Phase rule — Gibbs phase rule: Worked examples

Before starting, recall the three quantities (from the parent note):

$C$ = number of independent components (species minus reactions minus extra constraints),
$P$ = number of coexisting phases (distinct homogeneous regions),
$F$ = number of intensive knobs (like $T$ , pressure, composition) you can freely turn.

The full formula and its constant-pressure cousin (Condensed phase rule): $F = C - P + 2 and F_{condensed} = C - P + 1$

The scenario matrix

Every phase-rule problem lives in one of these cells. The worked examples below are each tagged with the cell they cover, and together they hit all of them.

Cell	What makes it tricky	Example
A. One component, sweeping $P$	Watch $F$ fall 2→1→0 as phases stack	Ex 1
B. Two components, single vs multi phase	Composition adds real freedom	Ex 2
C. Reaction reduces $C$	Must subtract reactions	Ex 3
D. Fixed pressure (condensed rule)	The $+ 2$ becomes $+ 1$	Ex 4
E. Degenerate $F = 0$ (invariant point)	Everything locked; solve for $P$ or $T$	Ex 5
F. Impossible / negative $F$	The formula forbids the scenario	Ex 6
G. Extra constraint (fixed composition ratio)	Subtract constraints beyond reactions	Ex 7
H. Real-world word problem	Translate words → $C, P$	Ex 8
I. Exam twist (immiscible liquids / azeotrope)	Miscounting phases	Ex 9

Example 1 — One component, sweeping through phases (Cell A)

Worked example Pure carbon dioxide, three situations

For pure $C O_{2}$ ( $C = 1$ ), find $F$ when it is (a) just gas, (b) liquid + gas on the boiling line, (c) solid + liquid + gas at the triple point.

Forecast: guess the three numbers before reading. (Hint: each extra phase eats one knob.)

Step 1 — Fix $C$ . Pure $C O_{2}$ is one chemical species, no reaction, so $C = 1$ . Why this step? $C$ is set by chemistry alone, before we count phases.

Step 2 — Plug each phase count.

(a) $P = 1$ : $F = 1 - 1 + 2 = 2$ .
(b) $P = 2$ : $F = 1 - 2 + 2 = 1$ .
(c) $P = 3$ : $F = 1 - 3 + 2 = 0$ . Why this step? Only $P$ changes between the three situations; $C$ and the $+ 2$ stay fixed.

Step 3 — Read off the geometry (see figure): $F = 2$ is an area, $F = 1$ a line, $F = 0$ a point.

Verify: the numbers $2, 1, 0$ decrease by exactly 1 each time $P$ rises by 1 — consistent with the $- P$ in the formula. ✅

The picture is the same shape as the water diagram in the parent — see Phase diagrams of pure substances. The lines are described by the Clausius–Clapeyron equation.

Example 2 — Two components: freedom hiding in composition (Cell B)

Worked example Salt water, three phase counts

A mixture of $NaCl$ and $H_{2} O$ ( $C = 2$ ). Find $F$ for (a) a single liquid solution, (b) solution + water vapour, (c) solution + solid salt + vapour.

Forecast: with one MORE component than water, do you expect more or fewer knobs than Example 1?

Step 1 — Fix $C$ . Two independent species, no reaction, so $C = 2$ . Why this step? $NaCl$ and $H_{2} O$ can be dialled independently — nothing links their amounts.

Step 2 — Plug each $P$ .

(a) $P = 1$ : $F = 2 - 1 + 2 = 3$ .
(b) $P = 2$ : $F = 2 - 2 + 2 = 2$ .
(c) $P = 3$ : $F = 2 - 3 + 2 = 1$ . Why this step? Same machine, but $C = 2$ shifts every answer up by one relative to Example 1.

Step 3 — Interpret (a). Three free knobs: $T$ , pressure, and the salt concentration. That is why you can heat, pressurise, and dilute a solution all independently.

