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CHEMICAL EQUILIBRIA

It involves the study of reversible chemical reactions, a dynamic equilibrium established

within the reversible reactions and the conditions that allow such reactions to attain

equilibrium.

Reversible reactions proceed in both the forward and backward direction.i.e

The reaction proceeds in both the forward and backward direction such that a given

moment, the concentration of both the reactants and products is constant. At this moment

the reaction is said to have reached equilibrium. the rate of the forward reaction is equal

to the rate of the backward reaction i.e the extent to which reactants are used up is equal

to the extent at which products combine to form back the reactants

Reversible reactions may involve molecules in equilibrium only or molecules in

equilibrium with its ions in solution. The later is called ionic equilibria, while the former

is called molecular equilibria

CHARACTERISTICS OF CHEMICAL EQUILIBRIUM ( EQUILIBRIUM STATE)

Chemical equilibrium is attained at a constant(fixed) temperature

The equilibrium state is reversible and the rate at which the forward reaction

proceeds is the same as that at which the backward reaction proceeds

It is dynamic i.e. opposing changes are always taking place at molecular level

It can be established in closed systems

MOLECULAR CHEMICAL EQUILIBRIA

This can be;

Homogenous gaseous equilibrium if it contains if it contains reactants and products

in gaseous state only.

Homogenous liquid equilibrium if it contains reactants and products in liquid or

aqueous sate only.

Heterogeneous equilibrium if it contains reactants and products in different

physical states

THE LAW OF MASS ACTION

This states that the rate of a chemical reaction at a given temperature is proportional to

the molar concentration of reactants raised to appropriate powers obtained from a

balanced equation.

This law is used to write the equilibrium constant in terms of molar concentration

denoted as Kc or in terms of partial pressure denoted as Kp;

General equation;

𝐴 + 𝐵 𝐶 𝐹𝑜𝑤𝑎𝑟𝑑

𝐵𝑎𝑐𝑘𝑤𝑎𝑟𝑑 + 𝐷

𝑎𝐴 + 𝑏𝐵 𝑐𝐶 𝐹𝑜𝑤𝑎𝑟𝑑

𝐵𝑎𝑐𝑘𝑤𝑎𝑟𝑑 + 𝑑𝐷

2

Equilibrium constant (𝑲𝒄)

By definition,

Equilibrium concentration KC is the ratio of the product of molar concentration of

products each raised to appropriate powers to the product of the molar concentration of

reactants each raised to appropriate powers obtained from a balanced equation.

Therefore 𝐾𝑐 = [𝐴]a[𝐵]b

[𝐶]c[𝐷]d

The units of kc are calculated from the expression by substituting the units of

concentration (𝑚𝑜𝑙 𝑑𝑚−3) or (𝑚𝑜𝑙 𝑙−1) Example

Sulphur trioxide decomposes according to the equation below;

2𝑆𝑂3(𝑔) 2𝑆𝑂2(𝑔) + 𝑂2(𝑔)

Kc expression Units

𝐾𝑐 = [𝑆𝑂2]

2[𝑂2]

[𝑆𝑂3]2 𝐾𝑐 =

(𝑚𝑜𝑙 𝑑𝑚−3)2(𝑚𝑜𝑙 𝑑𝑚−3)

(𝑚𝑜𝑙 𝑑𝑚−3)2

= 𝑚𝑜𝑙 𝑑𝑚−3 Note; In case of heterogeneous equilibrium reactions where solids exist in presence of

liquids or gases, the solids do not appear in the Kc expression since there concentration is

assumed to be relatively constant e.g

𝐶𝑎𝐶𝑂3(𝑠) 𝐶𝑎𝑂(𝑠) + 𝐶𝑂2(𝑔)

𝐾𝑐 = [𝐶𝑂2] Exercise 1;

Write the expression for the equilibrium constant Kc and its units for each of the

following reactions.

(a) 𝐻2(𝑔) + 𝐼2(𝑔) 𝐻𝐼(𝑔)

(b) 3𝐻2(𝑔) + 𝑁2(𝑔) 2𝑁𝐻3(𝑔)

(c) 𝐹𝑒(𝑠) + 𝐻2𝑂(𝑔) 4𝐻2(𝑔) + 𝐹𝑒3𝑂4(𝑠)

(d) 𝐶𝐻3𝐶𝑂𝑂𝐻(𝑙) + 𝐶𝐻3𝐶𝐻2𝑂𝐻(𝑙) 𝐶𝐻3𝐶𝑂𝑂𝐶𝐻2𝐶𝐻3(𝑙) + 𝐻2𝑂(𝑙)

Equilibrium constant in terms of partial pressure (Kp)

By definition,

Equilibrium concentration Kp is the ratio of the product of the partial of products each

raised to appropriate powers to the product of the partial of reactants each raised to

appropriate powers obtained from a balanced equation.

