1
Chemistry 30 - Equilibrium
Objectives:
1.
2.
3.
4.
5.
6.
7.
8.
Discuss what is meant by a reversible reaction, including the relationship between enthalpy, entropy and
the spontaneity of reactions (notes, 18-1).
Explain what equilibrium is and the conditions which must exist for equilibrium to become established
(notes, 12-3).
Discuss the difference between dynamic equilibrium and steady state (notes, 18-1).
Discuss the types of equilibria (notes).
Explain Le Chatelier's principle, understand why it works and use it to predict how various equilibrium
systems will shift in response to external stress (notes, 12-3).
Discuss the factors which affect equilibrium (notes, 12-3, 18-2).
Discuss the common ion effect (18-2).
Be able to write the equilibrium constant expression for a chemical reaction using the general equation
(18-1):
a A(aq) + b B(aq)
c C(aq) + d D(aq)
9.
10.
Make calculations based on the equilibrium constant expression (18-1).
Recognize the factors which affect Keq values (18-2).
11.
Make determinations of solubility based on Ksp values (18-4).
Vocabulary
equilibrium
chemical equilibrium
reversible reaction
Keq
steady state
equilibrium constant
solubility product constant
Ksp
liquid-vapour equilibrium
Le Chatelier's principle
common ion effect
solution equilibrium
Equilibrium
- a number of chemical reactions have a ΔH and ΔS that are both positive or both negative; one force
favours the reaction, while the other opposes it:
energy + H2O(l) → H2O(g)
ΔH = +

ΔS = +

- even at room temperature this reaction can still proceed; a few molecules will have the energy to
evaporate, until the air above is saturated. (relative humidity is the amount of water vapour in the
air, in %, relative to the amount the air can hold at that temperature. The air is saturated if the relative
humidity is 100%.)
- even at 100% humidity evaporation does not stop. How does the humidity stay at 100%?
- since the ΔH and ΔS are both positive, this reaction is reversible:
H2O(g) → energy + H2O(l)
ΔH = 
ΔS = 
2
- as a water molecule evaporates, another molecule of water vapour condenses. The rate of the
evaporation reaction equals the rate of the condensation reaction; the reactions are at equilibrium:
energy + H2O(l)
H2O(g)
How do we recognize a system at equilibrium?
i)
the reaction must be reversible
ii)
the system must be closed
iii)
both reactant and product must be present
iv)
concentrations of reactant and product do not change
v)
conditions of temperature and pressure must be held constant
What is a closed system?
- no inputs or outputs
- a steady-state system can appear like equilibrium, but has inputs and outputs (for example, your
weight does not change for a year. You appear to be at equilibrium, but you have inputs (food, water,
air) and outputs (wastes))
What are the types of equilibria?
1)
Liquid - Vapour (Phase)
A liquid and its vapour are present in a closed container. At equilibrium the rate of condensation equals the rate
of evaporation.
energy + H2O(l)
2)
H2O(g)
Solution - Solute (Solubility)
A saturated solution is present over undissolved crystals of the same solute. At equilibrium the rate of
crystallization equals the rate of dissolution. The container should be closed to prevent evaporation.
NaCl(s)
3)
Na1+(aq) + Cl1-(aq)
Chemical
The reactants and products of a chemical system are present in the same container at the same time. The rate
of the forward reaction equals the rate of the reverse reaction.
2 NO2(g)
N2O4 (g)
3
If the concentrations of the reactants and products in an equilibrium were measured over time and
graphed, this is what it would look like:
The container starts with NO2 and with time, N2O4 is produced. When the concentrations of each
compound are constant, equilibrium has been achieved.
Le Chatelier’s Principle

since reactions continue in both directions during equilibrium any factor which changes the rate
of one or both reactions will alter the relative amounts of reactants and products.

we can predict how an equilibrium system will respond to a given change by using Le Chatelier’s
Principle:
“When a system at equilibrium is disturbed by application of a stress, it attains a new equilibrium
position that minimizes the stress.”

