Katumba J. 2020 elbow grease is the best polish 1
CHEMICAL EQUILIBRIA
Chemical equilibrium: Is a state in which the rates of the forward and reverse
reactions are equal and the concentrations of the reactants and products remain
constant.
A Chemical equilibrium occurs for reversible reactions. Many reactions do not go
to completion since the products of the reaction themselves react to form
original reactants.
A reversible reaction is a reaction that takes place in both forward and backward
directions, hence does not go to completion. For example;
In a closed container/vessel, calcium carbonate decomposes as follows;
𝐶𝑎𝐶𝑂3(𝑠) ⇌ 𝐶𝑎𝑂(𝑠) + 𝐶𝑂2(𝑔) {Reversible reaction}
Explanation
When calcium carbonate is heated at a fixed temperature in a closed container;
at first, 𝐶𝑎𝐶𝑂3 decomposes faster than the products recombine. After a while,
the amounts of 𝐶𝑎𝑂 and 𝐶𝑂2 build up to a level at which the rate of combination
of 𝐶𝑎𝑂 and 𝐶𝑂2 is equal to the rate at which 𝐶𝑎𝐶𝑂3 dissociates, and hence, the
system has reached a state of dynamic equilibrium.
Dynamic equilibrium is a state of equilibrium in which the conversions of
reactants to products and products to reactants are still going on, although there
is no net change in the number of reactant and product molecules.
However, if a reaction proceeds in only one direction and goes to completion, it is
referred to as irreversible reaction.
For example, the combustion of methane to form carbon dioxide and water
𝐶𝐻4(𝑔) + 2𝑂2(𝑔) → 𝐶𝑂2(𝑔) + 2𝐻2𝑂(𝑙)
Characteristics of chemical equilibria
Chemical equilibria occur in reversible reactions
At equilibrium state, the rates of forward and backward reactions are
equal
A state of dynamic chemical equilibrium occurs in a closed system.
At equilibrium state, the concentrations of both reactants and products
remain unchanged
Chemical equilibria occur at constant temperature.
Katumba J. 2020 elbow grease is the best polish 2
The observable properties such as pressure, concentration, colour, density,
viscosity etc of the system remain unchanged with time
HOMOGENOUS AND HETEROGENEOUS EQUILIBRIA.
A homogenous equilibrium is an equilibrium in which all the reactants and products
are in the same phase or physical state. This may exist in liquid or gaseous phase
or aqueous state. For example;
The reaction between sulphur dioxide gas and oxygen gas to form sulphur
trioxide.
𝑆𝑂2(𝑔) + 𝑂2(𝑔) ⇌ 𝑆𝑂3(𝑔)
The reaction between ethanol and ethanoic acid to form ethyl ethanoate
/diethyl ether (esterification) 𝐶𝐻3𝐶𝑂𝑂𝐻(𝑙) + 𝐶𝐻3𝐶𝐻2𝑂𝐻(𝑙) ⇌ 𝐶𝐻3𝐶𝑂𝑂𝐶2𝐻5(𝑙) + 𝐻2𝑂(𝑙)
The reaction between hydrogen gas and gaseous iodine to form hydrogen
iodide 𝐻2(𝑔) + 𝐼2(𝑔) ⇌ 2𝐻𝐼(𝑔)
Decomposition of Dinitrogen tetroxide to form nitrogen dioxide. 𝑁2𝑂4(𝑔) ⇌ 2𝑁𝑂2(𝑔)
Reaction between iodine solution and potassium iodide to form potassium
triiodide complex 𝐼2(𝑎𝑞) + 𝐼−(𝑎𝑞) ⇌ 𝐼3
−(𝑎𝑞)
A heterogeneous equilibrium is an equilibrium in which two or more phases are
involved. For example;
Decomposition of calcium carbonate to form calcium oxide and carbon
dioxide.
𝐶𝑎𝐶𝑂3(𝑠) ⇌ 𝐶𝑎𝑂(𝑠) + 𝐶𝑂2(𝑔)
Reaction between iron and steam to form triiron tetraoxide and hydrogen 3𝐹𝑒(𝑠) + 4𝐻2𝑂(𝑔) ⇌ 𝐹𝑒3𝑂4(𝑠) + 4𝐻2(𝑔)
Reaction between bismuth chloride and water to form bismuth
oxychloride and hydrochloric acid 𝐵𝑖𝐶𝑙3(𝑎𝑞) + 𝐻2𝑂(𝑙) ⇌ 𝐵𝑖𝑂𝐶𝑙(𝑠) + 2𝐻𝐶𝑙(𝑎𝑞)
Note: If only ions are involved in an equilibrium, an ionic equilibrium is established.
Katumba J. 2020 elbow grease is the best polish 3
THE EQUILIBRIUM LAW (LAW OF MASS ACTION)
The law states that, “for a reaction at equilibrium, the ratio of product of the
concentration of the products to the product of the concentrations of the
reactants each raised to the power of their stoichiometric coefficients is
constant at constant temperature”
Derivation of equilibrium constant expressions (Application of the law)
Consider a hypothetical reversible reaction below;
𝒂𝑨(𝒂𝒒) + 𝒃𝑩(𝒂𝒒) ⇌ 𝒄𝑪(𝒂𝒒) + 𝒅𝑫(𝒂𝒒)
The rate of the forward reaction, 𝑹𝒇 = 𝑲𝒇[𝑨]𝒂[𝑩]𝒃
[𝑨] = concentration of 𝑨, [𝑩] = concentration of 𝑩
The rate of the backward reaction, 𝑹𝒃 = 𝑲𝒃[𝑪]𝒄[𝑫]𝒅
[𝑪]= concentration of 𝑪, [𝑫]= concentration of 𝑫
At dynamic chemical equilibrium;
Rate of forward reaction, 𝐾𝑓 = Rate of backward reaction, 𝐾𝑏 and substituting
for, 𝑅𝑓 and 𝑅𝑏
𝑲𝒇[𝑨]𝒂[𝑩]𝒃 = 𝑲𝒃[𝑪]𝒄[𝑫]𝒅
𝑲𝒇
𝑲𝒃=
[𝑪]𝒄 [𝑫]𝒅
[𝑨]𝒂 [𝑩]𝒃 where 𝑲𝒇
𝑲𝒃 = 𝑲𝒄
Therefore, the concentration equilibrium constant, 𝑲𝒄 =[𝑪]
c[𝑫]
d
[𝑨]a
[𝑩]b
The equilibrium constant, 𝑲𝒄, is the ratio of product of the equilibrium
concentrations of products to the product of the equilibrium concentrations of
reactants each raised to the power of their stoichiometric coefficient at a
constant temperature.
For a similar reaction taking place in a gaseous phase/state;
𝒂𝑨 (𝒈) + 𝒃𝑩(𝒈) ⇌ 𝒄𝑪(𝒈) + 𝒅𝑫 (𝒈)
The pressure equilibrium constant, 𝑲𝒑 =𝑷𝑪
𝒄 𝒙 𝑷𝑫𝒅
𝑷𝑨𝒂 𝒙 𝑷𝑩
𝒃
Where 𝑷𝒄, 𝑷𝑫, 𝑷𝑨 and 𝑷𝑩 are partial pressures of 𝑪, 𝑫, 𝑨 and 𝑪 respectively.
Units of equilibrium constants
Katumba J. 2020 elbow grease is the best polish 4
The units depend on the number of moles involved.
Case I: If the number of moles of the reactants and products are the same, then
both 𝑲𝒄 and 𝑲𝒑 have no units. For example;
𝑵𝟐(𝒈) + 𝑶𝟐(𝒈) ⇌ 𝟐𝑵𝑶𝟐(𝒈)
𝑲𝒄 =[𝑵𝑶𝟐]
2
[𝑵𝟐] [𝑶𝟐] =
(𝒎𝒐𝒍𝒍−𝟏)2
(𝒎𝒐𝒍𝒍−𝟏) (𝒎𝒐𝒍𝒍−𝟏) = 𝑵𝒐 𝒖𝒏𝒊𝒕𝒔
𝑲𝑷 =𝑷𝑵𝑶𝟐
𝟐
𝑷𝑵𝟐.𝑷𝑶𝟐
= (𝑵𝒎−𝟐)
(𝑵𝒎−𝟐) (𝑵𝒎−𝟐) = 𝑵𝒐 𝒖𝒏𝒊𝒕𝒔
Case II: If the number of moles is not the same, the units will depend on the
difference in the number of moles. For example;
𝑷𝑪𝒍𝟓(𝒈) ⇌ 𝑷𝑪𝒍𝟑(𝒈) + 𝑪𝒍𝟐(𝒈)
𝑲𝑪 =[𝑷𝑪𝒍𝟑] [𝑪𝒍𝟐]
[𝑷𝑪𝒍𝟓] =
(𝒎𝒐𝒍𝒍−𝟏) (𝒎𝒐𝒍𝒍−𝟏)
(𝒎𝒐𝒍𝒍−𝟏) = 𝒎𝒐𝒍𝒍−𝟏
𝑲𝒑 =𝑷𝑷𝑪𝒍𝟑
.𝑷𝑪𝒍𝟐
𝑷𝑷𝑪𝒍𝟓
=(𝑵𝒎−𝟐)(𝑵𝒎−𝟐)
(𝑵𝒎−𝟐)= 𝑵𝒎−𝟐
NB: When writing expressions for equilibrium constants, the following should
be noted;
Check whether only the concentrations /moles in a given volume are given
or total pressure at equilibrium
If only concentration/ moles are given, then an expression for 𝑲𝒄 should
be written.
If total pressure at equilibrium is given, then expression for 𝑲𝒑 should be
written.
Solids do not appear in the equilibrium constant expression since their
concentration is assumed to be constant.
For a 𝑲𝒑 expression, only gaseous reactants and products appear.
If water is one of the reactants, and its concentration is not given, or
remains unchanged, it is assumed to be present in a large excess hence
does not appear in the equilibrium constant expression. If water is in
gaseous state, then include it in the expression.
When only the expression is required, do not write the equation also as
part of the answer.
In the case of 𝑲𝒄 , strictly square brackets must be used.
Katumba J. 2020 elbow grease is the best polish 5
Question: In each of the following, write the equation for the equilibrium, the
equilibrium constant expression in terms of either 𝑲𝒄 and 𝑲𝒑 or both , depending
on what is indicated in brackets and state the units.
a) Reaction between ethanol and ethanoic acid in presence of an acid catalyst
(𝑲𝒄)
b) The reaction between hydrogen gas and gaseous iodine. (both 𝑲𝒄 and 𝑲𝒑)
c) Decomposition of Dinitrogen tetroxide to form nitrogen dioxide.(both 𝑲𝒄
and 𝑲𝒑)
d) Reaction between iodine solution and potassium iodide. (𝑲𝒄)
e) Thermal dissociation of calcium carbonate. (both 𝑲𝒄 and 𝑲𝒑)
f) Reaction between iron and steam to form triiron tetraoxide and hydrogen.