𝑭𝒓𝒐𝒎 𝑃𝑉 = 𝑛𝑅𝑇

𝑃 =𝑛

𝑉𝑅𝑇 Where

𝑛

𝑉= C (𝑚𝑜𝑙 𝐿−1)

Therefore 𝑃 = 𝐶𝑅𝑇 From the equation, it’s clear that partial pressure of a gas is directly proportional to

its concentration at constant temperature since R is also constant

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This also shows that Kp is directly proportional to Kc (Kp∝ 𝐾𝑐) Kp only applies to reactions involving gaseous components

Consider the general equation below;

𝑎𝐴(𝑔) + 𝑏𝐵(𝑔) 𝑐𝐶(𝑔) + 𝑑𝐷(𝑔)

𝐾𝑝 = (PC)

c.(PD)d

(PA)a.(PB)

b

The pressure exerted by a gas in a mixture of gases at equilibrium is equal to its partial

pressure and it depends on its mole fraction as seen from Dalton’s law i.e

𝑷𝑨 = 𝑿𝑨. 𝑷𝒕𝒐𝒕𝒂𝒍 (Dalton’s law) Where 𝑿𝑨 = 𝒏𝑨

𝒏𝒕𝒐𝒕𝒂𝒍 and 𝑷𝒕𝒐𝒕𝒂𝒍 = 𝒕𝒐𝒕𝒂𝒍 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆

Example

Write down the equilibrium expression in terms of Partial pressure (Kp) and state its

units for the equation below;

2𝑆𝑂2(𝑔)+ 𝑂2(𝑎𝑞) 2𝑆𝑂3(𝑔)

𝐾𝑝 = (PSO3)

2

(PSO2)2.(PO2)

Exercise 2

Write the expression for the equilibrium constant Kp and its units for each of the

following reactions. (Assuming pressure was measured in atmospheres)

(a) 𝐻2(𝑔) + 𝐼2(𝑔) 𝐻𝐼(𝑔)

(b) 3𝐻2(𝑔) + 𝑁2(𝑔) 2𝑁𝐻3(𝑔)

(c) 𝐹𝑒(𝑠) + 𝐻2𝑂(𝑔) 4𝐻2(𝑔) + 𝐹𝑒3𝑂4(𝑠)

Calculations involving Kc and Kp

There are three main types of calculations that we shall consider;

Calculations in which the moles of a reactant or product at equilibrium is given.

Calculations in which the percentage composition of a reactant or product in the

equilibrium mixture is given

Calculations in which the degree of dissociation (∝ ) or % decomposition of a

reactant is given.

All these calculations follow the same method as illustrated below using the equation

below;

Let X be moles of C formed at equilibrium

𝐴(𝑔) + 𝐵(𝑔) 𝐶(𝑔) + 𝐷(𝑔)

Initially 1 1 - -

Reacted X X X X (mole ratio 1:1:1:1)

Eq≡ 𝑚moles 1-X 1-X X X

[𝐸𝑞 ≡ 𝑚] 1−𝑋

𝑉

1−𝑋

𝑉

𝑋

𝑉

𝑋

𝑉

𝐾𝑐 = [𝐶] [𝐷]

[𝐴] [B]

Units: pressure is measured in Pa(𝑵𝑴−𝟐) 𝒐𝒓 𝒂𝒕𝒎 or mmHg

𝐾𝑝 = (atm)2

(atm)2.(atm) =𝑎𝑡𝑚−1

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𝐾𝑐 = (𝑋

𝑉).(𝑋

𝑉)

(1−𝑋

𝑉).(

1−𝑋

𝑉)

Note:

If the value of X is known then we would simplify this expression or if moles of

any reactant or product at equilibrium is known then would calculate X and use it

to determine Kc

initial moles of the reactants may be given in the question, however if not given

then we use the stoichiometric values like (A=1 and B=1 from the equation above)

V is the volume of the reaction vessel which can be assumed to be 1litre or 1𝑑𝑚3if it has not been mentioned in the question.

Calculations in which the moles of a reactant or product at equilibrium is

given.

Example

1) When 30moles of hydrogen iodide were put in a 1 litre vessel at 50℃, it decomposed

and at equilibrium, it was found to contain 10 moles of hydrogen iodide. Calculate the

KC for the reaction

Solution

In this example the initial moles of 𝑯𝑰 were given as 30 moles

Let X be moles of hydrogen formed

2𝐻𝐼(𝑔) 𝐻2(𝑔) + 𝐼2(𝑔)

Initially 30 - -

Reacted 2X X X (mole ratio 2:1:1)

Eq≡ 𝑚moles 30-2X X X

[𝐸𝑞 ≡ 𝑚] 30−2𝑋

1

𝑋

1

𝑋

1

But at equilibrium 30-2X= 10; Threfore X=10

[𝐻𝐼] at eq≡m = 10𝑚𝑜𝑙𝑙−1, [𝐻2] at eq≡m = 10𝑚𝑜𝑙𝑙−1, and [𝐼2] at eq≡m = 10 𝑚𝑜𝑙𝑙−1

From;

𝐾𝑐 = [𝐼2][𝐻2]

[𝐻𝐼]2;

Substituting the equilibrium concentration in the expression

𝐾𝑐 = 10𝑋10

102 = 1

2 Phosphorous pent chloride decomposes at high temperature according to the

following equation;

𝑃𝐶𝑙5(𝑔) 𝑃𝐶𝑙3(𝑔) + 𝐶𝑙2(𝑔).

