Phases and Phase Equilibrium
Types of Phases
- Solid phase: fixed shape and volume; particles in fixed lattice positions.
- Liquid phase: fixed volume, variable shape; particles mobile but in contact.
- Gas/vapour phase: variable volume and shape; particles widely separated.
- A system may have multiple solid phases (allotropes, polymorphs) e.g. graphite and diamond are two solid phases of carbon.
- There is only ever one liquid phase for a single component (miscible liquids count as one phase; immiscible liquids count as two).
- There is only ever one gas phase (all gases mix completely).
Conditions for Phase Equilibrium
For two phases to be in equilibrium, they must have the same:
- Temperature (thermal equilibrium)
- Pressure (mechanical equilibrium)
- Chemical potential of each component (chemical equilibrium)
The chemical potential (μ) is essentially the molar Gibbs free energy. At equilibrium: μsolid = μliquid = μvapour.
Homogeneous vs Heterogeneous Equilibria
Homogeneous Equilibrium
All species involved in the equilibrium are in the same phase. Example: N2(g) + 3H2(g) ⇌ 2NH3(g) — all gaseous.
Kc or Kp expressions include all species.
Heterogeneous Equilibrium
Species are in different phases. Example: CaCO3(s) ⇌ CaO(s) + CO2(g).
Concentrations of pure solids and liquids are constant → they are omitted from K expressions. Kp = p(CO2).
The Gibbs Phase Rule
where: F = degrees of freedom (number of intensive variables — T, P, composition — that can be independently varied without changing the number of phases); C = number of components; P = number of phases present.
Understanding the Variables
- Components (C): minimum number of chemically independent species needed to describe all phases. For pure water: C = 1.
- Phases (P): number of distinct phases present simultaneously.
- Degrees of freedom (F): number of variables (T, P, concentration) you can change independently while keeping the same phases.
| System (C=1, e.g. water) | Phases (P) | F = 1 − P + 2 | Meaning |
|---|---|---|---|
| Single phase region (e.g. liquid only) | 1 | F = 2 | Both T and P can vary — bivariant |
| Two-phase equilibrium (e.g. liquid + vapour) | 2 | F = 1 | Fix T → P fixed (or vice versa) — univariant |
| Triple point (solid + liquid + vapour) | 3 | F = 0 | T and P both fixed — invariant |
Applying the Phase Rule
Apply the Gibbs phase rule to: (a) liquid water only; (b) water at its boiling point (liquid + vapour); (c) water at its triple point.
One-Component Phase Diagrams
General Features of a One-Component Phase Diagram
A one-component (single substance) phase diagram plots pressure (P) on the y-axis and temperature (T) on the x-axis. It shows which phase (solid, liquid, gas) is stable at each combination of T and P.
Key Features
- Phase regions (areas): single-phase stable regions (F=2). Solid, liquid, or gas/vapour.
- Phase boundary curves (lines): two-phase equilibria (F=1). Three important curves:
- Sublimation curve (solid ⇌ gas): below triple point.
- Vapour pressure/vaporisation curve (liquid ⇌ gas): from triple point to critical point.
- Fusion/melting curve (solid ⇌ liquid): from triple point upward.
- Triple point (Tt, Pt): all three phases in equilibrium (F=0). Unique T and P.
- Critical point (Tc, Pc): above this, liquid and gas phases are indistinguishable (supercritical fluid). End of vaporisation curve.
General one-component phase diagram (schematic)
Reading the Phase Diagram
- The normal melting point is the T at which the fusion curve intersects P = 1 atm.
- The normal boiling point is the T at which the vaporisation curve intersects P = 1 atm.
- At T and P above the critical point, the substance exists as a supercritical fluid — cannot be liquefied by increasing pressure alone.
- Moving horizontally (constant P) across phase boundaries = heating/cooling transitions.
- Moving vertically (constant T) across phase boundaries = pressure-induced transitions.