in other words, for every action there is an equal and opposite reaction; whatever you do to a
reaction system, the system will respond in the opposite direction.
Factors that can affect equilibrium:
1)
Concentration - Increasing the concentration of any component will cause the reaction system to try
to decrease it, by favouring the opposite side.
2)
Temperature - Increasing the temperature of a system will favour the endothermic direction;
decreasing will favour the exothermic direction.
3)
Pressure - Increasing the pressure of a gaseous system favour the side with fewer particles. Solids,
liquids, and aqueous systems are not affected.
4)
Volume - In a closed system, increasing the volume decreases the pressure, and vice versa.
5)
Catalysts and inhibitors - Will increase or decrease the rate at which equilibrium is achieved; has no
effect on the relative amounts of reactants or products.
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Consider the following example:
N2(g) + 3 H2(g)
2 NH3(g) + 92 kJ
In this system, suppose I add some nitrogen gas. The system will try to consume the added nitrogen by
reacting it with some of the hydrogen gas that is present. The amount of hydrogen would go down and
the equilibrium will shift in favour of the ammonia, as its concentration increases. The same would
happen if I tried to add hydrogen; it would react, using up some nitrogen, again producing more
ammonia. If I added ammonia, the equilibrium would shift to favour the reactants; ammonia would be
consumed and nitrogen and hydrogen would be produced.
If energy is applied to the system above, the system will try to absorb energy by favouring the
production of reactants. If I cool the system, the system will respond by favouring the direction that gives
off energy, in this case the products.
If we increase the pressure on a vessel holding this equilibrium reaction, the increase affects both sides.
However, if the system changes its equilibrium concentrations to make more of the product and less
reactant, the number of total moles in the system decreases, thus decreasing the pressure. If pressure
is increased on this system, equilibrium will shift in favour of the products. More ammonia will
be made and less nitrogen and hydrogen will be present. If the pressure is decreased, the opposite is
true; the system will react to make more nitrogen and hydrogen, decreasing the amount of ammonia.
There will be more moles, so the pressure will increase. If the pressure is decreased, the
equilibrium will shift in favour of the reactants. By The Way: Changing the volume of a gaseous
system will have an effect on the pressure. Because the system is closed, decreasing the volume will
increase the pressure; the gases will be compressed. Conversely, if the volume is increased, the pressure
will decrease.
The Common Ion Effect

is related to solubility equilibria.

adding an ionic substance with an ion in common with an aqueous equilibrium will affect the equilibrium
in the same way as changing concentration:


NaCl(s)
Na1+(aq) + Cl1-(aq)
if NaNO3 (s) is added to this system, it breaks into Na1+(aq) and NO31-(aq). It is just like adding Na1+ ions to
the NaCl system. That system responds by favouring the formation of solid NaCl; the concentration of
Cl1- ions decreases.
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Mathematics in Equilibrium Expressions
despite the fact that changing the concentration of a reactant or product will shift the equilibrium
of a reaction, the ratio of the products over the reactants will stay the same. It is called:
The Equilibrium Constant
For the reaction:
aA(g) + bB(g)
cC(g) + dD(g)
-
where the lower-case letters represent number of moles (the numbers used to balance the equation) and
the upper-case letters represent the chemical species.
-
At equilibrium a mathematical relationship exists where a constant result from the equation:
[C]c[D]d
Keq =
[A]a[B]b
-
where [ ] represents the concentration. Equilibrium constants are always calculated as a ratio of the
products over the reactants.
-
only species which are gases or aqueous are included in the equilibrium constant. Solids and liquids
are not included (because changes in these concentrations do not affect the equilibrium).
e.g.
2 CaCO3
2Ca(s) + 2 CO2 (g) + O2 (g)
(s)
Keq =
e.g.
