CHAPTER 17
Equilibrium
1.
Answer depends on student choice of reaction.
2.
Four C–H bonds in the CH4 molecule and four Cl–Cl bonds in the four Cl2 molecules must be
broken. Four C–Cl bonds in CCl4 and four H–Cl bonds in the four HCl molecules must form.
3.
The collision model shows chemical reactions as taking place only when the reactant molecules p
hysically collide with one another, with enough energy to break bonds in the reactant molecules.
Not all collisions possess enough energy to break bonds in the reactant molecules. A minimum en
ergy, the activation energy (Ea), is needed for a collision to result in reaction. If molecules do not
possess this minimum energy when they collide, they bounce off one another without reacting. A
simple reaction is illustrated in Figure 17.2.
4.
The symbol Ea represents the activation energy of the reaction. The activation energy is the mini
mum energy two colliding molecules must possess in order for the collision to result in reaction. I
f molecules do not possess energies equal to or greater than Ea, a collision between these molecul
es will not result in a reaction.
5.
A catalyst provides an alternative pathway, with a lower activation energy barrier, by which the r
eaction can take place.
6.
Enzymes are biochemical catalysts that accelerate the complicated biochemical reactions in cells t
hat would ordinarily be too slow to sustain life at normal body temperatures.
7.
In an equilibrium system, two opposing processes are going on at the same time and speed. Ther
e is no net change in a system at equilibrium with time. Every time one process occurs, the opposi
te process occurs at the same time elsewhere in the system. A simple equilibrium might exist for t
he populations of two similar size towns connected by highway. Assuming there is no great attra
ction in one town compared to the other, we might assume the populations of the two towns woul
d remain constant with time as individual people drive between them, but in such a way that new
people are arriving in the first town from the second town as residents of the first town leave for t
he second town.
8.
A state of equilibrium is attained when two opposing processes are exactly balanced so there is n
o further observable net change in the system.
9.
We use a double-ended arrow (pointing both to the left and to the right) to indicate that the reacti
on is reversible and comes to equilibrium:
10.
Chemical equilibrium occurs when two opposing chemical reactions reach the same speed in a cl
osed system. When a state of chemical equilibrium has been reached, the concentrations of reacta
nts and products present in the system remain constant with time, and the reaction appears to “sto
p.” A chemical reaction that reaches a state of equilibrium is indicated by using a double arrow (
). The points of the double arrow point in opposite directions, to indicate that two opposite pr
ocesses are going on.
379
Chapter 17: Equilibrium
11.
The word dynamic is used to describe physical and chemical states of equilibrium to emphasize th
at the system has not “stopped”, but rather has reached the point where one process is exactly bala
nced by the opposing process. Once a chemical system has reached equilibrium the net concentrat
ion of product no longer increases because molecules of product already present react to form the
original reactants.
12.
a.
The green line is H2 because hydrogen is initially present in the greatest concentration. T
he blue line is N2 because some nitrogen is initially present but not as much as the H2 (a t
hird of the amount). The pink line is NH3 because at first no product is present but then
N2 and H2 react to form NH3.
b.
The concentrations of both N2 and H2 decrease at first because they react to form NH3
(which then causes the concentration of NH3 to go up). None of the concentrations
become zero over time because eventually some of the NH3 shifts back to form N2 and H2
again. Eventually the concentration of each species remains constant because the rate of
the forward reaction equals the rate of the backward reaction (equilibrium is reached).
c.
Equilibrium is reached when the lines become straight (the concentration
over time does not change). As stated in b, the rate of the forward reaction equals the rate
of the backward reaction.
13.
The equilibrium constant represents a ratio of the concentration of products present at the point of
equilibrium to the concentration of reactants present, with the concentration of each species raised
to the power of its coefficient in the balanced chemical equation for the reaction. For a general rea
ction
aA + bB
cC + dD
the equilibrium constant, K, has the algebraic form
[C] [D]
[A] [B]
c
d
a
b
K
The square brackets indicate the concentrations of the substances in moles per liter (molarity, M).
14.
The equilibrium constant is a ratio of concentration of products to concentration of reactants, wit
h all concentrations measured at equilibrium. Depending on the amount of reactant present at the
beginning of an experiment, there may be different absolute amounts of reactants and products pr
esent at equilibrium, but the ratio will always be the same for a given reaction at a given temperat
ure. For example, the ratios (4/2) and (6/3) are different absolutely in terms of the numbers involv
ed, but each of these ratios has the value of 2.
