the form of concentrations of products over
concentrations of reactants, each raised to the appropriate power.
Plug
in the concentrations of all three species to evaluate K_{c}.

Solution:

Diagram (a):
Diagram (b):
Diagram (c):
Diagram (d):
For diagrams
(a), (c), and (d), K_{c}
= 3. Therefore, diagram (b) is not at equilibrium.



15.41

Strategy:

The equilibrium constant K_{c} is given, and we start with a mixture of CO and
H_{2}O. From the
stoichiometry of the reaction, we can determine the concentration of H_{2}
at equilibrium. Knowing the
partial pressure of H_{2} and the volume of the container, we can
calculate the number of moles of H_{2}.

Solution:

The balanced equation shows that one mole of water
will combine with one mole of carbon monoxide to form one mole of
hydrogen and one mole of carbon dioxide. (This is the reverse reaction.) Let x
be the depletion in the concentration of either CO or H_{2}O at
equilibrium (why can x serve to
represent either quantity?). The
equilibrium concentration of hydrogen must then also be equal to x.
The changes are summarized as shown in the table.
H_{2} +
CO_{2} H_{2}O
+ CO
Initial (M): 0 0 0.0300 0.0300
Equilibrium (M): x x (0.0300 
x) (0.0300

x)
The equilibrium constant
is:
Substituting,
Taking the square root of both
sides, we obtain:
x
= 0.0173 M
The number of moles of H_{2} formed is:
0.0173
mol/L ´
10.0 L = 0.173 mol H_{2}



15.42

Since the reaction started with
only pure NO_{2}, the equilibrium concentration of NO must be twice
the equilibrium concentration of O_{2}, due to the 2:1 mole ratio
of the balanced equation. Therefore,
the equilibrium partial pressure of NO
is (2 ´
0.25 atm) = 0.50 atm.
We can find the
equilibrium NO_{2} pressure by rearranging the equilibrium
expression, then substituting in the known values.


15.43

Notice that the balanced
equation requires that for every two moles of HBr consumed, one mole of H_{2}
and one mole of Br_{2} must be formed. Let 2x
be the depletion in the concentration of HBr at equilibrium. The equilibrium concentrations of H_{2}
and Br_{2} must therefore each be x. The changes are shown
in the table.
H_{2} +
Br_{2} 2HBr
Initial
(M): 0
0 0.267
Equilibrium
(M): x x (0.267 
2x)
The equilibrium constant relationship is given by:
Substitution of the
equilibrium concentration expressions gives
Taking the square root of both sides we obtain:
x
= 1.80 ´
10^{4}
The equilibrium concentrations
are:
[H_{2}] = [Br_{2}] = 1.80 ´
10^{4} M
[HBr] =
0.267 
2(1.80 ´
10^{4}) = 0.267 M
Think About It:

If the depletion in the
concentration of HBr at equilibrium were defined as x, rather than 2x,
what would be the appropriate expressions for the equilibrium
concentrations of H_{2 }and Br_{2}? Should the final answers be different
in this case?



15.44

We follow the
procedure outlined in Section 15.4 of the text to calculate the equilibrium
concentrations.
Step 1:
The initial concentration of I_{2} is 0.0456 mol/2.30 L
= 0.0198 M. The stoichiometry of the problem shows 1
mole of I_{2} dissociating to 2 moles of I atoms. Let x
be the amount (in mol/L) of I_{2} dissociated. It follows that the equilibrium
concentration of I atoms must be 2x. We summarize the changes in
concentrations as follows:
I_{2}(g) 2I(g)
Initial
(M): 0.0198
0.000
Equilibrium (M): (0.0198  x) 2x
Step 2: Write the equilibrium expression in terms of the equilibrium
concentrations. Knowing the value of
the equilibrium constant, solve for x.
4x^{2} + (3.80 ´
10^{5})x  (7.52 ´
10^{7}) =
0
The above equation is a
quadratic equation of the form ax^{2}
+ bx + c = 0. The solution for a quadratic equation is
Here, we have a = 4, b =
3.80 ´
10^{5},
and c = 7.52
´
10^{7}. Substituting into the above equation,
x
= 4.29 ´
10^{4}
M or x = 4.39
´
10^{4}
M
The second
solution is physically impossible because you cannot have a negative
concentration. The first solution is
the correct answer.
Step 3: Having solved for x, calculate the equilibrium concentrations of all species.
[I] = 2x
= (2)(4.29 ´
10^{4}
M) = 8.58 ´
10^{4} M
[I_{2}] =
(0.0198 
x) =
[0.0198 
(4.29 ´
10^{4})]
M
= 0.0194 M
Think About It:

We could have simplified this problem by assuming that
x was small compared to
0.0198. We could then assume that
0.0198 
x »
0.0198. By making this assumption,
we could have avoided solving a quadratic equation.



15.45

Since equilibrium pressures are desired, we calculate K_{P}.
K_{P} = K_{c}(0.0821 T)^{Dn} = (4.63 ´ 10^{3})(0.0821
´
800)^{1} = 0.304
COCl_{2}(g)
CO(g)
+ Cl_{2}(g)
Initial
(atm): 0.760 0.000 0.000
Equilibrium
(atm): (0.760  x) x x
x^{2} + 0.304x  0.231 =
0
x
= 0.352 atm
At
equilibrium:
P_{CO} =


15.46

a.

The equilibrium constant, K_{c}, can be found by
simple substitution.

b.

The
magnitude of the reaction quotient Q_{c}
for the system after the concentration of CO_{2} becomes 0.50 mol/L, but before
equilibrium is reestablished, is:
The value of Q_{c}
is smaller than K_{c};
therefore, the system will shift to the right, increasing the
concentrations of CO and H_{2}O and decreasing the concentrations
of CO_{2} and H_{2}.
Let x be the depletion
in the concentration of CO_{2} at equilibrium. The stoichiometry of the balanced
equation then requires that the decrease in the concentration of H_{2}
must also be x, and that the
concentration increases of CO and H_{2}O be equal to x as well. The changes in the original
concentrations are shown in the table.
CO_{2 }+ H_{2} CO + H_{2}O
Initial
(M): 0.50 0.045 0.050 0.040
Equilibrium
(M): (0.50  x) (0.045 
x) (0.050 + x) (0.040 + x)
The
equilibrium expression is:
0.52(x^{2} 
0.545x + 0.0225) =
x^{2} + 0.090x + 0.0020
0.48x^{2} + 0.373x 
(9.7 ´
10^{3}) =
0
The
positive root of the equation is x
= 0.025.
The
equilibrium concentrations are:
[CO_{2}] =
(0.50 
0.025) M =
0.48 M
[H_{2}] =
(0.045 
0.025) M =
0.020 M
[CO]
= (0.050 + 0.025) M
= 0.075 M
[H_{2}O] =
(0.040 + 0.025) M =
0.065 M



