FAD1018 W5-W6 — Phase Equilibria

Weeks 5–6 lectures covering phase equilibria, colligative properties, Raoult's law, and fractional distillation. Source files: W5 (1).pdf, W5 (2).pdf, W6 (1).pdf.

Lecturer: Puan Zuraini Kadir (zuraini81@um.edu.my)

References:

Chemistry, R. Chang
Comprehensive College Chemistry
Principles of Chemistry, Tro

Summary

Three lectures on phase equilibria: (1) phase definitions, colligative properties, and one-component phase diagrams; (2) ideal/non-ideal solutions and Raoult's law; (3) fractional distillation and azeotropes.

Learning Outcomes

Define phase and component
Define colligative properties
Perform calculations on colligative properties
Define triple point and critical point for single-component systems
Sketch phase diagrams of compounds similar to H₂O and CO₂
Describe phase changes with respect to temperature and pressure
State properties of ideal and non-ideal solutions for two-component systems
Define and apply Raoult's law
Define azeotrope
Determine composition of an azeotropic mixture
Sketch phase diagrams for two-component systems: ideal, positive deviation, and negative deviation from Raoult's law
Explain principles involved in fractional distillation
Determine residue and distillate from boiling point–composition phase diagrams

Lecture 1 — Phase, Components & Colligative Properties (W5)

Phase & Component Definitions

Phase: A homogeneous part of a system separated by distinct physical boundaries (solid, liquid, gas).
Component: A chemically independent constituent of a system.

System	Phase	Component	Description
Mixture of O₂, N₂, H₂ gases	1	3	Gases well mixed; no visible boundary
Oil + water (unmixed)	2	2	Boundary between two liquids
Alcohol + water (mixed)	1	2	No boundary; miscible
Salt solution	1	2	Salt + water
Saturated CuSO₄ in closed bottle	3	2	Solid, liquid, gas (water vapour)
Steel	1	2	Fe + C

O=O
N#N
[H][H]
[Na+].[Cl-]
[Cu+2].[O-]S([O-])(=O)=O

Types of Phase Changes

graph LR
    S[Solid] -->|Fusion| L[Liquid]
    L -->|Freezing| S
    L -->|Vaporization| G[Gas]
    G -->|Condensation| L
    S -->|Sublimation| G
    G -->|Deposition| S

Colligative Properties

Properties that depend on solute particle concentration, not identity:

Freezing point depression: $ΔT_f = K_f m$
Boiling point elevation: $ΔT_b = K_b m$
Vapor pressure lowering: $ΔP = X_2 P_1^o$
Osmotic pressure: $Π = MRT$

Where:

$K_f$ = freezing-point depression constant
$K_b$ = boiling point elevation constant
$m$ = molality (mol solute / kg solvent)
$M$ = molarity (mol / L)
$X$ = mole fraction
$R$ = 0.0821 L atm mol⁻¹ K⁻¹

Vapor Pressure Lowering (Raoult's Law)

Partial pressure of solvent over solution: $$P_A = X_A P_A^o$$

For 2-component system: $$X_A + X_B = 1$$ $$P_A = (1 - X_B) P_A^o$$ $$ΔP = P_A^o - P_A = X_B P_A^o$$

[!example] Example: Glucose solution 218 g glucose (RMM = 180.2) dissolved in 460 mL water at 30°C. $P_{water}^o = 31.82$ mmHg.

$n_{water} = 460 / 18 = 25.5$ mol $n_{glucose} = 218 / 180.2 = 1.21$ mol $X_{glucose} = 1.21 / (25.5 + 1.21) = 0.0453$ $ΔP = 0.0453 × 31.82 = 1.44$ mmHg New vapor pressure = $31.82 - 1.44 = 30.38$ mmHg

C([C@@H]1[C@H]([C@@H]([C@H](C(O1)O)O)O)O)O

Freezing Point Depression

$$ΔT_f = K_f m = T_{solution} - T_{solvent}$$

[!example] Example: Naphthalene in benzene 1.60 g naphthalene (C₁₀H₈) in 20.0 g benzene. $K_f$ (benzene) = 4.3 °C m⁻¹. Pure benzene fp = 5.5°C.