Phase Diagram of Water
Key Data for Water
| Feature | Value | Notes |
|---|---|---|
| Triple point | T = 0.0098°C, P = 611 Pa (0.006 atm) | Solid + liquid + vapour coexist |
| Normal melting point | 0°C at 1 atm | Fusion curve at 1 atm |
| Normal boiling point | 100°C at 1 atm | Vaporisation curve at 1 atm |
| Critical point | T = 374°C, P = 218 atm | Above this: supercritical water |
Anomalous Feature: Negative Slope of Fusion Curve
The fusion curve (ice–water boundary) has a negative slope — uniquely different from most substances. This means that increasing pressure at constant temperature melts ice (converts solid to liquid).
Reason: Ice (solid) is less dense than liquid water (density: ice = 0.917 g/cm³, water = 1.00 g/cm³). Ice has an open hexagonal lattice with H-bonds; liquid water is more compact. By Le Chatelier's principle, increasing pressure favours the denser phase (liquid) → melting occurs.
Most substances have a positive slope fusion curve (solid denser than liquid).
Phase diagram of water (schematic — not to scale)
Interpreting the Water Phase Diagram
Using the water phase diagram: (a) What phase(s) exist at 50°C and 1 atm? (b) What happens when water at 25°C is put under very high pressure? (c) Why does water boil at <100°C on a mountaintop?
Phase Diagram of CO2
Key Data for CO2
| Feature | Value | Notes |
|---|---|---|
| Triple point | T = −56.6°C, P = 5.11 atm | All three phases coexist |
| Sublimation at 1 atm | −78.5°C | CO2 sublimates directly at normal pressure |
| Normal melting point | Does not exist at 1 atm (triple point P > 1 atm) | Liquid CO2 cannot exist at 1 atm |
| Critical point | T = 31.1°C, P = 73.8 atm | Supercritical CO2 above this |
Key Feature: Triple Point Above 1 atm
The triple point of CO2 occurs at 5.11 atm, which is above normal atmospheric pressure (1 atm). Therefore, at 1 atm, the solid–gas boundary is crossed directly — solid CO2 (dry ice) sublimates at −78.5°C without melting. Liquid CO2 can only exist at pressures above 5.11 atm.
Positive Slope of Fusion Curve (Normal)
Unlike water, CO2 has a positive slope on its fusion curve — solid CO2 is denser than liquid CO2. Increasing pressure at constant T favours the solid phase (Le Chatelier).
Supercritical CO2
Above 31.1°C and 73.8 atm, CO2 becomes a supercritical fluid — it has properties between a gas (low viscosity, high diffusivity) and a liquid (dissolves substances). Applications: supercritical CO2 extraction of caffeine from coffee, decaffeination; dry cleaning; extraction of essential oils; in HPLC (chromatography).
| Property | Water | CO2 |
|---|---|---|
| Triple point pressure | 0.006 atm (<1 atm) | 5.11 atm (>1 atm) |
| Behaviour at 1 atm on heating solid | Melts (solid → liquid → gas) | Sublimates (solid → gas directly) |
| Fusion curve slope | Negative (anomalous) | Positive (normal) |
| Solid denser than liquid? | No (ice floats) | Yes (normal) |
| Critical point | 374°C, 218 atm | 31.1°C, 73.8 atm |
Two-Component Phase Diagrams
Vapour Pressure–Composition Diagrams (Raoult's Law)
For an ideal binary mixture of two miscible liquids A and B, the vapour pressure of each component follows Raoult's Law:
On a P–x diagram, the total pressure varies linearly with composition for an ideal mixture.
Boiling Point–Composition Diagram (T–x Diagram)
At constant pressure (usually 1 atm), the boiling point–composition diagram shows:
- Liquid curve (bubble point curve): compositions and temperatures at which the liquid begins to boil.
- Vapour curve (dew point curve): compositions and temperatures at which vapour begins to condense.
- The region between the two curves = two-phase region (liquid + vapour in equilibrium).
T–x diagram for an ideal binary mixture (A more volatile, lower B.P.; B less volatile, higher B.P.)
The Lever Rule
At a point in the two-phase region, the relative amounts of liquid and vapour are given by the lever rule:
Azeotropes (Non-Ideal Mixtures)
Real mixtures may deviate from Raoult's law, forming azeotropes — mixtures with constant boiling point at a fixed composition:
- Minimum-boiling azeotrope (positive deviation): e.g. ethanol–water (95.6% ethanol, B.P. 78.1°C). Cannot be separated beyond azeotrope composition by simple distillation.
