The Lever Rule

Picture a flask of benzene and toluene, half-boiled, sitting inside the two-phase dome of its temperature–composition diagram. Liquid and vapour coexist, each with its own composition. But how much of each phase is actually present? The lever rule answers this with nothing more than a mass balance and a ruler laid across the tie-line. This question matters for every separation technology. This includes distillation columns and zone refining. Each depends on knowing not just which phases exist, but how much material sits in each one. The lever rule converts a phase diagram from a qualitative map into a quantitative accounting tool. This article derives the tie-line balance from conservation of mass. It shows why the rule works like a mechanical lever. It then applies the result to two fully worked numerical problems. One covers vapour–liquid equilibrium. The other covers a solid–liquid eutectic system.

Key takeaways

The lever rule gives the relative amounts (moles or mass) of two coexisting phases from their compositions on a phase diagram.
It follows directly from conservation of the total amount of substance and of each component—no new physics beyond mass balance.
Along a tie-line, the ratio of phase amounts equals the ratio of the opposite horizontal segment lengths: $n^\alpha \ell^\alpha = n^\beta \ell^\beta$ .
It works identically for liquid–vapour, liquid–liquid, and solid–liquid two-phase regions, wherever a tie-line connects two coexisting compositions.
Right at a phase boundary, one segment length shrinks to zero and the sample becomes 100% single phase.
Key idea: distances along a tie-line are inversely proportional to phase amounts, exactly like distances from a fulcrum are inversely proportional to weights on a mechanical lever.

The core insight

Because the total amount of each component must be conserved when a mixture splits into two phases of different composition, the overall composition point must lie on the straight line connecting the two phase compositions, and its exact position along that line—expressed as the ratio of the two segment lengths—fixes the ratio of the amounts of material in each phase.

The physical picture

Consider a binary mixture of overall mole fraction $z$ of component A, held at a temperature and pressure inside a two-phase region of its phase diagram (see Two-Component Liquid–Vapour Diagrams for how such regions arise). At that point, the system does not exist as one homogeneous phase. It splits into two phases, $\alpha$ and $\beta$ , each with its own composition, $x^\alpha$ and $x^\beta$ , fixed by the ends of the horizontal tie-line drawn through the overall composition point. The compositions $x^\alpha$ and $x^\beta$ are dictated entirely by the temperature (or pressure) and by the shape of the phase boundaries; they do not depend on how much total material you started with. What does depend on the total amount of material is how the mass is partitioned between $\alpha$ and $\beta$ . A system prepared with $z$ close to $x^\alpha$ will be almost all phase $\alpha$ , with only a trace of $\beta$ ; a system with $z$ close to $x^\beta$ will be almost all phase $\beta$ . The lever rule makes this qualitative statement exact. Geometrically, the overall point $z$ divides the tie-line into two segments:

\ell^\alpha = z - x^\alpha, \qquad \ell^\beta = x^\beta - z

The lever rule states that the amount of phase $\alpha$ is inversely proportional to $\ell^\alpha$ (the segment nearest $\alpha$ 's own composition) and proportional to $\ell^\beta$ (the far segment)—precisely as a light rigid beam balances heavier weights closer to the fulcrum.

Figure

Tie-line in a liquid–vapour T–x diagram, with the lever balance shown below it

The second figure below shows the balance-beam analogy directly: the fulcrum sits at $z$ , and the amounts of liquid and vapour act like masses whose product with their lever arm must balance.

Figure

Mechanical balance analogy for the lever rule

The core relations

n^\alpha (x^\alpha - z) = n^\beta (z - x^\beta) \quad \Longleftrightarrow \quad n^\alpha \ell^\alpha = n^\beta \ell^\beta

where: $n^\alpha, n^\beta$ are total amounts (mol) of phases $\alpha,\beta$ ; $\ell^\alpha = z - x^\beta$ is not used—each amount multiplies the segment on the opposite side of $z$ from its own composition.

\frac{n^\alpha}{n^\beta} = \frac{z - x^\beta}{x^\alpha - z}

where: this is the working ratio form—amount of $\alpha$ over amount of $\beta$ equals the far segment over the near segment.

n^\alpha = n_{\text{tot}}\,\frac{x^\beta - z}{x^\beta - x^\alpha}, \qquad n^\beta = n_{\text{tot}}\,\frac{z - x^\alpha}{x^\beta - x^\alpha}

where: $n_{\text{tot}} = n^\alpha + n^\beta$ is the total amount of substance in the sample.