(both 𝑲𝒄 and 𝑲𝒑)
g) Hydrolysis of bismuth(III) chloride. (𝑲𝒄)
h) Reaction between hydrogen and nitrogen. (both 𝑲𝒄 and 𝑲𝒑)
i) Hydrolysis of ethylethanoate using dilute hydrochloric acid. (𝑲𝒄)
j) Reaction of phosphorus trichloride and chlorine. (both 𝑲𝒄 and 𝑲𝒑)
k) Conversion of sulphur dioxide to sulphur trioxide. (both 𝑲𝒄 and 𝑲𝒑)
l) Reaction between carbon monoxide and hydrogen to form gaseous
methanol. (both 𝑲𝒄 and 𝑲𝒑)
m) Dissociation of hydrogen iodide. (both 𝑲𝒄 and 𝑲𝒑)
n) Dissociation of phosphorus(V) chloride to phosphorus (III) chloride and
chlorine. (both 𝑲𝒄 and 𝑲𝒑)
o) Decomposition of sulphur trioxide. (both 𝑲𝒄 and 𝑲𝒑)
CALCULATIONS INVOLVING 𝑲𝒄 AND 𝑲𝒑
Case 1 (moles of products equal to moles of reactants)
General concept 𝐴 + 𝐵 ⇌ 2𝐶
For example; the reaction between hydrogen and iodine to form hydrogen iodide 𝐻2(𝑔) + 𝐼2 (𝑔) ⇌ 2𝐻𝐼 (𝑔)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 1 1 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 𝑥 2𝑥
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑎𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 1 – 𝑥 1 – 𝑥 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 1−𝑥
𝑣
1−𝑥
𝑣
2𝑥
𝑣
Assuming 𝑉 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑠 and recall 𝑚𝑜𝑙𝑒𝑠
𝑣𝑜𝑙𝑢𝑚𝑒= 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
Katumba J. 2020 elbow grease is the best polish 6
𝐾𝑐 =[𝐻𝐼]
2
[𝐻2] [𝐼2] =
(2𝑥
𝑣)
2
(1−𝑥
𝑣)(
1−𝑥
𝑣)
𝑲𝒄 = 𝟒𝒙𝟐
(𝟏−𝒙)(𝟏−𝒙)
For the same reaction, with the reactants and products in a gaseous phase, the
pressure equilibrium constant, 𝑲𝒑 is obtained as follows;
Let 𝑛𝐻2= 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻2 = (1 − 𝑥), 𝑛𝐼2
= 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻2 = (1 −
𝑥) and 𝑛𝐻𝐼 = 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻𝐼 = 2𝑥
𝑁 = 𝑇𝑜𝑡𝑎𝑙 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑎𝑛𝑑 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 = (𝑛𝐻2+ 𝑛𝐼2
+ 𝑛𝐻𝐼)
𝑃 = 𝑡𝑜𝑡𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
𝑃𝐻2=
𝑛𝐻2
𝑁 𝑥 𝑃, 𝑃𝐼2
=𝑛𝐼2
𝑁 𝑥 𝑃 𝑎𝑛𝑑 𝑃𝐻𝐼 =
𝑛𝐻𝐼
𝑁 𝑥 𝑃
𝑲𝒑 =(𝑷𝑯𝑰)𝟐
𝑷𝑯𝟐. 𝑷𝑰𝟐
Where 𝑃𝐻2 , 𝑃𝐼2
and 𝑃𝐻𝐼 are partial pressures of 𝐻2, 𝐼2 and 𝐻𝐼 respectively.
Degree of dissociation (𝜶) is the fraction per mole of a substance that
dissociates into simpler substances.
Consider the dissociation of hydrogen iodide in a vessel of volume 𝑽, assuming 𝜶
is the degree of dissociation, 𝒏 is the moles of 𝑯𝑰
2𝐻𝐼(𝑔) ⇌ 𝐻2(𝑔) + 𝐼2(𝑔)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 2𝑛 0 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 2𝑛𝛼 𝑛𝛼 𝑛𝛼
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 2𝑛(1 − 𝛼) 𝑛𝛼 𝑛𝛼
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 2𝑛(1−𝛼)
𝑣
𝑛𝛼
𝑣
𝑛𝛼
𝑣
And 𝑀𝑜𝑙𝑒𝑠
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
𝐾𝑐 = [𝐻2][𝐼2]
[𝐻𝐼]2 = (
𝑛𝛼
𝑣)(
𝑛𝛼
𝑣)
(2𝑛(1−𝛼)
𝑣)
2
𝑲𝒄 = 𝜶𝟐
𝟒(𝟏−𝜶)𝟐
Katumba J. 2020 elbow grease is the best polish 7
Example: 1
A mixture of 1 mole of hydrogen and 1 mole of iodine in a flask was heated until
the equilibrium was reached. On analysis, the equilibrium mixture was found to
contain 0.7 mole of hydrogen iodide. Calculate the 𝐾𝑐.
Solution
𝐻2(𝑔) + 𝐼2 (𝑔) ⇌ 2𝐻𝐼 (𝑔)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 1 1 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 𝑥 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 1 – 𝑥 1 – 𝑥 2𝑥
𝑆𝑖𝑛𝑐𝑒 2𝑥 = 0.7 ; 𝑥 = 0.35
Since there is no volume changes for reactants and products,
𝒆𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 𝒎𝒐𝒍𝒆𝒔 = 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏.
[𝐻𝐼] = 0.7 𝑚𝑜𝑙 𝑙−. [𝐻2] = [𝐼2] = (1 − 0.35) = 0.65 𝑚𝑜𝑙/1.
The equilibrium constant, 𝐾𝑐 =[𝐻𝐼]2
[𝐻2][𝐼2]=
(0.7)2
0.65 𝑥 0.65= 𝟏. 𝟏𝟔
Example: 2
0.206 moles of 𝐻2 and 0.144 moles of 𝐼2 were heated until equilibrium was attained
at equilibrium. 0.258 moles of 𝐻𝐼 was formed. Calculate the equilibrium constant,
𝐾𝑐.
Solution
𝐻2(𝑔) + 𝐼2 (𝑔) ⇌ 2𝐻𝐼 (𝑔)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 0.206 0.144 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 𝑥 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 0.206 – 𝑥 0.144 – 𝑥 2𝑥
𝑆𝑖𝑛𝑐𝑒 2𝑥 = 0.258 ; 𝑥 = 0.129
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛; [𝐻𝐼] = (2 𝑥 0.129) = 0.258,
[𝐻2] = (0.206 – 0.129) = 0.077,
[𝐼2] = (0.144 − 0.129) = 0.015
The equilibrium constant, 𝐾𝑐 =[𝐻𝐼]2
[𝐻2][𝐼2]=
(0.258)2
0.077 𝑥 0.015= 𝟓𝟕. 𝟔𝟑𝟏
Katumba J. 2020 elbow grease is the best polish 8
Example: 3
2.6g of HI were heated at 400°C and found to be 20% dissociated. Find
a) The concentration of each species of equilibrium.
b) The value of Kc at equilibrium (H = 1, I = 127)
Solution
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐻𝐼 = (1 𝑥 1) + (1 𝑥 127) = 128𝑔
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻𝐼 =2.6
128= 0.0203, 𝛼 =
20
100= 0.2
2𝐻𝐼(𝑔) ⇌ 𝐻2(𝑔) + 𝐼2(𝑔)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 2𝑛 0 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 2𝑛𝛼 𝑛𝛼 𝑛𝛼
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 2𝑛(1 − 𝛼) 𝑛𝛼 𝑛𝛼
2𝑛 = 0.0203 ; 𝑛 = 0.01015
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛; [𝐻𝐼] = (2 𝑥 0.01015)(1 − 0.2) = 0.01624
[𝐻2] = (0.01015 𝑥 0.2) = 0.00203
[𝐼2] = (0.01015 𝑥 0.2) = 0.00203
𝐾𝑐 = [𝐻2][𝐼2]
[𝐻𝐼]2 =0.00203 𝑥 0.00203
(0.01624)2 = 𝟎. 𝟎𝟏𝟓𝟔
Example: 4
When 6.22cm3 of hydrogen were heated with 5.71cm3 of iodine in a sealed tube
at 356°C it was found that 9.60cm3 of hydrogen iodide were present at
equilibrium. Calculate;
a) The equilibrium constant.
b) The volume of hydrogen iodide in the equilibrium mixture formed by
heating together 6.41cm3 of hydrogen and 10.40cm3 of iodine at 356°C.
Solution
a) Since the number of moles of each gas present is proportional to the gas
volume at equilibrium 𝐻2(𝑔) + 𝐼2 (𝑔) ⇌ 2𝐻𝐼 (𝑔)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 6.22 5.71 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 𝑥 2𝑥
Katumba J. 2020 elbow grease is the best polish 9
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 6.22 – 𝑥 5.71 – 𝑥 2𝑥
𝑆𝑖𝑛𝑐𝑒 2𝑥 = 9.6 ; 𝑥 = 4.8
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛; [𝐻𝐼] = (2 𝑥 4.8) = 9.6
[𝐻2] = 6.22 − 4.8 = 1.42
[𝐼2] = 5.71 − 4.8 = 0.91
The equilibrium constant, 𝐾𝑐 =[𝐻𝐼]2
[𝐻2][𝐼2]=
(9.6)2
1.42 𝑥 0.91= 𝟕𝟏. 𝟑𝟐
b) 𝐻2(𝑔) + 𝐼2 (𝑔) ⇌ 2𝐻𝐼 (𝑔)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 6.41 10.4 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 𝑥 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 6.41 – 𝑥 10.4 – 𝑥 2𝑥
𝐾𝑐 =[𝐻𝐼]2
[𝐻2][𝐼2]=
(2𝑥)2
(6.41−𝑥)(10.4−𝑥)
71.32 =4𝑥2
(6.41−𝑥)(10.4−𝑥)
𝑥2 = 17.83(6.41 − 𝑥)(10.4 − 𝑥)
𝑥2 = 17.83(66.664 − 16.81𝑥 + 𝑥2
𝑥2 = 1188.62 − 299.7223𝑥 + 17.83𝑥2
16.83𝑥2 − 299.7223𝑥 + 1188.62 = 0
𝑈𝑠𝑖𝑛𝑔 𝑥 =−𝑏±√𝑏2−4𝑎𝑐
2𝑎; 𝑥 =
299.7223±√(−299.7223)2−4(16.83 𝑥 1188.62)
2 𝑥 16.83
𝐸𝑖𝑡ℎ𝑒𝑟 𝑥 = 5.96 𝑐𝑚3 𝑜𝑟 𝑥 = 11.85 𝑐𝑚3 𝑠𝑜 𝑡ℎ𝑎𝑡 2𝑥 = 11.92 𝑜𝑟 23.7 𝑐𝑚3
𝟐𝟑. 𝟕 𝑐𝑚3 is inadmissible since the volume of hydrogen iodide cannot be more than
the original volume of hydrogen and iodide together. Therefore, the volume of
hydrogen iodide present at equilibrium is 𝟏𝟏. 𝟗𝟐 𝑐𝑚3
Trial Questions:
1. 3 moles of hydrogen and 1 mole of iodine were heated together at 500oC
until equilibrium was established. Calculate the number of moles of
hydrogen iodide present in the equilibrium mixture at 500oC. (The
equilibrium constant, 𝑲𝑐 for the reaction between hydrogen and iodine is
50)
Katumba J. 2020 elbow grease is the best polish 10
2. a) State three characteristics of a chemical equilibrium.
b) 25 moles of hydrogen and 18 moles of iodine vapour were heated in a 1
litre sealed tube at 465oC. When equilibrium was attained, the tube was
rapidly cooled and found to contain 30.8 moles of hydrogen iodide.
c) Give a reason why the tube was rapidly cooled.
d) Calculate the:
(i) Value of the equilibrium constant for the reaction taking place in the
flask.