If 40.2g of chlorine were formed at equilibrium;

(a) Calculate the amount of phosphorous pent chloride and phosphorous tri-chloride at

equilibrium in 𝑀𝑜𝑙𝑙−1

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(b) Write the equilibrium expression in terms of concentration for the reaction

(c) Calculate the equilibrium constant KC and its units

Solution

In this example the initial moles of 𝑷𝑪𝒍𝟓 were not given so we shall take the

moles in the balanced stoichiometric equation to be the initial moles of the

reactants and we shall assume volume to be 1litre

(a) Let X be moles of chlorine formed

𝑃𝐶𝑙5(𝑔) 𝑃𝐶𝑙3(𝑔) + 𝐶𝑙2(𝑔)

Initially 1 - -

Reacted X X X (mole ratio 1:1:1)

Eq≡ 𝑚 moles 1-X X X

[𝐸𝑞 ≡ 𝑚] 1−𝑋

1

𝑋

1

𝑋

1

But X is equal to moles of chlorine at eq≡m = 40.2

(35.5𝑥2) =0.574

[𝑃𝐶𝑙5] at eq≡m = (1-0.574)= 0.426𝑚𝑜𝑙𝑙−1,

[𝑃𝐶𝑙3] at eq≡m = 0.574𝑚𝑜𝑙𝑙−1, and [𝐶𝑙2] at eq≡m = 0.574 𝑚𝑜𝑙𝑙−1

(b) 𝐾𝑐 = [𝑃𝐶𝑙3][𝐶𝑙2]

[𝑃𝐶𝑙5]

(c) 𝐾𝑐 = (0.574 ).(0.574 )

(0.426) (𝑚𝑜𝑙𝑙−1).(𝑚𝑜𝑙𝑙−1)

(𝑚𝑜𝑙𝑙−1)

Kc = 0.773𝑚𝑜𝑙𝑙−1

3 Sulphur dioxide reacts with oxygen according to the equation

2𝑆𝑂2(𝑔)+ 𝑂2(𝑎𝑞) 2𝑆𝑂3(𝑔)

At equilibrium it was found that there was 0.4 moles of Sulphur dioxide, 0.2 moles

of oxygen and 0.6 moles of sulphur trioxide at 700℃ and a pressure of 2 atm

(a) Calculate the partial pressure of each gas in the reaction

(b) Calculate the KP for the reaction and its units

Solution

In this questions the moles at equilibrium were given, therefore no need to

follow the general method which finds out the moles at equilibrium

(a)

Partial pressure = 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑥 𝑡𝑜𝑡𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒

Total number of moles = 0.2+ 0.4+0.6 = 1.2moles

Partial pressure of 𝑆𝑂2 = 0.4𝑋2

1.2 = 0.667 𝑎𝑡𝑚𝑠

Partial pressure of 𝑂2 = 0.2𝑋2

1.2 = 0.333 𝑎𝑡𝑚𝑠

Partial pressure of 𝑂2 = 0.6𝑋2

1.2 = 1𝑎𝑡𝑚

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But 𝐾𝑝 = 𝑃𝑆𝑂32 (𝑎𝑡𝑚)2

𝑃𝑆𝑂22 𝑃𝑂2 (𝑎𝑡𝑚)

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𝐾𝑝 = 12

0.6672𝑋 0.333 = 6.750 𝑎𝑡𝑚−1

Exercise 3

1)A mixture of 1.9 moles of hydrogen and 2.3 moles of iodine were allowed to react and

equilibrium attained at 440oC.Analysis of the equilibrium mixture, found out that it

contained 3moles of hydrogen Iodide.

Calculate the equilibrium constant KC.

Write it’s units

2) 3moles of nitrogen mono oxide and 1.5 moles of oxygen were put in 1litre vessel

which was heated to 400 oC until equilibrium was established; at equilibrium the vessel

was found to contain 0.5 moles of oxygen. If nitrogen monoxide and oxygen react

according to the equation

2𝑁𝑂(𝑔)+ 𝑂2(𝑎𝑞) 2𝑁𝑂2(𝑔)

Calculate the value of KC with its units

DEGREE OF DESSOCIATION (α)

This is the fraction of one mole of the original substance that has dissociated

Example

When hydrogen iodide dissociated to give iodine and hydrogen gas and the degree of

dissociation is α, the following information can be obtained

2𝐻𝐼(𝑔)) 𝐻2(𝑔) + 𝐼2(𝑔) Initially 1 ……. ……

Reacted α α1

2 α

1

2 (2:1:1)

Equilibrium (a- α) α1

2 α

1

2

EXAMPLE

1 If 2.6 moles of hydrogen iodide were heated at 400 oC and found to be 20%

dissociated. Calculate

The concentration of hydrogen iodine and hydrogen iodide present at equilibrium

The equilibrium constant, KC

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Solution

Initial moles of HI n=2.6 Degree of dissociation = 20

100 = 0.2

2𝐻𝐼(𝑔) 𝐻2(𝑔) + 𝐼2(𝑔) Initially 2.6 ……. ……

Reacted 2.6α 2.6α

2

2.6α

2 (2:1:1)

Equilibrium 2.6(1- α) 2.6α

2

2.6α

2

But α=0.2

Moles at equilibrium

𝐻𝐼=2.6(1-0.2)=2.08, 𝐻2 =2.6x0.2

2 =0.26, 𝐼2 =

2.6x0.2

2 =0.26

𝐾𝑐 = [𝐼2][𝐻2]

[𝐻𝐼]2=

0.26𝑋0.26

2.082 = 0.0156

Exercise 4

1 The degree of dissociation of 2.4 moles of hydrogen iodide at 450 oC was found to be

22%. Calculate the

(a) the number of moles of hydrogen iodide, iodine ad hydrogen at equilibrium.