- Maximum-boiling azeotrope (negative deviation): e.g. HCl–water (20.2% HCl, B.P. 108.6°C). Mixture boils above both pure components.
Applications of Phase Diagrams
| Application | Phase Diagram Used | Details |
|---|---|---|
| Fractional distillation of petroleum | Multi-component T–x diagram | Separation based on boiling point differences in complex mixture |
| Freeze-drying (lyophilisation) | Water phase diagram | Reduce P below triple point → ice sublimates directly to vapour without passing through liquid phase. Used for food and pharmaceutical preservation. |
| Supercritical CO2 extraction | CO2 phase diagram | Operate above critical point (31.1°C, 73.8 atm). Supercritical CO2 dissolves caffeine, essential oils etc. without toxic solvents. |
| Pressure cookers | Water phase diagram | Increase P above 1 atm → boiling point rises above 100°C → food cooks faster. |
| Autoclave sterilisation | Water phase diagram | Steam at 121°C (2 atm) kills all microbes; used in hospitals and labs. |
| Alloy solidification | Two-component solid–liquid diagram | Eutectic point gives minimum melting temperature mixture (e.g. tin–lead solder). |
| Fractional distillation (ethanol–water) | T–x diagram with azeotrope | Maximum ethanol concentration by distillation = 95.6% (azeotrope). Pure ethanol requires molecular sieves or chemical drying. |
Freeze-Drying Using the Water Phase Diagram
Explain, with reference to the water phase diagram, how freeze-drying preserves food.
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Exercises
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Define: (a) a phase; (b) the triple point; (c) the critical point; (d) degrees of freedom (F) in the Gibbs phase rule.
(a) A phase is a physically distinct, homogeneous region of a system separated from other regions by definite boundaries, with uniform composition and physical properties throughout.
(b) The triple point is the unique temperature and pressure at which solid, liquid, and gas phases of a substance coexist in equilibrium (F = 0; C=1, P=3).
(c) The critical point is the temperature and pressure above which the liquid and gas phases are indistinguishable (supercritical fluid); the end of the vaporisation curve.
(d) Degrees of freedom (F) = the number of intensive variables (T, P, composition) that can be independently varied without changing the number of phases present. From Gibbs: F = C − P + 2. -
Apply the Gibbs phase rule (F = C − P + 2) to: (a) a single gas; (b) liquid–vapour equilibrium for a pure substance; (c) the triple point of CO2; (d) an ideal binary liquid mixture (two components, one phase).
(a) C=1, P=1: F = 1−1+2 = 2. Both T and P can be varied independently.
(b) C=1, P=2: F = 1−2+2 = 1. Fix T → P fixed (vapour pressure curve). Univariant.
(c) C=1, P=3: F = 1−3+2 = 0. Invariant; unique T=−56.6°C, P=5.11 atm.
(d) C=2, P=1: F = 2−1+2 = 3. T, P, and composition can all be independently varied. -
Explain why the fusion curve of water has a negative slope while that of CO2 has a positive slope.
The slope of the fusion curve is governed by the Clausius–Clapeyron equation for melting: dP/dT = ΔHfus / (TΔVfus).
Water: ice is less dense than liquid water (ΔVfus = Vliquid − Vsolid < 0). Since ΔHfus > 0 and ΔVfus < 0, dP/dT < 0 → negative slope. Increasing pressure at constant T melts ice (favours denser liquid).
CO2: solid CO2 is denser than liquid CO2 (ΔVfus > 0). dP/dT > 0 → positive slope. Increasing pressure at constant T solidifies liquid CO2. -
Why does solid CO2 (dry ice) sublime at atmospheric pressure rather than melting?
The triple point of CO2 is at 5.11 atm, which is higher than atmospheric pressure (1 atm). At 1 atm, it is impossible to have liquid CO2 — the isobar at 1 atm crosses the solid–gas (sublimation) boundary directly without passing through the liquid region. Therefore, solid CO2 converts directly to gas (sublimates) at −78.5°C without melting. -
State Raoult's Law and explain what is meant by an ideal binary mixture. What is an azeotrope and why is it important in distillation?