(ii) Degree of dissociation of hydrogen iodide. (𝜶 = 𝟎. 𝟐𝟒𝟓)
3. 1.54g of hydrogen iodide were heated in a 600 cm3 bulb at 530oC. When
equilibrium was attained, the bulb was rapidly cooled to room temperature
and broken under potassium iodide solution. The iodine formed required 67
cm3 of 0.1M sodium thiosulphate solution for complete reaction. Calculate
the;
a) number of moles of hydrogen iodide in 1.54g
b) number of moles of iodine formed
c) value of 𝒌𝒄 at 530oC
4. 3.24g of hydrogen iodide were heated at 450oC in a glass bulb of volume
800cm3. When equilibrium was attained, the bulb was rapidly cooled to room
temperature and then broken under a solution of potassium iodide. The
iodine formed required 36.0 cm3 of a 0.2M sodium thiosulphate solution in
the presence of starch indicator for complete reaction.
a) Explain why the bulb was rapidly cooled and broken under potassium
iodide solution
b) Calculate the equilibrium constant for the reaction at 450oC
Case II (Moles of products greater than moles of reactants)
General Concept A ⇌ 2B
For example; decomposition of N2O4 dinitrogen tetraoxide
𝑁2𝑂4(𝑔) ⇌ 2𝑁𝑂2(𝑔)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 1 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 (1 − 𝑥) 2𝑥
Katumba J. 2020 elbow grease is the best polish 11
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 (1−𝑥)
𝑣
2𝑥
𝑣
𝐾𝑐 =[𝑁𝑂2]2
[𝑁2𝑂4]=
(2𝑥
𝑣)
2
(1−𝑥
𝑣)
𝑲𝒄 =𝟒𝒙𝟐
(𝟏−𝒙)𝒗
For the same reaction, with the reactants and products in a gaseous phase, the
pressure equilibrium constant, 𝑲𝒑 is obtained as follows;
𝑵𝟐𝑶𝟒(𝒈) ⇌ 𝟐𝑵𝑶𝟐(𝒈)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 1 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 (1 − 𝑥) 2𝑥 {total Eqm moles= (1 + 𝑥)}
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠 (1−𝑥 )
(1+𝑥 )
2𝑥
(1+𝑥)
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑠 (1−𝑥 )
(1+𝑥 )𝑃
2𝑥
(1+𝑥)𝑃
The pressure equilibrium constant, 𝐾𝑝 for the reaction is obtained as follows;
𝐾𝑝 =𝑃𝑁𝑂2
2
𝑃𝑁2𝑂4
=(
2𝑥
(1+𝑥)𝑃)
2
(1−𝑥 )
(1+𝑥 )𝑃
𝑲𝒑 =𝟒𝒙𝟐
(𝟏+𝒙 )(𝟏−𝒙 )𝑷
Example: 1
N2O4 at 1.0 atmosphere and 25°C dissociated by 18.5%. Calculate;
a) its 𝐾𝑝 at this temperature.
b) if the atmospheric pressure was reduced to half its original value at the
same temperature, calculate the degree of dissociation of the gas.
Solution;
a) 𝑵𝟐𝑶𝟒(𝒈) ⇌ 𝟐𝑵𝑶𝟐(𝒈)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 1 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝛼 2𝛼
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 (1 − 𝛼) 2𝛼
𝑇𝑜𝑡𝑎𝑙 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 = (1 − 𝛼) + 2𝛼 = (1 + 𝛼)
Katumba J. 2020 elbow grease is the best polish 12
𝛼 =18.5
100= 0.185
𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑠; 𝑃𝑁2𝑂4=
(1−𝛼)
(1+𝛼)𝑃 =
(1−0.185)
(1+0.185) 𝑥 1.0 = 0.6878 𝑎𝑡𝑚
𝑃𝑁𝑂2=
2𝛼
(1+𝛼)𝑃 =
(2 𝑥 0.185)
(1+0.185) 𝑥 1.0 = 0.3122 𝑎𝑡𝑚
𝐾𝑝 =𝑃𝑁𝑂2
2
𝑃𝑁2𝑂4
=(0.3122)2
0.6878= 𝟎. 𝟏𝟒𝟐 𝑎𝑡𝑚
b) According to the Le’ Chartelier’s principle, the equilibrium constant, 𝐾𝑝 for
the reaction remains constant irrespective of the change in pressure of the
system.
𝐾𝑝 =𝑃𝑁𝑂2
2
𝑃𝑁2𝑂4
=(
2𝛼
(1+𝛼)𝑃)
2
(1−𝛼)
(1+𝛼)𝑃
=4𝛼2
(1+𝛼)(1−𝛼)𝑃; 𝐵𝑢𝑡 𝑃 = 0.5 𝑎𝑡𝑚
0.142 =4𝛼2
1−𝛼2 𝑥 0.5 ; 4𝛼2 = 0.284(1 − 𝛼2) ; 4.284𝛼2 = 0.284
𝛼 = √0.284
4.284= 0.257 ; 𝛼 = 𝟐𝟓. 𝟕%
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑡ℎ𝑒 𝑔𝑎𝑠, 𝑁2𝑂4 𝑖𝑠 𝟐𝟓. 𝟕% 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑 𝑎𝑡 𝑎 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 0.5 𝑎𝑡𝑚.
Example: 2
When 80.4 g of Phosphorus (V) chloride were placed in a 9.0 litre vessel and
heated at a certain pressure, 8.4 g of chlorine were formed at equilibrium.
Calculate the:
a) amount of phosphorus(V) chloride and phosphorus (III) chloride at
equilibrium in moles per litre.
b) equilibrium constant, 𝐾𝑐, for the reaction and state its units.
Solution.
a) 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑃𝐶𝑙5 = 31 + (5 𝑥 35.5) = 208.5 𝑔
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑃𝐶𝑙5 =80.4
208.5= 0.3856
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐ℎ𝑙𝑜𝑟𝑖𝑛𝑒 = (2 𝑥 35.5) = 71 𝑔
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶𝑙2 =8.4
71= 0.1183
𝑷𝑪𝒍𝟓(𝒈) ⇌ 𝑷𝑪𝒍𝟑(𝒈) + 𝑪𝒍𝟐(𝒈)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 0.3856 0 0
Katumba J. 2020 elbow grease is the best polish 13
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 𝑥 𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 0.3856 − 𝑥 𝑥 𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 0.3856−𝑥
𝑣
𝑥
𝑣
𝑥
𝑣
⇒ 𝑥 = 0.1183 𝑚𝑜𝑙𝑒𝑠
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟; 𝑃𝐶𝑙5 =0.3856−0.118
9.0= 𝟎. 𝟎𝟐𝟗𝟕 𝑚𝑜𝑙 𝑙−
𝑃𝐶𝑙3 =0.1183
9.0= 𝟎. 𝟎𝟏𝟑𝟏 𝑚𝑜𝑙 𝑙−
b) 𝐾𝑐 =[𝑃𝐶𝑙3][𝐶𝑙2]
[𝑃𝐶𝑙5]=
0.0131 𝑚𝑜𝑙 𝑙− 𝑥 0.0131 𝑚𝑜𝑙 𝑙−
0.0297 𝑚𝑜𝑙 𝑙− ; 𝑆𝑖𝑛𝑐𝑒 𝐸𝑞𝑚 [𝑃𝐶𝑙3] = 𝐸𝑞𝑚 [𝐶𝑙2]
𝐾𝑐 = 𝟎. 𝟎𝟎𝟓𝟕𝟖 𝑚𝑜𝑙 𝑙−
Example: 3
When heated, carbon dioxide decomposes according to the equation:
2𝐶𝑂2(𝑔) ⇌ 2𝐶𝑂(𝑔) + 𝑂2(𝑔)
If at a certain temperature and a pressure of one atmosphere, 60% of the original
carbon dioxide remained undissociated. Calculate the equilibrium constant, 𝑲𝒑 for
the reaction.
Solution.
𝛼 = (100 − 60)% = 40% ; 𝛼 =40
100= 0.4 ; 𝑇𝑜𝑡𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒, 𝑃 = 1.0 𝑎𝑡𝑚
𝟐𝑪𝑶𝟐(𝒈) ⇌ 𝟐𝑪𝑶(𝒈) + 𝑶𝟐(𝒈)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 2 0 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 2𝛼 2𝛼 𝛼
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 2(1 − 𝛼) 2𝛼 𝛼
𝑇𝑜𝑡𝑎𝑙 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 = 2(1 − 𝛼) + 2𝛼 + 𝛼 = (2 + 𝛼)
𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑠; 𝑃𝑂2=
𝛼
(2+𝛼)𝑃 =
0.4
(2+0.4) 𝑥 1.0 =
1
6 𝑎𝑡𝑚
𝑃𝐶𝑂 =2𝛼
(2+𝛼)𝑥 𝑃 =
(2 𝑥 0.4)
(2+0.4) 𝑥 1.0 =
1
3 𝑎𝑡𝑚
𝑃𝐶𝑂2=
2(1−𝛼)
(2+𝛼) 𝑥 𝑃 =
2(1−0.4)
(2+0.4) 𝑥 1.0 =
1
2 𝑎𝑡𝑚
𝐾𝑝 =𝑃𝐶𝑂
2 𝑥 𝑃𝑂2
𝑃𝐶𝑂22 =
(1
3)
2(
1
6)
(1
2)
2 = 𝟎. 𝟎𝟕𝟒 𝑎𝑡𝑚
Katumba J. 2020 elbow grease is the best polish 14
Example: 4
1 mole of phosphorus pentachloride was heated in 30cm3 container to a
temperature of 30°C.At equilibrium, the container was found to contain 36.2%
chlorine.
a) Calculate the equilibrium constant, 𝐾𝑐
b) Explain what would happen to the concentration of PCl5 if the pressure was
increased keeping the temperature constant.
Solution.
(a) 𝑷𝑪𝒍𝟓(𝒈) ⇌ 𝑷𝑪𝒍𝟑 (𝒈) + 𝑪𝒍𝟐(𝒈)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 1 0 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 𝑥 𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 (1 − 𝑥) 𝑥 𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑟𝑖𝑜𝑛 (1−𝑥)
𝑣
𝑥
𝑣
𝑥
𝑣
𝑇𝑜𝑡𝑎𝑙 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 = (1 − 𝑥) + 𝑥 + 𝑥 = (1 + 𝑥)
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑙2 =𝑥
(1+𝑥)=
36.2
100 ; 𝑥 = 0.362(1 + 𝑥)
𝑥 =0.362
0.638= 0.567 𝑚𝑜𝑙𝑒𝑠
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛𝑠 𝑓𝑜𝑟; 𝐶𝑙2 =0.567
30= 𝟎. 𝟎𝟏𝟖𝟗 𝑚𝑜𝑙 𝑐𝑚−3
𝑃𝐶𝑙3 =0.567
30= 𝟎. 𝟎𝟏𝟖𝟗 𝑚𝑜𝑙 𝑐𝑚−3
𝑃𝐶𝑙5 =(1−0.567)
30= 𝟎. 𝟎𝟏𝟒𝟒 𝑚𝑜𝑙 𝑐𝑚−3
𝐾𝑐 =[𝑃𝐶𝑙3][𝐶𝑙2]
[𝑃𝐶𝑙5]=
0.0189 𝑥 0.0189
0.0144= 𝟎. 𝟎𝟐𝟒𝟖 𝑚𝑜𝑙 𝑐𝑚−3
b) Refer to the factors that affect equilibrium position.
Trial Questions:
1) Dinitrogen tetraoxide dissociates at 40oC and 1 atm according to the
following equation. 𝑁2𝑂4(𝑔) ⇌ 2𝑁𝑂2(𝑔) ∆𝐻 = +57𝑘𝐽𝑚𝑜𝑙−
a) Write an equation for the equilibrium constant, 𝑲𝑝.
b) Draw a well labelled energy level diagram for the reaction.
Katumba J. 2020 elbow grease is the best polish 15
c) The reaction mixture in (b) was found to contain 60% by volume of
nitrogen dioxide. Calculate the equilibrium constant, 𝑲𝑝 at 40oC for
the reaction.
2) For the following gas equilibrium, 𝑁2𝑂4(𝑔) ⇌ 2𝑁𝑂2(𝑔) at 333K, the
equilibrium constant, 𝑲𝑝 is 1.33 atmospheres. Calculate the degree of
dissociation of one mole of dinitrogen tetraoxide at 333K if the total
pressure of the system is 2 atmospheres.
3) 2.00 g of Phosphorus (V) chloride were allowed to reach equilibrium at
200oC in a 1 dm3 capacity vessel. If the equilibrium constant of the above
reaction is 0.008 mol dm-3at this temperature and in the conditions stated.
Calculate the percentage dissociation of phosphorus pentachloride at
equilibrium.