(b) The equilibrium constant Kc at, assuming the volume of the container is 600cm3

2 At a certain temperature and a total pressure of 1 atmosphere, di nitrogen tetra oxide is

50% dissociated according to the equation below;

𝑁2𝑂4(𝑔) 2𝑁𝑂2(𝑔) (a) Calculate the equilibrium constant Kp for the reaction

(b) The degree of dissociation of dinitrogen tetra oxide at a pressure of 10

atmospheres at the same temperature

Calculations in which the percentage amount of a reactant or product at

equilibrium is given.

Example

Carbon monoxide reacts with steam according to the following equation;

C𝑂(𝑔) +𝐻2𝑂(𝑔) 𝐶𝑂2(𝑔) + 𝐻2(g)

(a) Write the equilibrium constant Kc for the reaction

(b) Equal moles of carbon monoxide and steam were reacted in one litre vessel. When

equilibrium was established at 750℃, the vessel contained 26.7% carbon dioxide.

Calculate the equilibrium constant Kc for the reaction at 750℃.

Solution

(a)

𝐾𝑐 = [𝐶𝑂2] [𝐻2]

[CO] [𝐻2O]

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(b)

Let X moles of 𝐶𝑂2 formed at equilibrium

C𝑂(𝑔) + 𝐻2𝑂(𝑔) 𝐶𝑂2(𝑔) + 𝐻2(g)

Initially 1 1 - -

Reacted X X X X (mole ratio 1:1:1:1)

Eq≡ 𝑀 1-X 1-X X X

Total number of moles at equilibrium = 1-X+1-X+ X+ X= 2

Then we can express the carbon dioxide at equilibrium as a percentage to get the value of

X as below;

𝑀𝑜𝑙𝑒𝑠 𝑜𝑓𝐶𝑂2𝑎𝑡 𝑒𝑞𝑢𝑖𝑏𝑟𝑖𝑢𝑚

𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑎𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚𝑥 100 = 26.7

𝑋

2𝑥 100 = 26.

X= 0.534

Equilibrium concentrations

[𝐶𝑂2] =0534

1=0.534, [𝐻2] =

0534

1=0.534,

[𝐶𝑂 ] =(1-0.534)=0.466, [𝐻2𝑂] =(1-0.534)=0.466

𝐾𝑐 = 0.534 x 0.534

0.466 x0.466 = 1.313

Exercise 5

1 At a given temperature, 3.2 moles of phosphorus pent chloride dissociate so that

at equilibrium, it was found to 20% chlorine. Calculate the value of Kc.

2 Nitrogen monoxide combines with oxygen at 80℃ and 200atm to form Nitrogen

(iv) oxide (𝑁𝑂2).

(a) Write the expression of the equilibrium constant Kp for the reaction that take

place

(b) Calculate the Kp, if the mixture contained 67% Nitrogen (iv) oxide at

equilibrium.

HETEROGENEOUS SYSTEMS

These are systems where the reactants and products are not in the same phase e.g. some

may be solids others gases or liquid

Note: the concentration of solids is always taken to be constant and does not change

during the reaction therefore it does not appear in the expression of KC and KP e.g.

𝐶𝑎𝐶𝑂3(𝑠) 𝐶𝑎(𝑎𝑞)2+ + 𝐶𝑂3(𝑎𝑞)

2−

𝐾𝐶 = [𝐶𝑎2+][ 𝐶𝑂3(𝑎𝑞)2− ]

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When water in liquid state is one of the reactants, its concentration is regarded to be in

excess therefore does not appear in the KC expressions e.g.

𝐶𝐻3𝐶𝑂𝑂𝐶𝐻3(𝑙)+𝐻2𝑂(𝑙) 𝐶𝐻3𝐶𝑂𝑂𝐻(𝑙)+𝐶𝐻3𝑂𝐻(𝑙)

KC = [𝐶𝐻3𝐶𝑂𝑂𝐻][𝐶𝐻3𝑂𝐻]

[𝐶𝐻3𝐶𝑂𝑂𝐶𝐻3]

However if water is one of the products, then it appears in the KC expression e.g.

𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞)+𝐶𝐻3𝑂𝐻(𝑎𝑞) 𝐶𝐻3𝐶𝑂𝑂𝐶𝐻3(𝑎𝑞)+𝐻2𝑂(𝑙)

KC = [𝐶𝐻3𝐶𝑂𝑂𝐶𝐻3][𝐻2𝑂]

[𝐶𝐻3𝐶𝑂𝑂𝐻][𝐶𝐻3𝑂𝐻]

Example

Calcium hydride reacts with steam to produce calcium hydroxide and hdroge at 1

atmosphere as follows;

C𝑎𝐻2(𝑠) + 2𝐻2𝑂(𝑔) 𝐶𝑎(𝑂𝐻)2(𝑠) + 2𝐻2(g)

Calculate the equilibrium constant Kp if the mixture at equilibrium contained 58.62%

hydrogen gas.

