Unit 8: Chemical Equilibrium — Long Questions

Ans. (i) Reversible reactions
Definition: The reactions in which reactants are converted into products and products are converted back into reactants, and it proceeds in both directions without change in concentrations of reactants and products are called reversible reactions.
Examples

N₂₍ₐ₎ + 3H₂₍ₐ₎ ⇌ 2NH₃₍ₐ₎

2NO₂₍ₐ₎ ⇌ N₂O₄₍ₐ₎

PCl₅₍ₐ₎ ⇌ PCl₃₍ₐ₎ + Cl₂₍ₐ₎

Explanation
• Reversible reactions are represented by double arrow (⇌).
• Proceed in both forward and reverse directions.
• Do not go to completion; instead, they reach a state of balance.
• Occur in a closed system where no substance enters or leaves.
• The concentrations of reactants and products remain constant.
• The rate of the forward reaction equals the rate of the reverse reaction, when they achieve equilibrium.
• This state does not mean the reactions have stopped, only that the changes balance out.

(ii) Irreversible reactions
Definition: The reactions in which the reactants are completely consumed and converted into products are called irreversible reactions.
Example Combustion of methane CH₄ + 2O₂ ⟶ CO₂ + 2H₂O + Heat
Formation of water 2H₂ + O₂ ⟶ 2H₂O

Explanation
• Irreversible reactions are represented by single arrow (⟶).
• Reactants are completely converted to products.
• The reaction stops when the limiting reactant is used.
• They are represented by single arrow.

(iii) Macroscopic events
Definition: Macroscopic events refer to the phenomena that can be observed with the naked eye without considering the individual particles or molecules involved in the process.
Examples of macroscopic properties/events
• Color change: Rusting of iron.
• Evolution or absorption of heat: Exothermic reactions release heat, endothermic reactions absorbs it.
• Formation of a precipitate: Solid formed from two solutions, mixing silver nitrate and sodium chloride.
• Gas evolution: Bubbles or fizzing, vinegar reacting with baking soda.
• Change in volume or pressure: Noticeable in reactions involving gases.
• Change in composition: A substance transforms into new products, burning of wood.
These macroscopic observations are often used to infer what's happening on a molecular (microscopic) level.

(iv) Microscopic events
Definition: Microscopic events refer to the phenomena that cannot be observed with the naked eye.
Examples of microscopic events
• Molecular collisions: Particles interact and transfer energy.
• Breaking and forming of chemical bonds: Fundamental to chemical reactions.
• Rearrangement of atoms: Leads to new substances being formed.
• Electron transfer: In redox reactions, electrons move from one species to another.
• Changes in molecular structure or geometry: Affects properties and behavior of substances.

Ans. Dynamic equilibrium
Definition: The state in which rate of forward and rate of reverse reaction are continuous in both directions with similar rates is called dynamic equilibrium.

Dynamic nature of chemical equilibrium
When a reversible reaction reaches equilibrium, it might appear static from a macroscopic point of view concentrations of reactants and products no longer change visibly. At the microscopic level, the system is still highly dynamic:
• Reactant molecules oontinue to collide and form products.
• Product molecules simultaneously revert back to reactants.
• The forward and reverse reactions occur at equal rates, so there's no net change in concentration.

Reaction of steam and carbon monoxide
Forward Reaction: At the start maximum number of collisions between them will occurs as microscopic events.
• Equal moles of steam and CO with high frequency of collisions.
• Bonds in CO and H₂O are broken, forming H₂ and CO₂.
• The forward reaction rate is maximum initially due to high reactant concentrations.
H₂O₍ₐ₎ + CO₍ₐ₎ ⟶ H₂₍ₐ₎ + CO₂₍ₐ₎
• Reactant concentrations decrease, so forward rate slows down as the reaction proceeds.
• Product concentrations increase, so the reverse reaction begins to speed up.
• After the product formation reverse reaction starts so bonds are formed and broken at the same time.
• This ongoing molecular activity at equilibrium is define as dynamic, not static.

