Maximum vs Minimum Boiling Azeotropes — Easy Trick to Remember! ⚗️

 

❓ Question

Match List - I with List - II

List - I	List - II
(A) Solution of chloroform and acetone	(I) Minimum boiling azeotrope
(B) Solution of ethanol and water	(II) Dimerizes
(C) Solution of benzene and toluene	(III) Maximum boiling azeotrope
(D) Solution of acetic acid in benzene	(IV) ΔVₘᵢₓ = 0

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🖼️ Question Image

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✍️ Short Solution

Let’s match each solution with its correct behavior using the concept of ideal and non-ideal liquid mixtures.

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🔹 Step 1 — Recall: Types of Solutions Based on Raoult’s Law

1. Ideal Solutions:

   • Obey Raoult’s Law completely.

   • No enthalpy or volume change on mixing.

   • Example: Benzene + Toluene.

   • ⇒ ΔHₘᵢₓ = 0, ΔVₘᵢₓ = 0.

2. Non-Ideal Solutions:

   • Deviate from Raoult’s Law due to differences in molecular forces.

   • Can show positive or negative deviation.

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🔹 Step 2 — Identify the Type of Deviation

Type	Interaction Strength	Vapor Pressure	Boiling Point	Example
Positive Deviation	A–B < A–A or B–B	Higher than expected	Minimum boiling azeotrope	Ethanol + Water
Negative Deviation	A–B > A–A or B–B	Lower than expected	Maximum boiling azeotrope	Chloroform + Acetone

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🔹 Step 3 — Explain Each Pair

(A) Chloroform + Acetone → (III) Maximum Boiling Azeotrope

• Strong H-bonding forms between chloroform (H–CCl₃) and acetone (C=O).

• This results in a negative deviation from Raoult’s Law.

• Vapor pressure decreases → boiling point increases → maximum boiling azeotrope.

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(B) Ethanol + Water → (I) Minimum Boiling Azeotrope

• Hydrogen bonding between ethanol and water is weaker than between pure components.

• So, molecules escape easily → higher vapor pressure → positive deviation → minimum boiling azeotrope.

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(C) Benzene + Toluene → (IV) ΔVₘᵢₓ = 0

• Both are non-polar hydrocarbons with similar molecular sizes and forces.

• They form ideal solutions, showing no enthalpy or volume change on mixing.

• Hence, ΔVₘᵢₓ = 0.

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(D) Acetic Acid + Benzene → (II) Dimerizes

• Acetic acid molecules associate via hydrogen bonding in benzene.

• It forms dimers (CH₃COOH)₂, leading to molecular association rather than deviation from Raoult’s Law.

• Hence, Dimerizes is correct.

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🧮 Image Solution

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✅ Conclusion & Video Solution

✅ Final Answer:

List - I	List - II
(A)	(III)
(B)	(I)
(C)	(IV)
(D)	(II)

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📘 Concept Recap:

• Ideal solution: ΔHₘᵢₓ = 0, ΔVₘᵢₓ = 0

• Positive deviation: A–B < A–A or B–B → Min boiling azeotrope

• Negative deviation: A–B > A–A or B–B → Max boiling azeotrope

• Association: Dimer formation (as in acetic acid)

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