Solution;

𝐾𝑝 = (P𝐻2)

2

(P𝐻2𝑂)2

Let n be moles of steam that reacted

C𝑎𝐻2(𝑠) + 2𝐻2𝑂(𝑔) 𝐶𝑎(𝑂𝐻)2(𝑠) + 2𝐻2(g)

Initially 2 -

Reacted 2n 2n (mole ratio 2:2)

Eq≡ 𝑀 2(1-n) 2n

Total moles at equilibrium = 2-2n+2n= 2

𝑀𝑜𝑙𝑒𝑠 𝑜𝑓𝐻2𝑎𝑡 𝑒𝑞𝑢𝑖𝑏𝑟𝑖𝑢𝑚

𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑎𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚𝑥 100 = 58.62

2𝑛

2𝑥 100 = 58.62

n = 0.5862

Partial pressure of steam and hydrogen at equilibrium

P𝐻2= 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 ℎ𝑑𝑟𝑜𝑔𝑒𝑛

𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠𝑥 𝑡𝑜𝑡𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑟𝑒

P𝐻2=2𝑥0.5862

2𝑥 1 =0.5862

P𝐻2𝑂=2(1−0.5862)

2𝑥 1 =0.4138

𝐾𝑝 = 0.58622

0.4132 =2.015

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FACTORS THAT AFFECT EQUILIBRIUM REACTIONS

There are several factors that affect equilibrium reactions through in different ways and

these include

Temperature

Pressure

A catalyst

Addition of an inert gas

Concentration

The above factors may affect equilibrium reactions by either;

Changing the value of equilibrium constant (𝑲𝑪or𝑲𝑷)

This involves the factor or condition increasing or decreasing the value of 𝑲𝑪 or 𝑲𝑷 by

changing the concentration or partial pressure of reactants or products

However it should be noted that the equilibrium constant value is only temperature

dependent.

Shifting/changing the equilibrium position to either the left or right.

Equilibrium position refers to the proportion of reactants to products at equilibrium i.e.

If the concentration of reactants is greater than the concentration of the products then the

equilibrium position is said to shifted to the left however if the concentration of the

reactants is less than that of concentration of products then the equilibrium position is to

the right

Increasing or decreasing the rate of attainment of equilibrium.

This simply means how fast both the forward and backward reaction proceed and attain

equilibrium. Increase in rate of attainment means equilibrium is attained faster and vise

versa.

To explain how the equilibrium constant and position of equilibria of a given chemical

system is affected by the factors above, one must know about the Lechateliers’ principle

which states that “If a system is at equilibrium and subjected to external stress or

change, it will shift in a direction that tends to cancel or nullify the effect of the

stress”

A) CONCENTRATION

The concentration affects both the rate of reaction and position of equilibrium

High concentration of reactants increase the rate of react because of the high chances of

collision between the reacting particles

Generally,

𝒂𝑨 + 𝒃𝑩 𝒄𝑪 + 𝒅𝑫

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𝑲𝑪 = [𝑪]𝒄[𝑫]𝒅

[𝑨]𝒂[𝑩]𝒃

(i) Changing concentration of reactants (A and B)

If more of A is added to the equilibrium mixture the concentration of A will increase and

this attempts to change the equilibrium constant, this favours the forward reaction

therefore B reacts with the extra A added to form more of C and D shifting the

equilibrium position from left to the right while keeping the value of KC or Kp constant.

Similarly if more B was added to the equilibrium mixture, its concentration would

increase, favours the forward therefore A reacts with the extra B added to form C and D

so as to maintain equilibrium constant but this shifts the equilibrium position to the right.

(ii) Changing the concentration of products ( C and D)

If more of C is added to the equilibrium mixture the concentration of C will increase and

this attempts to change the equilibrium constant, this favours the Backward reaction

therefore D reacts with the extra C added to form more of A and B shifting the

equilibrium position from right to the left while keeping the value of KC or Kp constant.

Similarly if more D was added to the equilibrium mixture, its concentration would

increase, favours the backward therefore C reacts with the extra D added to form A and B

so as to maintain equilibrium constant but this shifts the equilibrium position to the left.

Exercise 6:

1(a) State and explain what would happen to the equilibrium constant and equilibrium

position of the general reaction above if some A was removed

(b) State and explain what would happen to the equilibrium constant and equilibrium

position of the general reaction above if some C was removed immediately its

formed.

B) TEMPERATURE

Generally increasing the temperature for any reaction increases the rate of the reaction

and therefore the equilibrium will be attained faster but this does not imply that more

products will be obtained or equilibrium constant increased

The effect of temperature on equilibrium constant and equilibrium position depends on

whether the reaction is exothermic or endothermic

For exothermic reaction e.g

𝐴 + 𝐵 𝐶 + 𝐷 𝐷𝐻 = –𝑉𝑒

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The forward reaction proceeds with release of heat, therefore increasing the temperature

will not favour the forward reaction instead favours the backward reaction such that more

of A and B will be formed .This increases the concentration or partial pressure of A and

B and decreases that of C and D shifting the equilibrium position to the left and decreases

the equilibrium position

However when the temperature is decreased, the forward reaction is favored such that

more of C and D are formed ,This increases the concentration or partial pressure of C

and D and decreases that of that of A and B shifting the equilibrium position to the right

and increasing the equilibrium constant

Example

Explain the effect of increasing and decreasing the temperature on the equilibrium

constant and position of the reaction below

𝑁2(𝑔)+ 𝐻2(𝑔) 𝑁𝐻3(𝑔)𝐷𝐻 = −92𝐾𝑔𝑚𝑜𝑙−1AT25 oC

Solution

The forward reaction proceeds with release of heat, therefore increasing the temperature

will not favour the forward reaction instead favours the backward reaction such that

ammonia dissociates to form more of nitrogen and hydrogen .This increases the

concentration or partial pressure of nitrogen and hydrogen and decreases that of ammonia

shifting the equilibrium position to the left and decreases the equilibrium position

However when the temperature is decreased, the forward reaction is favored such that

more of ammonia is formed ,This increases the concentration or partial pressure of

ammonia and decreases that of that of nitrogen and hydrogen shifting the equilibrium

position to the right and increasing the equilibrium constant

For endothermic reaction e.g.