Molar mass naphthalene = 128.17 g/mol $m = (1.60 / 128.17) / 0.0200 = 0.624$ mol/kg $ΔT_f = 4.3 × 0.624 = 2.68$°C Freezing point = $5.5 - 2.68 = 2.82$°C

c1ccc2ccccc2c1
c1ccccc1

Boiling Point Elevation

$$ΔT_b = K_b m = T_{b,solution} - T_{b,solvent}$$

[!example] Example: Ethylene glycol as antifreeze 651 g EG in 2505 g water. RMM EG = 62. $K_f = 1.86$ °C/m, $K_b = 0.52$ °C/m.

$n_{EG} = 651 / 62 = 10.5$ mol $m = 10.5 / 2.505 = 4.19$ mol/kg $ΔT_f = 1.86 × 4.19 = 7.79$°C → fp = $-7.79$°C $ΔT_b = 0.52 × 4.19 = 2.18$°C → bp = $102.18$°C

OCCO

Osmotic Pressure

$$ΠV = nRT \quad \text{or} \quad Π = MRT$$

[!example] Example: Glycerin solution 46.0 g glycerin (C₃H₈O₃) per liter at 0°C. RMM = 92.

$n = 46 / 92 = 0.5$ mol $Π = (0.5 / 1.0) × 0.0821 × 273 = 11.21$ atm

OCC(O)CO

[!example] Example: Polystyrene molecular weight 5.0 g polystyrene/L, $Π = 0.0100$ atm at 25°C.

$n = ΠV / RT = (0.0100 × 1) / (0.0821 × 298) = 4.09 × 10^{-4}$ mol $M = 5.0 / 4.09 × 10^{-4} = 1.22 × 10^4$ g/mol

C=CC1=CC=CC=C1

One-Component Phase Diagrams

Water:

Triple point: 0.01°C, 0.006 atm
Critical point: 374°C, 218 atm
Normal bp: 100°C (1 atm)
Normal fp: 0°C (1 atm)
Solid-liquid line has negative slope (ice less dense than water)

Carbon Dioxide:

Triple point: −56.6°C, 5.11 atm
Critical point: 31.1°C, 73 atm
Sublimes at 1 atm (dry ice)
Solid-liquid line has positive slope (solid denser than liquid)

O
O=C=O

Lecture 2 — Solutions & Raoult's Law (W5)

Molecular View of Solution Process

Break A–A and B–B interactions (requires energy, $E_1$)
Form A–B interactions (releases energy, $E_2$)
$ΔH_{solution} = E_1 - E_2$

Types of Solutions

Type	Condition	$ΔH_{soln}$	$ΔV$	Example
Ideal	A–A ≈ B–B ≈ A–B	0	0	Benzene–toluene
Positive deviation	A–A, B–B > A–B	+ve (endothermic)	+ve	Ethanol–water
Negative deviation	A–B > A–A, B–B	−ve (exothermic)	−ve	HCl–water

c1ccccc1
Cc1ccccc1
CCO
O
Cl

Raoult's Law

For a two-component miscible liquid mixture:

$$P_A = X_A P_A^o$$ $$P_B = X_B P_B^o$$

By Dalton's law: $$P_{total} = P_A + P_B = X_A P_A^o + X_B P_B^o$$

Where $X_A + X_B = 1$.

Ideal solution: obeys Raoult's law exactly ($P_1 = X_1 P_1^o$)
Positive deviation: $P_{actual} > P_{calculated}$
Negative deviation: $P_{actual} < P_{calculated}$

[!example] Example: Raoult's law verification Pure A: 60 kPa; Pure B: 30 kPa. Mixture: $X_A = 0.3$, $X_B = 0.7$, $P_{total} = 39$ kPa.

$P_{total,calc} = 0.3(60) + 0.7(30) = 18 + 21 = 39$ kPa Therefore, the mixture obeys Raoult's law (ideal).

[!example] Example: CS₂–acetone mixture 3.95 g CS₂ + 2.43 g acetone at 35°C. $P°{CS₂} = 515$ torr, $P°{acetone} = 332$ torr. Molar masses: CS₂ = 76.15, acetone = 58.0 g/mol.

$n_{CS₂} = 3.95 / 76.15 = 0.0519$ mol $n_{acetone} = 2.43 / 58.0 = 0.0419$ mol $X_{CS₂} = 0.0519 / (0.0519 + 0.0419) = 0.553$ $X_{acetone} = 0.447$ $P_{CS₂} = 0.553 × 515 = 285$ torr $P_{acetone} = 0.447 × 332 = 148$ torr $P_{total} = 285 + 148 = 433$ torr

C(=S)=S
CC(=O)C

[!example] Example: Sucrose solution vapor pressure 99.5 g sucrose + 300.0 mL water at 25°C. $P°_{water} = 23.8$ torr.