15.47

The equilibrium expression
for the system is:
The total pressure can be
expressed as:
If we let the partial pressure
of CO be x, then the partial
pressure of CO_{2} is:
Substitution gives the
equation:
This can be rearranged to the
quadratic:
x^{2} + 1.52x  6.84 =
0
The solutions
are x = 1.96 and x = 3.48; only the
positive result has physical significance (why?). The equilibrium pressures are
P_{CO} = x
= 1.96 atm


15.48

The initial concentrations are [H_{2}] = 0.80
mol/5.0 L = 0.16 M and [CO_{2}]
= 0.80 mol/5.0 L = 0.16 M.
H_{2}(g) +
CO_{2}(g) H_{2}O(g)
+ CO(g)
Initial (M): 0.16 0.16 0.00 0.00
Equilibrium (M): 0.16

x 0.16

x x x
Taking
the square root of both sides, we obtain:
x
= 0.11 M
The
equilibrium concentrations are:
[H_{2}] =
[CO_{2}] = (0.16  0.11) M
= 0.05 M
[H_{2}O] =
[CO] = 0.11 M


15.49

Strategy:

We
are given the concentrations of all species in a biochemical reaction and
must determine whether the reaction will proceed in the forward direction
or the reverse direction in order to establish equilibrium. This requires calculating the value of Q and comparing it to the value of
K, which is given in the
problem.

Solution:

The
reaction quotient is
Q =
Because
K is 1.11, Q is greater than K. Thus, the reaction will proceed to the
left (the reverse reaction will occur) in order to establish
equilibrium. The forward reaction will not occur.



15.54

Strategy:

According to Le
Châtelier's principle, increasing the temperature of an equilibrium
mixture causes a shift toward the product side (to the right) for an
endothermic reaction, and a shift toward the reactant side (to the left)
for an exothermic reaction.

Setup:

Reactions
where ∆H is negative are
exothermic.

Solution:

Equilibria (a), (d), and (e) are exothermic and will shift to the
left when the temperature is increased.



15.55

Strategy:

According to Le
Châtelier's principle, increasing
the volume causes a shift toward the side with the larger number of moles of gas. Decreasing the volume of an equilibrium
mixture causes a shift toward the side of the equation with the smaller number of moles of
gas. For an equilibrium with equal
numbers of gaseous moles on both sides, a change in volume does not cause
the equilibrium to shift in either direction.

Setup:

Determine the
numbers of moles of gas in both the reactants and products for each
reaction.

Solution:

There are an
equal number of moles of gas in the reactants and products for reaction (b). A change in volume will not affect the
position of its equilibrium.



15.56

Strategy:

According to Le
Châtelier's principle, adding a reactant to an equilibrium mixture shifts
the equilibrium toward the product side (to the right) of the equation.

Setup:

Only
equilibria containing H_{2} as a reactant will shift to the right
when H_{2} is added.

Solution:

Equilibria (a) and (c) contain H_{2} as a reactant
and will shift to the right when more H_{2} is added.



15.57

Strategy:

Use Le
Châtelier's principle to determine which of the following scenarios will
cause the reaction equilibrium to shift to the right.

Solution:

a.

Decreasing
the volume of an equilibrium mixture causes a shift toward the side of
the equation with the smaller
number of moles of gas. The
equilibrium will shift to the
left.

b.

Increasing
the volume of an equilibrium mixture causes a shift toward the side
with the larger number of
moles of gas. The equilibrium
will shift to the right.

c.

Altering the
amount of a solid does not
change the position of the equilibrium because doing so does not
change the value of Q.

d.

The addition
of more reactant will shift the
equilibrium to the right.

e.

The removal
of a product will shift the
equilibrium to the right.

(b), (d), and (e) will cause the equilibrium to shift to
the right.



15.58

a.

Addition
of more Cl_{2}(g) (a
reactant): The equilibrium would
shift to the right.

b.

Removal of SO_{2}Cl_{2}(g) (a product): The equilibrium would shift to the
right.

c.

Removal of SO_{2}(g) (a reactant): The
equilibrium would shift to the left.



15.59

a.

Removal
of CO_{2}(g) from the
system: The equilibrium would
shift to the right.

b.

Addition
of more solid Na_{2}CO_{3}: The equilibrium would be unaffected. [Na_{2}CO_{3}] does not
appear in the equilibrium expression.

c.

Removal
of some of the solid NaHCO_{3}: The equilibrium would be unaffected. Same reason as (b).



15.60

a.

This
reaction is endothermic.
(Why?) According to Section
15.5, an increase in temperature favors an endothermic reaction, so the
equilibrium constant should become larger.

b.

This
reaction is exothermic. Such
reactions are favored by decreases in temperature. The magnitude of K_{c} should decrease.

c.

In
this system heat is neither absorbed nor released. A change in temperature should have no effect on the magnitude of the
equilibrium constant.



15.61

Strategy:

A
change in pressure can affect only the volume of a gas, but not that of a
solid or liquid because solids and liquids are much less compressible. The stress applied is an increase in
pressure. According to Le
Châtelier's principle, the system will adjust to partially offset this
stress. In other words, the system
will adjust to decrease the pressure.
This can be achieved by shifting to the side of the equation that
has fewer moles of gas. Recall
that pressure is directly proportional to moles of gas: PV
= nRT so P µ
n.

Solution:

a.

Changes
in pressure ordinarily do not affect the concentrations of reacting
species in condensed phases because liquids and solids are virtually
incompressible. Pressure change
should have no effect on
this system.

b.

No
effect. Same situation as
(a).

c.

Only the product is
in the gas phase. An increase in
pressure should favor the reaction that decreases the total number of
moles of gas. The equilibrium
should shift to the left, that is, the amount of B
should decrease and that of A should increase

d.

In
this equation there are equal moles of gaseous reactants and
products. A shift in either
direction will have no effect on the total number of moles of gas
present. There will be no effect on the position of
the equilibrium when the pressure is increased.

e.

A
shift in the direction of the reverse reaction (shift to the left) will have the result of decreasing the
total number of moles of gas present.




15.62

a.

A
pressure increase will favor the reaction (forward or reverse?) that
decreases the total number of moles of gas. The equilibrium should shift to the right, i.e., more I_{2}
will be produced at the expense of I.

b.

If
the concentration of I_{2} is suddenly altered, the system is no
longer at equilibrium. Evaluating
the magnitude of the reaction quotient Q_{c} allows us to predict the direction of the
resulting equilibrium shift. The
reaction quotient for this system is:
Increasing
the concentration of I_{2} will increase Q_{c}. The
equilibrium will be reestablished in such a way that Q_{c} is again equal to the equilibrium
constant. More I will form. The system shifts to the left to establish equilibrium.

c.

The
forward reaction is exothermic. A
decease in temperature will shift the system to the right to reestablish equilibrium.



15.63

Strategy:

(a) What does the sign of DH°
indicate about the heat change (endothermic or exothermic) for the
forward reaction? (b) The stress is the addition of Cl_{2}
gas. How will the system adjust to
partially offset the stress?
(c) The stress is the
removal of PCl_{3} gas.
How will the system
adjust
to partially offset the stress?
(d) The stress is an
increase in pressure. The system
will adjust to decrease the pressure.
Remember, pressure is directly proportional to moles of gas.
(e) What is the function of a
catalyst? How does it affect a
reacting system not at
equilibrium? at equilibrium?