𝐴 + 𝐵 𝐶+ 𝐷 𝐷𝐻 = + 𝑉𝑒

In this case the forward reaction is endothermic favoured by increase in temperature such

that more of C and D are formed , this increases the concentration or partial pressure of C

and D and decreases that of A and B shifting the equilibrium position to the right which

also increases the equilibrium constant

However decreasing the temperature favours the backward reaction such that more of A

and B this decreases the concentration or partial pressure of C and D and increases that of

A and B shifting the equilibrium position and to the left which also decreases the

equilibrium constant

Example

Nitrogen and oxygen according to the following equation below at 25 oC

𝑁2(𝑔)+ 𝑂2(𝑔) 2𝑁𝑂(𝑔) 𝐷𝐻 + 180 𝐾𝑔𝑚𝑜𝑙−1

Explain how the reaction is affected if it’s carried out at temperatures of;

(i) 18 oC

The forward reaction proceeds with absorption of heat (endothermic), therefore

decreasing temperature from 25 oC to18 oC, the backward reaction is favored such that

13

nitrogen monoxide dissociates to form more of nitrogen and oxygen, This increases the

concentration or partial pressure of nitrogen and oxygen while decreasing that of nitrogen

monoxide shifting the equilibrium position to the left and decreasing the equilibrium

constant

(ii) 40 oC

The forward reaction proceeds with absorption of heat (endothermic), therefore

increasing the temperature 25 oC to40 oC, will favour the forward reaction such that

nitrogen reacts with oxygen to form more nitrogen monoxide .This increases the

concentration or partial pressure of nitrogen monoxide and decreases that of nitrogen and

oxygen shifting the equilibrium position to the right and increases the equilibrium

constant.

C )PRESSURE

Generally;

Pressure has negligible effect on the volumes of the liquids and solids since they are not

compressible

It affects only reactions in which gases are involved

Increasing pressure increases the rate of equilibrium reaction involving gases

Pressure has no effect on the equilibrium constant since it is only temperature dependent

The effect of pressure on equilibrium position depends on whether the reaction proceeds

with increase, decrease or no change in volume

Change in volume is determined by noting the total number of moles on the reactant and

product sides for example

If a reaction proceeds with decrease in volume e.g

𝑁2(𝑔)+ 3𝐻2(𝑔) 2𝑁𝐻3(𝑔)

This reaction proceeds with decrease in volume, increasing the pressure favours the

forward reaction, nitrogen reacts with hydrogen to form more ammonia, this increases its

partial pressure and decreases that of nitrogen and hydrogen shifting the equilibrium

position to the right.

The reverse is true if pressure is lowered.

The equilibrium constant Kp or Kc remains constant as long as temperature is constant.

If a reaction proceeds with increase in volume e.g

2𝑁𝑂2(𝑔) 2𝑁𝑂(𝑔)+ 𝑂2(𝑔)

Reactant side =2 moles

Low volume side

High pressure side

Product side =3 moles

High volume side

Low pressure side

Reactant side =4 moles

High volume side

Low pressure side

Reactant side =2 moles

Low volume side

High pressure side

14

This reaction proceeds with increase in volume, increasing the pressure favours the

backward reaction, nitrogen monoxide reacts with oxygen to form more nitrogen

dioxide, this increases its partial pressure and decreases that of nitrogen monoxide and

oxygen shifting the equilibrium position to the left .

The reverse is true if pressure is lowered.

The equilibrium constant Kp or Kc remains constant as long as temperature is constant.

If a reaction proceeds with No change in volume e.g

2𝐻𝐼(𝑔) 𝐼2(𝑔)+ 𝐻2(𝑔)

For all reactions that proceed with no change in volume, change in pressure does not

affect such reactions. Therefore the both equilibrium constant and position remain

constant

D) ADDITION OF AN INERT GAS (Ne, He, Ar) AT ACONSTANT PRESSURE

An inert gas does not take part in a chemical reaction.

The effect of an inert gas depends on the whether the reaction proceeds with increase,

decrease, or no change in volume.

For reactions that proceed with no change in volume e.g.

𝐼2(𝑔)+ 𝐻2(𝑔) 2𝐻𝐼(𝑔)

Addition of an inert gas to such a system will not affect the partial pressure of reactants

and products since volume is constant; therefore the equilibrium position remains

unchanged.

For reactions that proceed with increase in volume e.g.

2𝑁𝑂2(𝑔) 2𝑁𝑂(𝑔)+ 𝑂2(𝑔)

Addition of an inert gas at a constant pressure increases the total volume of the system;

this decreases the number of moles per unit volume of reactants and products which will

favour the forward reaction that proceeds with increase in number of moles hence

shifting the equilibrium position to the right.

For reactions that proceed with decrease in volume e.g

𝑁2(𝑔)+ 3𝐻2(𝑔) 2N𝐻3(𝑔)

Addition of an inert gas at a constant pressure increases the total volume of the system;

this decreases the number of moles per unit volume of reactants and products which will

Reactant side =3 moles

Product side =3 moles

15

favour the backward reaction that proceeds with increase in number of moles hence

shifting the equilibrium position to the left.