Reverse reaction
As the products hydrogen (H₂) and carbon dioxide (CO₂) accumulate during the forward reaction:
Their concentrations increase, leading to:
• More frequent collisions between H₂ and CO₂ molecules.
• These are microscopic events where molecules interact at the atomic level.
During these collisions
• Bonds in H₂ and CO₂ are broken.
• New bonds are formed to regenerate the reactants: CO and H₂O.
The forward reaction starts with maximum rate and gradually slows down, whereas at the start, the rate of the reverse reaction is low, gradually increases, and finally becomes constant.
H₂₍ₐ₎ + CO₂₍ₐ₎ ⟶ H₂O₍ₐ₎ + CO₍ₐ₎
• A time comes when both reactions proceed at the same rate.
• The reaction at this stage is said to be in chemical equilibrium and the concentration of reactants and products become constant.
H₂O₍ₐ₎ + CO₍ₐ₎ ⇌ H₂₍ₐ₎ + CO₂₍ₐ₎
Unless the system is somehow disturbed, no further changes in the concentrations will occur.

State of chemical equilibrium
Definition: The state of a reversible reaction at which composition of the reaction mixture does not change and forward and reverse rates are equal is called the state of chemical equilibrium.

Concentration time graphs in reversible reactions
Consider the reaction
A ⇌ B
The concentration of A (reactant) and B (product) is plotted over time.

Before equilibrium
• [A] (reactant) decreases over time.
• [B] (product) increases as it is formed from A.
At equilibrium
• The curves level off, meaning concentrations of both A and B become constant.
• No further net change, even though the forward and reverse reactions continue (dynamic equilibrium).

Different scenarios at equilibrium
A general reaction in which a reactant A is converted into product B. After attaining the chemical equilibrium, the graph shows three possibilities.

(i) Equal concentration: In this type of graph, the reactants and products are in equal concentration at the time of equilibrium.
• [A] = [B] at equilibrium.
• The system reaches a point where reactant and product are in equal concentrations.

(ii) Products are greater than reactants
• [A] < [B] at equilibrium.
• The reaction favors the products (more B formed) after achieving the equilibrium.

(iii) Reactants are greater than products
• Concentration of A and B are constant.
• [A] > [B] at equilibrium.
• The reaction favors the reactants (less B formed) after achieving the equilibrium.

• When a reversible reaction reaches equilibrium, the concentrations of reactants and products remain constant over time.
• This does not mean the reaction has stopped.

Microscopic activities
• Reactant molecules are still converting into products.
• Product molecules are simultaneously converting back into reactants.
• These processes are ongoing and balanced.

Equal reaction rates
• The rate of the forward reaction equals the rate of the reverse reaction.
• There is no net change in concentrations, even though reactions are still happening because these rates are equal.
• The system is not static, molecular collisions, bond breaking, and bond formation continues.
• Individual molecules are continuously reacting but the rates of forward and reverse reaction are equal.
• It's dynamic at the microscopic level, even though macroscopic properties (like color, pressure, or concentration) appear unchanged.
• This concept is essential in understanding how equilibrium systems behave under different conditions.

Ans. Relationship between microscopic and macroscopic events
• Macroscopic events are the observable changes in a system. These are the result of multiple simultaneous microscopic events at the molecular level.
• Importance of understanding microscopic events: Understanding microscopic events helps us to explain and predict the macroscopic changes that occur in an equilibrium system.
• Microscopic events in a chemical reaction: When there is a change in equilibrium, it is due to microscopic activities such as collisions between particles or molecules, formation and breaking of chemical bonds.
• Factors affecting reaction rates: These collisions affect the rates of the forward and reverse reactions. Reaction rates are influenced by, Activation energy, external factors such as temperature, pressure, and concentration.
Observable effects of microscopic changes: The combined effect of these microscopic events causes macroscopic changes. These changes can be observed by the naked eye, such as change in color, formation of precipitates, change in temperature.

Quick Check 8.1
(a) Differentiate between macroscopic and microscopic events.
Ans.
Macroscopic Events: Observable on a large scale (bulk properties), Examples: color change, pressure, volume, temperature, Measurable directly using instruments
Microscopic Events: Occur at the atomic or molecular level, Examples: collisions of molecules, bond breaking/forming, Not visible; explained using models and theories

(b) The equilibrium is dynamic in nature, explain in terms of microscopic events.
• At equilibrium, the rate of forward reaction is equal to rate of reverse reaction.
• Microscopically, both reactions are still occurring, but their effects cancel out.
• So, the concentrations remain constant even though molecules are still reacting.
• The system appears static macroscopically, but microscopic changes continue, making equilibrium dynamic in nature.

Ans. Dynamic Equilibrium
Definition: The state in which the rate of forward and reverse reaction are continuous in both directions with similar rates is called state of dynamic equilibrium.
• Dynamic equilibrium is a state in a reversible process where, the rate of change in one direction is equal to the rate of change in the opposite direction.
• Even though continuous changes occur at the microscopic level, there is no net change in the system as a whole.

