5b
Equilibrium
1. The Equilibrium State
In the previous (stoichiometry) section we assumed that
reactions proceeded until one or more of the reactants was
completely consumed; for many reactions this works well.
However, in some reactions there is a significant back
reaction. An example is the reaction of nitrogen and
hydrogen to form ammonia. Note the double arrow, which
indicates the occurrence of both a forward and back
reaction.
N2(g) + 3 H2(g)  2 NH3(g)
Here nitrogen and hydrogen are gradually consumed and
ammonia is formed until, as shown at the dashed line on
the right, the concentrations cease to change. At this point
the rates of the forward reaction and the reverse reaction
are equal. The system is then said to be at equilibrium.
Figure 1. A reaction reaching
equilibrium
2. The Equilibrium Constant
The extent to which a reaction proceeds before it reaches equilibrium is characterized by an expression
known as the mass action expression. This is set equal to a quantity known as K, the equilibrium
constant.
The mass action expression is a fraction with the products on the top and the reactants on the bottom,
each raised to an exponent equal to its coefficient in the balanced equation. Thus for the formation of
ammonia (above):
[NH3]2
K=
[N2][H2]3
In expressions like the one above, concentrations of gases are usually given as pressures, with units of
atmospheres. However the brackets shown above indicate that the concentrations are in moles per liter.
Brackets should not be used for pressures, but sometimes they are.
It bears repeating that equilibrium is a dynamic and not a static process. In the reaction A  B, A is
constantly being converted to B and B is constantly being converted to A. The equilibrium
concentrations are those at which the rates of the forward and back reactions are equal. What makes
equilibrium seem static is that since the rates of the forward and back reactions are equal, the reactant
(and product) concentrations do not change.
3. Gas Phase Equilibrium Calculations
Consider the ammonia equilibrium, shown above. Start with N2 = 0.35 atm; H2 = 0.25 atm; and NH3 =
0.40 atm. To help set up the calculation, we will use something called an ICE chart.
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An ICE chart gives the pressure or concentrations of each material present in the reaction. It gives the
initial concentration, the change in concentration and the concentration at equilibrium.
In most cases the Initial concentrations
N2
H2
NH3
are given; the Changes are obtained
from the stoichiometry of the reaction;
Initial
0.35 atm.
0.25 atm.
0.40 atm.
the Equilibrium values are obtained by
adding the Change to the Initial.
Change
-x
-3x
+2x
Thus the mass action expression would
be written as:
Equilibrium
0.35 - x
0.25 - 3x
0.40 + 2x
Of course this is mathematically rather messy. So let's try another equilibrium calculation which works
out more easily.
Consider the reaction
2 A + B  2 C.
The system starts with 0.50 atm. of
A, 0.50 atm. of B and 0.00 atm of C.
If the system ends up with 0.20 atm
of C, what is the value of K? First we
set up an ICE table. Here the changes
are determined from the fact that 0.20
atm. of C are formed.
A
B
C
Initial
0.50
0.50
0
Change
-0.20
-0.10
+0.20
Equilibrium
0.30
0.40
0.20
Now we use the equilibrium pressures from the ICE chart to calculate the value of K.
Now consider the reaction
H2(g) + I2(g)  2 HI(g)
The system starts with [H2] = [I2] = 0.50 atm. and no HI. If K = 0.0100 we calculate:
[HI]2
K = [H ][I ]
2 2
(2x)2
0.010 = (0.50 - x)(0.50 - x)
Take the square root of both sides
Initial
Change
Equilibrium
2x
0.10 = 0.50 - x
2x = 0.050 - 0.10 x, and x = 0.024
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H2(g)
I2(g)
HI(g)
0.50
0.50
0
-x
-x
+2x
0.50 - x
0.50 - x
2x
4. The Mathematics of Equilibrium Constants
Although units are not usually given in equilibrium constants, most do have units associated with them.