Note: Addition of an inert gas to a system at a constant volume will not a affect the

position of equilibrium because it will only increase the total pressure of the system but

the partial pressure of the reactants and products will not change.

E) EFFECT OF ADDING ACATALYST

Catalyst is defined as substance that alters or changes the rate of a chemical reaction but

remains unchanged in quantity at the end of the reaction or at equilibrium. This means

that a catalyst has no effect on equilibrium constant and equilibrium position.

Catalytic actions are homogeneous when the catalyst and the reacting substances are in

the same physical state but will be heterogeneous if the catalyst and the reacting

substances are in different physical state e.g.

In esterification of ethanol by ethanoic acid, sulphuric acid is homogeneous catalyst and

in the combination of hydrogen and oxygen to form water, platinum is a heterogeneous

catalyst

GENERAL PROPERTIES OF A CATALYST

A catalyst is unchanged at the end of the reaction

A catalyst alters the reaction rate but never the final position of the equilibrium;

catalyst may increase or decrease the rate of the reaction but has no effect on the

position of the equilibrium.

A small amount of a catalyst usually influences large amount of reacting

substances

Catalytic power is always more specific in character i.e. it catalyzes specific

reactions.

The efficiency of a catalyst is often increased or decreased by other substances

e.g. the catalyst used in Haber process (finally divided iron) is catalyzed by

addition of small amounts of Aluminum Oxide.

NOTE:

A catalyst which increase the rate of a reaction, achieve it by providing an alternative

pathway with lower activation energy than that without a catalyst

EXPERIMENT TO DETERMINE EQUILIBRIUM CONSTANT KC VALUE

16

(a) For etherification

An ester is formed when an alcohol (e.g Ethanol) is heated with a carboxylic acid

(e.g ethanoic acid) in presence of concentrated sulphuric acid.

𝐶𝐻3𝐶𝑂𝑂𝐻(𝑙+ 𝐶𝐻3𝐶𝐻2𝑂𝐻(𝑙) 𝐶𝐻3𝐶𝑂𝑂𝐶𝐻2𝐶𝐻3(𝑙)+ 𝐻2𝑂(𝑙)

Procedure

A known amount of ethanoic acid (a moles) and ethanol (b moles) are mixed in a dry

clean glass tube of a known volume (Vcm3). The mixture is gently shaken and a few

drops of concentrated sulphuric acid is added and the tube sealed.

The glass tube is then placed in a water bath at 70℃ and allowed to stand for several

hours so as to react up to equilibrium.

The mixture is cooled rapidly and the glass tube broken at ice cold to freeze the reaction

such that the position of equilibrium does not change.

A known volume of the equilibrium mixture is pipetted into a conical flask and titrated

with standard solution of sodium hydroxide using phenolphthalein indicator to determine

the amount of acid left at equilibrium.

Treatment of results.

Let x be the number of moles of acid (ethanoic acid) remaining at equilibrium

𝐶𝐻3𝐶𝑂𝑂𝐻(𝑙)+ 𝐶𝐻3𝐶𝐻2𝑂𝐻(𝑙) 𝐶𝐻3𝐶𝑂𝑂𝐶𝐻2𝐶𝐻3(𝑙)+ 𝐻2𝑂(𝑙)

Initially a b ………… .…….

Reacted (a-x) (a-x) (a-x) (a-x)

Equilibrium x b-(a-x) (a-x) (a-x)

Concentration 𝒙

𝒗

𝒃−(𝒂−𝒙)

𝒗

(𝒂−𝒙)

𝒗

(𝒂−𝒙)

𝒗

At equilibrium

From, KC = [𝐶𝐻3𝐶𝑂𝑂𝐶𝐻3][𝐻2𝑂]

[𝐶𝐻3𝐶𝑂𝑂𝐻][𝐶𝐻3𝑂𝐻] 𝐾𝑐=

(a−x)2

𝑥(𝑏−𝑎+𝑥)

17

(b) For a reaction between hydrogen and iodine to form hydrogen iodide

Procedure

A known amount of hydrogen (a moles) and iodine (b moles) are mixed and sealed in a

dry clean glass bulb of a known volume (Vcm3)

The glass tube is then heated at 450℃ for hours so as to react up to equilibrium.

The mixture is cooled rapidly and frozen so as to freeze the reaction such that the position

of equilibrium does not change.

The glass bulb is then broken under potassium iodide solution to form iodine solution.

A known volume of the equilibrium iodine solution is pipetted into a conical flask and

titrated with standard solution of sodium thiosulphate using starch indicator to determine

the amount of iodine left at equilibrium.

Equation

2𝑆2𝑂3(𝑎𝑞)2− + 𝐼2(𝑎𝑞)

→ 𝑆4𝑂6(𝑎𝑞)

2− + 2𝐼(𝑎𝑞)−

Treatment of results

Let X be moles iodine remained at equilibrium

Equation: 𝐻2(𝑔) + 𝐼2(𝑔) → 2𝐻𝐼(𝑔)

Initially a b …….