Dynamic equilibrium in phase changes: A reversible phase change also shows dynamic equilibrium between two physical states of a substance.

Example At 0 °C, ice and water coexist in equilibrium. Water freezes into ice and ice melts into water at the same rate.

Characteristics
• No net change is observed in the amount of each phase.
• The macroscopic properties of the system remain constant over time.

Relationship between dynamic equilibrium and vapor pressure
Studying the behavior of liquids and their vapor pressures is a significant concept in chemical equilibrium.

(i) Evaporation and condensation in a closed system
• When a liquid is placed in a closed container, some molecules at the surface gain enough kinetic energy to escape into the gaseous phase this is called evaporation.
• As more molecules evaporate, the vapor pressure (pressure exerted by gas molecules) increases at the walls of vessel.
• Simultaneously, gas molecules may collide with the liquid surface and return to the liquid phase this is called condensation.

(ii) Establishing dynamic equilibrium
• As evaporation and condensation continue, a point is reached where, the rate of evaporation is equal to rate of condensation. This state is called dynamic equilibrium.
• At this point, the number of molecules entering and leaving each phase becomes constant. The vapor pressure remains constant, until the temperature is constant and the system is undisturbed.

(iii) Effect of temperature on vapor pressure
• Increasing temperature raises the average kinetic energy of liquid molecules.
• Faster the evaporation, higher will be the vapor pressure.
• A new dynamic equilibrium is established at the higher temperature with a higher equilibrium vapor pressure.
• Dynamic equilibrium in vapor pressure describes the state at which evaporation and condensation occur at equal rates.
• Vapor pressure remains constant at constant temperature, and increases with temperature.

Liquid ⇌ Gas (with Evaporation above and Condensation below the arrow)
Evaporation

Quick Check 8.2
(a) Dynamic equilibrium exists between water and its vapour at 100°C. Justify.
Ans.
• At 100°C, water boils, so liquid water and water vapor coexist.
• Water molecules evaporate into vapor, and vapor molecules condense back into liquid.
• At equilibrium, the rate of evaporation = rate of condensation.
• Thus, the process is dynamic (both forward and reverse processes continue), but the amount of liquid and vapor remains constant.

(b) Do you think that dynamic equilibrium exists between ice and -water at 0°C? If yes, explain.
Ans. Yes, at 0°C, ice and water coexist.
• Ice melts to water, and water freezes back to ice simultaneously.
• The rates of melting and freezing become equal at equilibrium.
• So, the system is in dynamic equilibrium with constant amounts of ice and water.

Conditions for equilibrium
To study chemical equilibrium, the following conditions must be fulfilled:

(i) Reversibility of reaction
• Equilibrium occurs only in reversible reactions.
• The reaction must proceed in both forward and reverse directions.

(ii) Closed system
• The reaction must take place in a closed vessel.
• No reactants or products should be allowed to escape from the system.

Ans. Characteristics of chemical equilibrium
Important features of equilibrium are as follows:
(i) Concentration: At the stage of chemical equilibrium, the concentrations of reactants and products remain constant.
(ii) Starting point: The state of equilibrium in a reversible reaction can be approached from either side whether we start with reactants or products.
(iii) Role of catalyst: A catalyst does not change the equilibrium position and the equilibrium constant of the reaction. It helps to attain the equilibrium earlier.
(iv) Dependence of Kc on temperature: The value of equilibrium constant does not depend upon the initial concentrations of reactants, rather it is constants and depends on temperature only.

Keep in mind!
S.Q. What is the effect of the concentration of pure solid or liquid on equilibrium constant?
Ans. If pure solids or pure liquids are involved in an equilibrium system; their concentrations are not included in the equilibrium constant expression. This is because the change in concentration of any pure solid or liquid has no effect on the equilibrium constant.

Types of equilibrium
With respect to the physical states of reactants and products, there are two types of chemical equilibrium.