Solution:

a.

The
stress applied is the heat added to the system. Note that the reaction is endothermic
(DH°
> 0). Endothermic reactions
absorb heat from the surroundings; therefore, we can think of heat as a
reactant.
heat + PCl_{5}(g) ⇄ PCl_{3}(g) + Cl_{2}(g)
The
system will adjust to remove some of the added heat by undergoing a
decomposition reaction (Shift to
the right)

b.

The
stress is the addition of Cl_{2} gas. The system will shift in the
direction to remove some of the added Cl_{2}. The system shifts to the left
until equilibrium is reestablished.

c.

The stress is
the removal of PCl_{3} gas.
The system will shift to replace some of the PCl_{3}
that was
removed. The system shifts to the right until
equilibrium is reestablished.

d.

The
stress applied is an increase in pressure. The system will adjust to remove the
stress by decreasing the pressure.
Recall that pressure is directly proportional to the number of moles
of gas. In the balanced equation
we see 1 mole of gas on the reactants side and 2 moles of gas on the
products side. The pressure can
be decreased by shifting to the side with the fewer moles of gas. The system will shift to the left
to reestablish equilibrium.

e.

The
function of a catalyst is to increase the rate of a reaction. If a catalyst is added to the
reacting system not at equilibrium, the system will reach equilibrium
faster than if left undisturbed.
If a system is already at equilibrium, as in this case, the
addition of a catalyst will not affect either the concentrations of
reactant and product, or the equilibrium constant. A
catalyst has no effect on equilibrium position.




15.64

a.

Increasing the temperature favors the
endothermic reaction so that the concentrations of SO_{2} and O_{2}
will increase while that of SO_{3} will decrease.

b.

Increasing the pressure favors the
reaction that decreases the number of moles of gas. The concentration of SO_{3}
will increase.

c.

Increasing the concentration of SO_{2}
will lead to an increase in the concentration of SO_{3} and a decrease
in the concentration of O_{2}.

d.

A catalyst has
no effect on the position of equilibrium.

e.

Adding an inert gas at constant volume has no effect on the
position of equilibrium.



15.65

No change. A catalyst has no effect on the position
of the equilibrium.


15.66

a.

If helium gas is added to the
system without changing the pressure or the temperature, the volume of
the container must necessarily be increased. This will decrease the partial
pressures of all the reactants and products. A pressure decrease will favor the
reaction that increases the number of moles of gas. The position of equilibrium will shift
to the left.

b.

If the volume remains
unchanged, the partial pressures of all the reactants and products will
remain the same. The reaction
quotient Q_{c} will
still equal the equilibrium constant, and there will be no change in the position of
equilibrium.



15.67

For this system,
This
means that to remain at equilibrium, the pressure of carbon dioxide must
stay at a fixed value as long as the temperature remains the same.
a.

If the volume is
increased, the pressure of CO_{2} will drop (Boyle's law,
pressure and volume are inversely proportional). Some CaCO_{3} will break down
to form more CO_{2} and CaO.
(Shift to right)

b.

Assuming that the amount
of added solid CaO is not so large that the volume of the system is
altered significantly, there should be no effect. If a huge
amount of CaO were added, this would have the effect of reducing the
volume of the container. What
would happen then?

c.

Assuming that the amount
of CaCO_{3} removed doesn't alter the container volume
significantly, there should be no
effect. Removing a huge amount
of CaCO_{3} will have the effect of increasing the container
volume. The result in that case
will be the same as in part (a).

d.

The pressure of CO_{2}
will be greater and will exceed the value of K_{P}. Some CO_{2}
will combine with CaO to form more CaCO_{3}. (Shift to left)

e.

Carbon dioxide combines
with aqueous NaOH according to the equation
CO_{2}(g) + NaOH(aq) ® NaHCO_{3}(aq)
This will have the
effect of reducing the CO_{2} pressure and causing more CaCO_{3}
to break down to CO_{2} and CaO.
(Shift to the right)

f.

Carbon dioxide does not react with
hydrochloric acid, but CaCO_{3} does.
CaCO_{3}(s) + 2HCl(aq) ® CaCl_{2}(aq) + CO_{2}(g)
+ H_{2}O(l)
The CO_{2}
produced by the action of the acid will combine with CaO as discussed in
(d) above.
(Shift to the left)

g.

This is a
decomposition reaction.
Decomposition reactions are endothermic. Increasing the temperature will favor
this reaction and produce more CO_{2} and CaO. (Shift to the right)



15.68


15.69

a.

2O_{3}(g) 3O_{2}(g) DH= 284.4
kJ/mol.

b.

Equilibrium
would shift to the left. The number
of O_{3} molecules would increase and the number of O_{2}
molecules would decrease.



15.70

Strategy:

Use Le Châtelier's principle to
predict the direction of shift for each case. A shift to the left will cause a
decrease in the amount of HbO_{2} and a shift to the right will
cause an increase in the amount of HbO_{2}.

Solution:

a.

Decreasing the temperature of an
equilibrium mixture causes a shift toward the product side for an
exothermic reaction (∆H
< 0). The equilibrium will
shift to the right and increase
the amount of HbO_{2}.

b.

According to Henry’s Law (Equation
13.3), the solubility of a gas in a liquid is proportional to the
pressure of the gas over the solution.
Increasing the pressure of O_{2} would increase its
concentration. The equilibrium
would shift to the right and increase
the amount of HbO_{2}.

c.

The removal of a reactant will shift
the equilibrium to the left. The
amount of HbO_{2} will decrease.




15.71

Strategy:

According to Le
Châtelier's principle, adding a product to an equilibrium mixture shifts
the equilibrium toward the reactant side (to the left) of the equation.

Solution:

The addition
of acid (product) will shift the equilibrium to the left. As a result, the concentration of oxyhemoglobin will decrease as acidosis
occurs.



15.72

The relevant
relationships are:
and
K_{P} = K_{c}(0.0821 T)^{Dn} = K_{c}(0.0821
T) Dn = +1
We set
up a table for the calculated values of K_{c}
and K_{P}.
_{ }200 56.9(0.0821
´
473) = 2.21
´ 10^{3}
300 3.41(0.0821
´
573) = 1.60
´ 10^{2}
400 2.10(0.0821
´
673) = 116
Since K_{c}
(and K_{P}) decrease with
temperature, the reaction is exothermic.


15.73

a.

The equation that relates K_{P} and K_{c}
is:
K_{P} =
K_{c}(0.0821 T)^{Dn}
For this reaction, Dn = 3  2 = 1

b.

A mixture of H_{2} and O_{2} can be kept at room
temperature because of a very large activation energy. The reaction of hydrogen with
oxygen is infinitely slow without a catalyst or an initiator. The action of a single spark on a
mixture of these gases results in the explosive formation of water.



