Q.What is meant by positive and negative deviations from Raoult's law and how is the sign of $\Delta_{mix}H$ related to positive and negative deviations from Raoult's law?

Concept: Deviations from Raoult’s Law and Enthalpy of Mixing

A solution shows positive deviation when its total vapour pressure is higher than predicted by Raoult’s law ($P > P_{ideal}$). This occurs when A–B interactions are weaker than A–A and B–B interactions, making it easier for molecules to escape into vapour. Conversely, negative deviation means the vapour pressure is lower than ideal ($P < P_{ideal}$), due to stronger A–B interactions (e.g., hydrogen bonding) that hold molecules in the liquid.

The sign of $\Delta_{mix}H$ follows directly from the strength of interactions: …

The Core Idea

Raoult’s law states that for an ideal solution, the partial vapour pressure of each component is proportional to its mole fraction: $p_A = p_A^* x_A$, where $p_A^*$ is the vapour pressure of pure A. Real solutions often deviate from this because the molecules of A and B interact differently with each other than with themselves.

The sign of $\Delta_{mix}H$ (enthalpy change on mixing) is a direct thermodynamic fingerprint of these deviations. It tells you whether the new A–B interactions are energetically favourable or unfavourable compared to the original A–A and B–B interactions.

Step-by-Step Reasoning

1. Understand the reference: an ideal solution

   In an ideal solution, A–A, B–B, and A–B interactions are all equally strong. Mixing is energetically neutral: $\Delta_{mix}H = 0$. The vapour pressure follows Raoult’s law exactly.

2. Positive deviation: weaker A–B interactions

   If A–B interactions are weaker than A–A and B–B, the molecules in the mixture “prefer” to stay with their own kind. This makes it easier for molecules to escape into the vapour phase, so the observed vapour pressure is higher than Raoult’s law predicts.

   To break the stronger A–A and B–B interactions and form weaker A–B bonds, the system must absorb energy from the surroundings. Hence $\Delta_{mix}H > 0$ (endothermic mixing).

   Example: Ethanol + acetone — the hydrogen bonding in pure ethanol is disrupted.

3. Negative deviation: stronger A–B interactions

   If A–B interactions are stronger than A–A and B–B, molecules are held more tightly in the liquid. Fewer escape, so vapour pressure is lower than Raoult’s law predicts.

   Forming these stronger bonds releases energy, so $\Delta_{mix}H < 0$ (exothermic mixing).

   Example: Chloroform + acetone — hydrogen bonding between the two is stronger than in either pure liquid.

4. The direct relationship …

Method: Comparing Intermolecular Interactions (A-B vs A-A/B-B)

Step 1: Recall Raoult's law for a binary liquid mixture

For a binary mixture of A and B:

An ideal solution obeys this exactly at every composition, with $\Delta_{mix}H = 0$ and $\Delta_{mix}V = 0$.

Step 2: Positive deviation

• Observed vapour pressure > ideal vapour pressure predicted by Raoult's law.
• Cause: A-B interactions are weaker than A-A and B-B interactions, so molecules escape into the vapour more easily.
• Examples: Ethanol + water, acetone + benzene, carbon tetrachloride + chloroform.

Step 3: Negative deviation

• Observed vapour pressure < ideal vapour pressure.
• Cause: A-B interactions are stronger than A-A and B-B interactions, so molecules are held back in the liquid.
• Examples: Acetone + chloroform (hydrogen bonding), phenol + aniline, nitric acid + water.

Step 4: Relate the deviation to $\Delta_{mix}H$

| Deviation Type | $\Delta_{\text{mix}}H$ sign | Reason |

|----------------|-----------------------------|--------| …

What Raoult's Law Says First

For an ideal solution, Raoult's law states:

Where:

• $P_A^0$, $P_B^0$ = vapour pressures of pure components
• $x_A$, $x_B$ = mole fractions in the liquid phase

Ideal solution: $\Delta_{mix}H = 0$, $\Delta_{mix}V = 0$, and obeys Raoult's law at all compositions.

Positive Deviation from Raoult's Law

Definition: The observed total vapour pressure is greater than that predicted by Raoult's law.

Why it happens:

A–B interactions are weaker than A–A and B–B interactions. Molecules escape more easily into vapour.

Examples:

• Ethanol + Water
• Acetone + Benzene
• Carbon tetrachloride + Chloroform

Negative Deviation from Raoult's Law

Definition: The observed total vapour pressure is less than that predicted by Raoult's law.

Why it happens:

A–B interactions are stronger than A–A and B–B interactions. Molecules are held back in the liquid.

Examples:

• Acetone + Chloroform (hydrogen bonding)
• Phenol + Aniline
• Nitric acid + Water

Relation Between $\Delta_{mix}H$ and Deviation Type

This is the most common exam point — and where mistakes happen.

Key logic:

• Weaker new interactions → need extra energy → $\Delta_{mix}H > 0$
• Stronger new interactions → release energy → $\Delta_{mix}H < 0$

Common Mistakes Students Make

✗ Mistake 1: Reversing the sign

Wrong: "Positive deviation means $\Delta_{mix}H < 0$"

Why it happens: Students think "positive" means "good" or "energy released" — but here positive deviation means higher vapour pressure, which comes from weaker interactions, which requires energy input.

✓ How to avoid:

Always connect:

Weaker A–B bonds → need energy to break stronger A–A/B–B → endothermic → $\Delta_{mix}H > 0$

✗ Mistake 2: Confusing deviation with ideal behaviour

Wrong: "If $\Delta_{mix}H = 0$, it's a positive deviation"

Correct: $\Delta_{mix}H = 0$ means ideal solution, not deviation.

✓ How to avoid:

Memorise the three cases:

• Ideal: $\Delta_{mix}H = 0$
• Positive deviation: $\Delta_{mix}H > 0$
• Negative deviation: $\Delta_{mix}H < 0$

✗ Mistake 3: Forgetting the "why" — just memorising sign

Students memorise "positive → positive" without understanding. Then in a twisted question (e.g., "If $\Delta_{mix}H$ is positive, what type of deviation?"), they panic.

✓ How to avoid:

Draw a mental picture: …