SKH Thermodynamics Labs

Lab 10 · Solid–liquid equilibrium with pure solids

When does a liquid mixture start to freeze?

A liquid mixture of two synthetic compounds is cooled at constant pressure. A melts at 350 K (Δhfus = 12 kJ mol⁻¹) and B at 400 K (Δhfus = 15 kJ mol⁻¹). Each crystallizes as a pure solid, and the liquid is ideal. The feed has xA = 0.50 and is cooled to 340 K.

Find which solid appears, how much, and the composition of the remaining liquid. Then locate the temperature below which no liquid remains.

2027 prototype
Review the student workflow

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Core: predict, calculate, read the phase diagram and explain the result. Steps 1–5.

Graduate: the core steps plus a cooling path, a liquid-model comparison and a fusion-property sensitivity study in Step 4.

Graduate route. Complete Steps 1–3 for the starting case. In Step 4, follow the feed as it cools, change the liquid nonideality, and test how uncertain fusion data move the eutectic. Go to the graduate investigation.

Choose a caseAll cases use the same pure-solid data.

Reading a eutectic phase diagram

Ideal liquid, pure solids, constant pressure
Temperature–composition diagram for the starting case Two freezing curves fall from the melting points of pure B at 400 K and pure A at 350 K and meet at the eutectic, 319.8 K and x_A = 0.677. Above the curves the mixture is liquid. Below the left curve, liquid coexists with solid B; below the right curve, with solid A. Below the eutectic line, two solids coexist. The feed x_A = 0.50 first forms crystals of B at 346.7 K. At 340 K the tie line joins pure solid B with a liquid of x_A = 0.549. LiquidLiquid + solid BL + solid ASolid A + solid B first crystals · 346.7 KE · 319.8 K tie line at 340 Kliquid xA = 0.549 00.51300320340360380400pure B meltspure A melts Mole fraction of A, xA T / K

Model-generated for the starting case. The dotted vertical line is the feed. On it, the open circle marks the first crystals and the filled dot the state at 340 K; the tie line runs from pure solid B to the saturated liquid.

The solubility equation for a pure solid ln of x_i gamma_i equals minus delta h fusion over R times one over T minus one over T melting. The left side belongs to the liquid. The right side contains only properties of pure solid i. For B at 340 K the right side is minus 0.796, so an ideal liquid holds x_B = 0.451. liquid sidepure solid i only ln(xᵢγᵢ)=−(Δhᵢ/R)(1/T − 1/Tₘ,ᵢ) Assumes Δcp ≈ 0 and no solid solution. B at 340 K: −(15 000/8.314)(1/340 − 1/400) = −0.796ideal liquid: xB = e−0.796 = 0.451, so xA = 0.549

The same equation, written for A, gives the right-hand curve. The eutectic is where both hold at once.

At fixed pressure, a binary liquid saturated with one pure solid has one degree of freedom, so temperature alone fixes the liquid composition. The feed composition sets only the amounts.

1. Set the feed and the solids

Basis · 1.00 mol feed
K
mol/mol
Pure-solid properties

A

B

Synthetic teaching values, not real-compound data.

Optional. Your choice does not change the calculation.

Add a calculated result to the comparison before changing a setting.

Model and validity

Each solid is pure (no solid solution, polymorph or compound). The solubility equation neglects the heat-capacity difference between liquid and solid. The liquid model must stay single-phase; a liquid that splits is refused (use Lab 04). Temperatures are limited to 250–450 K. The global state is chosen by comparing the Gibbs energy of every candidate phase set.

Open the current Lab 10 for your own compound names and sources, and for sublimation of a volatile solid in a closed vessel.

2. Read the result

Calculate to see the phases, their amounts and the liquid composition.

4. Check your understanding

Write first, then compare

Use Steps 1–3 to answer each question in your own words. Then open the explanation and compare. Your answers are saved with your study.

A. Calculate the liquid composition on paper

For the starting case, use the solubility equation for B to find xA of the liquid. Then find the amount of solid B from a balance on A.

Compare with the explanation

ln xB = −(15 000/8.314)(1/340 − 1/400) = −0.796, so xB = 0.451 and xA = 0.549. All of A stays in the liquid, so the liquid amount is L = 0.50/0.549 = 0.911 mol and the solid B is 0.089 mol.

The liquid composition came from T alone. A feed of 0.45 or 0.30 at 340 K gives the same liquid; only the amount of solid changes.

B. Which solid appears first, and why?

Compare the starting case with the A-rich feed. Read the temperature at which the first crystals form.

Compare with the explanation

At xA = 0.50, B crystallizes first at 346.7 K, below its own melting point of 400 K. At xA = 0.90, A crystallizes first at 341.3 K, below 350 K.

Dissolving another component lowers each freezing point. The solid that appears first depends on the feed composition, not only on which pure compound melts higher.

C. What happens at the eutectic?

Load “Cooler · 320 K”, then “Below the eutectic · 300 K”. Compare the amount of liquid.

Compare with the explanation

At 320 K, 0.26 mol of solid B coexists with liquid of xA = 0.676. The eutectic lies at 319.8 K and xA = 0.677. Below it, the feed is entirely solid: 0.5 mol A and 0.5 mol B.

At the eutectic, liquid, solid A and solid B coexist. With three phases at fixed pressure there are no degrees of freedom: T and xA are fixed, but the amounts are not, as for the three-phase state in Lab 06. They depend on how much heat has been removed.

4. Investigate the freezing behavior

Master's core work

Calculate a state in Step 1 before running an investigation.

A. Follow the feed as it cools

  1. Tabulate the phases, liquid composition and solid amount from 400 K to 300 K.
  2. Locate the first-crystal temperature and the eutectic.
  3. Explain why the liquid composition moves along the freezing curve while the feed composition is unchanged.

B. Change the liquid nonideality

  1. Keep the pure-solid data and compare Margules A = −1, 0, 1 and 1.5.
  2. Compare the eutectic, the first-crystal temperature and the solid at the chosen T.
  3. Explain the direction of each change using xᵢγᵢ = constant along a freezing curve.

C. How sensitive is the result to fusion data?

  1. Change Δhfus by ±10% and Tₘ of B by ±5 K, one at a time.
  2. Compare the eutectic, the first-crystal temperature and the solid amount at the chosen T.
  3. Identify which property controls the crystallization yield of B and which affects only the eutectic.

D. Defend a limited conclusion

The record check only confirms that the sections and calculations are present. Scientific reasoning still needs instructor review.

5. Compare and save

  1. Calculate and save the four named cases.
  2. Check that the liquid composition at 340 K is the same for zA = 0.50 and 0.30.
  3. Answer questions A–C in Step 4 using the saved comparison.
  4. Download the complete study, reopen it, and reproduce a saved case.
  1. Calculate and save the four named cases.
  2. Complete investigations A–C and write the graduate claims.
  3. Download the complete study, reopen it, and reproduce a saved case and its investigation record.

Your calculated cases will appear here after you add them.

Starting case loaded. Press Calculate the equilibrium.

Your inputs, comparisons and explanations are kept in this open page. Download one study file to keep the complete record and reopen it later. Results are recalculated on restore.