15.74

Using data from Appendix 2 we calculate
the enthalpy change for the reaction.
The enthalpy
change is negative, so the reaction is exothermic. The formation of NOCl will be favored by low temperature.
A pressure increase
favors the reaction forming fewer moles of gas. The formation of NOCl will be favored by high pressure.


15.75

a.

Calculate the value of K_{P} by substituting the
equilibrium partial pressures into the equilibrium expression.

b.

The total pressure is the sum
of the partial pressures for the two gaseous components, A and B. We can write:
P_{A}
+
P_{B} =
1.5 atm
and
P_{B} =
1.5 
P_{A}
Substituting into the
expression for K_{P}
gives:
Solving the quadratic
equation, we obtain:
P_{A} =
0.69 atm
and by difference,
P_{B} =
0.81 atm
Check that substituting
these equilibrium concentrations into the equilibrium expression gives
the equilibrium constant calculated in part (a).



15.76

a.

The balanced equation
shows that equal amounts of ammonia and hydrogen sulfide are formed in
this decomposition. The partial
pressures of these gases must just be half the total pressure, i.e.,
0.355 atm. The value of K_{P} is

b.

We find the number
of moles of ammonia (or hydrogen sulfide) and ammonium hydrogen sulfide.
From
the balanced equation the percent decomposed is

c.

If the temperature
does not change, K_{P}
has the same value. The total
pressure will still be 0.709 atm at equilibrium. In other words the amounts of ammonia
and hydrogen sulfide will be twice as great, and the amount of solid
ammonium hydrogen sulfide will be:
[0.1205

2(0.0582)]mol = 0.0041
mol NH_{4}HS



15.77

Total
number of moles of gas is:
0.020
+ 0.040 + 0.96 = 1.02 mol of gas
You can calculate the partial pressure of each gaseous
component from the mole fraction and the total pressure.
Calculate K_{P} by substituting the
partial pressures into the equilibrium expression.


15.78

Since the reactant is a solid, we can
write:
The
total pressure is the sum of the ammonia and carbon dioxide pressures.
From
the stoichiometry,
Therefore:
Substituting
into the equilibrium expression:
K_{P} =
(0.212)^{2}(0.106)
= 4.76 ´
10^{3}


15.79

Set up a table that contains the initial concentrations,
the change in concentrations, and the equilibrium concentration. Assume that the vessel has a volume of 1
L.
H_{2} + Cl_{2} 2HCl
Initial
(M): 0.47 0 3.59
Equilibrium
(M): (0.47 + x) x (3.59 
2x)
Substitute
the equilibrium concentrations into the equilibrium expression, then solve
for x. Since
Dn = 0, K_{c} = K_{P}_{.}
Solving the quadratic
equation,
x
= 0.103
Having solved for x, calculate the equilibrium
concentrations of all species.
[H_{2}] =
0.573 M [Cl_{2}] =
0.103 M [HCl] =
3.384 M
Since we assumed that the vessel had a volume of 1 L,
the above molarities also correspond to the number of moles of each
component.
From the mole fraction of each component and the total
pressure, we can calculate the partial pressure of each component.
Total number of moles
= 0.573 + 0.103 + 3.384 =
4.06 mol
P=0.28 atm
P=0.051 atm
P=1.67 atm


15.80

Set up a table that contains the initial concentrations,
the change in concentrations, and the equilibrium concentrations. The initial concentration of I_{2}(g) is 0.054 mol/0.48 L =
0.1125 M. The amount of I_{2} that
dissociates is (0.0252)(0.1125 M)
= 0.002835 M. We carry extra significant figures
throughout this calculation to minimize rounding errors.
I_{2} 2I
Initial
(M): 0.1125 0
Change (M): 0.002835 +(2)(0.002835)
Equilibrium
(M): 0.1097 0.005670
Substitute
the equilibrium concentrations into the equilibrium expression to solve for
K_{c}.
K_{P} = K_{c}(0.0821
T)^{Dn}
K_{P}
= (2.93 × 10^{4})(0.0821 × 860)^{1} = 0.021


15.81

This
is a difficult problem. Express the
equilibrium number of moles in terms of the initial moles and the change in
number of moles (x). Next, calculate the mole fraction of each
component. Using the mole fraction,
you should come up with a relationship between partial pressure and total
pressure for each component.
Substitute the partial pressures into the equilibrium expression to
solve for the total pressure, P_{T}.
The
reaction is:
N_{2 }+ 3 H_{2} 2 NH_{3} _{}
Initial
(mol): 1 3 0
Equilibrium
(mol): (1

x) (3  3x) 2x
x
= 0.35 mol
Substituting
x into the following mole
fraction equations, the mole fractions of N_{2} and H_{2}
can be calculated.
The partial pressures of each component are equal to the
mole fraction multiplied by the total pressure.
Substitute
the partial pressures above (in terms of P_{T}) into the equilibrium expression, and solve for P_{T}.
P_{T} =
5.0 ´
10^{1} atm


15.82

For the balanced equation:


15.83

We carry an additional significant figure throughout
this calculation to minimize rounding errors. The initial molarity of SO_{2}Cl_{2}
is:
The concentration of SO_{2} at equilibrium is:
Since there is a 1:1 mole
ratio between SO_{2} and SO_{2}Cl_{2}, the
concentration of SO_{2} at equilibrium (0.01725 M) equals the concentration of SO_{2}Cl_{2}
reacted. The concentrations of SO_{2}Cl_{2}
and Cl_{2} at equilibrium are:
SO_{2}Cl_{2}(g)
SO_{2}(g) + Cl_{2}(g)
Initial
(M):
0.02500 0 0
Change (M): 0.01725 +0.01725 +0.01725
Equilibrium
(M): 0.00775 0.01725 0.01725
Substitute the equilibrium
concentrations into the equilibrium expression to calculate K_{c}.
or 0.038


15.84

a.

The reaction is endothermic because A—B bonds are
broken and no new bonds form.
Therefore, when temperature is decreased, the equilibrium will shift to the left causing more AB
to form.

b.

Equilibrium will shift
to the right (toward the larger number of gaseous moles).

c.

Adding He will have no effect on the position of the equilibrium.

d.

Adding
a catalyst will have no effect
on the position of the equilibrium.



