Chapter 6: Reversible & Irreversible Reactions | Complete Guide with Key Notes

Chapter 6: Reversible & Irreversible Changes

Complete Guide with Detailed Key Notes & Solved Exercises
Chemical Equilibrium | Dynamic Equilibrium | Physical Changes
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6.1 Reversible and Irreversible Changes

In chemical reactions, reactants transform into products. The extent of this transformation determines whether a reaction is reversible or irreversible.

Irreversible Reaction Example:

NaCl + AgNO₃ → AgCl↓ + NaNO₃

This precipitation reaction in aqueous solution goes to completion immediately. When stoichiometric amounts are used, no reactants remain at the end. Such reactions proceed only in the forward direction.

Reversible Reaction Example:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Conditions: 400°C, 200 atm, Fe catalyst. This reaction (Haber process) never reaches completion. At any given time, all three species (N₂, H₂, NH₃) coexist in the reaction mixture. Ammonia continuously forms and decomposes.

COMPREHENSIVE KEY NOTES: Reversible vs Irreversible

IRREVERSIBLE REACTIONS

• Go to completion (100% conversion)

• Single direction only (→ symbol)

• Products cannot reform reactants under same conditions

Examples: Precipitation (AgCl), combustion, strong acid-base neutralization

Characteristic: One product may escape (gas) or precipitate

REVERSIBLE REACTIONS

• Never reach 100% completion

• Both forward and backward directions (⇌ symbol)

• Products can reform reactants under same conditions

Examples: Haber process, esterification, dissociation of weak electrolytes

Characteristic: All species present at equilibrium

CONDITIONS AFFECTING REVERSIBILITY

Open container: Gaseous products escape → irreversible

Closed container: Products remain → reversible

Product removal: Reaction forced to completion

Catalyst: Speeds up equilibrium attainment only

PHYSICAL CHANGES

• Can also be reversible

• Hydration/dehydration of salts:

CuSO₄·5H₂O (Blue) ⇌ CuSO₄ (White) + 5H₂O
CoCl₂·6H₂O (Pink) ⇌ CoCl₂ (Blue) + 6H₂O

• Color indicates hydration state

Activity: Heating Copper Sulphate

Procedure: Take blue CuSO₄·5H₂O in dry test tube. Heat gently then strongly. Note color change to white. Let cool. Add water dropwise and observe color returns to blue.

Observation: Reversible physical change through hydration/dehydration.

Memory Trick:

“IRREVERSIBLE = One-way street, REVERSIBLE = Two-way street”

Irreversible reactions are like driving on a one-way street (→ only). Reversible reactions are like a two-way street (⇌ both directions).

6.2 Dynamic Equilibrium

In reversible reactions, forward and backward reactions occur simultaneously. Initially, forward reaction dominates. As products accumulate, the reverse reaction rate increases until equilibrium is established.

COMPREHENSIVE KEY NOTES: Dynamic Equilibrium

DEFINITION & CHARACTERISTICS

Dynamic equilibrium: Forward rate = Backward rate

Macroscopic level: No observable change (concentrations constant)

Microscopic level: Reactions continue in both directions

Equilibrium constant: Ratio of product/reactant concentrations remains fixed at given temperature

HOW EQUILIBRIUM IS ESTABLISHED

1. Initially: Only forward reaction (rate high), backward rate = 0

2. Progress: Forward rate decreases (reactants ↓), backward rate increases (products ↑)

3. Equilibrium: Rates equalize, concentrations stabilize (not necessarily equal!)

4. Dynamic nature: Molecules keep reacting but net change = 0

TIME TO REACH EQUILIBRIUM

• Depends on reaction nature and conditions

Fast: Water electrolysis (2H₂O ⇌ 2H₂ + O₂) = 4-5 seconds

Moderate: Ammonia synthesis (400°C, catalyst) = minutes

Slow: Some reactions may take years

Factors: Temperature, catalyst, concentration, pressure

REAL-WORLD EXAMPLE

Coal gasification in Thar, Sindh:

C + H₂O → CO + H₂ (Water gas)
CO + 3H₂ ⇌ CH₄ + H₂O (Catalytic methanation)

• Reversible step allows methane production from coal

Equilibrium Time Examples:

Quick equilibrium (seconds): Dissociation of weak acids, water electrolysis

Moderate equilibrium (minutes): Industrial ammonia synthesis with catalyst

Slow equilibrium (hours/days): Some esterification reactions

Key point: Catalyst speeds up equilibrium attainment but doesn’t change equilibrium position

Memory Trick:

“Dynamic = Moving, Equilibrium = Balanced = Dynamic Equilibrium = Moving but Balanced”

Imagine two teams pulling a rope with equal force – lots of effort (dynamic) but no movement (equilibrium).

Solved Exercises with Detailed Explanations

1 Multiple Choice Questions (Tick the correct answer):

(i) What will happen if the rates of forward and reverse reactions are very high?

(a) The equilibrium point will reach very soon.
(b) The equilibrium point will reach very late.
(c) The reaction will not attain the state of dynamic equilibrium.
(d) The reaction will be practically irreversible.
Show Explanation

Explanation: High reaction rates mean molecules collide more frequently and react faster. This speeds up both forward and reverse reactions, allowing equilibrium to be established quickly. The equilibrium position (ratio of products to reactants) is determined by thermodynamics, not kinetics. High rates only affect how fast equilibrium is reached, not where it lies.

Key concept: Reaction rate (kinetics) vs equilibrium position (thermodynamics).

(ii) Predict which components of the atmosphere react in the presence of lightning.

(a) N₂ and H₂O
(b) O₂ and H₂O
(c) CO₂ and O₂
(d) N₂ and O₂
Show Explanation

Explanation: Lightning provides extremely high temperatures (up to 30,000°C) that break the strong triple bond in N₂ (bond energy 945 kJ/mol). The nitrogen atoms then react with O₂ to form nitrogen oxides (NO, NO₂):

N₂ + O₂ → 2NO (at high temperature)
2NO + O₂ → 2NO₂

This is why thunderstorms produce nitrates that fertilize soil.

(v) What condition should be met for the reversible reaction to achieve the state of equilibrium?

(a) All the reactants should be converted into the products.
(b) 50% of the reactants should be converted into products.
(c) The concentrations of all the reactants and the products should become constant.
(d) One of the products should be removed from the reaction mixture.
Show Explanation

Explanation: At equilibrium, the forward and backward reaction rates are equal, so concentrations of all species remain constant over time. This doesn’t mean concentrations are equal or that reactants are completely converted. The exact concentrations at equilibrium depend on the equilibrium constant (K). Removing a product would shift equilibrium (Le Chatelier’s principle) but isn’t required to establish equilibrium.

Key point: Constant concentrations, not specific values or complete conversion.

(vi) Why the gas starts coming out when you open a can of fizzy drink?

(a) Because the solubility of the gas increases
(b) Because the gas is insoluble in water
(c) Because the gas is dissolved under pressure hence it comes out when pressure is decreased
(d) Because the solubility of the gas decreases at high pressure
Show Explanation

Explanation: CO₂ is dissolved in fizzy drinks under high pressure (3-4 atm). According to Henry’s Law, gas solubility is proportional to pressure. When you open the can, pressure drops to atmospheric pressure (1 atm), so solubility decreases and CO₂ bubbles out. This is an application of Le Chatelier’s principle to the equilibrium:

CO₂(g) ⇌ CO₂(aq)

Decreasing pressure shifts equilibrium toward the gaseous side (more molecules).

(viii) When a reaction will become a reversible one?