Reacted a-X a-X 2(a-X)

Equibrium X b-(a-X) 2(a-X)

Concentration 𝑿

𝑽

𝒃−(𝒂−𝑿)

𝑽

2(𝒂−𝑿)

𝒗

at equilibrium

From, KC = [HI]2

[𝐻2][𝐼2] 𝐾𝑐=

(2(a−X))2

𝑋(𝑏−𝑎+𝑋)

APPLICATION OF EQUILIBRIUM REACTIONS AND LE CHATELIER’S

PRINCIPLE

They are applied in industrial processes like the Haber processes and contact process and

manufacture of nitric acid to predict optimum conditions for maximum yield of products

CONTACT PROCESS

This is the process that involves manufacture of sulphuric acid from sulphur trioxide

𝑆𝑂2(𝑔)+ 𝑂2(𝑔) 2𝑆𝑂3(𝑔)𝐷𝐻𝜃 = −197 𝐾𝑔𝑚𝑜𝑙−1

The reaction proceeds with decrease in volume therefore to maintain the equilibrium

position to the right, such more sulphur trioxide is produced the pressure must be high

The reaction is exothermic so increasing temperature does not favour formation of

sulphur trioxide however at low temperatures a high yield of sulphur trioxide will be

obtained .In practice a temperature of 450 oC -500oCis used which is low temperature for

an industrial process.

18

Since the reaction is favoured by low temperature a catalyst is used in this reaction,

vanadium (v) oxide which does not affect the amount of sulphur trioxide formed and

position of equilibrium but increases the rate at which sulphur trioxide is obtained.

Therefore the optimum conditions for maximum yield sulphur trioxide are;

High pressure since the reaction proceeds with decrease in temperature

Low temperature since reaction is exothermic

Catalyst to increase rate of formation of sulphur trioxide at low temperature.

Note

(i) Once sulphur trioxide is formed it’s dissolved in concentrated sulphuric acid to

form a liquid called oleum. This then diluted with a calculated amount of water

to form the required concentration of sulphuric acid.

Equations:

𝑆𝑂2(𝑔)+ 𝑂2(𝑔) 2𝑆𝑂3(𝑔)

𝑆𝑂3(𝑔) + 𝐻2𝑆𝑂4(l) 𝐻2𝑆2𝑆𝑂7(l)

𝐻2𝑆2𝑆𝑂7(l) + 𝐻2𝑂(l) 2𝐻2𝑆𝑂4(l)

(ii) The sulphur trioxide is not directly added to water to form sulphuric acid

because the reaction is extremely exothermic releasing a lot of heat and fumes

of sulphuric acid vapour (sulphuric acid spray) that corrode and injure the

person preparing.

HABER PROCESS

This is the process that involves manufacture of ammonia from nitrogen and hydrogen

3𝐻2(𝑔)+ 𝑁2(𝑔) 2𝑁𝐻3(𝑔)𝐷𝐻𝜃 = −46 𝐾𝑔𝑚𝑜𝑙−1

The reaction proceeds with decrease in volume therefore to maintain the equilibrium

position to the right, such that more of ammonia is produced; the pressure must be high at

about 200atmospheres

The reaction is exothermic so increasing temperature does not favour formation of

ammonia however at low temperatures a high yield of ammonia will be obtained .In

practice a temperature of 450 oC is used which is low temperature for an industrial

process.

Since the reaction is favoured by low temperature a catalyst is used in this reaction; finely

divided iron which does not affect the amount of ammonia formed and position of

equilibrium but increases the rate at which ammonia is obtained.

Therefore the optimum conditions for maximum yield ammonia are;

High pressure since the reaction proceeds with decrease in temperature

Low temperature since reaction is exothermic

Catalyst to increase rate of formation of ammonia at low temperature.

19

MANFACTURE OF NITRIC ACID

Nitric acid is manufactured from nitrogen and oxygen gas that react to form nitrogen

monoxide.

Principle equation;

𝑁2(𝑔)+ 𝑂2(𝑔) 2𝑁𝑂(𝑔)𝐷𝐻𝜃 = +43.2 𝐾𝑐𝑎𝑙

The reaction proceeds with no change in volume therefore changing pressure has no

effect on the equilibrium position of this reaction and can be carried out at normal

atmospheric conditions.

The reaction is endothermic so increasing temperature favour formation of nitrogen

monoxide thus in practice a high temperature of 3000 oC is used which is used to obtain

a high yield of nitric acid.

Since the reaction occurs at a high temperature a catalyst may not necessary since the rate

of reaction increases with increase in temperature.

Keeping a high concentration of the reactants (Nitrogen and oxygen) favours the

formation of nitrogen monoxide and shifts the position of equilibrium to the right

Therefore the optimum conditions for maximum yield ammonia are;

High temperature since reaction is endothermic

Maintaining a high concentration of oxygen and nitrogen

Constant removal of nitrogen monoxide immediately its formed

Note

(i) Once nitrogen monoxide is formed it’s rapidly cooled and combined with

oxygen from excess air to form nitrogen dioxide. This is then absorbed in hot

water in presence of more air to form nitric acid.

Equations:

𝑁2(𝑔)+ 𝑂2(𝑔) 2𝑁𝑂(𝑔)

2𝑁𝑂(𝑔) + 𝑂2(g) 2N𝑂2(g)

4𝑁𝑂2(g) + 2𝐻2𝑂(l) + 𝑂2(g) 4HN𝑂3(aq)

(ii) Nitrogen monoxide may as well be made by oxidation of ammonia obtained

from the Haber process using excess oxygen from excess air and platinum

catalyst, this is the main reason ammonia manufacturing plants also make nitric

acid.

Equation;

4𝑁𝐻3(𝑔)+ 5𝑂2(𝑔) 4𝑁𝑂(𝑔)+ 6𝐻2𝑂(l)

END (THANKS)