15.85

For
a 100% yield, 2.00 moles of SO_{3} would be formed (why?). An 80% yield means 2.00 moles ´
(0.80) 1.60 moles SO_{3} is formed.
The
amount of SO_{2} remaining at equilibrium =
(2.00 
1.60)mol = 0.40 mol
The
amount of O_{2} reacted
= ´ (amount of SO_{2}
reacted) = (´ 1.60)mol =
0.80 mol
The
amount of O_{2} remaining at equilibrium =
(2.00 
0.80)mol = 1.20 mol
Total moles at equilibrium =
moles SO_{2} + moles
O_{2} + moles SO_{3}
= (0.40 + 1.20 + 1.60)mol = 3.20 moles
P_{total} = 3.3×10^{2} atm


15.86

We carry an additional significant figure throughout
this calculation to minimize rounding errors.
If
there were no dissociation, then the pressure would be:
We construct a table to
determine how much I_{2} has dissociated.
I_{2}(g) 2I(g)
Initial
P (atm): 0.953 0
Change in P (atm): x +2x
Equilibrium
P (atm): 0.953  x 2x
Knowing
the total pressure, we can write
0.953

x + 2x = 1.51 atm
and solve
for x:
x = 0.557
Therefore,
the equilibrium pressures are P= 0.396 atm and P= 1.114 atm.
Using these pressures in the equilibrium expression gives
K_{P} = 3.1


15.87

Of the original 1.05 moles of Br_{2}, 1.20% has
dissociated. The amount of Br_{2}
dissociated in molar concentration is:
Setting up a table:
Br_{2}(g) 2Br(g)
Initial (M): 0
Change (M): 0.0129 +2(0.0129)
Equilibrium
(M): 1.06 0.0258


15.88

Panting decreases the concentration of
CO_{2} because CO_{2} is exhaled during respiration. This decreases the concentration of
carbonate ions, shifting the equilibrium to the left. Less CaCO_{3} is produced. Two possible solutions would be either to
cool the chickens' environment or to feed them carbonated water.


15.89

According
to the ideal gas law, pressure is directly proportional to the
concentration of a gas in mol/L if the reaction is at constant volume and
temperature. Therefore, pressure may
be used as a concentration unit.
The
reaction is:
N_{2}
+ 3H_{2 } 2NH_{3} _{}
Initial
(atm): 0.862 0.373 0
Equilibrium
(atm): (0.862 
x) (0.373  3x) 2x
At this point, we need to make two assumptions that 3x is very small compared to 0.373
and that x is very small compared
to 0.862. Hence,
0.373

3x » 0.373
and
0.862

x
» 0.862
Solving for x.
x =
2.20 ´
10^{3}
atm
The equilibrium pressures are:
Think About It:

Was the assumption valid that we made above? Typically, the assumption is considered
valid if x is less than 5
percent of the number that we said it was very small compared to. Is this the case?



15.90

a.

The sum of the mole fractions
must equal one.
and
According to the
hint, the average molar mass is the sum of the products of the mole
fraction of each gas and its molar mass.
(X_{CO} ´
28.01 g) + [(1  X_{CO}) ´ 44.01 g] = 35 g
Solving,
X_{CO} =
0.56 and

b.

Solving for
the pressures
P_{CO} =
X_{CO}P_{total} =
(0.56)(11 atm) = 6.2 atm



15.91

a.

The equation is:
fructose glucose
Initial
(M): 0.244 0
Change (M): 0.131
+0.131
Equilibrium
(M): 0.113 0.131
Calculating
the equilibrium constant,

b.




15.92

If you started with radioactive iodine in the solid phase, then you
should fine radioactive iodine in the vapor phase at equilibrium. Conversely, if you started with
radioactive iodine in the vapor phase, you should find radioactive iodine
in the solid phase. Both of these
observations indicate a dynamic equilibrium between solid and vapor phase.


15.93

a.

There is only one gas phase component, O_{2}. The equilibrium constant is simply

b.

From the ideal gas equation,
we can calculate the moles of O_{2} produced by the decomposition
of CuO.
From the
balanced equation,
(23%)

c.

If a 1.0 mol sample were used,
the pressure of oxygen would still be the same (0.49 atm) and it would be
due to the same quantity of O_{2}. Remember, a pure solid does not affect
the equilibrium position. The
moles of CuO lost would still be 3.7 ´
10^{2}
mol. Thus the fraction decomposed
would be:
(3.7%)

d.

If the number of moles of CuO were less than 0.037
mol, the equilibrium could not be established because the pressure of O_{2}
would be less than 0.49 atm.
Therefore, the smallest number of moles of CuO needed to establish
equilibrium must be slightly greater
than 0.037 mol.



15.94

If
there were 0.88 mole of CO_{2} initially and at equilibrium there
were 0.11 moles, then (0.88  0.11) moles = 0.77 moles reacted.
NO + CO_{2} NO_{2} + CO
Initial
(mol): 3.9
0.88 0 0
Change (mol): 0.77
0.77 +.077 +0.77
Equilibrium (mol): (3.9 
0.77) 0.11 0.77 0.77
Solving
for the equilibrium constant:
In
the balanced equation there are equal number of moles of products and
reactants; therefore, the volume of the container will not affect the
calculation of K_{c}. We can solve for the equilibrium constant
in terms of moles.


15.95

We first must find the initial concentrations of all the species in the
system.
Calculate the reaction quotient by substituting the
initial concentrations into the appropriate equation.
We find that Q_{c}
is less than K_{c}. The equilibrium will shift to the right,
decreasing the concentrations of H_{2} and I_{2} and
increasing the concentration of HI.
We set up the usual
table. Let x be the decrease in concentration of H_{2} and I_{2}.
H_{2} + I_{2} 2
HI
Initial
(M): 0.298 0.410 0.369
Equilibrium
(M): (0.298  x) (0.410 
x) (0.369 + 2x)
The equilibrium expression is:
This becomes the quadratic
equation
50.3x^{2} 
39.9x + 6.49 = 0
The smaller root is x = 0.228 M. (The larger root is
physically impossible.)
Having solved for x, calculate the equilibrium
concentrations.
[H_{2}] =
(0.298 
0.228) M = 0.07 M
[I_{2}] =
(0.410 
0.228) M = 0.18 M
[HI] =
[0.36 + 2(0.228)] M = 0.83 M


15.96

Since
we started with pure A, then any A that is lost forms equal amounts of B
and C. Since the total pressure is P, the pressure of B + C = P  0.14 P = 0.86 P. The pressure of B = C
= 0.43 P.


15.97

The gas cannot be (a) because the color became lighter
with heating. Heating (a) to 150°C
would produce some HBr, which is colorless and would lighten rather than
darken the gas.
The gas cannot be (b) because Br_{2} doesn't
dissociate into Br atoms at 150°C, so the color shouldn't
change.
The gas must be (c). N_{2}O_{4}(colorless) ® 2NO_{2}(brown) is consistent with
the observations. The reaction is
endothermic so heating darkens the color.
Above 150°C,
the NO_{2} breaks up into colorless NO and O_{2}:
2NO_{2}(g) ® 2NO(g)
+ O_{2}(g)
An increase in pressure shifts the equilibrium back to the left,
restoring the color by producing NO_{2}.


15.98

As ionic size
decreases, charge density increases.
Thus, Ba^{2+} has the lowest charge density and Mg^{2+}
has the highest charge density. The
higher the charge density, the greater the Coulombic attraction between
cation and anion.


15.99

Given the following:
a.

Temperature must have units
of Kelvin.
K_{P} =
K_{c}(0.0821 T)^{Dn}
K_{P} =
(1.2)(0.0821 ´ 648)^{(2}^{4)} =
4.2 ´ 10^{4}

b.