15.
a.
K =
2
5
2
6
2
[C H Cl][HCl]
[C H ][Cl ]
b.
K =
4
6
2
4
5
3
2
[NO] [H O]
[NH ] [O ]
c.
K =
3
2
5
[PCl ][Cl ]
[PCl ]
380
Chapter 17: Equilibrium
16.
a.
K =
b.
K =
c.
K =
17.
a.
K =
2
2
[ClNO ( )][NO( )]
[NO ( )][ClNO( )]
g
g
g
g
b.
K =
2
5
5
2
2
[BrF ( )]
[Br ( )][F ( )]
g
g
g
c.
K =
5
6
2
2
4
6
3
[N ( )] [H O]
[NH ( )] [NO( )]
g
g
g
18.
a.
K =
3
2
2
[CH OH]
[CO][H ]
b.
K=
2
2
2
2
[NO] [O ]
[NO ]
c.
K =
4
3
6
4
2
[PBr ]
[P ][Br ]
19.
PCl5(g)
PCl3(g) + Cl2(g)
K =
3
2
5
[PCl ][Cl ]
[PCl ]
=
[0.0302 ][0.0491 ]
[0.0711 ]
M
M
M
= 2.09 × 10–2
20.
N2(g) + 3H2(g)
2NH3(g)
21.
N2(g) + O2(g)
2NO(g)
2
4
2
-4
2
2
[NO]
[4.7 10 ]
6.9 10
[N ][O ] [0.041 ][0.0078 ]
M
K
M
M
22.
2N2O(g) + O2(g)
4NO(g)
K =
4
4
6
2
2
2
2
[NO]
[0.00341 ]
=
= 4.85 10
[N O] [O ]
[0.0293 ] [0.0325 ]
M
M
M
381
Chapter 17: Equilibrium
23.
A homogeneous equilibrium system is one which all the substances present are in the same physic
al state. An example is
N2(g) + O2(g)
2NO(g).
Heterogeneous equilibrium systems are those involving substances in more than one physical
state (e.g., mixtures of liquids and gases, gases and solids, etc.). Examples are:
BaCO3(s)
BaO(s) + CO2(g)
NH3(g) + HCl(g)
NH4Cl(s).
24.
Equilibrium constants represent ratios of the concentrations of products and reactants present at t
he point of equilibrium. The concentration of a pure solid or of a pure liquid is constant and is det
ermined by the density of the solid or liquid. For example, suppose you had a liter of water. Withi
n that liter of water are 55.5 mol of water (the number of moles of water that is contained in one li
ter of water does not vary).
25.
a.
K =
4
3
6
2
[PF ]
[F ]
b.
K =
2
2
1
[Xe][F ]
c.
K =
2
4
2
[O ]
[Cl ]
26.
a.
K = [H2O(g)][CO2(g)]
b.
K = [CO2(g)]
c.
K =
27.
a.
K =
2
2
[H ][CO]
[H O]
b.
K = [H2O(g)]
c.
K =
3
2
1
[O ]
28.
a.
K = [N2(g)][Br2(g)]3
b.
K =
c.
K =
3
2
1
[O ( )]
g
29.
Le Châtelier’s principle states that when a change is imposed on a system at equilibrium, the posit
ion of the equilibrium shifts in a direction that tends to reduce the effect of the change.
382
Chapter 17: Equilibrium
30.
When an additional amount of one of the reactants is added to an equilibrium system, the system
shifts to the right and adjusts so as to consume some of the added reactant. This results in a net in
crease in the amount of product, compared to the equilibrium system before the additional reacta
nt was added, and so the amount of CO2(g) in the system will be higher than if the additional CO
(g) had not been added. The numerical value of the equilibrium constant does not change when a
reactant is added: the concentrations of all reactants and products adjust until the correct value of
K is once again achieved.
31.
When the volume of an equilibrium system involving gaseous substances is decreased suddenly, t
he pressure in the container increases. Reaction will occur, shifting in the direction that gives the
smaller number of gas molecules, to reduce this increase in pressure.
32.
If heat is applied to an endothermic reaction (i.e., the temperature is raised), the equilibrium is shi
fted to the right. More product will be present at equilibrium than if the temperature had not been
increased. The value of K increases.