Verify: compare with Example 1 cell-by-cell — every value is exactly $1$ larger, matching the extra component in $F = C - P + 2$ . ✅

Example 3 — A reaction shrinks the component count (Cell C)

Worked example Ammonium chloride decomposing

$N H_{4} Cl (s) ⇌ N H_{3} (g) + HCl (g)$ , started from pure $N H_{4} Cl$ . Find $F$ .

Forecast: there are 3 species — but is $C = 3$ ?

Step 1 — Count species and reactions. Species $= 3$ ( $N H_{4} Cl, N H_{3}, HCl$ ); independent reactions $= 1$ . Why this step? The equilibrium constant of the reaction links the three amounts — one dependency.

Step 2 — Extra constraint from the starting material. Because we start from pure $N H_{4} Cl$ , the gas phase must have $N H_{3}$ and $HCl$ in a 1:1 ratio. That is one more constraint. Why this step? Stoichiometry of the source fixes the composition ratio in the vapour — Components and independent reactions calls this an added constraint.

Step 3 — Components. $C = 3 - 1 - 1 = 1$ .

Step 4 — Phases. Solid $N H_{4} Cl$ + one gas phase $\Rightarrow P = 2$ .

Step 5 — Phase rule. $F = 1 - 2 + 2 = 1$ .

Verify: $F = 1$ means picking $T$ fixes the decomposition pressure — experimentally true, one pressure per temperature. ✅ Contrast with the parent's $CaC O_{3}$ example ( $C = 2$ ) where the extra 1:1 ratio was absent because the solid product $CaO$ locks nothing about gas composition.

Example 4 — Fixed pressure: the condensed rule (Cell D)

Example 5 — The invariant point, everything locked (Cell E)

Worked example Solving for the state at a triple point

Pure sulfur ( $C = 1$ ) shows rhombic solid, monoclinic solid, and liquid all coexisting. How many degrees of freedom, and what does that mean physically?

Forecast: three solids-or-liquids of ONE substance — how many knobs?

Step 1 — Count. $C = 1$ , $P = 3$ (two solid phases + one liquid — different crystal structures do count separately). Why this step? Rhombic and monoclinic sulfur are distinct homogeneous solids, so they are two phases, not one.

Step 2 — Phase rule. $F = 1 - 3 + 2 = 0$ .

Step 3 — Meaning. $F = 0$ ⇒ invariant point: both $T$ and pressure are fixed by nature. You cannot choose either.

Verify: experimentally the rhombic–monoclinic–liquid triple point of sulfur sits at one fixed $(T, P)$ ≈ $(15 1^{\circ} C, 1290 atm)$ — a single dot, exactly as $F = 0$ predicts. ✅

Example 6 — When the formula forbids a scenario (Cell F)

Worked example Can four phases of pure water coexist?

Claim: ice + liquid + vapour + a fourth ice polymorph, all of ordinary $H_{2} O$ under normal conditions, coexist simultaneously with a free choice of $T$ . Test it.

Forecast: is this possible?

Step 1 — Plug $C = 1$ , $P = 4$ . $F = 1 - 4 + 2 = - 1$ Why this step? We just run the machine and inspect the sign.

Step 2 — Interpret the negative. $F$ counts free knobs; it cannot be negative. A negative $F$ means the system is over-constrained: there are more equations than unknowns, so no such equilibrium exists. Why this step? Negative freedom is the formula's way of shouting "impossible."

Verify: rearranging $F \geq 0$ gives $P \leq C + 2 = 3$ . So a one-component system caps at 3 coexisting phases — the four-phase claim is ruled out. ✅ (Exotic high-pressure ice polymorphs can make new triple points, but never four phases at once for fixed $C = 1$ .)

Example 7 — An extra constraint beyond the reaction (Cell G)

Worked example Water gas equilibrium with imposed composition

$CO (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g)$ in one gas phase, prepared so that $CO$ and $H_{2} O$ are fed in equal moles. Find $F$ .

Forecast: 4 species, 1 reaction — but there is one more subtraction. Spot it.

Step 1 — Species and reactions. Species $= 4$ ; independent reactions $= 1$ .

