Non-ideal Solutions

What the Figure Shows

The figure presents two side-by-side graphs, labelled (a) and (b). Each graph plots vapour pressure (on the vertical $y$-axis) against mole fraction (on the horizontal $x$-axis). The composition axis runs from pure component 2 at the left ($x_1 = 0$, $x_2 = 1$) to pure component 1 at the right ($x_1 = 1$, $x_2 = 0$). In both panels, three dashed straight lines represent the ideal behaviour predicted by Raoult’s law: the partial pressure of component 1 ($p_1$), the partial pressure of component 2 ($p_2$), and the total vapour pressure ($p_{\text{total}} = p_1 + p_2$). These ideal lines are straight because Raoult’s law gives $p_i = x_i p_i^*$, where $p_i^*$ is the vapour pressure of the pure component.

In panel (a), three solid curves show the actual (non-ideal) partial pressures and total vapour pressure. All three curves lie above their corresponding dashed ideal lines — this is positive deviation from Raoult’s law. The topmost solid curve, labelled “vapour pressure of solution”, is the actual total vapour pressure. In panel (b), the three solid curves lie below their corresponding ideal dashed lines — this is negative deviation. The actual total vapour pressure is still the upper envelope of the solid curves (it is the sum $p_1 + p_2$); what makes the deviation negative is that this solid total curve sags below the dashed ideal-total line.

Physical Idea Taught

The figure illustrates how real solutions deviate from the ideal behaviour assumed by Raoult’s law. The key idea is that the strength of intermolecular forces between unlike molecules (A–B) compared to like molecules (A–A and B–B) determines the direction of deviation.

• Positive deviation (panel a): A–B interactions are weaker than A–A and B–B interactions. Molecules escape more easily into the vapour phase, so the actual vapour pressure is higher than ideal. Examples: ethanol–acetone, acetone–carbon disulphide.
• Negative deviation (panel b): A–B interactions are stronger than A–A and B–B interactions. Molecules are held more tightly in the liquid, so the actual vapour pressure is lower than ideal. Examples: phenol–aniline, chloroform–acetone.

The figure also sets the stage for understanding azeotropes. Solutions with large positive deviation form minimum boiling azeotropes (e.g., ethanol–water at ~95% ethanol by volume). Solutions with large negative deviation form maximum boiling azeotropes (e.g., nitric acid–water at ~68% HNO₃ by mass, boiling at 393.5 K). At the azeotropic composition, the liquid and vapour have the same composition, so further separation by fractional distillation is impossible.

Key Formula Developed

The figure is built on Raoult’s law for an ideal solution:

where:

• $p_i$ = partial vapour pressure of component $i$ in the solution
• $x_i$ = mole fraction of component $i$ in the liquid phase
• $p_i^*$ = vapour pressure of pure component $i$ at the same temperature

For a binary ideal solution, the total vapour pressure is:

$p_{\text{total}} = p_1 + p_2 = x_1 p_1^* + x_2 p_2^*$ …

Redrawn from the NCERT page with the structures, printed labels (H3C, CH3, C=O, Cl) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so wha …