(i) Homogeneous equilibrium
Definition: An equilibrium system in which all of the reactants and products are in the same phase.
Example: The following are examples of homogeneous equilibria.
N₂₍ₐ₎ + 3H₂₍ₐ₎ ⇌ 2NH₃₍ₐ₎
2SO₂₍ₐ₎ + O₂₍ₐ₎ ⇌ 2SO₃₍ₐ₎
CH₃COOH₍ₗ₎ + C₂H₅OH₍ₗ₎ ⇌ CH₃COOC₂H₅₍ₐq₎ + H₂O₍ₗ₎

(ii) Heterogeneous equilibrium
Definition: An equilibrium in which the reactants and products are in more than one phases is called heterogeneous equilibrium.
Example: The following are examples of heterogeneous equilibrium.
CaCO₃₍s₎ ⇌ CaO₍s₎ + CO₂₍ₐ₎
C₍s₎ + H₂O₍ₐ₎ ⇌ CO₍ₐ₎ + H₂₍ₐ₎
3Fe₍s₎ + 4H₂O₍ₐ₎ ⇌ Fe₃O₄₍s₎ + 4H₂₍ₐ₎

Quick Check 8.3
(a) Differentiate between homogeneous and heterogeneous equilibria.
Ans.
Homogeneous Equilibrium: All reactants and products are in the same phase (all gases or all liquids or all solids), Example: N₂₍ₐ₎ + 3H₂₍ₐ₎ ⇌ 2NH₃₍ₐ₎
Heterogeneous Equilibrium: Reactants and products are in different phases (solid, liquid, gas), Example: CaCO₃₍s₎ ⇌ CaO₍s₎ + CO₂₍ₐ₎

(b) Why do forward reaction rates slow down when a reversible reaction approaches the equilibrium stage?
Ans. Initially, forward reaction rate is high; reverse reaction is slow. As products form, reverse reaction rate increases. Near equilibrium, forward and reverse rates become equal. So, the net rate of forward reaction slows down and balances with reverse.

(c) In capped soda water bottles, gaseous CO₂ is in equilibrium with the aqueous CO₂ (HCO₃⁻ and H⁺).
CO₂₍ₐ₎ + H₂O₍ₗ₎ ⇌ HCO₃₍ₐq₎ + H⁺₍ₐq₎

(i) To which direction will the equilibrium shift if the bottle is opened?
(ii) Condition for this equilibrium requires the closed cap of the bottle. Why?
(iii) When the cap of the bottle is removed, to which direction the equilibrium shifts?

Ans. (i) CO₂ escapes when bottle is opened. So direction of reaction is forward and equilibrium shift towards the reactant side (left).
(ii) The system needs to be closed to prevent escape of gases like CO₂.
If the cap is removed, gas escapes and the equilibrium is disturbed.
So, the closed cap maintains a constant volume and pressure, allowing equilibrium to establish.

(iii)
Removing the cap allows CO₂ gas to escape.
According to Le Chatelier's Principle, equilibrium shifts to produce more CO₂ gas to replace lost gases.
So, the equilibrium shifts towards the left (reactants) side.

Ans. Position of Equilibrium
• The position of equilibrium refers to the relative amounts of reactants and products present in an equilibrium mixture.
• It indicates whether the reactants or products are favored in a reversible reaction.

Law of Mass Action
Introduction: C. Guldberg and P. Waage, (1864) Norwegian chemists, observed that, for any reversible reaction.
Statement: The rate at which reaction proceeds is directly proportional to the product of active masses of the reactants.

Equilibrium constant Kc
Definition: The ratio of the product of equilibrium concentrations of products (raised to their stoichiometric coefficients) to the product of equilibrium concentrations of reactants (also raised to their coefficients) is constant under given conditions.

Consider a general reaction
[C]ᶜ[D]ᵈ
Kc = ─────────
[A]ᵃ[B]ᵇ

• This expression is known as the law of mass action.
• [ ] denotes molar concentration in mol dm⁻³, also called active mass.
• Active mass is the concentration of a substance involved in a reaction.
• The rate of a chemical reaction is directly proportional to the product of the active masses of the reactants, each raised to its respective coefficient in the balanced equation.

Let us consider a general reaction
A + B ⇌ C + D

According to the Law of mass action,
Rate of forward reaction ∝ [A][B]
= kf [A][B]
'kf' is the proportionality constant, and is known as forward rate constant.

Rate of reverse reaction ∝ [C][D]
= kr [C][D]
'kr' is the proportionality constant, and is known as reverse rate constant.

Rate of reverse reaction ∝ [C][D]
= kr [C][D]

'kr' is the proportionality constant and is known as reverse rate constant.

At the equilibrium stage, the forward and the reverse rates are equal. Hence,
kf [A][B] = kr [C][D]
kf [C][D]
── = ─────────
kr [A][B]

The left side of this equation is the ratio of two rate constants, so it gives another constant called the equilibrium constant (Kc).
kf
Kc = ──
kr

[C][D]
So, Kc = ──────────
[A][B]

This equation is known as equilibrium constant expression.

General Reaction
Consider the following reversible reaction,
aA + bB ⇌ cC + dD

Where a, b, c and d are the coefficients of balanced equation.