Recalling that,
Therefore,

c.

Since the equation
N_{2}(g) + H_{2}(g) NH_{3}(g)
is equivalent to
[N_{2}(g) + 3H_{2}(g) 2NH_{3}(g)]
then, for the
reaction:
N_{2}(g) + H_{2}(g) NH_{3}(g)
equals for the
reaction:
N_{2}(g) + 3H_{2}(g)
2NH_{3}(g)
Thus,

d.

For K_{P} in part (b):
K_{P} =
(0.83)(0.0821 ´ 648)^{+2} =
2.3 ´
10^{3}
and for K_{P} in part (c):
K_{P} =
(1.1)(0.0821 ´ 648)^{1} =
2.1 ´
10^{2}



15.100

a.

P_{NO} =
1.0 ´
10^{6}
atm

b.

P_{NO} =
2.6 ´
10^{16}
atm

c.

Since K_{P}
increases with temperature, it is endothermic.

d.

Lightening. The electrical energy promotes the
endothermic reaction.



15.101


15.102

The vapor pressure of
water is equivalent to saying the partial pressure of H_{2}O(g).


15.103

Potassium is more volatile than sodium. Therefore, its removal shifts the
equilibrium from left to right.


15.104

We can calculate the average molar mass of the gaseous
mixture from the density.
Let be the
average molar mass of NO_{2} and N_{2}O_{4}. The above equation becomes:
= 50.4 g/mol
The average molar mass is equal to the sum of the molar
masses of each component times the respective mole fractions. Setting this up, we can calculate the
mole fraction of each component.
We can now calculate the
partial pressure of NO_{2} from the mole fraction and the total
pressure.
We can calculate the partial
pressure of N_{2}O_{4} by difference.
Finally, we can calculate K_{P} for the dissociation
of N_{2}O_{4}.


15.105

In this problem, you are
asked to calculate K_{c}.
Step 1: Calculate the
initial concentration of NOCl. We
carry an extra significant figure throughout this calculation to minimize
rounding errors.
Step 2: Let's represent
the change in concentration of NOCl as 2x. Setting up a table:
2NOCl(g) 2NO(g)
+ Cl_{2}(g)
Initial
(M): 1.667 0 0
Equilibrium
(M): 1.667  2x 2x x
If 28.0 percent of the NOCl
has dissociated at equilibrium, the amount consumed is:
(0.280)(1.667
M) =
0.4668 M
In
the table above, we have represented the amount of NOCl that reacts as 2x.
Therefore,
2x
= 0.4668 M
x
= 0.2334 M
The equilibrium
concentrations of NOCl, NO, and Cl_{2} are:
[NOCl] =
(1.667 
2x)M = (1.667  0.4668)M
= 1.200 M
[NO] =
2x =
0.4668 M
[Cl_{2}] = x
= 0.2334 M
Step 3: The equilibrium constant K_{c} can be calculated by
substituting the above concentrations into the equilibrium expression.
or 3.5 ´
10^{2}


15.106

a.

Since both reactions are
endothermic (DH°
is positive), according to Le Châtelier’s principle the products would be favored at high
temperatures. Indeed, the
steamreforming process is carried out at very high temperatures (between
800°C
and 1000°C). It is interesting to note that in a
plant that uses natural gas (methane) for both hydrogen generation and
heating, about onethird of the gas is burned to maintain the high
temperatures.
In each reaction there are
more moles of products than reactants; therefore, we expect products to be favored at low pressures. In reality, the reactions are carried
out at high pressures. The
reason is that when the hydrogen gas produced is used captively (usually
in the synthesis of ammonia), high pressure leads to higher yields of
ammonia.

b.

(i) The relation between K_{c} and K_{P} is given by Equation
15.4 of the text:
K_{P} =
K_{c}(0.0821 T)^{Dn}
Since
Dn = 4 
2 = 2, we write:
K_{P} =
(18)(0.0821 ´ 1073)^{2} =
1.4 ´
10^{5}
(ii) Let x
be the amount of CH_{4} and H_{2}O (in atm) reacted. We write:
CH_{4} + H_{2}O CO + 3H_{2}
Initial (atm): 15 15 0 0
Change (atm): x x +x +3x
Equilibrium (atm): 15  x 15 
x x 3x
The equilibrium constant
is given by:
Taking the square root
of both sides, we obtain:
which can be expressed
as
5.2x^{2} + (3.7 ´
10^{2}x) 
(5.6 ´
10^{3}) = 0
Solving
the quadratic equation, we obtain
x
= 13 atm
(The other solution for x is negative and is physically
impossible.)
At equilibrium, the
pressures are:
P_{CO} =
13 atm



15.107

a.

Shifts to right.

b.

Shifts to right.

c.

No change.

d.

No change.

e.

No change.

f.

Shifts to left.



15.108

K_{P} =
(1.1)(1.1) = 1.2


15.109

a.

Assuming the selfionization of
water occurs by a single elementary step mechanism, the equilibrium
constant is just the ratio of the forward and reverse rate constants.

b.

The product can be written as:
[H^{+}][OH^{}] =
K[H_{2}O]
What is [H_{2}O]? It is the concentration of pure
water. One liter of water has a
mass of 1000 g
(density = 1.00 g/mL). The number
of moles of H_{2}O is:
The concentration of water is
55.5 mol/1.00 L or 55.5 M. The product is:
[H^{+}][OH^{}] =
(1.8 ´
10^{16})(55.5) =
1.0 ´
10^{14}
We assume
the concentration of hydrogen ion and hydroxide ion are equal.
[H^{+}][OH^{}] =
(1.0 ´
10^{14})^{1/2} =
1.0 ´
10^{7}
M



15.110

At equilibrium, the value of K_{c} is equal to the ratio of the forward rate
constant to the rate constant for the reverse reaction.
k_{f} =
(12.6)(5.1 ´ 10^{2}) =
0.64
The forward reaction is third order, so the units of k_{f} must be:
rate = k_{f}[A]^{2}[B]
k_{f} =
0.64 /M^{2}×s


15.111

The equilibrium is: N_{2}O_{4}(g)
2NO_{2}(g)
Volume
is doubled so pressure is halved.
Let’s calculate Q_{P}
and compare it to K_{P}.
Equilibrium will shift
to the right. Some N_{2}O_{4}
will react, and some NO_{2} will be formed. Let
x = amount of N_{2}O_{4}
reacted.
N_{2}O_{4}(g)
2NO_{2}(g)
Initial
(atm): 0.10 0.075
Equilibrium
(atm): 0.10 
x 0.075 + 2x
Substitute
into the K_{P} expression
to solve for x.
4x^{2} + 0.413x  5.68 ´
10^{3} =
0
x
= 0.0123
At equilibrium:
Think About It:

; close enough to 0.113



15.112

a.

Combine Ni with CO above 50°C. Pump away the Ni(CO)_{4} vapor
(shift equilibrium to right), leaving the solid impurities behind.

b.

