Exhaustive Guide: Raoult's Law & Non-Ideal Solutions | CBSE Class 12 Chemistry | ChemCA

Vapour Pressure & Raoult's Law

Module 3 | CBSE Class 12 Chemistry | Solutions Chapter

1. Concept of Vapour Pressure

When a liquid is kept in a closed vessel, it evaporates. The vapour molecules accumulate and start condensing back into the liquid. When the rate of evaporation equals the rate of condensation, a dynamic equilibrium is established.

Vapour Pressure: The pressure exerted by the vapour in equilibrium with its liquid (or solid) phase at a given temperature in a closed system is called the vapour pressure of the liquid.

Vapour pressure depends on the nature of the liquid (weaker intermolecular forces mean higher vapour pressure) and temperature (vapour pressure increases with temperature).

2. Vapour Pressure of Liquid-Liquid Solutions (Raoult's Law)

Consider a binary solution of two volatile liquids, Component 1 and Component 2. In a closed vessel, both components will evaporate and contribute to the total vapour pressure.

French chemist François-Marie Raoult (1886) gave a quantitative relationship between the partial vapour pressures and mole fractions of the components.

Raoult's Law: For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.

For component 1: p₁ ∝ x₁ &implies; p₁ = p₁⁰ × x₁

For component 2: p₂ ∝ x₂ &implies; p₂ = p₂⁰ × x₂

(Where p₁⁰ and p₂⁰ are the vapour pressures of pure components 1 and 2 respectively at the same temperature.)

According to Dalton's law of partial pressures, the total pressure (p_total) over the solution phase is the sum of the partial pressures:

p_total = p₁ + p₂
p_total = x₁p₁⁰ + x₂p₂⁰
p_total = p₁⁰ + (p₂⁰ - p₁⁰)x₂

Key Conclusion: The total vapour pressure over the solution can be related to the mole fraction of any one component. It varies linearly with the mole fraction of component 2 (or 1).

2.1 Composition in Vapour Phase (Highly Tested Concept)

The mole fractions calculated above (x₁ and x₂) are for the liquid solution. To find the composition in the vapour phase, we use Dalton's Law. Let y₁ and y₂ be the mole fractions of components 1 and 2 respectively in the vapour phase.

y₁ = p₁ / p_total and y₂ = p₂ / p_total

2.2 Raoult's Law as a Special Case of Henry's Law

According to Raoult's law, p_i = p_i⁰ x_i.
According to Henry's law for gases in liquids, p = K_H x.
If we compare both equations, they are identical in form. The only difference is the proportionality constant (p_i⁰ replaces K_H). Thus, Raoult's law becomes a special case of Henry's law in which K_H becomes equal to the vapour pressure of the pure component (p_i⁰).

3. Vapour Pressure of Solutions of Solids in Liquids

When a non-volatile solute (like salt, sugar, or urea) is added to a solvent, the vapour pressure of the solution is found to be lower than that of the pure solvent.

Why does Vapour Pressure Decrease? In a pure liquid, the entire surface is occupied by solvent molecules. When a non-volatile solute is added, the surface now has both solute and solvent molecules. The fraction of the surface covered by solvent molecules decreases, which reduces the number of solvent molecules escaping into the vapour phase.

Since the solute is non-volatile, the entire vapour pressure is due to the solvent alone.
By Raoult's Law: p₁ = x₁ p₁⁰

Since x₁ (mole fraction of solvent) is always less than 1 in a solution, p₁ is always less than p₁⁰. This leads to the concept of Relative Lowering of Vapour Pressure, which is a colligative property (covered in Module 4).

4. Ideal and Non-Ideal Solutions

Liquid-liquid solutions are classified into ideal and non-ideal solutions based on their adherence to Raoult's Law.

4.1 Ideal Solutions

Solutions that obey Raoult's law over the entire range of concentration are known as ideal solutions.

Conditions for an Ideal Solution:

p_total = p_A + p_B (Obeys Raoult's Law)
ΔH_mix = 0 (No heat is absorbed or evolved during mixing)
ΔV_mix = 0 (Total volume of solution is exactly equal to the sum of volumes of components)

Molecular Reason: In an ideal solution, the intermolecular attractive forces between the A-A and B-B molecules are nearly equal to the A-B interaction.
Examples (NCERT): n-hexane + n-heptane; bromoethane + chloroethane; benzene + toluene.

4.2 Non-Ideal Solutions: Positive Deviation from Raoult's Law

When a solution does not obey Raoult's law, it is a non-ideal solution. If its vapour pressure is higher than predicted by Raoult's law, it exhibits a positive deviation.

Reason: A-B intermolecular interactions are weaker than A-A and B-B interactions. The molecules escape more easily into the vapour phase.
Conditions: p_A > p_A⁰x_A; ΔH_mix > 0 (Endothermic); ΔV_mix > 0.

NCERT Example: Ethanol + Acetone. Pure ethanol molecules are hydrogen-bonded. When acetone is added, its molecules get in between the host molecules and break some of the hydrogen bonds. Thus, escaping tendency increases, leading to positive deviation.
Other Examples: Carbon disulphide + acetone.

4.3 Non-Ideal Solutions: Negative Deviation from Raoult's Law

If the vapour pressure is lower than predicted by Raoult's law, it exhibits a negative deviation.

Reason: A-B intermolecular interactions are stronger than A-A and B-B interactions. This restricts molecules from escaping into the vapour phase.
Conditions: p_A < p_A⁰x_A; ΔH_mix < 0 (Exothermic); ΔV_mix < 0.

