Lever Rule and Material Balance Workshop

Introduction

Phase diagrams answer two different questions: what is in each phase, and how much of each phase exists. Tie-line endpoints answer the first; material balance answers the second. The lever rule is a geometric shorthand for that balance. This workshop makes the algebra explicit so that the rule can be used correctly on vapour–liquid, liquid–liquid, solid–liquid and ternary diagrams.

Core explanation

Let a binary sample have total amount n and component-B fraction z B. At one equilibrium T,p, phases α and β have amounts n α,n β and endpoint compositions x B^α,x B^β. Define f α = n α/n and f β = n β/n, so f α+f β = 1. Conservation of B gives z B = f αx B^α + f βx B^β. Solving yields f β = (z B − x B^α)/(x B^β − x B^α) and f α = (x B^β − z B)/(x B^β − x B^α). Each phase fraction is the length of the opposite lever arm divided by the full tie-line length.

The overall z B must lie between the two endpoint compositions for nonnegative fractions. If it lies outside, the chosen tie line cannot describe that overall sample as a two-phase mixture, or the data have been misread. Endpoint compositions are fixed by equilibrium conditions at the stated T,p; changing the overall z within the tie line changes phase amounts, not the endpoint compositions. At a critical point where endpoints merge, the ordinary ratio becomes 0/0 and the distinction between two phases vanishes.

For a ternary system, use vector balance z i = f αx i^α + f βx i^β for i = A,B,C. One independent component balance plus f α+f β = 1 can determine the fractions if the overall point truly lies on the specified tie line; the other balances are consistency checks. A two-phase overall point must lie on the straight segment joining the two endpoint compositions in the triangular diagram.

The composition basis matters. If a binary metal diagram is plotted in mass fraction, lever distances give mass fractions of phases. If a vapour–liquid diagram uses mole fractions, they give mole fractions of total amount in each phase. Converting between bases requires molar masses or densities as appropriate. Do not mix an overall mass fraction with endpoint mole fractions in one balance equation.

Step-by-step reasoning

Read T,p and the composition basis. Identify the two coexistence endpoints before looking at the overall composition. Write f α+f β = 1 and one component material balance. Solve algebraically, then check that fractions lie between zero and one and sum to one. Use a second component or complementary A balance as a check, especially in ternary problems.

Visual explanation

Draw a horizontal binary tie line with α endpoint at 0.20, overall point at 0.35 and β endpoint at 0.80. Label the 0.15 segment from α to overall as controlling β's fraction and the 0.45 segment from overall to β as controlling α's fraction. Under it draw a ternary tie segment with the same opposite-arm rule inside a triangle.

Real-world analogy

A balanced seesaw has a lighter weight farther from the pivot and a heavier weight closer to it. In a tie line, a small amount of a far-composition phase can balance a larger amount of a near-composition phase in the overall average. The analogy visualises the opposite-arm ratios; conservation equations supply the exact result.

Real-world example

In a distillation flash drum, feed overall composition z splits into liquid x and vapour y at specified T,p. The phase-equilibrium model supplies x and y; a material balance supplies how much vapour forms. A diagram without the feed z cannot determine the vapour fraction, and a feed z without equilibrium endpoints cannot determine both phase compositions.

Why?

The overall composition is a weighted average of the two phase compositions. A point nearer one endpoint requires more of that endpoint phase to keep the average nearby. Algebraic conservation produces the lever ratio; the phase diagram is needed only to provide the equilibrium endpoint values.

Common misconception

The phase fraction is not proportional to the distance from the overall point to that same phase's endpoint; it uses the opposite distance. Another mistake is to apply the rule across a one-phase region or between endpoints at different temperatures. A tie line connects phases at the same equilibrium T,p.

Worked example

At one T,p, liquid has x B = 0.20, vapour has y B = 0.80 and feed has z B = 0.35 on a mole basis. Let f V be vapour mole fraction. Balance gives 0.35 = (1−f V)0.20 + f V0.80 = 0.20 + 0.60f V, so f V = 0.25 and liquid fraction is 0.75. Check: 0.75(0.20) + 0.25(0.80) = 0.15 + 0.20 = 0.35. The feed is closer to liquid composition, so more liquid is present.

Quick check

1. If overall composition equals the α endpoint of a two-phase tie line, what β fraction does the lever equation give? Answer: f β = (z − x α)/(x β − x α) = 0. The sample is entirely α at that boundary, apart from limiting coexistence interpretation.

Exam focus

Write the conservation equation before using geometric distances. Identify mole versus mass basis and do not swap x, y and z labels. Check 0 ≤ f ≤ 1 and that weighted endpoints reproduce z. For ternary diagrams, verify all three component balances rather than trusting one incorrectly drawn tie line.

Advanced insight

The lever rule gives equilibrium phase amounts but not kinetic path or microstructure. In a solidifying alloy, diffusion limits can cause local compositions to depart from equilibrium tie-line endpoints; a quenched sample may retain gradients. A phase diagram is a thermodynamic map, while material balances and transport determine actual outcomes.

Summary

The lever rule is component conservation for two coexisting phases: z = f αx α + f βx β. Endpoint compositions come from a tie line, and opposite segment lengths give phase fractions. The rule extends to ternary tie segments and must use one consistent mass or mole basis.

Practice questions

1. Binary endpoints are x α = 0.10 and x β = 0.70, with z = 0.25. Find f β. Answer: f β = (0.25 − 0.10)/(0.70 − 0.10) = 0.15/0.60 = 0.25, so f α = 0.75. 2. A claimed tie line has endpoints 0.20 and 0.60, but an overall sample has z = 0.80. Can that tie line describe its two-phase composition? Answer: No. A weighted average of 0.20 and 0.60 cannot be 0.80 with nonnegative phase fractions. The sample is outside that tie-line segment or the endpoints are wrong. 3. What extra information is required to turn mass fractions from an alloy diagram into mole fractions of phases? Answer: Molar masses and phase compositions are needed to convert masses to mole amounts. One cannot use mass-basis lever fractions as mole fractions without conversion.