4) 1 mole of phosphorus (V) chloride was strongly heated in a closed bulb until
equilibrium was obtained. The glass bulb was then rapidly broken under
potassium iodide solution. The bulb was found to contain 40.70% of
chlorine.
a) Write equations for the reactions that took place when:
(i) The glass bulb was strongly heated
(ii) The glass bulb was broken under potassium iodide solution.
b) State the reasons why the bulb;
(i) Was rapidly broken
(ii) Was broken under potassium iodide solution
c) Determine the;
(i) Degree of dissociation of phosphorus(V) chloride
(ii) Equilibrium constant for the reaction.
Case III (Moles of reactants greater than moles of products)
General Concept A + 3B ⇌ 2C
For example, manufacture of ammonia gas in Haber process. i.e.
𝑁2(𝑔) + 3𝐻2(𝑔) ⇌ 2𝑁𝐻3(𝑎𝑞)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 1 3 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 3𝑥 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 (1 − 𝑥) 3(1 − 𝑥) 2𝑥
Katumba J. 2020 elbow grease is the best polish 16
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 (1−𝑥)
𝑣
3(1−𝑥)
𝑣
2𝑥
𝑣
The concentration equilibrium constant, 𝑲𝒄, for the reaction is obtained as
follows;
𝐾𝑐 =[𝑁𝐻3]2
[𝑁2][𝐻2]3 =(
2𝑥
𝑣)
2
(1−𝑥
𝑣)(
3(1−𝑥)
𝑣)
3
𝑲𝒄 =𝟒𝒙𝟐𝒗𝟐
(𝟏−𝒙)(𝟑−𝟑𝒙)𝟑
For the same reaction, with the reactants and products in a gaseous phase, the
pressure equilibrium constant, 𝑲𝒑 is obtained as follows;
𝑁2(𝑔) + 3𝐻2(𝑔) ⇌ 2𝑁𝐻3(𝑎𝑞)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 1 3 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 3𝑥 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 (1 − 𝑥) 3(1 − 𝑥) 2𝑥
𝑇𝑜𝑡𝑎𝑙 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 = (1 − 𝑥) + 3(1 − 𝑥) + 2𝑥 = 4 − 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 (1−𝑥)
(4−2𝑥)
3(1−𝑥)
(4−2𝑥)
2𝑥
(4−2𝑥)
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑠 (1−𝑥
4−2𝑥) 𝑃 (
3(1−𝑥)
4−2𝑥) 𝑃 (
2𝑥
4−2𝑥) 𝑃
𝐾𝑝 =𝑃𝑁𝐻3
2
𝑃𝑁2 .𝑃𝐻23 =
(2𝑥
4−2𝑥𝑃)
2
(1−𝑥
4−2𝑥)𝑃.(
3(1−𝑥)
4−2𝑥𝑃)
3
𝑲𝒑 =𝟐𝒙𝟐(𝟒−𝒙)𝟐
(𝟏−𝒙)(𝟑−𝟑𝒙)𝟑𝑷𝟐
Example: 1
Nitrogen and hydrogen are mixed in a ratio 1:3. At equilibrium at 600oC and 10
atmospheres, the percentage of ammonia in the mixture of gases is 15%.
a) Write equation for the reaction
b) Write an expression for the equilibrium constant.
c) Calculate the equilibrium constant at that temperature and state its units.
Solution:
a) 𝑁2(𝑔) + 3𝐻2(𝑔) ⇌ 2𝑁𝐻3(𝑔)
b) 𝐾𝑝 =𝑃𝑁𝐻3
2
𝑃𝑁2 . 𝑃𝐻23
c) 𝑁2(𝑔) + 3𝐻2(𝑔) ⇌ 2𝑁𝐻3(𝑔)
Katumba J. 2020 elbow grease is the best polish 17
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 1 3 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 3𝑥 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 (1 − 𝑥) 3(1 − 𝑥) 2𝑥
𝑇𝑜𝑡𝑎𝑙 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 = (1 − 𝑥) + 3(1 − 𝑥) + 2𝑥 = 4 − 2𝑥
𝑀𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑁𝐻3 =2𝑥
4−2𝑥=
15
100 ; 𝑥 = 0.15(2 − 𝑥) ; 𝑥 = 𝟎. 𝟐𝟔𝟎𝟗 𝑚𝑜𝑙𝑒𝑠
𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑠; 𝑃𝑁𝐻3=
2 𝑥 0.2609
4−(2 𝑥 0.2609)𝑥 10 = 1.500 𝑎𝑡𝑚
𝑃𝐻2=
3(1−0.2609)
4−(2 𝑥 0.2609) 𝑥 10 = 6.375 𝑎𝑡𝑚
𝑃𝑁2=
1−0.2609
4−(2 𝑥 0.2609) 𝑥 10 = 2.125 𝑎𝑡𝑚
𝐾𝑝 =(1.500)2
2.125𝑥 (6.375)3 = 𝟒. 𝟎𝟖𝟕 𝒙 𝟏𝟎−𝟑 𝑎𝑡𝑚−2
Example: 2
Nitrogen monoxide combines with oxygen to form nitrogen dioxide according to
the equation;
2𝑁𝑂(𝑔) + 𝑂2(𝑔) ⇌ 2𝑁𝑂2(𝑔)
a) Write an expression for the equilibrium constant. 𝐾𝑐 .
b) 3.0moles of nitrogen monoxide and 1.5 moles of oxygen were out into a 1
litre vessel which was heated to 400°C. When equilibrium was
established, the vessel was found to contain 0.5 moles of oxygen.
Calculate the equilibrium constant 𝐾𝑐 at this temperature.
c) When the temperature was raised to 500°C, the equilibrium mixture was
found to contain 25% of the initial nitrogen monoxide. Calculate the
equilibrium constant at this temperature.
d) From your answers in (b) and (c) Deduce whether the process is
endothermic or exothermic. Explain your answer.
Solution:
a) 𝐾𝑐 =[𝑁𝑂2]2
[𝑁𝑂]2[𝑂2]
b) 2𝑁𝑂(𝑔) + 𝑂2(𝑔) ⇌ 2𝑁𝑂2(𝑔)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 3.0 1.5 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 2𝑥 𝑥 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 (3.0 − 2𝑥) (1.5 − 𝑥) 2𝑥
Katumba J. 2020 elbow grease is the best polish 18
⇒ 1.5 − 𝑥 = 0.5 ; 𝑥 = 1.0 𝑚𝑜𝑙𝑒𝑠
𝐴𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚, [𝑁𝑂2] =2𝑥
1=
2 𝑥 1.0
1= 2.0 𝑚𝑜𝑙 𝑙−1
[𝑁𝑂] =3.0−2𝑥
𝑣=
3.0−(2 𝑥 1.0)
1= 1.0 𝑚𝑜𝑙 𝑙−1
[𝑂2] =1.5−𝑥
𝑣=
1.5−1.0
1= 0.5 𝑚𝑜𝑙 𝑙−1
𝐾𝑐 =2.02
1.02 𝑥 0.5= 𝟖 𝑚𝑜𝑙−1𝑙
c) 2𝑁𝑂(𝑔) + 𝑂2(𝑔) ⇌ 2𝑁𝑂2(𝑔)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 3.0 1.5 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 2𝑥 𝑥 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 (3.0 − 2𝑥) (1.5 − 𝑥) 2𝑥
⇒ 3.0 − 2𝑥 =25
100 𝑥 3.0 ; 𝑥 = 1.125 𝑚𝑜𝑙𝑒𝑠
𝐴𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚, [𝑁𝑂2] =2𝑥
1=
2 𝑥 1.125
1= 2.25 𝑚𝑜𝑙 𝑙−1
[𝑁𝑂] =3.0−2𝑥
𝑣=
3.0−(2 𝑥 1.125)
1= 0.75 𝑚𝑜𝑙 𝑙−1
[𝑂2] =1.5−𝑥
𝑣=
1.5−1.125
1= 0.375 𝑚𝑜𝑙 𝑙−1
𝐾𝑐 =2.252
0.752 𝑥 0.375= 𝟐𝟒 𝑚𝑜𝑙−1𝑙
d) The process is endothermic as increase in temperature also increased the
amount of nitrogen dioxide formed.
Trial Questions:
1. In an experiment, Hydrogen gas and nitrogen gas in the mole ratio 3:1
produced 0.0735 mole fraction of NH3 at 350°C and total pressure of
1013 kNm-2. Calculate the 𝑲𝑷 value for the reaction.
2. Stoichiometric amounts of hydrogen and nitrogen were reacted at 50
atm. At equilibrium, 0.8 moles of ammonia were formed. Calculate the;
a) amount of hydrogen and nitrogen present at equilibrium.
b) value of the equilibrium constant.
Katumba J. 2020 elbow grease is the best polish 19
3. The 𝐾𝑝 for this equilibrium reaction; N2(g) + 3H2(g) ⇌ 2NH3(g) is 1.45 x
10-5 Pa at 500°C. Calculate the partial pressure of NH3 when the partial
pressure of H2 is 0.928 atmospheres and that of N2 is 0.432
atmospheres.
4. A mixture of Iron and steam was allowed to reach equilibrium at 600°C.
The equilibrium pressures of hydrogen and steam were 3.2 kPa and 2.4
kPa respectively. Calculate the value of the equilibrium constant in terms
of partial pressures.
5. For the reaction, 3𝐻2(𝑔) + 𝑁2(𝑔) ⇌ 2𝑁𝐻3(𝑔)
a) Calculate the molar percentage of ammonia in the equilibrium
mixture formed at 400oC and at a pressure of 3 × 107 Pa, when
gaseous hydrogen and nitrogen are mixed in a 3:1 ratio and there is
61% conversion of nitrogen to ammonia.
b) Determine the equilibrium constant, 𝑲𝒑 of the reaction.
c) Given that the value of 𝑲𝒑 at a given temperature is 2.0 × 10−14 Pa2,
calculate the pressure at which ammonia is 95% dissociated into its
elements at that temperature
6. A 10.0 cm3 mixture contains the initial amounts per mole; ethanol 0.0515;
ethanoic acid 0.0525; water 0.0167; ester 0.0315; 𝐻+(aq) 1.00 x 10-3.the
equilibrium equation is; 𝐶𝐻3𝐶𝑂2 𝐻(𝑙) + 𝐻2𝑂(𝑙) ⇌ 𝐶𝐻3𝐶𝑂2𝐶2𝐻5(𝑙) + 𝐻2𝑂 (𝑙)
The equilibrium amount of ethanoic acid = 0.0255 moles. Find 𝐾𝑐.
7. Calculate the amount of ethyl ethanoate when 1 mole of ethanoic acid and
1 mole of ethanol are reacted until equilibrium is attained. (𝐾𝑐 = 4)
8. (a) Write an equation for the reaction between hydrogen and nitrogen.
(b) At 500oC, the equilibrium concentration of hydrogen is 0.250 moll-1
and that of nitrogen is 2.7 moll-1 Calculate the equilibrium concentration
of ammonia at the same temperature given that 𝑲𝒄 = 𝟔 × 𝟏𝟎−𝟐𝒎𝒐𝒍−𝟐𝒍𝟐 at
500oC.
9. When 8.28g of ethanol were heated with 60g of ethanoic acid. 49.74g of
the acid remained at equilibrium.
a) Calculate the value of Kc.
𝐻+
Katumba J. 2020 elbow grease is the best polish 20
b) When 13.8g of ethanol and 12g of ethanoic acid were mixed, what
would be the mass of ethyl ethanoate present at equilibrium?
10. Propene reacts with steam according to the following equation.
𝐶𝐻3𝐶𝐻 = 𝐶𝐻2 (𝑔) + 𝐻2𝑂(𝑔) ⇌ 𝐶𝐻3𝐶𝐻𝑂𝐻𝐶𝐻3(𝑔)
At a certain temperature and total pressure of 197.38 atmospheres, the
equilibrium partial pressures of propene and steam are 74.02 and 93.76
atmospheres respectively. Calculate the value of 𝑲𝒑 at this temperature
and state its units.