Consider, for example:
[NH3]2
K=
[N2][H2]3
Here K will have the units of atm-2 or kPa-2.
However, suppose that the concentrations were given in moles/L instead of atm. Now the value of K
must be in units of L2/mol2. Thus, a gas phase equilibrium reaction can have two different equilibrium
constants. The term Kp is used when the units are atmospheres (pressure) and the term Kc is used when
the units are moles/L (concentration). If your concentrations are in one set of units and your
equilibrium constant uses another, K will have to be changed. This can be done with the equation:
KP = KC (RT)Δn
Here the gas constant, R, is 0.08206 L atm/mol  K (or 8.31 L kPa/mol), T is the Kelvin temperature,
and Δn is the change in the number of moles of gas as the reaction is written. For the formation of
ammonia, Δn is -2. This formula is given in the AP Chemistry formula sheet.
In addition to changing the units of equilibrium constants, you must also know when and how to
combine them. Suppose we want to combine the reactions:
A+BC
K = 50
A + C  2D
K = 300
If we combine (add) the two reactions we get a third reaction:
2A+B2D
What is K for the combined reaction?
The rule is that when you add reactions, you multiply the equilibrium constants. So K for the third
reaction would be 15,000. If you subtract the second reaction from the first, then you would divide the
first K by the second. If you double a reaction, you square its K. If you reverse a reaction, the new K is
the reciprocal of the K for the forward reaction. Thus:
H2 + X2  2 HX
K = 2500
½ H2 + ½ X2  HX
K = 50
Although we say that the products are on the top of the mass action expression, an equilibrium reaction
does not have a reactant or product since it goes in both directions. What we actually mean is that the
compounds on the right side of the equation are on the top and the compounds on the left side of the
equation are on the bottom.
To summarize what is used in the mass action expression:
- the concentration of a solute is given in moles/L
- the concentration of a gas is either moles/L or atmospheres
- the concentration of an undissolved solid is exactly one.
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5. Le Chatelier's Principle
Le Chatelier's Principle says that "if a stress is applied to a system at equilibrium, the system adjusts to
reduce the stress." After you are accustomed to using Le Chatelier's Principle, it seems so obvious that
one wonders why it had to be given a name. Yet it often shows up on the multiple choice portion of
every AP Chemistry test.
Consider the reaction:
N2 + 3 H2  2 NH3 + heat
Adding ammonia to a mixture of nitrogen, hydrogen and ammonia at equilibrium will place a stress on
the system. The system will relieve this stress by converting some of the ammonia to nitrogen and
hydrogen.
Adding nitrogen to the system will cause the equilibrium to shift to the right, consuming some of the
added nitrogen as well as some of the hydrogen you started with.
If the temperature is raised, the system will remove some of the excess heat by shifting to the left. If the
volume of the system is decreased, the equilibrium will shift to the right, toward fewer moles of gas, in
order to remove some of the excess pressure.
A common "trick" is to ask about raising the pressure by adding a nonreactive gas (e.g. argon).
Although the total pressure is increased, the partial pressures of the reactants are not affected and the
equilibrium is not shifted. Except for changing the temperature, none of the changes we have discussed
affect the value of K. Yet it would require a change of K in order to alter the concentrations of any of
the reactants.
Consider the equilibrium in which an excess of solid naphthalene (moth balls) is in equilibrium with its
vapor. How is the vapor concentration affected by adding more naphthalene? It isn't! Whether a system
contains more or less solid, the concentration of the solid is still considered to be one. It hasn't changed.
6. Solubility Equilibria
When a soluble or slightly soluble salt is added to water it dissolves initially. Eventually, however, the
solution reaches its capacity and further dissolution stops. What has happened is that the rate at which
the salt dissolves has become equal to the rate at which the dissolved salt precipitates. We say that the
solution is now "saturated." The system has reached equilibrium.