Consider the reverse
reaction:
Ni(CO)_{4}(g) ® Ni(s)
+ 4CO(g)
DH° =
(4)(110.5
kJ/mol) 
(1)(602.9
kJ/mol) = 160.9 kJ/mol
The decomposition is endothermic, which is favored at high
temperatures. Heat Ni(CO)_{4}
above 200°C
to convert it back to Ni.



15.113

a.

Molar mass of PCl_{5} =
208.2 g/mol

b.

PCl_{5} PCl_{3} + Cl_{2}
Initial (atm) 1.03 0 0
Equilibrium (atm) 1.03  x x
x
x^{2} + 1.05x
 1.08
= 0
x
= 0.639
At equilibrium:

c.

P_{T} =
(1.03 
x) + x + x = 1.03 +
0.639 = 1.67 atm

d.

(62.0%)



15.114


15.115

a.

K_{P} = P_{Hg} = 0.0020 mmHg = 2.6 ´ 10^{6} atm = 2.6 ´
10^{6} (equil. constants are expressed without
units)

b.

Volume
of lab = (6.1 m)(5.3 m)(3.1 m) =
100 m^{3}
[Hg] =
K_{c}
_{ }
The concentration of mercury
vapor in the room is:
Yes. This concentration exceeds the safety
limit of 0.05 mg/m^{3}.



15.116

Initially,
at equilibrium: [NO_{2}] =
0.0475 M and [N_{2}O_{4}]
= 0.487 M. At the instant the volume is halved, the
concentrations double.
[NO_{2}] = 2(0.0475 M) = 0.0950 M and [N_{2}O_{4}] = 2(0.487 M) = 0.974 M. The system is no
longer at equilibrium. The system
will shift to the left to offset the increase in pressure when the volume
is halved. When a new equilibrium
position is established, we write:
N_{2}O_{4} 2
NO_{2}
0.974
M + x 0.0950 M – 2x
4x^{2}
– 0.3846x + 4.52 ´
10^{3} =
0
Solving
x =
0.0824 M (impossible) and
x =
0.0137 M
At the new equilibrium,
[N_{2}O_{4}] =
0.974 + 0.0137 = 0.988
M
[NO_{2}]
= 0.0950 – (2 ´
0.0137) = 0.0676
M
As we can see, the new
equilibrium concentration of NO_{2} is greater than the initial equilibrium concentration (0.0475 M).
Therefore, the gases should look darker!


15.117

There is a temporary dynamic equilibrium between the melting ice
cubes and the freezing of water between the ice cubes.


15.118

a.

A catalyst
speeds up the rates of the forward and reverse reactions to the same
extent.

b.

A catalyst would not change the energies of the reactant and
product.

c.

The first reaction is exothermic.
Raising the temperature would favor the reverse reaction,
increasing the amount of reactant and decreasing the amount of product at
equilibrium. The equilibrium
constant, K, would decrease.
The second reaction is endothermic.
Raising the temperature would favor the forward reaction,
increasing the amount of product and decreasing the amount of reactant at
equilibrium. The equilibrium
constant, K, would increase.

d.

A catalyst lowers the activation energy for the forward and
reverse reactions to the same extent. Adding a catalyst to a reaction
mixture will simply cause the mixture to reach equilibrium sooner. The same equilibrium mixture could be
obtained without the catalyst, but we might have to wait longer for
equilibrium to be reached. If the
same equilibrium position is reached, with or without a catalyst, then
the equilibrium constant is the same.



15.119

First, let's calculate the initial concentration of ammonia.
Let's set up a table to
represent the equilibrium concentrations.
We represent the amount of NH_{3} that reacts as 2x.
2NH_{3}(g) N_{2}(g) + 3H_{2}(g)
Initial
(M):
0.214 0 0
Equilibrium
(M): 0.214  2x x 3x
Substitute
into the equilibrium expression to solve for x.
Taking
the square root of both sides of the equation gives:
Rearranging,
5.20x^{2} + 1.82x  0.195 =
0
Solving
the quadratic equation gives the solutions:
x = 0.086 M and x = 0.44
M
The
positive root is the correct answer.
The equilibrium concentrations are:
[NH_{3}] =
0.214 
2(0.086) = 0.042
M
[N_{2}] = 0.086 M
[H_{2}] =
3(0.086) = 0.26
M


15.120

To determine DH°, we need to plot ln K_{P} versus 1/T
(y vs. x).
ln K_{P}

1/T

4.93

0.00167

1.63

0.00143

–0.83

0.00125

–2.77

0.00111

–4.34

0.00100

The
slope of the plot equals DH°/R.
DH° = 1.15
× 10^{5} J/mol = 115 kJ/mol


15.121

a.

From the balanced equation
N_{2}O_{4} 2NO_{2}
Initial
(mol): 1 0
Equilibrium (mol): (1 
x) 2x
The total moles in
the system = (moles N_{2}O_{4} + moles NO_{2}) =
[(1 
x) + 2x] = 1 + x. If the total pressure in the system is P, then:
and
K_{P} =

b.

Rearranging the K_{P} expression:
4x^{2}P = K_{P}

x^{2}K_{P}
x^{2}(4P + K_{P}) = K_{P}
K_{P} is a constant (at
constant temperature). If P increases, the fraction (and therefore x) must decrease.
Equilibrium shifts to the left to produce less NO_{2} and
more N_{2}O_{4} as predicted.



15.122

a.

We start by writing the van’t Hoff equation at two different
temperatures.
Assuming an endothermic reaction, DH°
> 0 and T_{2} > T_{1}. Then, < 0,
meaning that < 0 or
K_{1} < K_{2}. A larger K_{2} indicates that there are more products at
equilibrium as the temperature is raised.
This agrees with LeChatelier’s principle that an increase in
temperature favors the forward endothermic reaction. The opposite of the above discussion
holds for an exothermic reaction.

b.

Treating
H_{2}O(l)
H_{2}O(g) DH_{vap} = ?
as a heterogeneous equilibrium, .
We substitute into the equation derived in part (a) to solve for DH_{vap}.
1.067 =
DH°(2.466
× 10^{5})
DH° = 4.34
× 10^{4} J/mol = 434
kJ/mol



15.123

Initially, the pressure of SO_{2}Cl_{2} is 9.00 atm.
The pressure is held constant, so after the reaction reaches
equilibrium, . The
amount (pressure) of SO_{2}Cl_{2} reacted must equal the
pressure of SO_{2} and Cl_{2} produced for the pressure to
remain constant. If we let , then the pressure of SO_{2}Cl_{2}
reacted must be 2x. We set up a table showing the initial
pressures, the change in pressures, and the equilibrium pressures.
SO_{2}Cl_{2}(g) SO_{2}(g) + Cl_{2}(g)
Initial
(atm): 9.00 0 0
Equilibrium
(atm): 9.00  2x x x
Again, note that the change in pressure for SO_{2}Cl_{2}
(2x) does not match the
stoichiometry of the reaction, because we are expressing changes in
pressure. The total pressure is kept
at 9.00 atm throughout.
x^{2} + 4.10x  18.45 = 0
Solving the quadratic equation, x = 2.71 atm. At equilibrium,