Raoult's Law: The partial vapour pressure of a component in a mixture equals the mole fraction of that component multiplied by the vapour pressure of the pure component: pA = xA × PA*.
An ideal binary mixture obeys Raoult's law at all compositions. This occurs when the two components have similar intermolecular forces (A–B forces ≈ A–A and B–B forces). The total vapour pressure varies linearly with composition.
An azeotrope is a mixture with a fixed composition that boils at a constant temperature (the vapour and liquid have the same composition). Example: ethanol–water azeotrope (95.6% ethanol, B.P. 78.1°C). At this point, distillation cannot further concentrate the mixture because the vapour has the same composition as the liquid. To obtain pure ethanol, molecular sieves or chemical drying agents must be used. -
A binary mixture of A (B.P. 80°C) and B (B.P. 110°C) has an overall composition xB = 0.40. At a certain temperature, the liquid composition is xB(liquid) = 0.25 and the vapour composition is xB(vapour) = 0.60. Use the lever rule to find the mole fraction of the system that is vapour.
Lever rule: moles vapour / moles liquid = (xoverall − xliquid) / (xvapour − xoverall)
= (0.40 − 0.25) / (0.60 − 0.40)
= 0.15 / 0.20 = 0.75
So moles vapour = 0.75 × moles liquid. If total moles = 1: vapour/(vapour+liquid) = 0.75/1.75 = 0.429 (42.9% vapour).
Interactive Quiz
Unit 10 Quiz — Phase Diagrams
25 QuestionsA phase is defined as:
The Gibbs phase rule is F = C − P + 2. For a pure substance at its triple point, F equals:
What is the triple point of water?
The fusion curve of water has a negative slope because:
Why does dry ice (solid CO2) sublime at atmospheric pressure rather than melting?
Above the critical point, a substance exists as:
For a one-component system along the vapour pressure curve (liquid + vapour equilibrium), F equals:
The normal boiling point of a liquid is defined as the temperature at which:
In a heterogeneous equilibrium such as CaCO3(s) ⇌ CaO(s) + CO2(g), the equilibrium expression Kp is:
Raoult's Law states that the partial vapour pressure of a component equals:
In the T–x diagram for a binary mixture, the area between the liquid and vapour curves represents:
The ethanol–water azeotrope at 95.6% ethanol is an example of:
Water boils at a lower temperature on a mountain than at sea level because:
Freeze-drying works by:
Supercritical CO2 is used in coffee decaffeination because:
At the triple point of water, F = 0. This means:
For a binary mixture of two components in a single liquid phase, F equals:
In a T–x diagram, a tie line connects:
A pressure cooker works because:
The critical temperature is the temperature above which:
The equilibrium H2O(l) ⇌ H2O(g) is an example of:
For the ideal binary mixture of A and B, if xA = 0.6, PA* = 80 kPa and PB* = 40 kPa, the total vapour pressure is:
What does it mean if a binary mixture shows positive deviation from Raoult's Law?
Along the sublimation curve of a substance, the equilibrium is between:
The lever rule in a two-phase region gives the:
Unit Test
Section A — Short Answer
30 marksState the Gibbs phase rule and define each variable. Apply it to: (a) ice melting at 0°C and 1 atm; (b) liquid water at 60°C and 1 atm; (c) the triple point of water.
F = degrees of freedom; C = number of components; P = number of phases.
(a) Ice melting (solid + liquid): C=1, P=2 → F = 1−2+2 = 1. Univariant — fixing T (0°C) fixes P (1 atm) along the fusion curve.
(b) Liquid water only: C=1, P=1 → F = 1−1+2 = 2. Bivariant — both T and P can be changed independently.
(c) Triple point: C=1, P=3 → F = 1−3+2 = 0. Invariant — T (0.0098°C) and P (611 Pa) are fixed uniquely.
Sketch a labelled phase diagram for water (P vs T). On it, mark and label: (a) the three phase regions; (b) the triple point with its T and P values; (c) the critical point; (d) the normal melting point and normal boiling point; (e) the anomalous feature of the fusion curve and explain its significance.
Three regions: ICE (solid, top-left), LIQUID WATER (top-right), WATER VAPOUR (bottom-right).