$n_{sucrose} = 99.5 / 342.3 = 0.291$ mol $n_{water} = 300 / 18 = 16.67$ mol $X_{water} = 16.67 / (16.67 + 0.291) = 0.983$ $P_{solution} = 0.983 × 23.8 = 23.4$ torr

C([C@@H]1[C@H]([C@@H]([C@H]([C@H](O1)O[C@]2([C@H]([C@@H]([C@H](O2)CO)O)O)CO)O)O)O)O

Lecture 3 — Fractional Distillation & Azeotropes (W6)

Fractional Distillation

A procedure for separating liquid components based on different boiling points.

At the end of distillation:

Liquid with lower boiling point → collected in receiving flask (distillate)
Liquid with higher boiling point → left in distilling flask (residue)

Phase Diagram: Boiling Point vs Composition

For an ideal solution A–B:

Upper curve: vapor composition
Lower curve: liquid composition
Region between curves: liquid + vapor coexistence

Azeotrope

A mixture that distills at constant composition (cannot be separated by simple fractional distillation).

Positive Deviation Azeotropes

Azeotrope has minimum boiling point and maximum vapor pressure
Boiling point of mixture lower than either pure component

Ethanol–Benzene system:

Azeotrope: 32.4% ethanol
Starting < 32.4% ethanol → distillate = azeotrope, residue = benzene
Starting > 32.4% ethanol → distillate = azeotrope, residue = ethanol
Starting = azeotrope → only azeotrope distills over

Ethanol–Water system:

Azeotrope: 95.6% ethanol + 4.4% water
Bp = 78.2°C (lower than pure ethanol 78.4°C and water 100°C)

CCO
c1ccccc1

[!note] Characteristics of positive deviation

$P_{total} > P_{theoretical}$

Endothermic solution ($ΔH = +ve$)

$ΔV = +ve$ (expansion)

$T_{b,A-B} > T_{b,A}$ or $T_{b,B}$ — correction: actually $T_{b,azeotrope} < T_{b,pure}$

Negative Deviation Azeotropes

Azeotrope has maximum boiling point and minimum vapor pressure
Boiling point of mixture higher than either pure component

HCl–Water system:

Azeotrope: 20.2% HCl
Starting < 20.2% HCl → distillate = pure H₂O, residue = azeotrope
Starting > 20.2% HCl → distillate = pure HCl, residue = azeotrope

Nitric Acid–Water system:

Azeotrope: 68% HNO₃ + 32% water
Bp = 120.5°C (higher than pure HNO₃ 78°C and water 100°C)

Cl
O
O=[N+]([O-])O

[!note] Characteristics of negative deviation

$P_{total} < P_{theoretical}$

Exothermic solution ($ΔH = -ve$)

$ΔV = -ve$ (shrinkage)

Key Distillation Rules

Deviation	Starting Composition	Distillate	Residue
Positive	< azeotrope %	Azeotrope	Higher bp component
Positive	> azeotrope %	Azeotrope	Higher bp component
Negative	< azeotrope %	Lower bp pure component	Azeotrope
Negative	> azeotrope %	Higher bp pure component	Azeotrope

Key Equations

Property	Equation
Gibbs Phase Rule	$F = C - P + 2$
Raoult's Law	$P_A = X_A P_A^o$
Dalton's Law	$P_{total} = P_A + P_B$
Vapor pressure lowering	$ΔP = X_{solute} P°_{solvent}$
Boiling point elevation	$ΔT_b = K_b m$
Freezing point depression	$ΔT_f = K_f m$
Osmotic pressure	$Π = MRT$
Clausius-Clapeyron	$\ln\frac{P_2}{P_1} = -\frac{ΔH_{vap}}{R}(\frac{1}{T_2} - \frac{1}{T_1})$

Study Notes

[!note] Exam weightage Phase Equilibria appears in most papers with ~5–8% mark weight. Focus on:

Raoult's Law calculations (vapor pressure, mole fraction)

Colligative property calculations ($ΔT_b$, $ΔT_f$, $Π$)

Drawing and interpreting phase diagrams

Fractional distillation predictions from bp-composition diagrams

Identifying azeotrope composition and behavior

[!warning] Common errors

Confusing positive vs negative deviation characteristics

Forgetting that azeotropes in positive deviation have minimum bp

Mixing up distillate vs residue for non-ideal solutions

Using molarity instead of molality for colligative properties