(a) If the activation energy of the forward reaction is comparable to that of backward reaction
(b) If the activation energy of the forward reaction is higher than that of backward reaction
(c) If the activation energy of the forward reaction is lower than that of backward reaction
(d) If the enthalpy change of both the reactions is zero.
Show Explanation

Explanation: For a reaction to be reversible under given conditions, both forward and backward reactions must be feasible. This requires comparable activation energies so that both directions can occur at measurable rates. If Ea(forward) is much lower than Ea(backward), only forward reaction occurs (irreversible). If Ea(backward) is much lower, reverse dominates. Comparable Ea values allow both reactions to proceed, establishing equilibrium.

(ix) Is reversible reaction useful for preparing compounds on large scale?

(a) No
(b) Yes
(c) They are useful only when equilibrium lies far to the right side
(d) They are useful only when equilibrium lies far to the left side
Show Explanation

Explanation: Many important industrial processes use reversible reactions. They are economically viable when the equilibrium constant K is large (equilibrium lies far to the right, favoring products). Examples:

Haber process: N₂ + 3H₂ ⇌ 2NH₃ (K ~ 0.5 at 400°C, improved by Le Chatelier’s principle)

Contact process: 2SO₂ + O₂ ⇌ 2SO₃ (K large, high yield)

Esterification: Used despite moderate K because ester can be removed to shift equilibrium

If K is very small (equilibrium far left), the reaction isn’t useful industrially without continuous product removal.

(x) What will happen to the concentrations of the products if a reversible reaction at equilibrium is not disturbed?

(a) They will remain constant
(b) They will keep on increasing
(c) They will keep on decreasing
(d) They will remain constant for some time and then start decreasing
Show Explanation

Explanation: By definition, at equilibrium, forward rate = backward rate. This means product molecules form at the same rate they convert back to reactants. Therefore, product concentrations remain constant as long as conditions (temperature, pressure, volume) remain unchanged. This is the fundamental characteristic of equilibrium.

Common misconception: Students often think reactions stop at equilibrium. In reality, reactions continue (dynamic) but with no net change.

2 Questions for Short Answers:

i. How is dynamic equilibrium different from the static equilibrium?

Answer:
Dynamic Equilibrium Static Equilibrium
• Reactions continue in both directions • No movement or reactions occur
• Forward rate = Backward rate • All forces balanced, no motion
• Concentrations constant but not zero • Position constant
• Example: N₂ + 3H₂ ⇌ 2NH₃ • Example: Book on table, balanced seesaw

Key difference: Dynamic = continuous molecular activity, Static = complete rest.

iii. How can you get the maximum yield in a reversible reaction?

Answer:

To maximize yield (product formation) in reversible reactions, apply Le Chatelier’s principle:

1. Remove products as they form: Especially effective for gaseous products or those that can be distilled, precipitated, or otherwise removed.

2. Use excess of cheaper reactant: Shifts equilibrium toward products.

3. Adjust temperature: For exothermic reactions (ΔH negative), decrease temperature favors products. For endothermic reactions, increase temperature.

4. Adjust pressure (for gases): Increase pressure if fewer gas molecules on product side. Decrease pressure if more gas molecules on product side.

5. Use catalyst: Doesn’t increase yield but helps reach equilibrium faster, allowing more product in given time.

Example for Haber process: N₂ + 3H₂ ⇌ 2NH₃ (exothermic, fewer gas molecules on right)

• Use high pressure (200 atm)

• Moderate temperature (400°C) – compromise between yield and rate

• Remove NH₃ by liquefaction

• Use excess N₂ (cheaper than H₂)

iv. How can you decrease the time to attain the position of equilibrium in a reversible reaction?

Answer:

To reach equilibrium faster (increase reaction rate):

1. Increase temperature: Increases kinetic energy, more collisions with sufficient energy to overcome Ea.

2. Use catalyst: Provides alternative pathway with lower activation energy for both forward and reverse reactions.

3. Increase concentrations of reactants: More collisions per unit time.

4. Increase pressure (for gaseous reactions): Increases collision frequency.

5. Increase surface area (for heterogeneous reactions): Solid reactants in powdered form.

Important note: These factors speed up attainment of equilibrium but do NOT change the equilibrium constant or equilibrium position (except temperature, which changes K). The equilibrium concentrations will be the same regardless of how fast equilibrium is reached.