[C]ᶜ[D]ᵈ
Then Kc = ──────────
[A]ᵃ[B]ᵇ

Equilibrium constant expressions of some important reactions
(i) N₂₍ₐ₎ + 3H₂₍ₐ₎ ⇌ 2NH₃₍ₐ₎
[NH₃]²
Kc = ─────────────
[N₂][H₂]³

(ii) 2N₂O₅₍ₐ₎ ⇌ 4NO₂₍ₐ₎ + O₂₍ₐ₎
[NO₂]⁴[O₂]
Kc = ──────────────
[N₂O₅]²

Quick Check 8.4
(a) Write Kc expressions for the given reactions:
Ans. (i)
Sn²⁺₍ₐq₎ + 2Fe³⁺₍ₐq₎ ⇌ Sn⁴⁺₍ₐq₎ + 2Fe²⁺₍ₐq₎
[Sn⁴⁺][Fe²⁺]²
Kc = ─────────────────
[Sn²⁺][Fe³⁺]²

(ii) 4NH₃₍ₐ₎ + 5O₂₍ₐ₎ ⇌ 4NO₍ₐ₎ + 6H₂O₍ₐ₎
[NO]⁴[H₂O]⁶
Kc = ──────────────────
[NH₃]⁴[O₂]⁵

(iii) PCl₅₍ₐ₎ ⇌ PCl₃₍ₐ₎ + Cl₂₍ₐ₎
[PCl₃][Cl₂]
Kc = ──────────────
[PCl₅]

(b) Calculate the value of Kc for the following reaction using the information below:
H₂₍ₐ₎ + CO₂₍ₐ₎ ⇌ H₂O₍ₐ₎ + CO₍ₐ₎

Initial concentrations [H₂₍ₐ₎] = 10.00 mol/dm³ [CO₂₍ₐ₎] = 10.00 mol/dm³

Equilibrium concentration [CO₍ₐ₎] = 9.47 mol/dm³

Ans. Step 1 Determine change in concentration
• For every mole of CO formed, 1 mole of H₂ and 1 mole of CO₂ are consumed.
So, x = 9.47 mol/dm³

Step 2 Calculate equilibrium concentrations of reactants
[H₂]ₑq = 10.00 - 9.47 = 0.53 mol/dm³
[CO₂]ₑq = 10.00 - 9.47 = 0.53 mol
[H₂O]ₑq = x = 9.47 mol/dm³

Step 3 Write Kc expression:
[H₂O][CO] (9.47)(9.47)
Kc = ────────────── = ────────────
[H₂][CO₂] (0.53)(0.53)

Step 4 Calculate Kc
89.7
Kc = ─────── = 319.3
0.2809

Answer Kc = 319.3

Units of equilibrium constant (Kc)
(i) Units of equilibrium constant when moles are equal
• The units of Kc depend on the number of moles of reactants and products in the balanced chemical equation.
• When the total number of moles of reactants equals the total number of moles of products, the units cancel out.
• The equilibrium constant Kc becomes dimensionless (has no units).

Example Ester formation reaction
CH₃COOH₍ₐq₎ + C₂H₅OH₍ₐq₎ ⇌ CH₃COOC₂H₅₍ₐq₎ + H₂O₍ₗ₎
[CH₃COOC₂H₅][H₂O] mol dm⁻³ × mol dm⁻³
Kc = ─────────────────────────────────── = ───────────────────── = no units
[CH₃COOH₍ₐq₎][C₂H₅OH₍ₐq₎] mol dm⁻³ × mol dm⁻³

• Esterification occurs in solution state and follows this pattern:
The number of moles of reactants = number of moles of products.
• The value of Kc for this reaction has no unit.

(ii) Units of kc when moles of reactants and products are unequal
• When the number of moles of reactants and products are unequal, the units of the equilibrium constant (kc) do not cancel out.
• The value of kc depends on the units of concentration used (typically mol dm⁻³).