33.
a.
shifts left (system reacts to the left to get rid of excess H2)
b.
shifts right (system reacts to the right to replace carbon monoxide as it is removed)
c.
no effect (carbon is in the solid state)
34.
a.
shifts right (system reacts to the right to get rid of excess fluorine)
b.
no change (P is in the solid state)
c.
shifts left (system reacts to get rid of excess PF3)
35.
a.
no change (water is in the liquid state)
b.
Assuming the system is warm enough to convert the dry ice to the gaseous state, the
equilibrium will shift to the left.
c.
shifts to right
d.
shifts to right
36.
a.
no change (B is solid)
b.
shift right (system reacts to replace removed C)
c.
shift left (system reacts by shifting in direction of fewer mol of gas)
d.
shift right (the reaction is endothermic as written)
37.
As the reaction is exothermic, heat is effectively a product. An increase in temperature would basi
cally fight against the forward reaction as it tries to release energy, and would disfavor the produc
tion of hydrogen chloride.
38.
The answer is d. When hydrogen gas is added, equilibrium will shift away from the addition of re
actant and toward the product side, producing more water vapor. The value of K does not change.
39.
As the reaction is exothermic, heat is effectively a product of the reaction as written. Raising the t
emperature of the system (adding heat to the system) tends to disfavor the formation of products,
and shifts the equilibrium to the left (toward reactants).
383
Chapter 17: Equilibrium
40.
For an endothermic reaction, a decrease in temperature will shift the position of equilibrium to th
e left (toward the reactants).
41.
The production of dextrose will be favored.
42.
CO(g) + 2H2(g)
CH3OH(l)
add additional CO(g) or H2(g): the system will react in the forward direction to remove the excess
decrease the volume of the system: the system will react in the direction of fewer moles of gas
decrease the temperature: removal of heat favors products in an exothermic reaction
43.
A large equilibrium constant means that the concentration of products is large compared to the co
ncentration of remaining reactants. The position of the equilibrium lies far to the right. Reactions
with numerically large equilibrium constants are greatly favored as a source of product. When w
e calculated the theoretical yield for a reaction in earlier chapters, we tacitly assumed that the reac
tion had a large equilibrium constant.
44.
A small equilibrium constant implies that not much product forms before equilibrium is reached.
The reaction would not be a good source of the products unless Le Châtelier’s principle can be us
ed to force the reaction to the right.
45.
K =
2
9
2
11
5
5
2
2
5
2
2
[BrF ]
[1.01 10
]
1.18 10
[Br ][F ]
[2.41 10
][8.15 10 ]
M
M
M
46.
K =
5
7
–7
3
2
3
2
2
[SO ][NO] [4.99 10
][6.31 10
]
8.63 10
[SO ][NO ] [2.11 10
][1.73 10
]
M
M
M
M
47.
K =
2
1
2
7
2
2
4
2
3
2
[CO ]
[1.1 10
]
8.7 10
[CO] [O ] [2.7 10
] [1.9 10
]
M
M
M
48.
K = 5.21 × 10–3 =
3
3
2
2
2
2
2
[CO][H O] [4.73 10
][5.21 10
]
[CO ][H ]
[3.99 10
][H ]
M
M
M
[H2] = 0.119 M
49.
K = 2.1 × 103 =
2
2
2
2
[HF]
[HF]
[H ][F ] [0.0021 ][0.0021 ]
M
M
[HF]2 = 9.26 10–3
[HF] = 9.6 10–2 M
50.
2
2 2
3
2
2
2
2
1 2
2
[H ] [O ] [1.9 10 ] [O ]
2.4 10
[H O]
[1.1 10 ]
K
[O2] = 8.0 10–2 M
384
Chapter 17: Equilibrium
51.
a.
The extent of conversion of O2 to O3 is very small.
b.
K =
2
54
3
3
2
[O ]
1.12 10
[O ]
=
2
3
2
3
[O ]
[3.04 10 ]
M
[O3] = 5.61 × 10–30 M
52.
K = 8.1 × 10–3 =
2
2
2
2
4
2
4
[NO ]
[0.0021 ]
=
[N O ]
[N O ]
M
[N2O4] = 5.4 10–4 M
53.
When a crystal of ionic solute M+X– is placed in water, initially the crystal simply dissolves produ
cing M+(aq) ions and X–(aq) ions. As the concentration of the ions in solution increases, however,
the likelihood of oppositely charged ions colliding and reforming the solid increases. Eventually,
an equilibrium is reached when dissolving and reforming of the solid occur at the same speed. Pas
t this point in time, there is no further net increase in the concentration of the dissolved ions.
