What type of deviation from Raoult's law is shown by a mixture of phenol and aniline? Give reason.

What will happen to the boiling point of the solution on mixing phenol and aniline?

Q: What type of deviation from Raoult's law is shown by a mixture of phenol and aniline? Give reason. What will happen to the boiling point of the solution on mixing phenol and aniline?

Lead: A phenol–aniline mixture shows **negative deviation** from Raoult's law. **Reason:** - The –OH group of phenol and the –NH₂ group of aniline form **new intermolecular hydrogen bonds** (O–H···N). - These A–B attractions are **stronger** than the A–A and B–B attractions in the pure liquids. - The escaping tendency of the molecules falls, so the vapour pressure is **lower** than predicted by Raoult's law → negative deviation. **Effect on boiling point:** - Because the vapour pressure is lowered, the **boiling point increases** (rises above that of the individual components).

Q: Why does a phenol–aniline mixture show negative deviation from Raoult's law?

Because the acidic –OH of phenol and the basic –NH₂ of aniline form strong intermolecular O–H···N hydrogen bonds. These A–B attractions are stronger than the attractions in the pure liquids, so the escaping tendency and vapour pressure drop below the Raoult's-law prediction.

Q: Does the boiling point rise or fall for the phenol–aniline mixture?

It rises. Negative deviation means the vapour pressure is lower than expected, and a lower vapour pressure corresponds to a higher boiling point — often higher than that of either pure component.

Solutions

Non-ideal solutions: negative deviation

Official Answer

Lead: A phenol–aniline mixture shows negative deviation from Raoult's law.

Reason:

The –OH group of phenol and the –NH₂ group of aniline form new intermolecular hydrogen bonds (O–H···N).
These A–B attractions are stronger than the A–A and B–B attractions in the pure liquids.
The escaping tendency of the molecules falls, so the vapour pressure is lower than predicted by Raoult's law → negative deviation.

Effect on boiling point:

Because the vapour pressure is lowered, the boiling point increases (rises above that of the individual components).

negative deviationRaoult's lawphenol anilineintermolecular hydrogen bondingO-H...N bondlower vapour pressureboiling point increasesstronger A-B interactions

Marking Scheme

11 mark: identifying negative deviation from Raoult's law with correct reason (strong intermolecular hydrogen bonding between phenol –OH and aniline –NH₂ lowers vapour pressure).
21 mark: stating that the boiling point of the solution increases (rises) because vapour pressure is lowered.
3Accept equivalent wording: 'intermolecular H-bonding stronger than in pure components' and 'boiling point higher than either component'.

Hint

Ask whether new A–B forces (here O–H···N hydrogen bonds) are stronger than the forces in the pure liquids — stronger means negative deviation and a higher boiling point.

Quick Oral Answer

A phenol–aniline mixture shows negative deviation from Raoult's law because strong O–H···N hydrogen bonding between the –OH of phenol and the –NH₂ of aniline lowers the vapour pressure, so the boiling point of the solution increases.

Analysis & Explanation

Concept:

Deviations from Raoult's law depend on the relative strengths of solute–solvent (A–B) versus pure-component (A–A, B–B) interactions.

Why negative deviation here:

Phenol supplies an acidic –OH hydrogen; aniline supplies a basic nitrogen lone pair on –NH₂.
Together they form strong intermolecular hydrogen bonds that did not exist (or were weaker) in the pure liquids.
Stronger A–B forces reduce escaping tendency → observed vapour pressure < Raoult's-law prediction.

Consequences of negative deviation:

Vapour pressure is lower than expected.
Boiling point is higher than expected.
Mixing is exothermic ( $\Delta H < 0$ ) and volume contracts slightly ( $\Delta V < 0$ ).

Exam trap:

Students sometimes label it positive deviation by analogy with ethanol–water. Here the NEW O–H···N bonding is stronger than the interactions in the separate liquids, so it is firmly a negative deviation.

Real-world:

Strong negative-deviation pairs (like nitric acid–water) form maximum-boiling azeotropes, which is why such mixtures boil at a temperature higher than either pure component.

Common Mistakes

1Calling it positive deviation — the new O–H···N hydrogen bonds are stronger than the interactions in the pure liquids, giving negative deviation.
2Saying the boiling point decreases; lowered vapour pressure means the boiling point rises.
3Omitting the reason (hydrogen bonding) and stating only the type of deviation, losing half the marks.

Interesting Facts

Phenol–aniline is a textbook example of a negative-deviation pair precisely because an acidic O–H and a basic N combine to hydrogen bond strongly.

Negative-deviation mixtures release heat on mixing (exothermic) and contract in volume, the opposite of positive-deviation mixtures like ethanol–acetone.

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Frequently Asked Questions

Why does a phenol–aniline mixture show negative deviation from Raoult's law?

Because the acidic –OH of phenol and the basic –NH₂ of aniline form strong intermolecular O–H···N hydrogen bonds. These A–B attractions are stronger than the attractions in the pure liquids, so the escaping tendency and vapour pressure drop below the Raoult's-law prediction.

Does the boiling point rise or fall for the phenol–aniline mixture?

It rises. Negative deviation means the vapour pressure is lower than expected, and a lower vapour pressure corresponds to a higher boiling point — often higher than that of either pure component.