Example Synthesis of ammonia (Haber's process)
N₂₍ₐ₎ + 3H₂₍ₐ₎ ⇌ 2NH₃₍ₐ₎
[NH₃₍ₐ₎]² [mol dm⁻³]²
Kc = ─────────────────── = ──────────────────────────
[N₂₍ₐ₎][H₂₍ₐ₎]³ [mol dm⁻³][mol dm⁻³]³
1
= ───────────────
[mol dm⁻³]²

Kc = mol⁻² dm⁶

• Reactant side: 1 + 3 = 4 moles
• Product side: 2 moles
• Change in number of moles: -2
The number of moles are not equal, the units of Kc are mol⁻² dm⁺⁶

Ans. Relationship between equilibrium constants
There are four different types of quantities which may be used to calculate the equilibrium constants of reversible reaction. Let the general reaction be
aA + bB ⇌ cC + dD

(i) kc
When the concentrations of reactants and products are in mole dm⁻³, then equilibrium constant is written as:
[C]ᶜ[D]ᵈ
kc = ───────────────── (i)
[A]ᵃ[B]ᵇ

Square brackets [ ] are used for mole dm⁻³

(ii) kp
When the concentrations are expressed in terms of partial pressures (p) for gaseous reactants and products, then
p°c • p°D
kp = ──────────────── (ii)
p°A • p°B

(iii) kn
When the concentrations are expressed in terms of number of moles, then
n°c • n°D
kn = ──────────────── (iii)
n°A • n°B

(iv) kx
When the concentrations are expressed in terms of mole fractions, (X) then
X°c • X°D
kx = ──────────────── (iv)
X°A • X°B

The relationships between these equilibrium constants are as follows:
kp = kc(RT)^Δn
kp = kx(P)^Δn
kp = kn(N)^Δn
R = General gas constant
T = Absolute temperature of the system
P = Pressure of the system
N = Total number of moles of reactants and products
Δn = Difference of number of moles of products and reactants in the balanced chemical equation
• It depends upon the value of 'Δn' that which of the equilibrium constants is bigger or smaller than the other.
• If the number of moles of reactants and products in a balanced chemical equation are equal, and all the constants have equal values.
Δn = 0 then kp = kc = kx = kn

Whichever concentration units are used, the equilibrium constants are same, if change in number of moles of reactants and products are equal.

Position of equilibrium and reaction conditions
The position of equilibrium refers to the relative amounts of products and reactants present in an equilibrium mixture. If a system in equilibrium is disturbed.

(i) [Products] > [Reactants]
If the concentration of products is increased relative to the reactants, we say that the position of equilibrium has shifted to the left.

(ii) [Reactants] > [Products]
If the concentration of products is decreased relative to the reactants, we say that the position of equilibrium has shifted to the right.

Quick Check 8.5
(a) Compare the magnitudes of kc and kp for the following reactions
(i) Ammonia synthesis:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Ans. (i) Change in number of moles of gas:
Δn = moles of products – mole of reactants = 2 – (1 + 3) = - 2

Relationship kp = kc(RT)^Δn as kp < kc, (Since Δn = -2)
kp < kc for ammonia synthesis because change in number of moles is negative.

(ii) Dissociation of PC ℓ₅
PCℓ₅₍g₎ ⇌ PCℓ₃₍g₎ + Clℓ₂₍g₎

Ans. Le-Chatelier's principle

Statement It can be stated as follows: "If a system in equilibrium is disturbed, it behaves in such a way as to nullify the effect of that disturbance".
It describes what happens to a system when something momentarily takes it away from equilibrium.

Applications of Le-Chatelier's principle

The most common applications of this principle with reference to certain physical and chemical equilibria are discussed below:
(i) Effect of change in concentration
(ii) Effect of change in pressure
(iii) Effect of change in temperature
(iv) Effect of catalyst on equilibrium

1. The effect of change of concentrations
• When a system is at equilibrium, the concentrations of reactants and products remain constant.
• If a reactant or product is added or removed, the equilibrium position is disturbed.
• The system then adjusts itself to restore equilibrium this is explained by Le Chatelier's Principle.

There are four possibilities for changes in concentration
(i) Addition of a reactant
(ii) Addition of a product
(iii) Removal of a reactant
(iv) Removal of a product

Hydrolysis of BiCℓ₃

Consider reversible reaction proceed by combining BiC ℓ₃ with H₂O to give BiOC ℓ (insoluble white solid) and HC ℓ

BiCℓ₃ + H₂O ⇌ BiOCℓ + 2HCℓ

The kc expression for the reaction is:

kc = [BiOCℓ][HCℓ]² / [BiCℓ₃][H₂O]

BiCℓ ₃ reacts with water to form BiOC ℓ, a white insoluble precipitate, and hydrochloric acid ( HC ℓ ). If BiOC ℓ is a solid, its concentration is considered constant and may be omitted from the expression depending on the context.

• When water is added to BiC ℓ₃ solution, it becomes cloudy due to formation of BiOC ℓ precipitate.
• At equilibrium, a certain amount of BiOC ℓ and HC ℓ is formed, some BiC ℓ₃ remains unreacted.