54.
solubility product, Ksp
55.
The equilibrium represents the balancing of the dynamic processes of dissolving and of reformati
on of the solid. The amount of excess solid added in preparing a solution might affect the speed at
which the point of equilibrium is reached, but it will not affect the net amount of solute present in
solution at the point of equilibrium. A given amount of solvent can only “hold” a certain amount
of solute.
56.
Stirring or grinding the solute increases the speed with which the solute dissolves, but the ultimat
e amount of solute that dissolves is fixed by the equilibrium constant for the dissolving process, K
sp, which changes only with temperature. Therefore only the temperature will affect the solubility.
57.
a.
AgIO3(s)
Ag+(aq) + IO3–(aq)
Ksp = [Ag+(aq)][IO3–(aq)]
b.
Sn(OH)2(s)
Sn2+(aq) + 2OH–(aq)
Ksp = [Sn2+(aq)][OH–(aq)]2
c.
Zn3(PO4)2(s)
3Zn2+(aq) + 2PO43–(aq)
Ksp = [Zn2+(aq)]3[PO43–(aq)]2
d.
BaF2(s)
Ba2+(aq) + 2F–(aq)
Ksp = [Ba2+(aq)][F–(aq)]2
58.
a.
NiS(s)
Ni2+(aq) + S2–(aq)
Ksp = [Ni2+(aq)][S2–(aq)]
b.
CuCO3(s)
Cu2+(aq) + CO32–(aq)
Ksp = [Cu2+(aq)][CO32–(aq)]
c.
BaCrO4(s)
Ba2+(aq) + CrO42–(aq)
Ksp = [Ba2+(aq)][CrO42–(aq)]
d.
Ag3PO4(s)
3Ag+(aq) + PO43–(aq)
Ksp = [Ag+(aq)]3[PO43–(aq)]
59.
Cu(OH)2(s)
Cu2+(aq) + 2OH–(aq)
molar mass Cu(OH)2 = 97.57 g
Let x represent the solubility of Cu(OH)2 in mol/L. Then [Cu2+] = x and [OH–] = 2x from the
stoichiometry of the equation
Ksp = [Cu2+][OH–]2 = 2.2 10–20 = [x][2x]2 = 4x3
385
Chapter 17: Equilibrium
x = the molar solubility of Cu(OH)2 = 1.77 × 10–7 M (1.8 × 10–7 M)
1.77 × 10–7
mol
L
× 97.57
g
mol
= 1.7 × 10–5 g/L
60.
MgCO3(s)
Mg2+(aq) + CO32–(aq)
Molar mass MgCO3 = 84.32 g
Let x represent the solubility of MgCO3 in mol/L. Then [CO32–] = x and [Mg2+] = x from the
stoichiometry of the equation.
Ksp = [Mg2+][CO32–] = 3.5 × 10–8 = (x)(x) = x2
then the molar solubility of MgCO3 = x = 1.87 × 10–4 M (1.9 × 10–4 M)
1.87 × 10–4
mol
84.32 g
×
0.016g/L
L
1 mol
61.
NiS(s)
Ni2+(aq) + S2–(aq)
molar mass NiS = 90.77 g
-4
-6
g NiS
1moL NiS
3.6×10
×
= 3.97 × 10
L
90.77 g NiS
M
Ksp = [Ni2+][S2–] = (3.97 10–6 M)(3.97 10–6 M) = 1.6 10–11
62.
Ni(OH)2(s)
Ni2+(aq) + 2OH–(aq)
molar mass Ni(OH)2 = 92.71 g
let x represent the molar solubility of Ni(OH)2: then [Ni2+] = x and [OH–] = 2x.
Ksp = [Ni2+][OH–]2 = 2.0 × 10–15 = [x][2x]2 = 4x3
then the molar solubility of Ni(OH)2 = x = 7.9 × 10–6 M
gram solubility =
6
4
mol
92.71 g
7.98×10
×
= 7.4 ×10 g/L
L
1 mol
63.