Since salts break into their component ions when they dissolve, this must be seen in the mass action
expression. Consider, for example, the dissolution of lead (II) chloride in water:
PbCl2(s)  Pb2+ (aq) + 2 Cl–(aq)
(Note that you don't get Cl2 from PbCl2). The appropriate mass action expression would be:
[Pb+2][Cl-]2
K = [PbCl ]
2
However, since lead chloride is a solid, it will have a concentration of unity and thus can be removed as
a term in the equation. The equilibrium constant for a dissolution reaction is normally referred to as the
"solubility product" and is abbreviated Ksp. Thus: Ksp = [Pb+2][Cl-]2
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In addition to the solubility product, a second quantity is often used in solubility calculations. This is
the molar solubility, the solubility of the substance in moles per liter. The simplest solubility
calculations are those which involve the conversion of the molar solubility to solubility product and the
solubility product to the molar solubility. Be certain you know the difference!
Given that the Ksp of lead chloride is 1.6  10–5, what is its molar solubility? If the molar solubility of
lead chloride is s, then in a saturated solution the concentration of Pb2+ is s and the concentration of Cl–
must be 2s. Plugging these terms into the mass action expression gives:
Ksp = (s)(2s)2
1.6  10–5 = 4s3
solubility = s = 0.016 M
Be careful that when you calculate the value of (2s)2 you remember to square both the "2" and the "s."
An astonishing number of students make mistakes here.
Given that its molar solubility is 1.0  10–15, what is the Ksp of Bi2S3?
[Bi+3] = 2  1.0  10–15 = 2.0  10–15 M
[S-2] = 3  1.0  10–15 = 3.0  10–15 M
Therefore:
Ksp = [Bi+3]2[S-2]3 = [2.0  10–15]2[3.0  10–15]3 = 108  10–75 = 1.1  10–73
To make it easy for the grader to give you partial credit, always write out the mass action
expression and then plug in the numbers.
7. The Common Ion Effect
If a 0.10 M sodium fluoride solution is used to dissolve calcium fluoride, substantially less calcium
fluoride will dissolve than would have dissolved in pure water. This is caused by fluoride, an ion which
is common to both compounds. This phenomenon is therefore known as the common ion effect. Let us
calculate how much calcium fluoride actually dissolves.
If the molar solubility of calcium fluoride is s, we have:
[Ca+2] = s and [F-] = 0.10 + 2s
This gives the following mass action expression:
Ksp = 4.0  10–11 = (s)(0.10 + 2s)2
We can solve the cubic equation or we can make the simplifying assumption that 0.10 >> 2s. If we
make this assumption, as we usually do, the 2s can be neglected and the equation becomes:
4.0  10–11 = (s)(0.10)2
s = [Ca2+] = 4.0  10–9 M
Was neglecting 2s a valid assumption? The criterion we will use is called "the 5% rule." If a variable is
less than 5% of the constant to which it is added, the variable can be neglected. Is 2s = 2  4.0  10–9
less than 5% of 0.10? It certainly is!
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Most of the difficulties which you will have with solubility products can be summarized in the
following questions. Why did we double the s but not double 0.10? When do you double the fluoride
concentration? The answer is..............never. You are not doubling the fluoride concentration. You are
doubling s, the molar solubility, in order to get the fluoride concentration.
8. Selective Precipitation
In these problems two different compounds precipitate from the same solution. There are all sorts of
questions which can be asked. Consider, for example, a solution containing 0.010 M Ag+ solution and
0.010 M Ba2+. You are told that for silver chromate Ksp = 9.0  10–12 and for barium chromate Ksp =
8.5  10–11.
1) Na2CrO4 is added gradually and with stirring. Assuming that the chromate solution does not affect
the volume, which compound precipitates first? We will imagine the chromate starting at zero and
gradually rising until one compound — silver chromate or barium chromate — starts to precipitate. To
do this we calculate the chromate concentration at which the precipitation starts.