Step 2 — Added constraint. Feeding $CO$ and $H_{2} O$ equally imposes a fixed relationship among mole numbers — one extra constraint. Why this step? Any imposed composition ratio removes a degree of independence, just like the 1:1 ratio in Example 3.

Step 3 — Components. $C = 4 - 1 - 1 = 2$ .

Step 4 — Phases. All in one gas phase, $P = 1$ .

Step 5 — Phase rule. $F = 2 - 1 + 2 = 3$ .

Verify: three free knobs = $T$ , pressure, and one composition variable — a homogeneous reacting gas should indeed be freely adjustable in three ways. ✅

Example 8 — Real-world word problem (Cell H)

Worked example Espresso steam wand

A barista's boiler holds liquid water and steam in equilibrium. The barista wants to set the steam temperature. How many independent settings does thermodynamics allow, and what happens when they pick one?

Forecast: can they set both temperature and pressure freely?

Step 1 — Translate to symbols. Pure water: $C = 1$ . Liquid + vapour: $P = 2$ . Why this step? Word problems must be reduced to $C$ and $P$ before any formula.

Step 2 — Phase rule. $F = 1 - 2 + 2 = 1$ .

Step 3 — Interpret. Exactly one free knob. Choosing the temperature forces the pressure (the boiling-point relation), and vice versa. Why this step? $F = 1$ is a curve on the $P$ – $T$ plane — pick a point along it via one variable, the other is then determined.

Verify: real espresso boilers are rated by pressure OR temperature but not both independently — matching $F = 1$ . The link between the two is the Clausius–Clapeyron equation. ✅

Example 9 — Exam twist: don't miscount phases (Cell I)

Worked example Oil, water, and their shared vapour

Immiscible liquid oil and liquid water sit in a sealed jar with a common vapour containing both. Treating oil and water as two independent components, find $F$ .

Forecast: the trap is $P$ . How many phases: 2 or 3?

Step 1 — Count phases carefully. Oil (liquid) and water (liquid) do not mix — they are two distinct liquid phases, plus one vapour phase. So $P = 3$ , not 2. Why this step? Phases are physically distinct homogeneous regions; two immiscible liquids are two phases even though both are "liquid."

Step 2 — Components. Oil and water: $C = 2$ .

Step 3 — Phase rule. $F = 2 - 3 + 2 = 1$ .

Step 4 — Interpret. One free knob. In practice the mixture boils at a fixed temperature for a given pressure (the basis of steam distillation) — pick pressure, temperature is set.

Verify: if you had wrongly used $P = 2$ you'd get $F = 2$ and predict independent $T$ – $P$ control, contradicting the observed fixed steam-distillation temperature. So $P = 3 \Rightarrow F = 1$ is right. ✅ (The underlying equality of Chemical potential across all three phases supplies the constraints; see also Gibbs free energy and equilibrium.)

Recall Quick self-test

A three-component system shows two phases at fixed pressure. $F = ?$ ::: Condensed rule: $F = 3 - 2 + 1 = 2$ . Pure substance, $F = 0$ . How many phases? ::: $P = C + 2 = 1 + 2 = 3$ . Why can $F$ never be negative? ::: A negative $F$ means more constraints than variables — the equilibrium is impossible, so $P \leq C + 2$ . Two immiscible liquids + their vapour, two components: $F = ?$ ::: $F = 2 - 3 + 2 = 1$ .

Connections

Phase rule — Gibbs phase rule (index 2.4.7) — the parent note that derives $F = C - P + 2$ .
Condensed phase rule — the $+ 1$ version used in Examples 4.
Components and independent reactions — how reactions and constraints lower $C$ (Examples 3, 7).
Chemical potential — supplies every equilibrium constraint.
Phase diagrams of pure substances and Clausius–Clapeyron equation — the geometry behind $F = 2, 1, 0$ .
Gibbs free energy and equilibrium — why chemical potentials equalise.
2.4.07 Phase rule — Gibbs phase rule (Hinglish) — Hinglish companion.