CaCO3(s)
Ca2+(aq) + CO32–(aq)
Ksp = [Ca2+][CO32–] = 3.0 10–9
Let x represent the number of moles of CaCO3(s) that dissolve per liter, then [Ca2+] = x and
[CO32–] = x also from the stoichiometry of the reaction;
Ksp = [x][x] = x2 = 3.0 10–9
then x = [CaCO3] = 5.5 10–5 M (5.5 10–3 g/L)
64.
CaSO4(s)
Ca2+(aq) + SO42–(aq)
molar mass CaSO4 = 136.15 g
2.05
g
1 mol
L
136.15 g
= 1.506 × 10–2 M
386
Chapter 17: Equilibrium
If CaSO4 dissolves to the extent of 1.506 × 10–2 M, then [Ca2+] will be 1.506 × 10–2 M and [SO42–]
will be 1.506 × 10–2 M also.
Ksp = [Ca2+][SO42–] = [1.506 × 10–2 M][ 1.506 × 10–2 M] = 2.27 × 10–4
65.
Fe(OH)2(s)
= Fe2+(aq) + 2OH–(aq)
molar mass Fe(OH)2 = 89.87 g
1.5 10–3 g/L
2
1mol Fe(OH)
89.87g
= 1.67 10–5 mol Fe(OH)2/L
Ksp = [Fe2+][OH–]2
If 1.67 10–5 M of Fe(OH)2 dissolves then
[Fe2+] = 1.67 10–5 M and [OH–] = 2 (1.67 10–5 M) = 3.34 10–5 M
Ksp = (1.67 10–5 M)(3.34 10–5 M)2 = 1.9 10–14
66.
Cr(OH)3(s)
Cr3+(aq) + 3OH–(aq)
If Cr(OH)3 dissolves to the extent of 8.21 × 10–5 M, then [Cr3+] will be 8.21 × 10–5 M and [OH–]
will be 3(8.21 × 10–5 M) in a saturated solution.
Ksp = [Cr3+][OH–]3 = [8.21 × 10–5 M][2.46 × 10–4 M]3 = 1.23 × 10–15
67.
MgF2(s)
Mg2+(aq) + 2F–(aq)
molar mass MgF2 = 62.31 g
8.0 10–2 g MgF2/L
2
2
1 mol MgF
62.31 g MgF
= 1.28 10–3 M = 1.3 10–3 M
Ksp = [Mg2+][F–]2
If 1.28 10–3 M of MgF2 dissolves, then [Mg2+] = 1.28 10–3 M and [F–] = 2 1.28 10–3 M =
2.56 10–3 M.
Ksp = (1.28 10–3 M)(2.56 10–3 M)2 = 8.4 10–9
68.
PbCl2(s)
Pb2+(aq) + 2Cl–(aq)
Ksp = [Pb2+][Cl–]2
If PbCl2 dissolves to the extent of 3.6 10–2 M, then [Pb2+] = 3.6 10–2 M and [Cl–] = 2 (3.6
10–2) = 7.2 10–2 M.
Ksp = (3.6 10–2 M)(7.2 10–2 M)2 = 1.9 10–4
molar mass PbCl2 = 278.1 g
2
3.6×10 mol
278.1g
×
1L
1mol
= 10. g/L
69.
Hg2Cl2(s)
Hg22+(aq) + 2Cl–(aq)
Ksp = [Hg22+][Cl–]2
387
Chapter 17: Equilibrium
Let x represent the number of moles of Hg2Cl2 that dissolve per liter; then [Hg22+] = x and [Cl–] =
2x.
Ksp = [x][2x]2 = 4x3 = 1.3 10–18
x3 = 3.25 10–19
x = [Hg22+] = 6.9 10–7 M
70.
Fe(OH)3(s)
Fe3+(aq) + 3OH–(aq)
Ksp = [Fe3+][OH–]3 = 4 10–38
Let x represent the number of moles of Fe(OH)3 that dissolve per liter; then [Fe3+] = x.
The amount of hydroxide ion that would be produced by the dissolving of the Fe(OH)3 would
then be 3x, but pure water itself contains hydroxide ion at the concentration of 1.0 10–7 M (see
Chapter 17). The total concentration of hydroxide ion is then [OH–] = (3x + 1.0 10–7). As x
must be a very small number [because Fe(OH)3 is not very soluble], we can save ourselves a lot
of arithmetic if we use the approximation that
(3x + 1.0 10–7 M) = 1.0 10–7
Ksp = [x][1.0 10–7]3 = 4 10–38
x = 4 10–17 M
molar mass Fe(OH)3 = 106.9 g
17
4 10 mol 106.9 g
1 L
1 mol
= 4 10–15 g/L
71.
