2
EQUILIBRIUM
2.1 WHAT IS EQUILIBRIUM ?
In general terms equilibrium implies a situation that is unchanging or steady. This is generally
achieved through a balance of opposing forces. In chemistry equilibrium refers to the exact balancing
of two processes, with one being the opposite of the other. An example of this is a sealed bottle of
water. As long as the lid is left on the water level stays fairly constant, but if the lid is removed the
equilibrium is disturbed and the water evaporates. If you concluded that evaporation is not occurring
when the lid is on, you would be wrong. In fact it occurs at the same rate whether the lid is on or off.
The difference is that when the lid is off there is very little water condensing in the air above the liquid
and returning to the water. In the sealed container the two processes (evaporation and condensation)
are occurring at the same rate – hence there is no apparent change in the level of the water.
Many of the processes that occur in atmospheric, soil and water systems are controlled by
chemical equilibrium. These include solubilisation, precipitation, acid-base and oxidation-reduction
processes.
2.2 WHEN IS A SYSTEM AT EQUILIBRIUM ?
So far in this module you might assume that all reactions proceed in one direction only and that they
go till one of the reactants run out – called completion. The reality is that many reactions stop far short
of completion. These reactions reach a state of chemical equilibrium.
An equilibrium reaction is reversible. In a reversible reaction, the substances on the left-hand
side of the chemical equation produce those on the right-hand side, and the substances on the right
change back to those on the left.
The equilibrium is closed - no substance is added or removed, and the temperature and
pressure remain constant.
The equilibrium is dynamic. The reversible reactions occur continuously, but there is no
appearance of change. The concentrations of the components do not change at equilibrium. In an
equilibrium system, the rates of the forward and reverse reactions are equal. A double arrow is used
to indicate that the reaction is proceeding in both directions.
CHEMICAL EQUILIBRIUM
RATE OF FORWARD REACTION = RATE OF REVERSE REACTION
2.3 THE EQUILIBRIUM CONSTANT
The Law Of Chemical Equilibrium (formerly called the Law of Mass Action) states that, “when a
reaction reaches equilibrium, the product of the concentration of each substance on the right hand side
of the equation divided by the product of the concentration of each substance on the left hand side is a
constant, at a given temperature”. This is known as the equilibrium constant, K.
Chemistry 2
2. Equilibrium
Thus, for a general equilibrium reaction,
aA + bB
K
cC + dD
[C]c[D]d
=
[A]a[B]b
The square brackets indicate concentration in moles per litre.
We can use the change of ozone (O3) an important atmospheric pollutant to oxygen (O 2) as an
example of how to construct a chemical equilibrium expression.
The equation is