Three boundary curves: sublimation (solid-gas, below triple point), vaporisation (liquid-gas, up to critical point), fusion (solid-liquid, from triple point upward).
(a) Regions clearly labelled with phases.
(b) Triple point at T = 0.0098°C, P = 611 Pa (0.006 atm).
(c) Critical point at T = 374°C, P = 218 atm (end of vaporisation curve).
(d) Normal M.P. = 0°C at 1 atm (intersection of fusion curve with P=1 atm); Normal B.P. = 100°C at 1 atm (intersection of vaporisation curve with P=1 atm).
(e) Anomalous feature: fusion curve has a negative slope (leans slightly to the left — higher P gives lower melting T). Significance: ice is less dense than liquid water; increasing pressure favours the denser phase (liquid) → melting. Opposite to most substances.
Compare the phase diagrams of water and CO2, focusing on: (i) triple point pressure; (ii) behaviour at 1 atm on heating the solid; (iii) slope of fusion curve; (iv) properties of the solid vs liquid density. Explain one practical consequence of each difference.
Consequence: At 1 atm, water melts normally (solid→liquid→gas); dry ice sublimates (solid→gas directly). Dry ice useful for cooling without liquid mess.
(ii) Behaviour at 1 atm: Water: melts at 0°C then boils at 100°C. CO2: sublimates at −78.5°C (no liquid phase at 1 atm).
(iii) Fusion curve slope: Water: negative (anomalous). CO2: positive (normal).
Consequence: Increasing pressure on ice at ~0°C melts it (water pipe bursts in frost); increasing pressure on solid CO2 keeps it solid (used safely in pressurised fire extinguishers).
(iv) Density: Ice less dense than water (ice floats — critical for aquatic life to survive under ice). Solid CO2 denser than liquid CO2 (normal behaviour).
Write Kp expressions for the following heterogeneous equilibria and explain why pure solids and liquids are excluded:
(a) CaCO3(s) ⇌ CaO(s) + CO2(g)
(b) 3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g)
(c) NH4HCO3(s) ⇌ NH3(g) + H2O(g) + CO2(g)
(a) Kp = p(CO2)
(Both CaCO3 and CaO are pure solids.)
(b) Kp = [p(H2)]4 / [p(H2O)]4
(Fe and Fe3O4 are pure solids.)
(c) Kp = p(NH3) × p(H2O) × p(CO2)
(NH4HCO3 is a pure solid.)
For an ideal binary mixture of hexane (P* = 150 mmHg) and heptane (P* = 45 mmHg) at 25°C:
(a) Calculate the total vapour pressure of a mixture with mole fraction of hexane xhex = 0.70.
(b) Calculate the mole fraction of hexane in the vapour phase above this mixture.
(c) Is the vapour richer or poorer in hexane compared to the liquid? What does this mean for distillation?
phep = 0.30 × 45 = 13.5 mmHg
Ptotal = 105 + 13.5 = 118.5 mmHg
(b) yhex (mole fraction in vapour) = phex/Ptotal = 105/118.5 = 0.886
(c) The vapour (yhex = 0.886) is richer in hexane than the liquid (xhex = 0.70). This makes sense because hexane is more volatile (higher P*). In distillation, the vapour phase is enriched in the more volatile component → repeated distillation (fractionation) progressively purifies hexane from the mixture.
Explain the following, using phase diagrams in your answer:
(a) Why a pressure cooker allows food to cook faster at high altitude. [2]
(b) How supercritical CO2 is used in decaffeination of coffee. [2]
(c) Why mountaineers cannot make a proper cup of tea at very high altitude. [2]
(b) CO2 is brought to T > 31.1°C and P > 73.8 atm → supercritical fluid region on the CO2 phase diagram. Supercritical CO2 has gas-like diffusivity (penetrates coffee beans easily) and liquid-like solvating ability (dissolves caffeine). When pressure is released, CO2 returns to gas phase → leaves no toxic residue in coffee. Clean, efficient, selective extraction.
(c) At very high altitude (e.g. Everest base camp ~5300 m), atmospheric pressure ≈ 0.54 atm. The vaporisation curve of water is intersected at ~83°C. Tea requires >90°C to properly brew (extract flavour compounds, kill bacteria). Since water boils at only ~83°C, tea is under-extracted and does not taste right. A pressure cooker would solve this.