Collision between molecules is not the only prerequisite for a reaction. The molecules must also p
ossess sufficient energy to react with each other, and must have the proper spatial orientation for r
eaction.
72.
An increase in temperature increases the fraction of molecules that possess sufficient energy for a
collision to result in a reaction.
73.
activation
74.
catalyst
75.
balancing
76.
constant
77.
To say a reaction is reversible means that, to one extent or another, the reaction may occur in eith
er direction.
78.
When we say that a chemical equilibrium is dynamic, we are recognizing the fact that even thoug
h the reaction has appeared macroscopically to have stopped, on a microscopic basis the forward
and reverse reactions are still taking place, at the same speed.
79.
equals
80.
heterogeneous
388
Chapter 17: Equilibrium
81.
increase
82.
position
83.
pressure (and concentration)
84.
The balanced equation is H2O(g) + CO(g)
H2(g) + CO2(g). Initially, 8 H2O molecules are pre
sent and 6 CO molecules are present in the same 1.0-L container. The system reaches equilibriu
m by
H2O(g) + CO(g)
H2(g) + CO2(g)
Initial
8
6 0 0
Change
–x
–x
+x +x
Equilibrium
8–x 6–x
x x
To determine the value of x, use the equilibrium expression
Use the quadratic equation to solve for x. x = 24, 4. The value of x cannot be 24 or else a
negative equilibrium concentration for the reactants would result. Therefore, x must equal 4. The
number of each type of molecule in the container at equilibrium is
H2O = 8 – x = 8 – 4 = 4
CO = 6 – x = 6 – 4 = 2
H2 = x = 4
CO2 = x = 4
85.
The solubility product (Ksp) for a sparingly soluble salt is the equilibrium constant for the dynami
c equilibrium between solution and undissolved solute in a saturated solution of the salt. Consider
the sparingly soluble salt AgCl.
AgCl(s)
Ag+(aq) + Cl–(aq)
Ksp = [Ag+(aq)][Cl–(aq)]
86.
An equilibrium reaction may come to many positions of equilibrium, but at each possible position
of equilibrium, the numerical value of the equilibrium constant is fulfilled. If different amounts of
reactant are taken in different experiments, the absolute amounts of reactant and product present a
t the point of equilibrium reached will differ from one experiment to another, but the ratio that de
fines the equilibrium constant will be the same.
87.
alpha
beta
K =
[beta]
[alpha]
At equilibrium, [alpha] = 2 [beta]
K =
[beta]
2 [beta]
= 0.5
88.
PCl5(g)
PCl3(g) + Cl2(g)
389
Chapter 17: Equilibrium
K =
3
2
5
[PCl ][Cl ]
[PCl ]
= 4.5 10–3
The concentration of PCl5 is to be twice the concentration of PCl3: [PCl5] = 2 [PCl3]
K =
3
2
3
[PCl ][Cl ]
2 [PCl ]
= 4.5 10–3
K =
2
[Cl ]
2
= 4.5 10–3 and [Cl2] = 9.0 10–3 M
89.
density CaCO3(s) = 2.930 g/cm3
molar mass CaCO3 = 100.1 g
density CaO(s) = 3.30 g/cm3 (lime)
molar mass CaO = 56.08 g
90.
As all of the metal carbonates indicated have the metal ion in the +2 oxidation state, we can illustr
ate the calculations for a general metal carbonate, MCO3:
MCO3(s)
M2+(aq) + CO32–(aq)
Ksp = [M2+(aq)][CO32–(aq)]
If we then let x represent the number of moles of MCO3 that dissolve per liter, then [M2+(aq)] = x
and [CO32–(aq)] = x also because the reaction is of 1:1 stoichiometry. Therefore,
Ksp = [M2+(aq)][CO32–(aq)] = x2 for each salt. Solving for x gives the following results.
[BaCO3] = x = 7.1 10–5 M
[CdCO3] = x = 2.3 10–6 M
[CaCO3] = x = 5.3 10–5 M
[CoCO3] = x = 3.9 10–7 M
91.
Ca3(PO4)2(s)
3Ca2+(aq) + 2PO43–(aq)
Let x represent the number of moles of Ca3(PO4)2 that dissolve per liter; then [Ca2+] = 3x and
[PO43–] = 2x.