2O3 (g)
3O2(g)
Coefficient
Reactant
Product
To obtain the equilibrium expression, place the concentration of the product on the top, and the
concentration of the reactant on the bottom.
[O2]
Product
[O3]
Reactant
Then use the co-efficients as powers
K=
[O2]3
[O3]2
Co-efficients become powers
PRACTICE QUESTIONS
Write the equilibrium constant expressions for the following:
1.
2.
When nitrogen gas (N2) and hydrogen gas (H2) are reacted together, they form an equilibrium
with the product, ammonia gas (NH3).
The reaction of sulfur dioxide (SO2) with oxygen to produce sulfur trioxide (SO3).
2.4 THE M EANING OF THE E QUILIBRIUM CONSTANT
What does the numerical value of K mean? Because it is a ratio of concentrations, we can tell
something about the equilibrium position, and whether it favours the products or the reactants.
If K is large (> 1), then the top of the ratio is larger than the bottom, so the concentration of
the products must be greater than the reactants. The opposite case applies whether K is small (< 1).
Chemistry 2
2.2
2. Equilibrium
Calculating is a very simple exercise if you know the concentration of each component of the reaction,
because it is no more than substitution. If you don’t, you can’t! It is as simple as that.
EXAMPLE
Calculate the value of K for the ozone reaction above, given the following data: [O 3] = 1.81 x
10-5 M and [O2] = 6.90 x 10-4 M. Comment on the equilibrium position.
K
= [O3]2 ÷ [O2]3 = (1.81 x 10-5)2 ÷ (6.90 x 10-4)3
= 0.997
Since K is around 1, the reaction neither favours the reactants or products.
PRACTICE QUESTIONS
3.
Calculate K for the following reaction and comment on the equilibrium position:
(a)
(b)
2A + B  C
A + 2B  3C
[A] = 0.013, [B] = 0.003, [C] = 0.12
[A] = 0.0732, [B] = 0.000181, [C] = 0.000069
The value of K is fixed for a given reaction at a given temperature. Its value is not affected by
concentration or pressure.
2.5 SOLIDS AND WATER IN EQUILIBRIUM EXPRESSIONS
Solids
Limestone (calcium carbonate) is commonly used to produce lime (a soil additive) by the following
reaction
CaCO3(s)
CaO(s) + CO2(g)
If we simply write out an equilibrium expression as per the rules above we end up with
K = [CO2][CaO]
[CaCO3]
If we perform experiments to confirm this however, it can be established that the position of the
equilibrium does not depend on the amounts of pure solids or pure liquids present. This is because
the concentrations of pure solids or pure liquids do not change – that is they are constant. This means
that they can be omitted from the equilibrium constant expression. The equilibrium constant
expression therefore becomes.
K = [CO2]
Pure solids are omitted from equilibrium expressions, but ALL gases and solutions MUST be
included as their concentrations can vary.
Chemistry 2
2.3
2. Equilibrium
Water
In aqueous reactions (those involving water based solutions), the concentration of water, while a real
value (55.55M), is essentially a constant, and as above, can be incorporated into K. Thus for the
reaction of ammonia and water,

NH3(l) + H2O(1)
NH4+(aq)
+
OH-(aq)
K = [NH4+][OH-]
[NH3][H2O]
But since [H2O] can be included with K it becomes,
K = [NH4+][OH-]
[NH3]
PRACTICE QUESTIONS
Write the equilibrium constant expressions for the following:
4.
5.
Carbon dioxide (CO2) reversibly dissolves in water, forming carbonic acid (H2CO3). This is why
"fizzy" drinks are fizzy. When you open the can or bottle, you "open the closed system" and
change the equilibrium. Carbon dioxide is released from solution, creating the bubbles.
Sulfur trioxide gas (SO3) dissolves in water to form sulfuric acid H2SO4 – one of the
components of acid rain.
2.6 CHANGES TO AN EQUILIBRIUM SYSTEM
When iron (III) ions are mixed with thiocyanate ions (SCN-), a deep red colour forms. This is due to
the FeSCN2+ ion. The reaction exists as an equilibrium, and also gives off heat.
Fe3+ + SCN-

FeSCN2+
+
heat
Observe what happens to the amount of colour, when the equilibrium is disturbed in a number of ways
in the following exercise. Account for the changes by writing answers in the spaces provided.
The work above will have shown you that the equilibrium position is not fixed, and can be altered to
more or less of one side, if the conditions change.
2.7 LE CHATELIER’S PRINCIPLE
It is important to understand the factors that affect the position of any chemical equilibrium. When
chemicals are manufactured, the process is designed to maximise the yield of any desired product by
choosing the conditions that favour its formation. In this section we will examine how changes in
factors such as temperature and concentration affect the position of a chemical equilibrium.
We can predict the effects of changes on chemical equilibria by using Le Chateleier’s
Principle, which states that “when a system at equilibrium is subjected to change, a reaction will
occur that will tend to reduce the effect of the change”.