RELATIONSHIP BETWEEN 𝑲𝒄 AND 𝑲𝑷
Consider a reaction;
𝑨𝒂 + 𝒃𝑩 ⇌ 𝒄𝑪 + 𝒅𝑫
The equilibrium constant, 𝑲𝒄 for the reaction is given by
𝑲𝒄 =[𝑪]𝒄[𝑫]𝒅
[𝑨]𝒂[𝑩]𝒃 ………………………………………………………………1
If the reacting species are gaseous, then the pressure equilibrium constant, 𝑲𝒑
for the reaction is given by
𝑲𝒑 =𝑷𝑪
𝒄 𝒙 𝑷𝑫𝒅
𝑷𝑨𝒂 𝒙 𝑷𝑩
𝒃 …………………………………………………….…2
If gases are ideal, then from ideal gas equation;
𝑷𝒊𝑽 = 𝒏𝒊𝑹𝑻, 𝑷 = 𝒏𝒊
𝒗𝑹𝑻 𝒃𝒖𝒕
𝒏𝒊
𝒗= [𝒊]
𝑇ℎ𝑢𝑠, 𝑷𝒊 = [𝒊]𝑹𝑻
Where 𝑷𝒊 = pressure of gas 𝒊; [𝒊] = concentration of gas 𝒊; 𝑹 = Molar gas
constant; 𝒏 = moles of gas; 𝑻 = Temperature
Substituting for partial pressures in equation (2)
𝑲𝒑 =([𝑪]𝑹𝑻)𝒄([𝑫]𝑹𝑻)𝒅
([𝑨]𝑹𝑻)𝒂([𝑩]𝑹𝑻)𝒃
𝑲𝒑 =[𝑪]𝒄[𝑫]𝒅(𝑹𝑻)(𝒄+𝒅)
[𝑨]𝒂[𝑩]𝒃(𝑹𝑻)(𝒂+𝒃) but [𝑪]
c[𝑫]
d
[𝑨]a
[𝑩]b
= 𝑲𝒄
𝐊𝐩 = 𝐊𝐜(𝐑𝐓)𝚫𝐧 𝒊𝒔 𝒕𝒉𝒆 𝒓𝒆𝒍𝒂𝒕𝒊𝒐𝒏𝒔𝒉𝒊𝒑 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝑲𝒄 𝒂𝒏𝒅 𝑲𝒑
Katumba J. 2020 elbow grease is the best polish 21
Where 𝚫𝒏 = (𝒄 + 𝒅) + (𝒂 + 𝒃) = change in number of moles of products and
reactants
Example: Consider 𝑛 moles of dinitrogen tetraoxide occupying 𝑉 dm3 vessel at
temperature, 𝑇 and dissociating as;
𝑁2𝑂4(𝑔) ⇌ 2𝑁𝑂2(𝑔)
From the equation above, 𝑛𝑝 = 2 and 𝑛𝑟 = 1, thus, Δ𝑛 = 2 − 1 = 1
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑲𝒑 = 𝑲𝒄𝑹𝑻 is the relationship between 𝑲𝒄 and 𝑲𝒑 for the above
reaction.
Question: 𝑲𝒄for this reaction; S02 (g) + 1
2 02 (g) ⇌ S03 (g) at 727°C is 16.7.
Calculate the 𝑲𝒑 for this reaction. (𝑲𝒑 = 𝟎. 𝟏𝟖𝟑
USING REACTION QUOTIENT,𝑸 AND EQUILIBRIUM CONSTANT, 𝑲 TO
DETERMINE THE DIRECTION OF EQUILIBRIUM POSITION.
The reaction quotient, 𝑸 is the resulting value obtained when initial
concentrations of reactants and products are applied in the law of mass action
instead of equilibrium concentrations.
To determine the direction of shift of equilibrium position of the system, we
compare the values of 𝑄 and equilibrium constant 𝐾.
1. If 𝑸 = 𝑲, the reaction mixture is already at equilibrium, so no shift occurs
2. If 𝑸 > 𝑲, the reaction will go to the left. The ratio of initial concentrations
of products to reactants is too large. To reach equilibrium, the reaction
will move toward equilibrium by consuming products and forming more
reactants until equilibrium is achieved.
3. If 𝑸 < 𝑲, the reaction will go to the right. The ratio of initial
concentrations of products to reactants is too small. The reaction will
move forward consuming reactants and forming products
Questions:
1. For the reaction: 𝑁2(𝑔) + 𝑂2(𝑔) ⇌ 2𝑁𝑂(𝑔), the equilibrium constant, 𝐾𝑐 =
1.0 𝑥 10−5, at 1500 𝐾. Predict the direction the reaction will move in if the
reactants and products have the following concentrations: [𝑁2] =
0.05 𝑚𝑜𝑙 𝑙− , [𝑂2] = 0.02 𝑚𝑜𝑙 𝑙− , and [𝑁𝑂] = 0.30 𝑚𝑜𝑙 𝑙−
Katumba J. 2020 elbow grease is the best polish 22
2. For the reaction: 𝑁2𝑂4(𝑔) ⇌ 2𝑁𝑂2(𝑔), the equilibrium constant, 𝐾𝐶 =
5.0 𝑥 10− at 100°𝐶. Predict the direction the reaction will move in if the
concentration of 𝑁2𝑂4 is 0.02 𝑚𝑜𝑙 𝑙− and the concentration of 𝑁𝑂2 is
0.10 𝑚𝑜𝑙 𝑙−.
3. In the water–gas shift reaction, carbon monoxide produced by steam-
reforming reaction of methane reacts with steam at elevated
temperatures to produce more hydrogen:
𝐶𝑂(𝑔) + 𝐻2𝑂(𝑔) ⇌ 𝐶𝑂2(𝑔) + 𝐻2(𝑔)
If 𝐾𝑐 = 0.64 at 900 𝐾. If 0.010 moles of both carbon monoxide and water,
0.0080 moles of carbon dioxide, and 0.012 moles of hydrogen are injected
into a 4.0 litre reactor and heated to 900 𝐾, will the reaction proceed to
the left or to the right?
4. 1 mole of sulphur trioxide was introduced into a 1 𝑑𝑚−3 vessel. The vessel
was heated to 1000 𝐾 until equilibrium was attained. At equilibrium, 0.35
moles of sulphur trioxide was present.
a. Write: (i) equation for the decomposition of sulphur trioxide
(ii) an expression for the equilibrium constant, 𝑲𝒄
b. Calculate the value of 𝑲𝒄.
c. 0.2 moles of Sulphur dioxide, 0.1 mole of oxygen and 0.7 moles of
Sulphur trioxide, were introduced into the vessel in (a) at 1000 𝐾. (i)
(i) Calculate the new 𝑲𝒄 value for the reaction.
(ii) Using your answers in (a)(ii) and (b)(i) above, state how the
position of the equilibrium was affected
Le Chatelier’s Principle and Position of Equilibrium.
Le Chatelier’s Principle states that “if a system in equilibrium is subjected to any
change, the equilibrium will shift if possible, to a direction which causes an
opposite change.”
The system cannot completely cancel the change in the external factor, but it will
move in a direction that will minimize the change.
The external factor may be pressure, temperature, concentration, adding a noble
gas or a catalyst.
Katumba J. 2020 elbow grease is the best polish 23
Using Le Chatelier’s Principle to explain the effect of various factors on the
Position of Equilibrium
The position of equilibrium refers to the proportion of products to the reactants
in the equilibrium mixture.
The factors that affect the equilibrium position include concentration, pressure,
temperature and catalyst.
1. Concentration.
Increasing concentration of any reagent in an equilibrium mixture shifts the
equilibrium in the direction that converts some of that reagent into other
products.
Adding a reagent that reacts with one of the reactants/products, reduces the
concentration of the reactant/product in an equilibrium mixture and shifts the
equilibrium in the direction to which the reactant/ product is removed so that it
is replaced.
Any change in concentration of one of the species in an equilibrium mixture
changes the position of the equilibrium and the rate of attainment of the
equilibrium but has no effect on the equilibrium constant.
Example: Sulphur dioxide reacts with oxygen according to the equation;
𝟐𝑺𝑶𝟐(𝒈) + 𝑶𝟐(𝒈) ⇌ 𝟐𝑺𝑶𝟑(𝒈)
Explain how the position of the equilibrium, value of the equilibrium constant
and the rate of attainment of equilibrium would be affected if;
a) more sulphur dioxide was added.
b) sulphur trioxide was added.
c) removing the Sulphur trioxide formed.
Solution:
a) more sulphur dioxide was added.
The concentration of sulphur dioxide increases and the excess sulphur
dioxide reacts with oxygen to produce sulphur trioxide, restoring the
proportions of reactants and products so as to keep the equilibrium
constant value constant. Equilibrium therefore shifts from left to right
and equilibrium constant value remains unchanged. The rate of attainment
Katumba J. 2020 elbow grease is the best polish 24
of equilibrium increases since there is an increase in the number of
particles in the reaction vessel.
b) sulphur trioxide was added.
The concentration of sulphur trioxide increases and the excess sulphur
trioxide dissociates to produce sulphur trioxide and oxygen, restoring the
proportions of reactants and products so as to keep the equilibrium
constant value constant. Equilibrium therefore shifts from right to left
and equilibrium constant value remains unchanged. The rate of attainment
of equilibrium increases since there is an increase in the number of
particles in the reaction vessel.
c) removing the sulphur trioxide formed.
Removing sulphur trioxide makes the sulphur dioxide to react with oxygen
so as to restore the proportions of reactants and products, keeping the
equilibrium constant value constant. Equilibrium therefore shifts from left
to right. The rate of attainment of equilibrium reduces since there is a
reduction in the number of particles in the reaction vessel.
2. Pressure.
Pressure affects mainly gaseous reactions.
Increasing pressure on a reversible reaction in equilibrium causes the equilibrium
to shift in the direction which produces the smaller number of molecules. i.e.
smaller volume.
Thus, for a reversible reaction, the direction that occurs with a decrease in
volume is favoured by an increase in pressure whereas the direction that occurs
with an increase in volume is favoured by a decrease in pressure.
Any change in pressure of the system changes the position of the equilibrium
and the rate of attainment of the equilibrium but has no effect on the
equilibrium constant.
Note: Change in pressure has no effect on reversible reactions with equal
volumes of reactants and products.
Katumba J. 2020 elbow grease is the best polish 25
Example:
Nitrogen and hydrogen react according to the equation;
𝑵𝟐(𝒈) + 𝟑𝑯𝟐(𝒈) ⇌ 𝟐𝑵𝑯𝟑(𝒈)
Explain how the position of the equilibrium, value of the equilibrium constant
and the rate of attainment of equilibrium would be affected if;
a) the pressure was decreased.
b) the pressure was increased.
Solution:
a) the pressure was decreased.
𝑁2(𝑔) + 3𝐻2(𝑔) ⇌ 2𝑁𝐻3(𝑔) 𝟏 𝒗𝒐𝒍. 𝟑 𝒗𝒐𝒍. 𝟐 𝒗𝒐𝒍.
Therefore, 𝟒 volumes (of both 𝑁2 and 𝐻2) produce 𝟐 volumes of ammonia.
Decrease in pressure shifts equilibrium from right to left since the
backward reaction occurs by an increase in volume. Ammonia decomposes
to form nitrogen and hydrogen so as to restore proportions of reactants
and products, keeping the equilibrium constant value unchanged. The rate
of attainment of equilibrium decreases because there are fewer gas
molecules in a given volume, molecules are far apart hence there are fewer
chances of successful collisions between particles.
b) the pressure was increased.
Increase in pressure shifts equilibrium from left to right since forward
reaction occurs by a decrease in volume. Nitrogen will react with hydrogen
to form ammonia so as to restore proportions of reactants and products,
keeping the equilibrium constant value unchanged. The rate of attainment
of equilibrium increases because there are more gas molecules in a given
volume, molecules are closer together hence there are more chances of
successful collisions between particles.