For BaCrO4
For Ag2CrO4
Ksp = [Ba2+][CrO42–]
Ksp = [Ag+]2[CrO42–]
8.5  10–11 = (0.010)(s)
9.0  10–12 = (0.010)2(s)
2–
–9
s = [CrO4 ] = 8.5  10 M
s = [CrO42–] = 9.0  10–8 M
The answer is barium chromate because it starts to precipitate at a lower chromate concentration.
2) What is the concentration of the first cation to precipitate when the second cation has just started to
come out of solution?
Since the first cation to precipitate is barium, the second cation must be silver. We previously
calculated that when silver chromate starts to precipitate the chromate concentration will be 9.0  10–8
M. What barium concentration is consistent with that? We can calculate that as follows:
Ksp = [Ba2+][CrO42–]
8.5  10–11 = [Ba2+](9.0  10–8)
[Ba2+] = 9.4  10–4 M
9. Mixtures of Two Solutions and Bounce Back
A common type of solubility problem involves the mixing of two solutions which might or might not
form a precipitate. The first question you will have to answer is whether a precipitate forms.
Suppose, for example, we mix 40 mL of 6.5  10–3 M Pb(NO3)2 with 60 mL of 0.30 M KI solution.
Knowing that for PbI2 the solubility product, Ksp = 1.4  10–8, does a precipitate form? To answer that
question we must first calculate the Preliminary concentrations of Pb2+ and I– - that means the
concentration after dilution, but before any reaction occurs.
If we assume that the volumes of the two solutions are additive, which we always do, then the final
volume of the solution will be 100 mL. Thus, after dilution:
Then take the concentrations of the diluted ions and use them to calculate Q, the equilibrium quotient.
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You may recall the Q has the same form as K but consists of concentrations as they really are and not
the equilibrium values. So at equilibrium Q = K.
Q = [Pb2+][I–]2 = [2.6  10–3][0.18]2 = 8.42  10–7
Since Q > K, that is 8.42  10–7 > 1.4  10–8, a precipitate forms.
Next calculate the concentrations of ions after Reaction has occurred and the precipitate has been
removed. It would not work to set up an equation like:
1.4  10–8 = (2.6  10–3 - s)(0.18 – 2s)2
This gives s = 2.6  10–3 and, therefore, [Pb2+]= 0. Certainly the lead concentration is close to zero, but
it is not exactly zero and the exact value is what we need to calculate. The calculation is done by a
method called "bounce back."
In bounce back we assume that the concentration of limiting reactant, in this case Pb2+, really does go
all the way to zero. If necessary we correct the concentration of the other ion, iodide in this case. Then
we calculate how much of the slightly soluble salt re-dissolves. The concentration of iodide is
0.18 - 2(2.6  10–3) = 0.17 M
so we calculate the concentration of Pb2+ using the mass action expression. We can now use an ICE
chart to calculate the equilibrium concentrations.
PbI2(s)
Initial
Change
Equilibrium
-s
Pb+2(aq)
I-(aq)
0M
0.17 M
+s
+2s
s
0.17 + 2s
we will assume that the 2s <<0.17M, and can then calculate the
concentration of Pb+2:
“Bounce back,” which you
use when two solutions are
mixed to form a precipitate,
has three steps.
1. Dilution
2. Reaction
3. Equilibration
It is easiest to use a PRICE
chart
1.4  10–8 = [Pb2+][I–]2 = [s](0.17)2
[Pb2+] = s = 4.8  10–5 M
You may also be asked the concentrations of the other ions, K+
and NO–3 in this case. To calculate them, go back to the initial concentrations after dilution. If after
dilution but before reaction [Pb2+] = 2.6  10–3, then [NO–3] = 2  2.6  10–3 = 5.2  10–3. Similarly the
K+ concentration in KI solution will be equal to [I–] before precipitation occurs. Thus [K+] = 0.18 M.
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