Ksp = [Ca2+]3[PO43–]2 = 1.3 10–32
[3x]3[2x]2 = 1.3 10–32
108x5 = 1.3 10–32
x5 = 1.20 10–34
x = 1.64 10–7 M
[Ca2+] = 3x = 3 1.64 10–7 M = 4.9 10–7 M
390
Chapter 17: Equilibrium
92.
Although a small solubility product generally implies a small solubility, comparisons of solubility
based directly on Ksp values are only valid if the salts produce the same numbers of positive and n
egative ions per formula when they dissolve. For example, one can compare the solubilities of Ag
Cl(s) and NiS(s) directly using Ksp, because each salt produces one positive and one negative ion
per formula when dissolved. One could not directly compare AgCl(s) with a salt such as Ca3(PO4)
2, however.
93.
A higher concentration means there are more molecules present, which results in a greater freque
ncy of collision between molecules.
94.
At higher temperatures, the average kinetic energy of the reactant molecules is larger. At higher t
emperatures, the probability that a collision between molecules will be energetic enough for reacti
on to take place is larger. On a molecular basis, a higher temperature means a given molecule will
be moving faster.
95.
When a liquid is confined in an otherwise empty closed container, the liquid begins to evaporate,
producing molecules of vapor in the empty space of the container. As the amount of vapor increa
ses, molecules in the vapor phase begin to condense and reenter the liquid state. Eventually the op
posite processes of evaporation and condensation will be going on at the same speed: beyond this
point, for every molecule that leaves the liquid state and evaporates, there is a molecule of vapor
which leaves the vapor state and condenses. We know the state of equilibrium has been reached
when there is no further change in the pressure of the vapor.
96.
a.
K =
2
2
2
[HBr]
[H ][Br ]
b.
K =
2
2
2
2
2
[H S]
[H ] [S ]
c.
K =
2
2
2
2
[HCN]
[H ][C N ]
97.
a.
K =
3
2
2
3
[O ]
[O ]
b.
K =
2
2
2
2
4
2
[CO ][H O]
[CH ][O ]
c.
K =
2
4
2
2
4
2
[C H Cl ]
[C H ][Cl ]
391
Chapter 17: Equilibrium
98.
N2(g) + 3Cl2(g)
2NCl3(g)
99.
3
3
2
2
5
[PCl ][Cl ] [0.325 ][3.9 10
]
0.12
[PCl ]
[1.1 10
]
M
M
K
M
100.
a.
3
2
1
[O ]
K
b.
3
1
[NH ][HCl]
K
c.
2
1
[O ]
K
101.
a.
5
2
1
[O ]
K
b.
2
2
[H O]
[CO ]
K
c.
K = [N2O][H2O]2
102.
The second snapshot is the first to represent an equilibrium mixture because after this point, the
concentrations of the reactant and products remain constant. 6 molecules of A2B reacted initially.
2A2B(g)
2A2(g) + B2(g)
Initial
? 0 0
Change
–2x
+2x +x
Equilibrium
? – 2x = 2 2x = 4 x = 2
Therefore ? = 6.
103.
2NO(g) + O2(g)
2NO2(g)
a.
shifts to the right
b.
shifts to the right
c.
no effect (He is not involved in the reaction)
104.
The reaction is exothermic as written. An increase in temperature (addition of heat) will shift the r
eaction to the left (toward reactants).
105.
2
2
2
[CO ][H ]
[1.3 ][1.4
]
3.9
[CO][H O] [0.71 ][0.66
]
M
M
K
M
M
392
Chapter 17: Equilibrium
106.
2
2
2
3
3
3
3
2
2
[NH ]
[NH ]
1.3 10
[N ][H ]
[0.1M][0.1M]
K
[NH3]2 = 1.3 10–6
[NH3] = 1.1 10–3 M
107.
[Cl2] = 0.79 M
108.
a.
Cu(OH)2(s)
Cu2+(aq) + 2OH–(aq)
Ksp = [Cu2+][OH–]2
b.
Cr(OH)3(s)
Cr3+(aq) + 3OH–(aq)
Ksp = [Cr3+][OH–]3
c.
Ba(OH)2(s)
Ba2+(aq) + 2OH–(aq)
Ksp = [Ba2+][OH–]2
d.
Sn(OH)2(s)
Sn2+(aq) + 2OH–(aq)
Ksp = [Sn2+][OH–]2
109.