2 H2
(g)
+ O2 (g)
2 H2O (l)
Keq =
-
[CO2]2[O2]
1
[H2]2[O2]
remember, these are mathematical expressions and must be written as such
(i.e. always write “Keq = “)
Solving equilibrium problems using the equilibrium constant
If you are given any chemical reaction at equilibrium and the concentration of each species, you can
calculate the value of the equilibrium constant. For instance, if you are given the following equation:
N2(g) + 3 H2(g)
2 NH3(g) + 92 kJ
Suppose at equilibrium the concentration of N2 = 0.142 mol/L, the [H2] = 0.341 mol/L and the [NH3]
= 0.220 mol/L. You can calculate the value of the Keq by first writing the Keq expression:
Keq =
[NH3]2
[N2] [H2]3
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If we fill in the numbers we get:
Keq =
[NH3]2
[N2] [H2]3
=
(0.220 mol/L)2
(0.142 mol/L)(0.341 mol/L)3
Keq =
[NH3]2
[N2] [H2]3
=
(0.220 mol/L)2
(0.142 mol/L)(0.341 mol/L)3
Working this out gives
= 8.60
This means that products are slightly favoured over reactants, since the ratio is slightly over 1.
This same technique can be used for any problem involving equilibrium concentrations; if the K eq value is
given, for instance, the concentration of one of the substances in the equilibrium can be calculated.
ICE box problems
This type of problem gives you information about the initial situation, before equilibrium is established.
Then you get 1 or more pieces of information concerning the situation at equilibrium. You are required
to fill in the blanks.
For Example:
Consider the following reaction:
2 A(g) + B(g)
C(g) + 3 D(g)
0.100 mol/L each of A and B are placed in a flask and allowed to come to equilibrium. At equilibrium
0.022 mol/L of C is present. Calculate the value of the Keq. (note that M is given to mean mol/L, to save
space)
Step 1. Write down what you know:
2 A(g)
[Initial]
0.100 mol/L
+
B(g)
0.100 mol/L
C(g)
0.0
[Change]
[Equilibrium]
0.022 mol/L
+
3 D(g)
0.0
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Step 2. Determine the Change in Concentration:
2 A(g)
[Initial]
[Change]
+
0.100 mol/L
B(g)
0.100 mol/L
-(2/1)0.022 mol/L
= -0.044 mol/L
-(1/1)0.022 mol/L
= -0.022 mol/L
[Equilibrium]
C(g)
+
3 D(g)
0.0
0.0
+0.022 mol/L
+(3/1)0.022 mol/L
= +0.066 mol/L
0.022 mol/L
You know the change in concentration of C is +0.022 mol/L, since it started at 0.0 mol/L and ended at
0.022 mol/L at equilibrium. The changes of all other components is proportional to the change in [C].
This works exactly the same as mole ratio does in gravimetric stoichiometry. Show your work for the
change, as it is done here.
Step 3. Determine the Equilibrium Concentrations:
2 A(g)
[Initial]
[Change]
[Equilibrium]
0.100 mol/L
-(2/1)0.022 mol/L
= -0.044 mol/L
0.056 mol/L
+
B(g)
0.100 mol/L
-(1/1)0.022 mol/L
= -0.022 mol/L
0.078 mol/L
C(g)
0.0
+0.022 mol/L
0.022 mol/L
+
3 D(g)
0.0
+(3/1)0.022 mol/L
= +0.066 mol/L
0.066 mol/L
To get each equilibrium concentration subtract the change in each case from the initial concentration.
Step 4: Write the Keq expression and calculate the Keq:
Keq =
[C][D]3
[A]2[B]
=
(0.022 mol/L)(0.066 mol/L)2
(0.056 mol/L)2(0.078 mol/L)
Keq = 0.026
At this stage it is just like the calculation when you are given the equilibrium concentrations and are
asked to calculate the value of the K eq.
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Example 2. Consider the following reaction:
A(g) + B(g)
2 C(g)
If the initial concentration of C is 0.350 mol/L, calculate the equilibrium concentrations of all
three species if the value of the Keq is 145.