Section B — Extended Response
20 marks(a) Describe the one-component phase diagram of CO2. Sketch and label it, indicating: all three phase regions, the triple point (with values), the critical point (with values), and the sublimation curve. Explain why CO2 does not have a normal melting point. [5 marks]
(b) Using the Gibbs phase rule, calculate F for three different conditions on the CO2 phase diagram and explain what each value means physically. [3 marks]
(c) Explain what a homogeneous equilibrium is and give one example with its K expression. [2 marks]
Three regions: solid (top-left), liquid (top-middle-right), gas (bottom-right).
Triple point: −56.6°C, 5.11 atm. Critical point: 31.1°C, 73.8 atm.
Fusion curve: positive slope (normal — solid denser than liquid).
Sublimation curve: runs from bottom-left to triple point (solid–gas equilibrium below triple point).
No normal melting point: The triple point (5.11 atm) is above 1 atm. Therefore the 1 atm isobar does not cross the solid–liquid (fusion) boundary at any temperature. At 1 atm, moving from solid to gas crosses only the sublimation curve (−78.5°C). Liquid CO2 can only exist above 5.11 atm.
(b)
1. Solid CO2 only: C=1, P=1 → F = 2. Both T and P can be varied — bivariant.
2. Solid + gas (on sublimation curve): C=1, P=2 → F = 1. Fix T → P fixed (sublimation pressure) — univariant.
3. Triple point: C=1, P=3 → F = 0. T and P both fixed at −56.6°C, 5.11 atm — invariant.
(c) A homogeneous equilibrium is one where all reactants and products are in the same phase.
Example: N2(g) + 3H2(g) ⇌ 2NH3(g)
Kp = [p(NH3)]2 / (p(N2) × [p(H2)]3)
(a) Describe the T–x (boiling point–composition) diagram for an ideal binary mixture. Define and identify: the liquid curve, vapour curve, two-phase region, and a tie line. Explain what the lever rule gives and derive the formula. [5 marks]
(b) Explain what an azeotrope is, distinguishing between minimum-boiling and maximum-boiling azeotropes. Give one example of each and explain why the ethanol–water azeotrope limits the purity achievable by simple fractional distillation. [5 marks]
T–x diagram (at constant P, usually 1 atm): T on y-axis, mole fraction xB on x-axis from 0 (pure A) to 1 (pure B).
Liquid (bubble point) curve: the locus of temperatures at which a liquid of given composition begins to boil. Below this curve = single-phase liquid.
Vapour (dew point) curve: temperatures at which vapour of given composition begins to condense. Above this curve = single-phase vapour.
Two-phase region: between the two curves; liquid and vapour coexist in equilibrium. F = 2−2+1 = 1 (at constant P) — univariant.
Tie line: horizontal line at constant T within the two-phase region, connecting a point on the liquid curve (xL) to a point on the vapour curve (xV).
Lever rule: Moles(vapour) × (xV − xoverall) = Moles(liquid) × (xoverall − xL)
→ nV/nL = (xoverall − xL)/(xV − xoverall)
Gives the relative moles of vapour to liquid at any point in the two-phase region.
(b)
An azeotrope is a mixture of fixed composition at which the vapour and liquid have identical composition → distillation cannot change the composition further at that point.
Minimum-boiling azeotrope (positive deviation from Raoult's Law): vapour pressure curve has a maximum → boiling point curve has a minimum. The mixture boils at a temperature below both pure components. Example: ethanol–water (95.6% ethanol, B.P. 78.1°C; pure ethanol B.P. 78.4°C, water 100°C).
Maximum-boiling azeotrope (negative deviation): vapour pressure minimum → boiling point maximum. Mixture boils above both pure components. Example: HCl–water (20.2% HCl, B.P. 108.6°C).
Ethanol–water limitation: Starting from any dilute ethanol–water mixture and distilling, the distillate becomes progressively richer in ethanol until it reaches 95.6% ethanol (azeotrope). At this point, vapour = liquid composition → no further enrichment is possible by distillation. To get 100% ethanol, physical desiccants (molecular sieves, CaO) must be used.