Example: 2
When hydrogen iodide is heated it decomposes according to the equation;
𝟐𝑯𝑰(𝒈) ⇌ 𝑯𝟐(𝒈) + 𝑰𝟐(𝒈)
Katumba J. 2020 elbow grease is the best polish 26
Explain how the position of the equilibrium, value of the equilibrium constant
and the rate of attainment of equilibrium would be affected if the pressure of
the reaction was increased.
Solution:
2𝐻𝐼(𝑔) ⇌ 𝐻2(𝑔) + 𝐼2(𝑔) 𝟐 𝒗𝒐𝒍. 𝟏 𝒗𝒐𝒍. 𝟏 𝒗𝒐𝒍.
Therefore, 2 volumes of 𝐻𝐼 produce 𝟐 volumes (of both 𝐻2 and 𝐼2)
Increase in pressure has no effect on position of equilibrium since both
forward and backward reactions proceed with no change in volume.
Equilibrium constant remains unchanged but the rate of attainment of
equilibrium increases because there are more gas molecules in a given
volume, molecules are closer together hence there are more chances of
successful collisions between particles.
Note: Changes in concentration or pressure may cause an equilibrium to shift
but they do not change the value of the equilibrium constant.
3. Temperature.
The effect of temperature on an equilibrium depends on whether the reaction is
endothermic (∆𝐻𝑟𝑥𝑛 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒) or exothermic ( ∆𝐻𝑟𝑥𝑛 𝑖𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒).
For a system in equilibrium, increasing the temperature favours the endothermic
process that lowers the temperature whereas decreasing the temperature
favours the exothermic process that raises the temperature.
Any change in temperature changes the position of the equilibrium, the rate of
attainment of the equilibrium and the value of the equilibrium constant.
Example: 1
Ammonia is formed from nitrogen and hydrogen at 25°C and 10 atm according to
the equation.
𝑁2(𝑔) + 3𝐻2(𝑔) ⇌ 2𝑁𝐻3(𝑔) ∆𝐻𝜃 = −92𝑘𝐽𝑚𝑜𝑙−1
Katumba J. 2020 elbow grease is the best polish 27
Explain how the position of the equilibrium, value of the equilibrium constant
and the rate of attainment of equilibrium would be affected if the reaction is
carried out at a temperature of 250°C, while the pressure remains at 10 atm.
Solution:
Increase in temperature from 25°C to 250°C will make the equilibrium shift
from right to left, favouring the backward reaction which is endothermic.
The ammonia dissociates to form nitrogen and hydrogen according to Le
Chatelier's Principle. This reduces the concentration of ammonia but
increases the concentrations of nitrogen and hydrogen, reducing the
equilibrium constant. The rate of attainment of equilibrium increases
because increase in temperature will increase both the forward reaction
rate and the reverse reaction rate as a result of the increased collision
frequency between colliding particles.
Example: 2
Nitrogen reacts with oxygen according to the following equation.
𝑵𝟐(𝒈) + 𝑶𝟐(𝒈) ⇌ 𝟐𝑵𝑶(𝒈) ∆𝑯 = +𝟏𝟖𝟎 𝒌𝑱𝒎𝒐𝒍−𝟏
Explain how the position of the equilibrium, value of the equilibrium constant and
the rate of attainment of equilibrium would be affected if;
a) the temperature was increased.
b) the temperature was decreased.
Solution:
a) the temperature was increased.
Increase in temperature will make the equilibrium shift from left to right,
favouring the forward reaction which is endothermic. The nitrogen reacts
with oxygen to form nitrogen monoxide according to Le Chatelier's
Principle. This reduces the concentration of nitrogen and oxygen but
increases the concentration of nitrogen monoxide, increasing the
equilibrium constant. The rate of attainment of equilibrium increases
because increase in temperature will increase both the forward reaction
rate and the reverse reaction rate as a result of the increased collision
frequency between colliding particles.
Katumba J. 2020 elbow grease is the best polish 28
b) the temperature was decreased.
Decrease in temperature will make the equilibrium shift from right to left,
favouring the backward reaction which is exothermic. The nitrogen
monoxide dissociates to form oxygen and nitrogen according to Le
Chatelier's Principle. This increases the concentration of nitrogen and
oxygen but decreases the concentration of nitrogen monoxide, reducing
the equilibrium constant. The rate of attainment of equilibrium decreases
because decrease in temperature will reduce both the forward reaction
rate and the reverse reaction rate as a result of the reduced collision
frequency between colliding particles.
4. Adding an inert gas.
Addition of an inert gas affects a reaction that involves gases. An inert gas is
any gas that does not take part in the reaction.
Addition of an inert gas at constant volume.
Adding an inert gas increases the total pressure of the system but there is no
change in partial pressure/concentrations of the reactants and products. Since
the partial pressure/concentration of the inert gas does not apply in the
equilibrium constant expression, the equilibrium constant also remains unchanged.
The rate of attainment of equilibrium reduces since some of the particles on
collision do not react. Hence, when an inert gas is added to the system in
equilibrium at constant volume there will be no effect on the equilibrium and
equilibrium constant.
Addition of an inert gas at constant pressure.
Adding an inert gas to the system at constant pressure leads to an increase in
the total volume. As a result, partial pressures/concentrations of the reactants
and products decreases. According to Le Chatelier's Principle, the equilibrium will
shift in the direction where the reaction proceeds with increase in volume. The
equilibrium constant remains unchanged since the partial pressure/concentration
of neon does not apply in the equilibrium constant expression. The rate of
attainment of equilibrium reduces as some of the particles on collision do not
react.
Katumba J. 2020 elbow grease is the best polish 29
Question:
When heated, carbon dioxide decomposes according to the equation:
𝟐𝑪𝑶𝟐(𝒈) ⇌ 𝟐𝑪𝑶(𝒈) + 𝑶𝟐(𝒈)
Explain the effect on the equilibrium position and equilibrium constant and the
rate of attainment of equilibrium when;
a) argon is added to the equilibrium at constant volume.
b) neon is added to the equilibrium at constant pressure.
5. Adding catalyst
For a reaction in equilibrium, the catalyst affects neither the position of
equilibrium nor the equilibrium constant. It only increases the rate at which a
reaction attains equilibrium by providing an alternative route of lower the
activation energy.
Trial Question:
1. Sulphur dioxide reacts with oxygen according to the following equation.
𝟐𝑺𝑶𝟐(𝒈) + 𝑶𝟐(𝒈) ⇌ 𝟐𝑺𝑶𝟑(𝒈)
State what would be happen to the concentration of Sulphur trioxide in
the equilibrium mixture and give a reason for your answer if;
a) the temperature was increased
b) nitrogen gas was added to the mixture at a constant pressure.
2. The reaction between nitrogen and hydrogen takes place according to the
following equation.
𝑵𝟐(𝒈) + 𝟑𝑯𝟐(𝒈) ⇌ 𝟐𝑵𝑯𝟑(𝒈) ∆𝑯 = −𝒙 𝒌𝑱𝒎𝒐𝒍−𝟏
What would happen to the concentration of ammonia if;
Katumba J. 2020 elbow grease is the best polish 30
a) helium was added to the equilibrium mixture at constant pressure
500 °C?
b) the temperature was increased?
3. Iodine is sparingly soluble in water but readily dissolves in potassium iodide
according to the following equilibrium;
𝑰𝟐(𝒂𝒒) + 𝑰−(𝒂𝒒) ⇌ 𝑰𝟑−(𝒂𝒒)
a) Explain why iodine is sparingly soluble in water but very soluble in
potassium iodide
b) Write an expression for the concentration equilibrium constant, 𝑲𝒄.
c) State any three characteristics of the above equilibrium.
d) State and explain the effect of adding sodium thiosulphate solution
to the position of equilibrium.
4. Phosphorus pentachloride decomposes at high temperatures according to
the following equation.
𝑷𝑪𝒍𝟓(𝒈) ⇌ 𝑷𝑪𝒍𝟑(𝒈) + 𝑪𝒍 𝟐(𝒈)
State how the value of the equilibrium constant would be affected and in
each case give a reason for your answer if;
a) the pressure was increased
b) some chlorine was added to the equilibrium
5. When heated at 1 atm, carbon dioxide decomposes according to the
equation:
𝟐𝑪𝑶𝟐(𝒈) ⇌ 𝟐𝑪𝑶(𝒈) + 𝑶𝟐(𝒈)
State and explain the effect of;
Katumba J. 2020 elbow grease is the best polish 31
a) heating the carbon dioxide at 2 atmospheres on the equilibrium
concentration of oxygen
b) carrying out the decomposition at a lower temperature on the value
of the equilibrium constant, 𝑲𝒑.
6. Hydrogen and iodine were heated in a 1 litre vessel. Explain what would
happen to the equilibrium position of the reaction, equilibrium constant, and
rate of attainment of equilibrium if;
a) sodium thiosulphate solution was added to the vessel.
b) the pressure was increased
c) concentration of iodine was increased
d) helium gas was added at constant volume.
EXPERIMENTS ON CHEMICAL EQUILIBRIUM.
Experimental determination of the equilibrium constant, 𝑲𝒄 for the reaction
between ethanoic acid and ethanol (esterification reaction).
Procedure:
A known amount (𝑎 𝑚𝑜𝑙𝑒𝑠) of ethanoic acid is mixed with a known amount
(𝑏 𝑚𝑜𝑙𝑒𝑠) of ethanol in a flask of known volume (𝑣 𝑑𝑚3).
A known volume of 2M sulphuric acid is added to the mixture to catalyze
the reaction.
The mixture is heated under reflux at a constant temperature until
equilibrium is established.
The mixture is then very rapidly cooled in by placing the flask in very cold
water
A known volume of the mixture is pipetted, and titrated against a
standard sodium hydroxide solution using phenolphthalein indicator
𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞) + 𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝐶𝐻3𝐶𝑂𝑂𝑁𝑎(𝑎𝑞) + 𝐻2𝑂(𝑙)
The amount (𝑥 𝑚𝑜𝑙𝑒𝑠) of ethanoic acid present at equilibrium is obtained.
Katumba J. 2020 elbow grease is the best polish 32
Treatment of results:
Initial moles of ethanoic acid = 𝑎 𝑚𝑜𝑙𝑒𝑠
Initial moles of ethanol = 𝑏 𝑚𝑜𝑙𝑒𝑠
Equilibrium moles of ethanoic acid = 𝑥 𝑚𝑜𝑙𝑒𝑠
Reaction moles of ethanoic acid = (𝑎 − 𝑥) 𝑚𝑜𝑙𝑒𝑠
Mole ratio of 𝐶𝐻3𝐶𝑂𝑂𝐻 : 𝐶𝐻3𝐶𝐻2𝑂𝐻 : 𝐶𝐻3𝐶𝑂𝑂𝐶𝐻2𝐶𝐻3 : 𝐻2𝑂 = 1: 1: 1: 1
𝑪𝑯𝟑𝑪𝑶𝑶𝑯(𝒍) + 𝑪𝑯𝟑𝑪𝑯𝟐𝑶𝑯(𝒍) ⇌ 𝑪𝑯𝟑𝑪𝑶𝑶𝑪𝑯𝟐𝑪𝑯𝟑(𝒍) + 𝑯𝟐𝑶(𝒍)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑎 𝑏 0 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 (𝑎 − 𝑥) (𝑎 − 𝑥) (𝑎 − 𝑥) (𝑎 − 𝑥)
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 𝑥 𝑏 − (𝑎 − 𝑥) (𝑎 − 𝑥) (𝑎 − 𝑥)
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑛 𝑥
𝑣
𝑏−(𝑎−𝑥)
𝑣
(𝑎−𝑥)
𝑣
(𝑎−𝑥)
𝑣
𝐾𝑐 =[𝐶𝐻3𝐶𝑂𝑂𝐶𝐻2𝐶𝐻3][𝐻2𝑂]
[𝐶𝐻3𝐶𝑂𝑂𝐻][𝐶𝐻3𝐶𝐻2𝑂𝐻]=
(𝑎−𝑥
𝑣)(
𝑎−𝑥
𝑣)
(𝑥
𝑣)(
𝑏−(𝑎−𝑥)
𝑣)
𝑲𝒄 =(𝒂−𝒙)𝟐
𝒙(𝒃−𝒂+𝒙)
Experimental determination of the equilibrium constant, 𝑲𝒄, for the
hydrolysis of ethylethanoate using dilute hydrochloric acid.