Chemistry 2
An easy way to remember Le Chatelier’s Principle is to note that when a reactant or
product is added to a system at equilibrium, the system shifts away from the added
component. What actually happens is that one of the reactions – forward or reverse –
increases in rate for a time until a new equilibrium is reached.
2.4
2. Equilibrium
Imposing a change on an equilibrium that favours the products is known “shifting the
equilibrium to the right”. A change that favours the reactants is known as “shifting the equilibrium to
the left”
Concentration
Adding a component to a reaction at equilibrium will cause the system to shift to reduce some of the
added component. Therefore, adding more reactant causes the forward reaction to dominate for a
while and more product will be formed. Removing a component has the opposite effect – the
equilibrium shifts to make more of this component.
Pressure
When a gas is compressed (its volume decreases), and the pressure in its container increases. Hence
when the volume of any gaseous system at equilibrium is placed under increased pressure, then Le
Chatelier’s Principle predicts that the system will shift in the direction that reduces the pressure. Since
1 mole of any gas occupies the same volume, the effect of changing the pressure of a gaseous
equilibrium can be predicted by examining which side of the equation has the least number of moles
of gas.
Case I. Forward reaction produces more gas molecules – therefore an increase in system pressure
will force the equilibrium to the left.
H2CO3(aq)  H2O(l) + CO2(g)
Case 2. Forward reaction produces less gas molecules – therefore an increase in system pressure
will force the equilibrium to the right.
3H2(g)
+
N2(g)

2NH3(g)
Case 3. Forward reaction does not change the number of gas molecules – therefore has no effect
on the equilibrium.
H2(g)
+
Cl2(g)
 2HCl(g)
Temperature
Like pressure, temperature also affects the position of chemical equilibria. Many chemical reactions
either produce or consume heat. Heat can be thought of as a product or reactant. So if we consider the
reaction of sulfur dioxide (a common pollutant gas) with oxygen (from the atmosphere) the following
reaction can be written.
2SO2(g) + O2(g)
 2 SO3(g) +
heat
As it is an reaction that produces heat. Le Chatelier’s Principle still applies here, hence if we heat this
reaction up it will tend to go to the left, whilst cooling it down will force the reaction to the right.
Only temperature affects the value of K, so changing the temperature not only disturbs the
equilibrium for a time, it alters the position.
Chemistry 2
2.5
2. Equilibrium
PRACTICE QUESTIONS
Use Le Chatelier’s Principle to predict the effects of changes in the following equilibrium systems.
6.
The Haber process is used to manufacture ammonia, by the following equilibrium reaction,
N2
+ 3H2 + heat
2NH3
Which reaction (forward or reverse) will be favoured by the following changes?
(a) adding more N2
(b) removing NH3 as it forms
(c) increasing the temperature
(d) decreasing the temperature
(e) increasing the container volume
(f) decreasing the container volume
2.8 GENERAL IONIC SOLUBILITY RULES
There are some general rules that apply to the solubility of various common ions when they are
present with other ions. Table 2.1 provides these rules.
TABLE 2.1 Solubility rules
The following compounds are SOLUBLE
Except
Ammonium (NH4+) salts
Alkali metal salts (Group lA)
Nitrates (NO3-)
Halides (Group 7A)
Ethanoates (formerly Acetates)(CH3 COO-)
Chlorates (C1O3-)
Sulfates (SO42-)
Some Li+
Ag+, Hg22+, Pb2+
Ag+, A13+
Ba2+, Sr2+, Ca2+, Pb2+, Ag+, Hg2+, Hg22+
The following compounds are INSOLUBLE
Carbonates (CO32-)
Phosphates (PO43-)
Hydroxides (OH-)
Sulfides (S2-)
Except
NH4+ and alkali metals
Na+, K+ and NH4+
Ba2+ and alkali metals
NH4+ and alkali metals
EXAMPLE
What is the solubility of (a) sodium sulfate, (b) barium sulfate and (c) ammonium carbonate?
(a) Sodium is an alkali metal, and almost all these are soluble. Sulfate is not an exception, so
it is soluble.
(b) Barium is one of the exceptions to the sulfates are soluble rule, which means that is
insoluble.
(c) Most carbonates are insoluble, but ammonium is one of the exceptions, so it is soluble.
Chemistry 2
2.6
2. Equilibrium
PRACTICE QUESTIONS
10.








Predict the solubilities of the following compounds:
iron (III) phosphate
sodium carbonate
magnesium sulfate
lead (II) chloride
silver ethanoate
barium hydroxide
ammonium carbonate
lead nitrate
WHAT YOU NEED TO BE ABLE TO DO







explain the meaning of equilibrium
define terms associated with equilibrium
write equilibrium constants
calculate and interpret values of the equilibrium constant
list factors affecting the equilibrium position
perform solubility equilibria calculations
apply the solubility rules
Chemistry 2
2.7