(i) Addition of BiC ℓ₃ (reactant)
• Equilibrium position is disturbed by the addition of BiC ℓ₃ .
• Equilibrium shifts to the right (forward direction)
• More BiOC ℓ and HC ℓ are formed to re-establish the equilibrium.
• kc remains constant, but equilibrium position will change.

(ii) Addition of BiOC ℓ or HC ℓ (product)
• Equilibrium will be disturbed by the addition of BiOC ℓ.
• Equilibrium shifts to the left (reverse direction)
• More BiC ℓ ₃ is produced to re-establish the equilibrium.
• kc remains constant, but the position of equilibrium changes.

(iii) Removal of BiC ℓ₃ (reactant)
• Equilibrium will be disturbed by the removal of BiC ℓ₃ .
• Equilibrium shifts to the left (reverse direction).
• Reaction produces more BiC ℓ₃ to compensate, and to re-establish the equilibrium.
• A new equilibrium is established, kc stays constant.

(iv) Removal of BiOC ℓ or HC ℓ (product)
• Equilibrium will be disturbed by the removal of BiOC ℓ.
• Equilibrium shifts to the right (forward direction).
• More BiOC ℓ and HC ℓ are produced to re-establish the equilibrium.
• A new equilibrium is reached, with kc unchanged.

2. The effect of change in pressure or volume

The effect of pressure or volume change does not apply when:

(i) No gaseous components are involved
• If no gases are involved in the reaction, pressure or volume changes have no effect on equilibrium.

(ii) Equal moles of gaseous reactants and products
• In gaseous homogeneous equilibrium, if the number of gaseous moles of reactants and products are equal.

n_reactants = n_products

• Change in pressure or volume has no effect on the equilibrium position.

Effect applies only when gaseous moles of reactants and products are unequal.

n_reactants ≠ n_products

Synthesis of Ammonia (Haber's process)

N₂ + 3H₂(g) ⇌ 2NH₃(g)

• Total moles of reactants are equal to 4.
• Total moles of products are equal to 2.
• So, number of moles of reactants are not equal to number of mole of products.

(i) Increase in pressure / decrease in volume
• The number of moles and volume decrease for the forward reaction.
• At equilibrium volume of mixture is less than the total volume at initial stage.
• If pressure increase the system shifts to the side with fewer gas moles.
• In this case the reaction is shifted to forward direction.

(ii) Decrease in pressure / increase in volume
• If pressure increase or volume is decreased, the system shifts to the side where the reaction mixture occupy more volume.
• Synthesis of ammonia is shifted to reverse direction
• More N₂ and H₂ are formed, ammonia formation decreases.

Ans. Application of Le-Chatelier's principle

1. Effect of Change in temperature
• Le-Chatelier's principle helps predict the direction of a chemical reaction when temperature is changed.
• Temperature is the only factor that changes the value of equilibrium constant (kc).

Changing temperature can disturb equilibrium in two types of reactions:

(i) Exothermic Reactions

Heat is released ΔH° is negative

Example

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH° = -198 kJ

The increase in temperature will shift the reaction from right to left, according to the Le-Chatelier's principle.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

(a) Increase in temperature
• The equilibrium shifts left to right in reverse direction.
• [SO₃] decreases, and [SO₂] [O₂] increases.
• The value of kc decreases.

(b) Decrease in Temperature
• The equilibrium shifts right to left in forward reaction.
• [SO₃] increases, and [SO₂] [O₂] decreases.
• The value of kc increases.
• Production of SO₃ is favoured at lower temperatures.

kc values: The preparation of SO₃ is not favored at high temperatures.
• At 1000 K kc = 2.8 × 10²
• At 298 K kc = 1 × 10²⁶

(ii) Endothermic reactions

Heat is absorbed ΔH° is positive

Example The equation for the conversion of N₂O₄ to NO₂ is given below:

N₂O₄(g) ⇌ 2NO₂(g) ΔH° = +57.2 kJ

The increase in temperature will shift the reaction from left to right, according to the Le-Chatelier's principle.

N₂O₄(g) ⇌ 2NO₂(g)

(a) Increase in temperature
• The equilibrium shifts right (forward reaction).
• [NO₂] increases, [N₂O₄] decreases.
• The value of kc increases
• Product formation is favored at higher temperatures

(b) Decrease in temperature
• The equilibrium shifts left (reverse reaction).
• [NO₂] decreases, [N₂O₄] increases.
• The value of kc decreases.