Step 1. Write down what you know:
A(g)
[Initial]
+
B(g)
0.0
2 C(g)
0.0
0.350 mol/L
[Change]
[Equilibrium]
Step 2. Determine the Change in Concentration:
A(g)
+
B(g)
[Initial]
0.0
0.0
[Change]
+x
+x
2 C(g)
0.350 mol/L
- 2x
[Equilibrium]
When you don’t know something in math, you call it ‘x’. In this case you don’t know what the change in
concentration is, so we can call it ‘x’. The change in concentration of ‘C’ reflects the stoichiometry of the
equation. The reactants are positive x because since they start at zero, they have no where to go but
up. That means the change in C has to be negative (one side goes up, the other goes down.)
Step 3. Determine the Equilibrium Concentrations:
A(g)
+
B(g)
[Initial]
0.0
0.0
[Change]
+x
+x
[Equilibrium]
x
2 C(g)
0.350 mol/L
- 2x
x
0.350 mol/L - 2x
Step 4: Write the Keq expression and calculate the value of x:
Keq =
145 =
[C]2
[A][B]
(0.350 - 2x)2
(x)(x)
=
(0.350 - 2x)2
x2
9
Take the square root of both sides:
√(0.350 - 2x)2
√x2
√145 =
x = 0.0249 mol/L
Step 5. Solve for the concentration of each species.
A(g)
+
B(g)
[Initial]
0.0
0.0
[Change]
+x
+x
[Equilibrium]
x
x
0.0249 mol/L
0.0249 mol/L
2 C(g)
0.350 mol/L
- 2x
0.350 mol/L - 2x
0.350 mol/L – 2(0.0249 mol/L)
= 0.300 mol/L
Example 3. Consider the following reaction:
A(g) + B(g)
C(g)
If the initial concentration of C is 0.350 mol/L, calculate the equilibrium concentrations of all
three species if the value of the Keq is 145.
Step 1. Write down what you know:
A(g)
[Initial]
+
0.0
B(g)
0.0
C(g)
0.350 mol/L
[Change]
[Equilibrium]
Step 2. Determine the Change in Concentration:
A(g)
+
B(g)
C(g)
[Initial]
0.0
0.0
0.350 mol/L
[Change]
+x
+x
-x
[Equilibrium]
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Step 3. Determine the Equilibrium Concentrations:
A(g)
+
B(g)
[Initial]
0.0
0.0
[Change]
+x
+x
[Equilibrium]
x
C(g)
0.350 mol/L
-x
x
0.350 mol/L - x
Step 4: Write the Keq expression and calculate the value of x:
Keq =
[C]
[A][B]
145 =
(0.350 - x)
(x)(x)
=
(0.350 - x)
x2
145x2 = 0.350 - x
145x2 + 1 x - 0.350 = 0
a
b
c
You will notice this problem set up just like the previous one, but instead of resolving neatly, this problem
ended up in a quadratic. Use the quadratic formula to solve it:
x = -b ± √ b2 – 4ac
2a
x = 0.0458 mol/L
Step 5. Solve for the concentration of each species.
A(g)
[Equilibrium]
x
0.0458 mol/L
+
B(g)
x
0.0458 mol/L
C(g)
0.350 mol/L - x
0.350 mol/L – 0.0458 mol/L
= 0.304 mol/L
11
The Solubility Product
-
is a variation on the equilibrium constant for a solute-solution equilibrium.
-
remember that the solubility equilibrium is based on a saturated solution.
CaCl2 (s)
-
Ca2+(aq) + 2 Cl1-(aq)
the Keq would be:
Keq =
-
but the CaCl2 is a solid, so it is not included in the expression and is replaced by a 1:
Keq =
-
[Ca2+][Cl1-]2
[CaCl2]
[Ca2+][Cl1-]2
1
the resulting expression is the Ksp:
Ksp =
[Ca2+][Cl1-]2
Calculations using the Ksp
e.g.
A saturated solution of CuBr has a concentration of 2.0 x 10-4 mol/L. Calculate the Ksp.