As the calculations for each of the silver halides would be similar, we can illustrate the calculatio
ns for a general silver halide, AgX:
AgX(s)
Ag+(aq) + X–(aq)
Ksp = [Ag+(aq)][X–(aq)]
If we then let x represent the number of moles of AgX that dissolve per liter, then [Ag+(aq)] = x
and [X–(aq)] = x also because the reaction is of 1:1 stoichiometry. Therefore,
[Ag+(aq)][X–(aq)] = x2 = Ksp for each salt. Solving for x gives the following results.
[AgCl] = x = 1.3 10–5 M
[AgBr] = x = 7.1 10–7 M
[AgI] = x = 9.1 10–9 M
110.
molar mass AgCl = 143.4 g
9.0 10–4 g AgCl/L
1 mol AgCl
143.4g AgCl
= 6.28 10–6 mol AgCl/L
AgCl(s)
Ag+(aq) + Cl–(aq)
Ksp = [Ag+][Cl–] = (6.28 10–6 M)(6.28 10–6 M) = 3.9 10–11
111.
HgS(s)
Hg2+(aq) + S2–(aq)
Ksp = [Hg2+][S2–] = 1.6 10–54
Let x represent the number of moles of HgS that dissolve per liter; then [Hg2+]= x and [S2–] = x.
Ksp = [x][x] = x2 = 1.6 10–54
393
Chapter 17: Equilibrium
x = 1.26 10–27 M
molar mass HgS = 232.7 g
1.26 10–27 M
232.7 g
1 mol
= 2.9 10–25 g/L
112.
molar mass Ni(OH)2 = 92.71 g
2
2
0.14 g Ni(OH)
1 mol
1 L
92.71 g Ni(OH)
= 1.510 10–3 M
Ni(OH)2(s)
Ni2+(aq) + 2OH–(aq)
Ksp = [Ni2+][OH–]2
If 1.510 10–3 M of Ni(OH)2 dissolves, then [Ni2+] = 1.510 10–3 M and [OH–] = 2 (1.510
10–3 M) = 3.020 10–3 M.
Ksp = (1.510 10–3 M)(3.020 10–3 M)2 = 1.4 10–8
113.
The nitrogen–nitrogen triple bond in N2 and the three hydrogen–hydrogen bonds (in the three H2
molecules) must be broken. Six nitrogen–hydrogen bonds must form (in the two ammonia molec
ules).
114.
The activation energy is the minimum energy two colliding molecules must possess in order for t
he collision to result in reaction. If molecules do not possess energies equal to or greater than Ea,
a collision between these molecules will not result in a reaction.
115.
Living cells contain biological catalysts called enzymes. Such enzymes speed up the complicated
biochemical processes that must occur in cells (which would be too slow to sustain life without th
e enzyme’s action).
116.
Once a system has reached equilibrium the net concentration of product no longer increases becau
se molecules of product already present react to form the original reactants. This is not to say that
the same product molecules are necessarily always present.
117.
N2(g) + 3H2(g)
2NH3(g)
2
2
10
3
3
4
3
3
2
2
[NH ]
[0.34
]
2.5 10
[N ][H ]
[4.9 10 ][2.1 10 ]
M
M
M
118.
119.
A small equilibrium constant means that the concentration of products is small, compared to the c
oncentration of reactants. The position of equilibrium lies far to the left. Reactions with very smal
l equilibrium constants are generally not very useful as a source of the products unless the equilib
rium position can be shifted by application of Le Châtelier’s principle.
394
Chapter 17: Equilibrium
120.
3H2(g) + N2(g)
2NH3(g)
121.
A2(g) + 2B(g)
2AB(g)
122.
3O2(g)
2O3(g)
123.
H2(g) + I2(g)
2HI(g)
124.
H2(g) + F2(g)
2HF(g)
125.
The correct answer is a. For an endothermic reaction, an increase in temperature will shift the
position of equilibrium to the right (toward the products), which will increase K.
395
Chapter 17: Equilibrium
126.
N2
H2
NH3
Add N2
Increase
Decrease
Increase
Remove H2
Increase
Decrease
Decrease
Add NH3
Increase
Increase
Increase
Add Ne
No change
No change
No change
Increase T
Increase
Increase
Decrease
Decrease V
Decrease
Decrease
Increase
Add catalyst
No change
No change
No change
396