NCERT Example: Phenol + Aniline. The intermolecular hydrogen bonding between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bonding between similar molecules.
Other Examples: Chloroform + Acetone (New H-bond forms between them).

5. Azeotropes (Constant Boiling Mixtures)

Azeotropes: These are binary mixtures having the same composition in liquid and vapour phases, and boil at a constant temperature. In such cases, it is not possible to separate the components by fractional distillation.

Type of Azeotrope	Deviation from Raoult's Law	NCERT Example
Minimum Boiling Azeotrope (Boils at a temp lower than both components)	Large Positive Deviation	Ethanol + Water (approx 95% ethanol by volume). Fractional distillation of sugar fermentation stops here.
Maximum Boiling Azeotrope (Boils at a temp higher than both components)	Large Negative Deviation	Nitric Acid + Water (68% HNO₃ and 32% H₂O by mass, boiling point 393.5 K).

6. NCERT Solved Examples (Step-by-Step)

NCERT Example 2.5: Vapour pressure of chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) at 298 K are 200 mm Hg and 415 mm Hg respectively.
(i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at 298 K.
(ii) Mole fractions of each component in vapour phase.

Solution:
Step 1: Calculate moles
Molar mass CH₂Cl₂ = 12 + 2 + 71 = 85 g mol^-1
Molar mass CHCl₃ = 12 + 1 + 106.5 = 119.5 g mol^-1
Moles of CH₂Cl₂ = 40 / 85 = 0.47 mol
Moles of CHCl₃ = 25.5 / 119.5 = 0.213 mol
Total moles = 0.47 + 0.213 = 0.683 mol

Step 2: Mole fractions in solution (x)
x(CH₂Cl₂) = 0.47 / 0.683 = 0.688
x(CHCl₃) = 1 - 0.688 = 0.312

Step 3: Total Vapour Pressure (Raoult's Law)
p_total = p⁰(CH₂Cl₂) × x(CH₂Cl₂) + p⁰(CHCl₃) × x(CHCl₃)
p_total = (415 × 0.688) + (200 × 0.312) = 285.5 + 62.4 = 347.9 mm Hg

Step 4: Mole fraction in Vapour Phase (y)
y(CH₂Cl₂) = p(CH₂Cl₂) / p_total = 285.5 / 347.9 = 0.82
y(CHCl₃) = p(CHCl₃) / p_total = 62.4 / 347.9 = 0.18
Note: The vapour phase is richer in the more volatile component (dichloromethane).

7. Previous Year Questions (PYQs) & Exhaustive Question Bank

Part A: Conceptual (1-2 Marks)

[CBSE 2018, 2020]

Q1. What type of deviation from Raoult's law is shown by a mixture of ethanol and acetone? Give reason.

Answer: A mixture of ethanol and acetone shows a positive deviation from Raoult's law.
Reason: Pure ethanol possesses strong intermolecular hydrogen bonding. When acetone is added, its molecules interpose between ethanol molecules and break some of these hydrogen bonds. This weakens the overall intermolecular forces (A-B interactions are weaker than A-A interactions), making it easier for molecules to escape into the vapour phase, hence increasing vapour pressure.

[CBSE 2016, 2019]

Q2. What is meant by azeotropic mixture? What type of azeotrope is formed by negative deviation from Raoult's law?

Answer: Azeotropic mixtures (or Azeotropes) are binary liquid mixtures which have the same composition in the liquid and vapour phase and boil at a constant temperature. They cannot be separated by fractional distillation.
A mixture showing a large negative deviation from Raoult's law forms a maximum boiling azeotrope (e.g., 68% Nitric acid and water).

Part B: Assertion-Reason Type (1 Mark)

[CBSE 2023]

Q3. Assertion (A): When NaCl is added to water, a depression in the freezing point is observed. However, the vapour pressure of the solution is found to be lowered.
Reason (R): Lowering of vapour pressure of a solution causes depression in the freezing point.

Answer: Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion. The addition of a non-volatile solute (NaCl) decreases the escaping tendency of water molecules, lowering the vapour pressure. Freezing occurs when the vapour pressure of the liquid equals that of the solid phase, which now happens at a lower temperature.

[CBSE Sample Paper]

Q4. Assertion (A): A mixture of chloroform and acetone forms a solution with negative deviation from Raoult's law.
Reason (R): Chloroform and acetone molecules form hydrogen bonds with each other.

Answer: Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion. The new hydrogen bond (C-H...O) formed between chloroform and acetone makes the A-B interaction stronger than pure A-A and B-B interactions, reducing escaping tendency and lowering vapour pressure.

Part C: Numerical Problems (3 Marks)

[CBSE 2017, NCERT Intext 2.8]

Q5. The vapour pressures of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Answer:
Given: p_A⁰ = 450 mm Hg; p_B⁰ = 700 mm Hg; p_total = 600 mm Hg.

Step 1: Composition in liquid phase (x)
p_total = p_A⁰ + (p_B⁰ - p_A⁰)x_B
600 = 450 + (700 - 450)x_B
150 = 250 x_B &implies; x_B = 150 / 250 = 0.60
x_A = 1 - 0.60 = 0.40

Step 2: Partial Pressures
p_A = x_A × p_A⁰ = 0.40 × 450 = 180 mm Hg
p_B = x_B × p_B⁰ = 0.60 × 700 = 420 mm Hg

Step 3: Composition in vapour phase (y)
y_A = p_A / p_total = 180 / 600 = 0.30
y_B = p_B / p_total = 420 / 600 = 0.70

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Exhaustive Guide: Raoult's Law & Non-Ideal Solutions | CBSE Class 12 Chemistry