Cu1+(aq)
CuBr(s)
2.0 x 10-4 mol/L
Ksp
Br1-(aq)
+
1(2.0 x 10-4 mol/L)
= 2.0 x 10-4 mol/L
1(2.0 x 10-4 mol/L)
= 2.0 x 10-4 mol/L
= [Cu1+][Br1-]
= (2.0 x 10-4 mol/L)( 2.0 x 10-4 mol/L)
= 4.0 x 10-8
e.g.
Calculate the Ksp for Bi2S3, which has a solubility of 1.36 x 10-15 mol/L at 25°C.
Bi2S3 (s)
2 Bi3+(aq)
1.36 x 10-15 mol/L
2(1.36 x 10-15 mol/L)
= 2.72 x 10-15 mol/L
Ksp
+
3 S2-(aq)
3(1.36 x 10-15 mol/L)
= 4.08 x 10-15 mol/L
= [Bi3+]2[S2-]3
= (2.72 x 10-15 mol/L)2(4.08 x 10-15 mol/L)3
= 5.02 x 10-73
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We can also use a known value of Ksp to find out the solubility of a substance, since Ksp is based on a
saturated solution.
e.g.
The Ksp value for Cu(IO3)2 is 1.4 x 10-7 at 25°C. Calculate its solubility at this
temperature.
Cu(IO3)2 (s)
x
Cu2+(aq)
+
2 IO31-(aq)
1(x)
Ksp
2(x)
= [Cu2+][IO31-]2
= (x)(2x)2
1.4 x 10-7
x
= 4x3
= 3 1.4 x 10-7
4
= 3.3 x 10-3 mol/L
The concentration of a saturated solution of copper (II) iodate is 2.2 x 10 -3 mol/L
Ksp values can also be used to predict if mixing solutions will produce a precipitate.
e.g. A solution is prepared by adding 750.0 mL of 4.00 x 10-3 mol/L Ce(NO3)3 to 300.0 mL of
2.00 x 10-2 mol/L KIO3. Will Ce(IO3)3 precipitate from this solution ? The Ksp of Ce(IO3)3 is 1.9 x 10-10.
Step 1. Write a dissociation equation for each substance and determine the concentration of each ion in
solution.
Ce(NO3)3 (aq)
Ce3+(aq)
+
3 NO31-(aq)
4.00 x 10-3 mol/L
1(4.00 x 10-3 mol/L)
= 4.00 x 10-3 mol/L
KIO3 (aq)
K1+(aq)
2.00 x 10-2 mol/L
1(2.00 x 10-2 mol/L)
= 2.00 x 10-2 mol/L
3(4.00 x 10-3 mol/L)
= 1.20 x 10-2 mol/L
+
IO31-(aq)
1(2.00 x 10-2 mol/L)
= 2.00 x 10-2 mol/L
Step 2. Use the dilution equation to determine the concentration of each of the significant species. In
this case they are Ce3+(aq) and IO31-(aq)
[Ce3+] = csvs = (4.00 x 10-3 mol/L)(750.0 mL)
vd
750.0 mL + 300.0 mL
= 2.86 x 10-3 mol/L
[IO31-] = csvs = (2.00 x 10-2 mol/L)(300.0 mL)
vd
750.0 mL + 300.0 mL
= 5.71 x 10-3 mol/L
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Step 3. Write the Ksp expression for the substance you are testing and solve for the K sp using the
numbers you just calculated. This is a trial Ksp
Substance: Ce(IO3)3
Ksp
= [Ce3+][IO31-]3
Trial Ksp = (2.86 x 10-3 mol/L)( 5.71 x 10-3 mol/L)3
= 5.52 x 10-10
Step 4. Compare the trial Ksp with the actual Ksp from the question:
-
if the trial Ksp is larger than the actual Ksp, there is more solute than can dissolve; there
will be a precipitate.
-
if the trial Ksp is smaller than the actual Ksp, there is less solute than can dissolve; there
will not be a precipitate.
In this case 5.52 x 10-10 is larger than 1.9 x 10-10; there will be a precipitate.