Example: In Haber process, Fe catalyst allows equilibrium to be reached in minutes instead of hours at 400°C.

3 Constructed Response Questions:

1. Why are some reactions irreversible while others are reversible?

Answer:

Irreversible reactions occur when:

1. Products are removed from system: Gas escapes (open container), precipitate forms and settles, volatile product evaporates.

2. Large energy difference: Highly exothermic reactions (large negative ΔH) release so much energy that reverse reaction is highly unfavorable.

3. Products are very stable: CO₂ and H₂O from combustion are among the most stable compounds.

4. Irreversible under conditions: Some reactions require extreme conditions to reverse (e.g., decomposition of organic compounds).

Reversible reactions occur when:

1. Closed system: All species remain in contact.

2. Small energy difference: ΔH is small enough that both directions are feasible.

3. Comparable activation energies: Both forward and reverse reactions can occur at measurable rates.

4. No product removal: Products can interact with each other to reform reactants.

Examples:

Irreversible: NaCl + AgNO₃ → AgCl↓ + NaNO₃ (precipitate removes AgCl)

Reversible: N₂ + 3H₂ ⇌ 2NH₃ (all gases remain in closed container)

Key concept: Reversibility depends on whether products can reform reactants under the given conditions.

ii. Why are combustion reactions generally irreversible?

Answer:

Combustion reactions (burning fuels) are generally irreversible because:

1. Highly exothermic: Release large amounts of energy (ΔH very negative). For example:

CH₄ + 2O₂ → CO₂ + 2H₂O ΔH = -890 kJ/mol

Reverse reaction would require absorbing 890 kJ/mol, which doesn’t occur under normal conditions.

2. Products are very stable: CO₂ and H₂O are among the most thermodynamically stable compounds. Their formation releases maximum energy.

3. Gaseous products escape: In open air, CO₂ and H₂O vapor disperse into atmosphere, preventing reverse reaction.

4. Entropy increase: Combustion increases disorder (ΔS positive), making reverse reaction even less favorable.

5. Kinetic factors: Activation energy for reverse reaction is extremely high under normal conditions.

Exception: Some combustion-like reactions can be reversed with extreme conditions. For example, photosynthesis reverses the combustion of glucose but requires sunlight and chlorophyll as catalyst:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ (photosynthesis)

This shows that with sufficient energy input (sunlight) and specific conditions, even combustion-like reactions can be reversed.

iii. Can you make an irreversible reaction reversible and vice versa?

Answer:

YES, by changing conditions, we can often convert irreversible reactions to reversible and vice versa.

A. Making irreversible reaction reversible:

Method: Prevent product removal or change conditions to allow reverse reaction.

Example 1: CaCO₃ decomposition

CaCO₃(s) → CaO(s) + CO₂(g) (irreversible in open container)
CaCO₃(s) ⇌ CaO(s) + CO₂(g) (reversible in closed container)

In closed container, CO₂ builds up and reacts with CaO to reform CaCO₃.

Example 2: Electrolysis of water

2H₂O → 2H₂ + O₂ (irreversible with continuous current)
2H₂O ⇌ 2H₂ + O₂ (reversible in fuel cell)

B. Making reversible reaction irreversible:

Method: Continuously remove products or change conditions to favor one direction.

Example: Haber process

N₂ + 3H₂ ⇌ 2NH₃ (reversible in closed system)

By continuously removing NH₃ (liquefying it), the reaction is forced to completion, effectively becoming irreversible in practice.

Limitations: Some reactions are intrinsically irreversible due to large energy differences. For example, combustion of hydrocarbons to CO₂ and H₂O is essentially irreversible under normal conditions because reverse reaction (forming hydrocarbons from CO₂ and H₂O) requires enormous energy input and specific conditions (like photosynthesis).