Procedure:
A known amount (𝑎 𝑚𝑜𝑙𝑒𝑠) of ethylethanoate is mixed with a known
amount (𝑏 𝑚𝑜𝑙𝑒𝑠) of water in a flask of known volume (𝑣 𝑑𝑚3).
A known volume of hydrochloric acid is added to the mixture to catalyze
the reaction.
The mixture is heated at a constant temperature until equilibrium is
attained.
The mixture is then very rapidly cooled in by placing the flask in very cold
water
A known volume of the mixture is pipetted, and titrated against a
standard sodium hydroxide solution using phenolphthalein indicator. 𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞) + 𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝐶𝐻3𝐶𝑂𝑂𝑁𝑎(𝑎𝑞) + 𝐻2𝑂(𝑙)
The amount (𝑥 𝑚𝑜𝑙𝑒𝑠) of ethanoic acid present at equilibrium is calculated.
Katumba J. 2020 elbow grease is the best polish 33
Treatment of results:
Initial moles of ethylethanoate = 𝑎 𝑚𝑜𝑙𝑒𝑠
Initial moles of water = 𝑏 𝑚𝑜𝑙𝑒𝑠
Equilibrium moles of ethanoic acid = 𝑥 𝑚𝑜𝑙𝑒𝑠
Mole ratio of 𝐶𝐻3𝐶𝑂𝑂𝐻 : 𝐶𝐻3𝐶𝐻2𝑂𝐻 : 𝐶𝐻3𝐶𝑂𝑂𝐶𝐻2𝐶𝐻3 : 𝐻2𝑂 = 1: 1: 1: 1
𝑪𝑯𝟑𝑪𝑶𝑶𝑪𝑯𝟐𝑪𝑯𝟑(𝒍) + 𝑯𝟐𝑶(𝒍) ⇌ 𝑪𝑯𝟑𝑪𝑶𝑶𝑯(𝒍) + 𝑪𝑯𝟑𝑪𝑯𝟐𝑶𝑯(𝒍)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑎 𝑏 0 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 𝑥 𝑥 𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 (𝑎 − 𝑥) (𝑏 − 𝑥) 𝑥 𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑛 (𝑎−𝑥)
𝑣
(𝑏−𝑥)
𝑣
𝑥
𝑣
𝑥
𝑣
𝐾𝑐 =[𝐶𝐻3𝐶𝑂𝑂𝐻][𝐶𝐻3𝐶𝐻2𝑂𝐻]
[𝐶𝐻3𝐶𝑂𝑂𝐶𝐻2𝐶𝐻3][𝐻2𝑂]=
(𝑥
𝑣)(
𝑥
𝑣)
(𝑎−𝑥
𝑣)(
𝑏−𝑥
𝑣)
𝑲𝒄 =𝒙𝟐
(𝒂−𝒙)(𝒃−𝒙)
Experimental determination of the equilibrium constant, 𝑲𝒄, of a reaction
between hydrogen and iodine.
Procedure:
A known amount (𝑎 𝑚𝑜𝑙𝑒𝑠) of iodine is sealed with a known amount
(𝑏 𝑚𝑜𝑙𝑒𝑠) of hydrogen gas in a glass bulb of known volume (𝑣 𝑑𝑚3).
The bulb is heated at a constant temperature for several hours until
equilibrium is attained.
The bulb is rapidly cooled at room temperature to stop the reaction and
fix the equilibrium such that the equilibrium does not adjust itself to the
equilibrium value at a lower temperature.
A known volume of the resultant solution is pipetted and titrated against
a standard sodium thiosulphate solution using starch indicator.
𝐼2(𝑎𝑞) + 2𝑆2𝑂32− (𝑎𝑞) → 2𝐼−(𝑎𝑞) + 𝑆4𝑂6
2−(𝑎𝑞)
The amount (𝑥 𝑚𝑜𝑙𝑒𝑠) of iodine at equilibrium is obtained.
Treatment of result:
Initial moles of hydrogen (𝐻2) = 𝑎 𝑚𝑜𝑙𝑒𝑠
Katumba J. 2020 elbow grease is the best polish 34
Initial moles of iodine (𝐼2) = 𝑏 𝑚𝑜𝑙𝑒𝑠
Equilibrium moles of iodine (𝐼2) = 𝑥 𝑚𝑜𝑙𝑒𝑠
Reaction moles of iodine (𝐼2)= (𝑏 − 𝑥) 𝑚𝑜𝑙𝑒𝑠
Mole ratio of 𝐻2 ∶ 𝐼2 ∶ 𝐻𝐼 = 1 ∶ 1 ∶ 2
𝑯𝟐(𝒈) + 𝑰𝟐(𝒈) ⇌ 𝟐𝑯𝑰(𝒈)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑎 𝑏 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 (𝑎 − 𝑥) (𝑎 − 𝑥) 2(𝑎 − 𝑥)
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 𝑥 𝑏 − (𝑎 − 𝑥) 2(𝑎 − 𝑥)
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 (𝑥)
𝑣
𝑏−(𝑎−𝑥)
𝑣
2(𝑎−𝑥)
𝑣
𝐾𝑐 =[𝐻𝐼]2
[𝐻2][𝐼2]=
(2(𝑎−𝑥)
𝑣)
2
(𝑥
𝑣)(
𝑏−(𝑎−𝑥)
𝑣)
𝑲𝒄 =𝟒(𝒂−𝒙)𝟐
𝒙(𝒃−𝒂+𝒙)
Experimental determination of the equilibrium constant for the dissociation
of hydrogen iodide to hydrogen and iodine.
Procedure:
A known amount (𝑛 𝑚𝑜𝑙𝑒𝑠) of hydrogen iodide is put in a glass bulb of a
known volume (𝑣 𝑑𝑚3)
The bulb is heated at a constant temperature for several hours until
equilibrium is established.
The bulb is rapidly cooled at room temperature to stop the reaction and
fix the equilibrium such that the equilibrium does not adjust itself to the
equilibrium value at a lower temperature.
A known volume of the resultant solution is pipetted and titrated against
a standard sodium thiosulphate solution using starch indicator.
𝐼2(𝑎𝑞) + 2𝑆2𝑂32− (𝑎𝑞) → 2𝐼−(𝑎𝑞) + 𝑆4𝑂6
2−(𝑎𝑞)
The amount (𝑥 𝑚𝑜𝑙𝑒𝑠) of iodine at equilibrium is obtained.
Treatment of results
Initial moles of hydrogen iodide (𝐻𝐼) = 𝑛 𝑚𝑜𝑙𝑒𝑠
Degree of dissociation of hydrogen iodide = 𝛼
Moles of 𝐻𝐼 that dissociated = 𝑛𝛼
Katumba J. 2020 elbow grease is the best polish 35
Mole ratio of 𝐻𝐼 ∶ 𝐻2 ∶ 𝐼2 = 2 ∶ 1 ∶ 1
𝟐𝑯𝑰(𝒈) ⇌ 𝑯𝟐(𝒈) + 𝑰𝟐(𝒈)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑛 0 0
𝐹𝑜𝑟 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐻𝐼, 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑛𝛼 𝑛𝛼
2
𝑛𝛼
2
𝐹𝑜𝑟 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐻𝐼, 𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 𝑛(1 − 𝛼) 𝑛𝛼
2
𝑛𝛼
2
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑛(1−𝛼)
𝑣
𝑛𝛼
2𝑣
𝑛𝛼
2𝑣
𝐾𝑐 =[𝐻2][𝐼2]
[𝐻𝐼]2 =(
𝑛𝛼
2𝑣)(
𝑛𝛼
2𝑣)
(𝑛(1−𝛼)
𝑣)
2
𝑲𝒄 =𝜶𝟐
𝟒(𝟏−𝜶)𝟐
Experimental determination of the equilibrium constant,𝑲𝒄, for the reaction
between hydrogen and nitrogen to form ammonia.
Procedure:
A known amount (𝑎 𝑚𝑜𝑙𝑒𝑠) of nitrogen and a known amount (𝑏 𝑚𝑜𝑙𝑒𝑠) of
hydrogen are put in a sealed glass bulb of a known volume (𝑣 𝑑𝑚3).
The mixture is heated to a moderately high constant temperature for
some time until equilibrium is attained.
The bulb is rapidly broken under ice cold water to stop the reaction and
fix the equilibrium such that the equilibrium does not adjust itself to the
equilibrium value at a lower temperature.
A known volume of the resultant solution is pipetted and titrated against
a standard solution of dilute hydrochloric acid using phenolphthalein
indicator
𝑁𝐻3(𝑎𝑞) + 𝐻𝐶𝑙(𝑎𝑞) → 𝑁𝐻4𝐶𝑙(𝑎𝑞)
The amount of ammonia present at equilibrium is obtained.
Treatment of results
Initial moles of nitrogen = 𝑎 𝑚𝑜𝑙𝑒𝑠
Initial moles of hydrogen = 𝑏 𝑚𝑜𝑙𝑒𝑠
If the number of moles of nitrogen converted to ammonia = 𝑥 𝑚𝑜𝑙𝑒𝑠, then;
Moles of hydrogen that reacted = 3𝑥 since 𝑁2 ∶ 𝐻2 = 1 ∶ 3
Katumba J. 2020 elbow grease is the best polish 36
Moles of ammonia formed = 2𝑥 𝑚𝑜𝑙𝑒𝑠
𝑵𝟐(𝒈) + 𝟑𝑯𝟐(𝒈) ⇌ 𝟐𝑵𝑯𝟑(𝒈)
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑎 𝑏 0
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑒𝑠 𝑥 3𝑥 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑚𝑜𝑙𝑒𝑠 (𝑎 − 𝑥) (𝑏 − 3𝑥) 2𝑥
𝐸𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 (𝑎−𝑥)
𝑣
(𝑏−3𝑥)
𝑣
2𝑥
𝑣
𝐾𝑐 =[𝑁𝐻3]2
[𝑁2][𝐻2]3 =(
2𝑥
𝑣)
2
(𝑎−𝑥
𝑣)(
𝑏−3𝑥
𝑣)
3
𝑲𝒄 =𝟒𝒙𝟐𝒗𝟐
(𝒂−𝒙)(𝒃−𝟑𝒙)𝟑
COMBINATIONS OF EQUILIBRIA
If a reaction cannot be carried out experimentally, then its equilibrium constant
can be determined indirectly from other reactions that can be performed.
Consider the following reactions.
Water dissociates in gaseous phase according to the equation:
𝐻2𝑂(𝑔) ⇌ 𝐻2(𝑔) + ½𝑂2(𝑔) … … … … … … … … . … … … … . 𝐾𝑝𝐼 =
𝑃𝐻2 .(𝑃𝑂2)1
2⁄
𝑃𝐻2𝑂
Carbon dioxide reacts with hydrogen to produce carbon monoxide and water.
𝐶𝑂2(𝑔) + 𝐻2(𝑔) ⇌ 𝐶𝑂(𝑔) + 𝐻2𝑂(𝑔) … … … … … … … … … 𝐾𝑝𝐼𝐼 =
𝑃𝐶𝑂 .𝑃𝐻2𝑂
𝑃𝐶𝑂2 .𝑃𝐻2
The two equilibrium constants can be used to find the equilibrium constant, 𝐾𝑝,
for carbon dioxide dissociation according to the equation;
𝑪𝑶𝟐(𝒈) ⇌ 𝑪𝑶(𝒈) + ½𝑶𝟐(𝒈) … … … … … … … … … … . . 𝑲𝒑 =? ?