15.124

Using Equation 14.8 of the text, we can calculate k_{1}.
Then, we can calculate k_{1}
using the expression
(see Section 15.2 of the text)
k_{1} = 6.5 × 10^{4} s^{1}
k_{1} = 6.4 × 10^{8} s^{1}


15.125

We start with a table.
A_{2} + B_{2} 2AB
Initial
(mol): 1 3 0
Change (mol): +x
Equilibrium (mol): x
After
the addition of 2 moles of A,
A_{2} + B_{2} 2AB
Initial
(mol): x
Change (mol): +x
Equilibrium (mol): 3  x 3  x 2x
We write two different
equilibrium constants expressions for the two tables.
We equate the equilibrium expressions
and solve for x.
6x + 9 = 8x + 12
x
= 1.5
We substitute x back into one of the equilibrium
expressions to solve for K.
Substitute x
into the other equilibrium expression to see if you obtain the same value
for K. Note that we used moles rather than
molarity for the concentrations, because the volume, V, cancels in the equilibrium expressions.


15.126

a.

First, we calculate the moles
of I_{2}.
Let x be the number of
moles of I_{2} that dissolves in CCl_{4}, so (1.26 × 10^{4}  x)mol remains
dissolved in water. We set up
expressions for the concentrations of I_{2} in CCl_{4}
and H_{2}O.
Next, we substitute these concentrations
into the equilibrium expression and solve for x.
83(1.26 × 10^{4}  x) =
6.67x
x
= 1.166 × 10^{4}
The fraction of I_{2} remaining in the aqueous phase is:

b.

The first extraction leaves only 7% I_{2} in the
water. The next extraction with
0.030 L of CCl_{4} will leave only (0.07)(0.07) = 5 × 10^{3}. This is the fraction remaining after
the second extraction which is only 0.5%.

c.

A single
extraction using 0.060 L of CCl_{4} gives:
The concentration of I_{2} is the same as in part
(a).
Substituting these concentrations into the
equilibrium expression and solving for x gives:
83(1.26 × 10^{4}  x) =
3.33x
x
= 1.211 × 10^{4}
The fraction of I_{2} remaining in the aqueous phase is:



15.127

Strategy:

The concentrations of solids and pure liquids do not appear in K_{C}. So, even though the
given equilibrium is heterogeneous, Equation 15.4 still applies.

Solution:

There are 3 moles of gas molecules on the product side and 6
moles on the reactant side. So,
Dn = 3 – 6 = –3.



15.128

According
to Le Châtelier’s principle, an exothermic equilibrium can be shifted to
the right by lowering the temperature. Also, raising the pressure will
favor the production of product since the equilibrium will respond to this
stress by minimizing the number of gas particles. From the standpoint of optimizing the yield, temperature should be
low and pressure should be high. To optimize the rate of production,
both the forward and reverse reactions should be very rapid to ensure that
equilibrium is established quickly. From
a kinetic standpoint, temperature and pressure (concentration) should both
be high.


15.129

Strategy:

According to Equation
15.4, K_{P} and K_{C} are the same when Dn = 0:
So, look for reactions that show no change in the number of gas
molecules as the reaction occurs. Also, note that Equation 15.4 does not
apply to heterogeneous
equilibria if liquid phases are solutions.

Solution:

a.

K_{P} ¹ K_{C}
(Dn ¹ 0)

b.

K_{P} ¹ K_{C}
(Dn ¹ 0)

c.

Equation 15.4 is not applicable (some species are aqueous).

d.

K_{P} = K_{C} (Dn = 0)

e.

K_{P} ¹ K_{C}
(Dn ¹ 0)

f.

Equation 15.4 is not
applicable (some species
are aqueous).

g.

K_{P} = K_{C} (Dn = 0)

h.

K_{P} ¹ K_{C}
(Dn ¹ 0)




15.130

a.

Strategy:

Use the law
of mass action to write the equilibrium expression.

Setup:

The
equilibrium expression has the form of the concentrations of products
over the concentration of reactants, each raised to a power equal to
its stoichiometric coefficient in the balanced chemical equation.

Solution:



b.

Strategy:

To evaluate K_{c}, plug in the
equilibrium concentrations of all three species into the equilibrium
expression derived in part ‘a’.

Solution:



c.

Strategy:

Begin by
writing the equilibrium expression for the precipitation reaction. Take the sum of the two reactions and
write the net reaction. The
overall equilibrium constant is the product of the individual
constants.

Setup:

The equilibrium expression for the aqueous reaction
was calculated in part (a):
The equilibrium expression for the
precipitation reaction is
Taking the sum of the two reactions gives:
OD1A(aq) + OF2A(aq) OD17X(aq)
OD17X(aq) + A771A(aq) OD17XA77(s)
OD1A(aq) + OF2A(aq) +
A771A(aq) OD17XA11(s)

Solution:



d.

Strategy:

Use the
concentrations provided to calculate Q_{c}, and then compare Q_{c} with K_{c}.

Setup:


Solution:

The
calculated value of Q_{c}
is greater than K_{c}. Therefore,
the reaction is not at equilibrium and must proceed to the left to
establish equilibrium.


e.

Strategy:

Construct an
equilibrium table to determine the equilibrium concentration of each
species in terms of an unknown (x);
solve for x, and use it to
calculate the equilibrium molar concentrations.

Setup:

Insert the
starting concentrations that we know into the equilibrium table:
OD1A + OF2A OD17X
Initial (M): 1.0 1.0 0
Equilibrium (M):

Solution:

We define
the change in concentration of one of the reactants as x. Because there is no product at the
start of the reaction, the reactant concentration must decrease; that
is, this reaction must proceed in the forward direction to reach
equilibrium. According to the
stoichiometry of the chemical reaction, the reactant concentrations
will both decrease by the same amount (x), and the product concentration will increase by that
amount (x). Combining the initial concentration
and the change in concentration for each species, we get expressions
(in terms of x) for the
equilibrium concentrations:
OD1A + OF2A OD17X
Initial (M): 1.0 1.0 0
Equilibrium (M): (1.0  x) (1.0  x) x
Next,
we insert these expressions for the equilibrium concentrations into the
equilibrium expression and solve for x.
Collecting
terms we get
This is a
quadratic equation of the form ax^{2}
+ bx + c = 0. The solution
for the quadratic equation (see Appendix 1) is
Here
we have a = 3.8 × 10^{2},
b = –7.61 × 10^{2}, c = 3.8 × 10^{2}, so
x = 1.05 or x = 0.95
Only the
second of these values, 0.95, makes sense because concentration cannot
be a negative number (1 – 1.05 = –0.5).
Using the calculated values of x, we can determine the equilibrium concentration of each
species as follows:
[OD17X] = 0.95 M
[OD1A] = [OF2A] = (1 – 0.95) = 0.05 M