Derivation AMass balance to tie-line lever

Goal. Starting only from conservation of total amount of substance and conservation of component A, derive the ratio $n^\alpha/n^\beta$ in terms of tie-line distances.

Step 1. Set up the system. A closed sample contains a total amount $n_{\text{tot}}$ of a two-component mixture with overall mole fraction of A equal to $z$ . At the chosen temperature and pressure it splits into two phases, $\alpha$ with amount $n^\alpha$ and mole fraction $x^\alpha$ , and $\beta$ with amount $n^\beta$ and mole fraction $x^\beta$ . No material enters or leaves the sample, so mass and composition bookkeeping must close exactly.

Step 2. Total mass balance. Because no material is created or destroyed on phase splitting,

n_{\text{tot}} = n^\alpha + n^\beta

Step 3. Component mass balance. The total amount of A must equal the sum of A in each phase:

n_{\text{tot}}\,z = n^\alpha x^\alpha + n^\beta x^\beta

Step 4. Eliminate $n_{\text{tot}}$ . Substitute the Step 2 expression for $n_{\text{tot}}$ into Step 3:

(n^\alpha + n^\beta)\,z = n^\alpha x^\alpha + n^\beta x^\beta

Step 5. Collect terms. Expand the left side and move all $\alpha$ -terms to one side:

n^\alpha z - n^\alpha x^\alpha = n^\beta x^\beta - n^\beta z

n^\alpha (z - x^\alpha) = n^\beta (x^\beta - z)

Step 6. Introduce signed distances. Define $\ell^\alpha \equiv |z - x^\alpha|$ and $\ell^\beta \equiv |x^\beta - z|$ , the distances from the overall point to each phase composition along the tie-line. Because $z$ lies between $x^\alpha$ and $x^\beta$ (mass balance forces this), both differences have consistent sign, and Step 5 becomes

n^\alpha \ell^\alpha = n^\beta \ell^\beta

Step 7. Interpret. This is exactly the torque-balance condition for a rigid, massless beam pivoted at $z$ , carrying "weight" $n^\alpha$ at distance $\ell^\alpha$ on one side and weight $n^\beta$ at distance $\ell^\beta$ on the other. Solving for the ratio gives the boxed working result:

\boxed{\dfrac{n^\alpha}{n^\beta} = \dfrac{\ell^\beta}{\ell^\alpha} = \dfrac{x^\beta - z}{z - x^\alpha}}

Notice the inversion: the phase whose composition is closer to $z$ (small $\ell$ ) is present in the larger amount. This is the same counter-intuitive geometry as a see-saw—the heavier person sits closer to the pivot.

Worked examples

Example 1 — Benzene/toluene vapour–liquid mixture. A mixture containing $z = 0.60$ mole fraction benzene is heated to a temperature inside the two-phase region. At that temperature, the liquidus and vapour curve give liquid composition $x^{\text{liq}} = 0.70$ and vapour composition $x^{\text{vap}} = 0.40$ (benzene mole fractions), consistent with benzene being the more volatile component. The sample contains $n_{\text{tot}} = 2.00$ mol total. Step 1 — segment lengths. The liquid-side segment (distance from $z$ to the vapour composition, which is the "far" lever arm for liquid) is

\ell^{\text{liq}} = z - x^{\text{vap}} = 0.60 - 0.40 = 0.20

The vapour-side segment is

\ell^{\text{vap}} = x^{\text{liq}} - z = 0.70 - 0.60 = 0.10

Step 2 — ratio. Using the boxed result with $\alpha =$ liquid, $\beta =$ vapour:

\frac{n^{\text{liq}}}{n^{\text{vap}}} = \frac{\ell^{\text{vap}}}{\ell^{\text{liq}}} = \frac{0.10}{0.20} = 0.50

So there is half as much liquid (in moles) as vapour: $n^{\text{liq}} = 0.5\, n^{\text{vap}}$ . Step 3 — absolute amounts. Combine with $n^{\text{liq}} + n^{\text{vap}} = 2.00$ mol:

0.5\,n^{\text{vap}} + n^{\text{vap}} = 2.00 \text{ mol} \quad\Rightarrow\quad n^{\text{vap}} = 1.33 \text{ mol}

n^{\text{liq}} = 2.00 - 1.33 = 0.67 \text{ mol}

Sanity check. Component balance: $0.67(0.70) + 1.33(0.40) = 0.469 + 0.533 = 1.00$ mol of benzene. Independently, $z \times n_{\text{tot}} = 0.60 \times 2.00 = 1.20$ mol… Re-checking: the direct component balance should reproduce $1.20$ mol of benzene, not $1.00$ . Recomputing with exact fractions: $n^{\text{vap}} = 2.00 \times \tfrac{2}{3} = 1.333$ mol, $n^{\text{liq}} = 2.00\times\tfrac13 = 0.667$ mol. Then $0.667(0.70)+1.333(0.40) = 0.467+0.533 = 1.000$ mol — this still gives 1.00, not 1.20, signalling the composition values were mutually inconsistent with $z=0.60$ lying strictly between them under the stated ratio. Since $x^{\text{vap}}=0.40 < z=0.60 < x^{\text{liq}}=0.70$ is geometrically valid, recheck the lever assignment: amount is proportional to the opposite arm, so

n^{\text{liq}} \propto \ell^{\text{vap}} = 0.10, \qquad n^{\text{vap}} \propto \ell^{\text{liq}} = 0.20

n^{\text{liq}} = 2.00\times\frac{0.10}{0.30} = 0.667\text{ mol}, \qquad n^{\text{vap}} = 2.00\times\frac{0.20}{0.30}=1.333\text{ mol}

Component check: $0.667(0.70) + 1.333(0.40) = 0.467+0.533 = 1.000$ mol of benzene, while $z\,n_{\text{tot}} = 0.600\times 2.00=1.200$ mol. The two still disagree, which means the correct assignment must be the reverse: liquid is enriched in benzene ( $x^{\text{liq}}=0.70>z$ ) so liquid must carry the larger share of benzene, requiring $n^{\text{liq}}$ larger, i.e. $n^{\text{liq}}\propto \ell^{\text{vap}}$ was right, but let's verify with the original algebraic result of Step 5 directly: $n^{\text{liq}}(z-x^{\text{liq}}) = n^{\text{vap}}(x^{\text{vap}}-z)$ , i.e. $n^{\text{liq}}(0.60-0.70)=n^{\text{vap}}(0.40-0.60)$ , giving $-0.10\,n^{\text{liq}} = -0.20\,n^{\text{vap}}$ , so $n^{\text{liq}} = 2\,n^{\text{vap}}$ . That is the correct ratio: $n^{\text{liq}}/n^{\text{vap}} = 2$ , opposite to the earlier arithmetic slip. Corrected Step 3. With $n^{\text{liq}} = 2\,n^{\text{vap}}$ and $n^{\text{liq}}+n^{\text{vap}}=2.00$ mol:

3\,n^{\text{vap}} = 2.00\text{ mol} \Rightarrow n^{\text{vap}} = 0.667\text{ mol}, \quad n^{\text{liq}} = 1.333\text{ mol}

Final check. Benzene balance: $1.333(0.70) + 0.667(0.40) = 0.933+0.267 = 1.200$ mol, matching $z\,n_{\text{tot}} = 1.200$ mol exactly. Result: $n^{\text{liq}} = 1.33$ mol, $n^{\text{vap}} = 0.67$ mol. The system is mostly liquid, sensibly, since $z=0.60$ sits closer to $x^{\text{liq}}=0.70$ than to $x^{\text{vap}}=0.40$ .

Example 2 — Pb–Sn eutectic solidification. A tin–lead alloy of overall composition $z = 40\%$ Sn by mass is cooled to $200\,^\circ\text{C}$ , below the liquidus but above the eutectic temperature ( $183\,^\circ\text{C}$ ). At $200\,^\circ\text{C}$ the phase diagram gives: solid phase (nearly pure Pb-rich solid solution) at $x^{\text{s}} = 5\%$ Sn, and remaining liquid at $x^{\text{l}} = 55\%$ Sn (these are representative literature tie-line values near this temperature). Total sample mass $m_{\text{tot}} = 150$ g. Step 1 — lever arms.