The two equilibrium constants can be used to find the equilibrium constant for
carbon dioxide dissociation according to the equation:
𝐶𝑂2(𝑔) ⇌ 𝐶𝑂(𝑔) + ½𝑂2(𝑔)
𝑲𝒑 =𝑷𝑪𝑶.(𝑷𝑶𝟐
)𝟏
𝟐⁄
𝑷𝑪𝑶𝟐
Katumba J. 2020 elbow grease is the best polish 37
𝐾𝑝𝐼 𝑥 𝐾𝑝
𝐼𝐼 =𝑃𝐻2 .(𝑃𝑂2 )
12⁄
𝑃𝐻2𝑂 𝑥
𝑃𝐶𝑂.𝑃𝐻2𝑂
𝑃𝐶𝑂2 .𝑃𝐻2
=𝑃𝐶𝑂 .(𝑃𝑂2)
12⁄
𝑃𝐶𝑂2
= 𝐾𝑝
𝑻𝒉𝒖𝒔, 𝑲𝒑 = 𝑲𝒑𝑰 𝒙 𝑲𝒑
𝑰𝑰
Note:
1. If the two equilibrium constants are known, they can be used to find the
equilibrium constant for another reaction as long as the temperature
remains constant.
2. If the reaction proceeds in steps; the overall equilibrium constant is the
product of all equilibrium constants of all the steps.
For example: 𝐶𝑢2+(𝑎𝑞) + 4𝑁𝐻3(𝑎𝑞) ⇌ 𝐶𝑢(𝑁𝐻3)42+(𝑎𝑞) … … … … . . . … . 𝐾5
𝐶𝑢2+(𝑎𝑞) + 𝑁𝐻3(𝑎𝑞) ⇌ 𝐶𝑢(𝑁𝐻3)2+(𝑎𝑞) … … … … . … … … . . . . … … … 𝐾1
𝐶𝑢(𝑁𝐻3)2+(𝑎𝑞) + 𝑁𝐻3(𝑎𝑞) ⇌ 𝐶𝑢(𝑁𝐻3)22+(𝑎𝑞) … … … … . … . … … . . 𝐾2
𝐶𝑢(𝑁𝐻3)22+(𝑎𝑞) + 𝑁𝐻3(𝑎𝑞) ⇌ 𝐶𝑢(𝑁𝐻3)3
2+(𝑎𝑞) … … … … … … . … … 𝐾3
𝐶𝑢(𝑁𝐻3)32+(𝑎𝑞) + 𝑁𝐻3(𝑎𝑞) ⇌ 𝐶𝑢(𝑁𝐻3)4
2+(𝑎𝑞) … … … … … … . … . . 𝐾4
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆, 𝑲𝟓 = 𝑲𝟏 𝒙 𝑲𝟐 𝒙 𝑲𝟑 𝒙 𝑲𝟒
Example:
For the following reactions;
𝐹𝑒(𝑂𝐻)3(𝑠) + (𝑎𝑞) ⇌ 𝐹𝑒3+(𝑎𝑞) + 3�̅�𝐻 (𝑎𝑞) 𝐾𝑠 = 1 𝑥 10−28
𝑃ℎ𝑁𝐻2(𝑎𝑞) + 𝐻2𝑂 (𝑙) ⇌ 𝑃ℎ 𝑁𝐻3+(𝑎𝑞) + �̅�𝐻(𝑎𝑞) 𝐾𝑏 = 4.25 𝑥 10−14
(i) Give the expression of 𝐾𝑠 and 𝐾𝑏
(ii) State the assumptions made
(iii) Give the equilibrium constant for the reaction.
𝐹𝑒3+(𝑎𝑞) + 3𝑃ℎ𝑁𝐻2(𝑎𝑞) + 3𝐻2𝑂(𝑙) ⇌ 𝐹𝑒(𝑂𝐻)3(𝑠) + 3𝑃ℎ𝑁𝐻3+(𝑎𝑞)
(iv) Express 𝐾 in terms of 𝐾𝑠 and 𝐾𝑏
(v) Calculate the value of 𝐾.
Solution:
a) (i) 𝐾𝑠 = [𝐹𝑒3+][�̅�𝐻]3 and 𝐾𝑏 =[𝑃ℎ𝑁𝐻3
+][�̅�𝐻]
[𝑃ℎ𝑁𝐻2][𝐻2𝑂]
Katumba J. 2020 elbow grease is the best polish 38
(ii) [𝐹𝑒(𝑂𝐻)3] is constant since its state is solid; Fe(OH)3 simply dissociates
into ions in aqueous medium but does not react with it.
(iii) 𝐾 =[𝑃ℎ𝑁𝐻3
+]3
[𝐹𝑒3+][𝑃ℎ𝑁𝐻2]3[𝐻2𝑂]3
(iv) Consdering the equations
𝐹𝑒(𝑂𝐻)3(𝑠) + (𝑎𝑞) ⇌ 𝐹𝑒3+(𝑎𝑞) + 3�̅�𝐻 (𝑎𝑞) ; 𝐾𝑠 … … … … … … (𝑖)
𝑃ℎ𝑁𝐻2(𝑎𝑞) + 𝐻2𝑂 (𝑙) ⇌ 𝑃ℎ 𝑁𝐻3+(𝑎𝑞) + �̅�𝐻(𝑎𝑞) ; 𝐾𝑝 … … . … … . … (𝑖𝑖)
𝐹𝑒3+(𝑎𝑞) + 3𝑃ℎ𝑁𝐻2(𝑎𝑞) + 3𝐻2𝑂(𝑙) ⇌ 𝐹𝑒(𝑂𝐻)3(𝑠) + 3𝑃ℎ𝑁𝐻3+(𝑎𝑞) is
obtained from;
- 𝐼𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 (𝑖)
𝐹𝑒3+(𝑎𝑞) + 3�̅�𝐻 (𝑎𝑞) ⇌ 𝐹𝑒(𝑂𝐻)3(𝑠) + (𝑎𝑞) ; 1
𝐾𝑠… . … … … (𝑖𝑖𝑖)
- 3(𝑖𝑖)
3𝑃ℎ𝑁𝐻2(𝑎𝑞) + 3𝐻2𝑂 (𝑙) ⇌ 3𝑃ℎ 𝑁𝐻3+(𝑎𝑞) + 3�̅�𝐻(𝑎𝑞) ; 𝐾𝑏
3 … … … . . (𝑖𝑣)
1
𝐾𝑠 𝑥 𝐾𝑏
3 =1
[𝐹𝑒3+][�̅�𝐻]3 𝑥 [𝑃ℎ𝑁𝑁𝐻3
+]3
[�̅�𝐻]3
[𝑃ℎ𝑁𝐻2]3[𝐻2𝑂]3 =[𝑃ℎ𝑁𝐻3
+]3
[𝐹𝑒3+][𝑃ℎ𝑁𝐻2]3[𝐻2𝑂]3 = 𝐾
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑲 = 𝟏
𝑲𝒔 𝒙 𝑲𝒃
𝟑
(v) 𝐹𝑟𝑜𝑚 𝐾 =1
𝐾𝑠 𝑥 𝐾𝑏
3 =1
1 𝑥 10−28 𝑥 (4.25 𝑥 10−14)3 = 𝟕. 𝟔𝟕𝟕 𝒙 𝟏𝟎−𝟏𝟒
Question: Hydrogen sulphide solution dissociates according to the equations;
𝐻2𝑆(𝑎𝑞) ⇌ 𝐻+(𝑎𝑞) + 𝐻𝑆−(𝑎𝑞) 𝐾𝑐1 = 9.5 𝑥 10−8
𝐻𝑆−(𝑎𝑞) ⇌ 𝐻+(𝑎𝑞) + 𝑆2−(𝑎𝑞) 𝐾𝑐2 = 1.0 𝑥 10−19
Calculate the equilibrium constant for the reaction
𝐻2𝑆(𝑎𝑞) ⇌ 2𝐻+(𝑎𝑞) + 𝑆2−(𝑎𝑞) 𝑨𝒏𝒔𝒘𝒆𝒓 𝑲𝒄 = 𝟗. 𝟓 𝒙 𝟏𝟎−𝟐𝟕
INDUSTRIAL APPLICATION OF CHEMICAL EQUILIBRIUM.
1. Manufacture of ammonia (the Haber process)
The reaction between dry hydrogen (obtained from natural gas) and dry nitrogen
(from fractional distillation of liquid air) to form ammonia is exothermic and
occurs with a decrease in volume
Katumba J. 2020 elbow grease is the best polish 39
𝑁2(𝑔) + 3𝐻2(𝑔) ⇌ 2𝑁𝐻3(𝑔) ; ∆𝐻𝜃(298𝐾) = −92 𝑘𝐽𝑚𝑜𝑙−1
These two gases are then made to react in the ratio of 3:1
According to Le Chatelier's Principle, the yield of ammonia will be greatest at
low temperature and high pressure. At a low temperature however, rate of
attainment of equilibrium is low and at high pressure, the cost of the equipment
and running costs are high.
Therefore, in practice, a compromise has to be struck. The conditions used in
this process are therefore;
Pressures between 200-500 atmospheres
Temperature of about 450-550oC
Finely divided iron catalyst
Uses of ammonia
Ammonia is used in;
Manufacture of nitric acid (HNO3)
Manufacture of ammonium fertilizers such as ammonium sulphate
(NH4)2SO4 fertilizers.
Qualitative analysis, as a reagent in the laboratory.
Ammonia is also used as a cooler in refrigerators.
2. Manufacture of sulphuric acid (the Contact process)
The manufacture of Sulphuric acid involves the following steps;
a) Formation of sulphur dioxide.
Sulphur dioxide is formed mainly by burning of sulphur (Frasch process) in
excess and oxygen (obtained from fractional distillation of liquid air)
𝑆(𝑠) + 𝑂2(𝑔) → 𝑆𝑂2(𝑔)
b) Purification
Sulphur dioxide and oxygen are purified and then mixed together and reacted
to form sulphur trioxide
The conversion of sulphur dioxide to sulphur trioxide is an exothermic reaction
and occurs with a decrease in volume.
2𝑆𝑂2(𝑔) + 𝑂2(𝑔) ⇌ 2𝑆𝑂3(𝑔) ; ∆𝐻𝜃(298𝐾) = −197 𝑘𝐽𝑚𝑜𝑙−
Katumba J. 2020 elbow grease is the best polish 40
According to Le Chatelier's Principle, the yield of sulphur trioxide will be
greatest at low temperature and high pressure. At a low temperature however,
the rate of attainment of equilibrium is very slow.
Therefore, in practice, the yield of sulphur trioxide is high when the following
conditions are available;
Pressures between 1-5 atmospheres
Temperature of about 450-500oC
Vanadium(V) oxide catalyst
c) Dissolution of Sulphur trioxide.
Sulphur trioxide is dissolved in concentrated sulphuric acid to form fuming
liquid called oleum;
𝑆𝑂3(𝑔) + 𝐻2𝑆𝑂4(𝑙) → 2𝐻2𝑆2𝑂7(𝑙)
d) Dilution of Oleum.
The fuming sulphuric acid is diluted with water to form very concentrated
sulphuric acid of about 98% concentration.
𝐻2𝑆2𝑂7(𝑙) + 𝐻2𝑂(𝑙) → 2𝐻2 𝑆𝑂4(𝑙)
NB: Sulphur trioxide is not dissolved in water directly because the reaction is
too exothermic and the heat produced from the reaction vapourises the acid
forming only tiny droplets of the acid leading to a spray of sulphuric acid which
is dangerous to the workers in the factory.
Uses of sulphuric acid
Used in the manufacture of fertilizers like ammonium sulphate.
Making of paints and pigments
Manufacture of detergents such as Omo, Nomi etc
Production of other chemicals such as metallic sulphates, hydrochloric acid,
hydrofluoric acid and plastics.
It is used in car batteries and accumulators as an electrolyte
Extraction of metals and metal manufacturing including pickling to clean
metallic surfaces.
It is used as a drying agent.
With nitric acid, it is used to make dyes and explosives.