CHEMICAL EQUILIBRIA Chemical equilibrium reversible reactions · A Chemical equilibrium occurs for...

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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; 3 () ⇌ () + 2 () {Reversible reaction} Explanation When calcium carbonate is heated at a fixed temperature in a closed container; at first, 3 decomposes faster than the products recombine. After a while, the amounts of and 2 build up to a level at which the rate of combination of and 2 is equal to the rate at which 3 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 4 () + 2 2 () → 2 () + 2 2 () 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.

Transcript of CHEMICAL EQUILIBRIA Chemical equilibrium reversible reactions · A Chemical equilibrium occurs for...

Page 1: CHEMICAL EQUILIBRIA Chemical equilibrium reversible reactions · A Chemical equilibrium occurs for reversible reactions. Many reactions do not go to completion since the products

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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.

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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.

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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

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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.

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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

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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

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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

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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

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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)

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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

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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)

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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

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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

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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.

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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

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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)

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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

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⇒ 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.

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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+

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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

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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−

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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.

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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

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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.

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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)

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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

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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.

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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.

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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;

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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;

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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.

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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.

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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

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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

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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

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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

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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]

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(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

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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−

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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.