2. Effect of catalyst on equilibrium

Catalyst

Definition A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the reaction.

• Does not change the position of equilibrium.
• Does not affect the equilibrium constant (kc).
• Only speeds up the rate at which equilibrium is achieved.
• Yield remains the same, but equilibrium is achieved faster

Working of a Catalyst
• Provides a new reaction pathway with lower activation energy (Ea).
• This means more reactant molecules have enough energy to overcome the activation barrier
• Reaction proceeds faster in both forward and reverse directions.

Energy profile diagram
• X-axis: Reaction coordinate
• Y-axis: Potential energy
• The peak of the curve (activation energy) is lowered in the presence of a catalyst.

Comparison

Property Activation Energy (Ea), Uncatalyzed Reaction: High, Catalyzed Reaction: Lower
Property: Reaction Speed, Uncatalyzed Reaction: Slow, Catalyzed Reaction: Faster
Property: Equilibrium Constant, Uncatalyzed Reaction: Unchanged, Catalyzed Reaction: Unchanged
Property: Equilibrium Position, Uncatalyzed Reaction: Unaffected, Catalyzed Reaction: Unaffected

Ans. Synthesis of Ammonia by Haber's process

The maximum yield of product is to be achieved in chemistry is the basic need of a reaction. Le-Chatelier's principle helps to have maximum yield in industrial synthesis of ammonia. Ammonia synthesis from nitrogen and hydrogen is a reversible process.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH° = -46.11 kJ mol⁻¹

Industrial importance Ammonia is an important compound used in fertilizers, explosives, and chemicals.

Yield of ammonia

The maximum yield of ammonia can be achieved by following ways.

(i) Removal of Ammonia (product removal)
• Removal of ammonia from the reaction vessel time to time decreases its concentration, according to Le Chatelier's principle.
• The equilibrium shifts forward direction, producing more ammonia.

(ii) Effect of Pressure
• Four moles of reactants combine to give two moles of products, reaction happens with the decreasing volume.
• High pressure will shift the equilibrium position to the right to give more and more ammonia.

(iii) Effect of Temperature
• The reaction is exothermic (releases heat).
• Lower temperature favors forward reaction (more ammonia).
• The equilibrium shifts forward direction, according to Le-Chatelier's principle.

Explanation From the table 8.1 given below and the figure, then it becomes clear that the effect of pressure and temperature on the yield of ammonia is very prominent. The most complete conversion is 98.3% at 473 K (200°C) and 1000 atmospheric pressure.

• The yield is being favoured at 200°C, but the rate of reaction becomes very slow and the process becomes uneconomical.
• The temperature is raised to a moderate level i.e. 400°C and a catalyst is used to increase the rate.
• If we want to achieve the same rate without a catalyst then we require much higher temperature, which lowers the yield.

Optimum industrial conditions

The most suitable to get maximum yield of ammonia are:
• Pressure between 200-300 atmospheres
• Temperature around 673 K (400°C)
• Pieces of iron crystals present in a fused mixture of MgO, Al₂O₃ and SiO₂ as catalyst

Preparation of sulphur trioxide

To manufacture H₂SO₄, sulphur trioxide gas is produced from SO₂ and O₂ in a reversible process.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH° = -97.9 kJ mol⁻¹

This is a reversible and exothermic reaction.

Table 8.2 Effect of temperature on equilibrium position of SO₃ formation

Temperature (°C) 200, 300, 400, 500, 600, 700
Kc: 5500, 690, 160, 55, 25, 13
Mole% of SO₃: 98, 91, 75, 61, 46, 31

Sulphur trioxide is a key intermediate in the manufacture of sulphuric acid (H₂SO₄).

Effect of temperature and pressure
• High pressure will shift equilibrium forward, increasing SO₃ yield.
• Low temperature will favours formation of SO₃ but slows the reaction.
• High temperature will increase rate but reduces SO₃ yield.

Effect of Temperature at 1 atm Pressure

Observation As temperature increases, kc decreases, and SO₃ yield drops.

Optimum conditions for SO₃ formation
• A mixture of SO₂ and O₂ (air) is passed over a solid catalyst (V₂O₅).
• The exothermic reaction raises the gas temperature to about 600°C.
• To increase yield, the gas is recycled at a lower temperature (400-500°C).

Conclusion For industrial production of SO₃
• Use a catalyst (V₂O₅) to increase the rate.
• Maintain moderate temperature (400-500°C).
• Operate at high pressure for better yield.
• Recycle the gas to maximize efficiency and conserve energy.