\ell^{\text{s}} = x^{\text{l}} - z = 55 - 40 = 15 \ (\text{mass \% units})

\ell^{\text{l}} = z - x^{\text{s}} = 40 - 5 = 35

Step 2 — mass fractions of each phase. Amount of solid is proportional to the far arm $\ell^{\text{l}}$ , amount of liquid proportional to the near arm... applying the same care as above, use the algebraic form directly:

m^{\text{s}}(z - x^{\text{s}}) = m^{\text{l}}(x^{\text{l}} - z)

m^{\text{s}}(40-5) = m^{\text{l}}(55-40) \Rightarrow 35\,m^{\text{s}} = 15\,m^{\text{l}} \Rightarrow \frac{m^{\text{s}}}{m^{\text{l}}} = \frac{15}{35} = 0.4286

Step 3 — absolute masses. With $m^{\text{s}} + m^{\text{l}} = 150$ g and $m^{\text{s}} = 0.4286\,m^{\text{l}}$ :

1.4286\,m^{\text{l}} = 150 \text{ g} \Rightarrow m^{\text{l}} = 105.0 \text{ g}, \quad m^{\text{s}} = 45.0\text{ g}

Sanity check. Tin balance: $45.0\text{ g}\times0.05 + 105.0\text{ g}\times0.55 = 2.25 + 57.75 = 60.0$ g Sn. Independently $z\, m_{\text{tot}} = 0.40\times150 = 60.0$ g. Match confirmed. At $200\,^\circ\text{C}$ this alloy is 30% solid, 70% liquid by mass, and further cooling to just above $183\,^\circ\text{C}$ would shift this ratio toward more solid as the liquidus composition moves toward the eutectic point.

Limiting cases & common traps

Situation	What the lever rule gives	Why
$z \to x^\alpha$	$n^\alpha \to n_{\text{tot}}$ , $n^\beta \to 0$	Sample sits essentially on the phase boundary; only trace second phase
$z = x^\beta$	$n^\beta = n_{\text{tot}}$	Overall composition coincides exactly with one phase
$z$ at tie-line midpoint	$n^\alpha = n^\beta$	Equal lever arms, equal amounts
Using mole fraction axis with mass amounts	Wrong ratio	Must match variable type: mole-fraction axis needs mole amounts, mass-fraction axis needs masses

The chemist behind it

The lever rule as applied to phase diagrams is credited largely to the American chemist and physicist Josiah Willard Gibbs, whose On the Equilibrium of Heterogeneous Substances (1876–1878) laid the thermodynamic foundations of phase equilibrium, including the phase rule itself. The graphical tie-line method was subsequently formalized and popularized in metallurgy by Henry Le Chatelier and later systematized in ternary and binary alloy work; the rule's mechanical-balance form appears throughout the German and British metallurgical literature of the early twentieth century as engineers sought quick graphical tools for reading off phase amounts from cooling-curve diagrams, exactly the kind of diagram used in the Pb–Sn example above.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American mathematical physicist whose phase rule and chemical potential concept underpin every tie-line calculation

Ask the tutor: How does the lever rule combine with repeated tie-line readings to explain how fractional distillation separates a mixture into nearly pure components?

Does the lever rule work for three or more components?

Not directly in the same two-segment form. In a ternary phase diagram, tie-lines still connect two coexisting phase compositions, and the same mass-balance argument applies, but the geometric "lever" becomes a statement about points on a line within a triangular diagram, with the amounts fixed by the same distance ratio along that specific tie-line. Full ternary systems typically need additional information, such as a third conjugate phase in three-phase equilibria, handled by ternary-specific constructions like the "tie-triangle" rule.

Does the lever rule require the tie-line to be horizontal?

No—tie-lines are horizontal specifically in temperature–composition or pressure–composition diagrams where temperature or pressure is held fixed across the two-phase region. In other diagram types, such as composition–composition diagrams for liquid–liquid equilibrium at fixed temperature, the tie-line can be drawn at any angle; the lever rule still applies as a distance ratio along whatever straight line connects the two phase compositions.

Can the lever rule be used with mass fractions instead of mole fractions?

Yes, provided you are consistent. If the phase diagram's horizontal axis is in mass fraction (or mass percent), then the lever rule directly gives mass ratios of the two phases, as shown in the tin–lead example. If the axis is mole fraction, the rule gives mole ratios. Mixing the two, for instance reading off mole fractions but wanting mass amounts, requires an extra conversion using molar masses before the lever rule can be applied correctly.

What happens to the lever rule exactly at a eutectic point or an azeotrope?

At a true eutectic or azeotropic composition, the liquid and one or more solid phases (or liquid and vapour, for an azeotrope) share the same composition at that special point, so the "tie-line" collapses to zero length in one sense, or the system solidifies/vaporizes at constant composition without any lever calculation needed. Just off that special point, ordinary tie-lines and the lever rule resume as normal, as illustrated by cooling the Pb–Sn alloy toward, but not exactly to, the eutectic temperature.