Engineering & Technology · Reaction Engineering, Separations & Transport
Phase equilibria and separation driving forces
A liquid and its vapour can have different compositions while they are in equilibrium. Separation can exploit that difference. Equilibrium describes a limiting relationship between phases; it does not say that a real contact reaches it instantly.
Use Process Balances & Design for amount, flow, energy and boundary accounts. Review material structure and properties. Chemistry’s reaction rates and equilibrium explain the molecular and thermodynamic ideas further. Exponentials are evaluated explicitly; each model states its scope before calculation.
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Interpret an equilibrium composition relationship and combine it with an amount balance for a simple phase split.
Equal tendency to transfer need not mean equal composition
At phase equilibrium there is no net tendency for a component to transfer between the phases under the specified conditions. Molecules can still exchange. The corresponding thermodynamic statement concerns chemical potential, rather than equality of gas and liquid concentration. Temperature, pressure, composition and mixture identity determine the relationship.
Let x be the mole fraction of component A in a binary liquid, and y its mole fraction in the equilibrium vapour. For an idealised constant-relative-volatility model, y=α x/[1+(α−1)x]. Here α compares the vapour-to-liquid partition tendencies of A and B. We supply α=2 for an imaginary mixture over this exercise; it is not property data for a real solvent pair.
A composition gap enables enrichment
With α=2, the curve is y=2x/(1+x). At x=0.40, y≈0.5714: the vapour is enriched in A but contains substantial B. The diagonal y=x represents no enrichment. At pure-component endpoints the curve and diagonal meet. When α=1 throughout, this model gives no composition difference to exploit by that phase split.
The figure also marks an original flash balance. One mole of feed with overall A fraction z=0.50 separates into 0.50 mol liquid and 0.50 mol vapour. Total and A balances give 1.00=0.50+0.50 and 0.50=0.50x+0.50y. The operating line y=1−x intersects the supplied equilibrium curve at x≈0.4142 and y≈0.5858.
Equilibrium and equipment have separate constraints
A flash is one contact followed by separation of the phases. An equilibrium-stage calculation assumes the departing phases satisfy the relationship; an actual device needs sufficient contact and disengagement. The energy balance and phase-equilibrium data must also support the chosen vapour fraction. Supplying half vapour here teaches composition accounting, not how to set a heater or pressure.
Real mixtures can be nonideal. Their volatility relationship may change with composition or create an azeotrope, where vapour and liquid compositions coincide at a non-pure mixture under given conditions. Repeated ordinary distillation cannot be assumed to cross that limit. Use appropriate measured or validated thermodynamic models, identify their range and retain uncertainty before promising purity or recovery.
Close the component balance
Substitute y=1−x into y=2x/(1+x). Rearrangement gives x²+2x−1=0. Only x=√2−1≈0.414214 lies between 0 and 1; the other root is not a physical mole fraction. Then y=2−√2≈0.585786.
Liquid contains 0.50×0.414214≈0.207107 mol A and vapour contains 0.50×0.585786≈0.292893 mol A. Their sum is 0.500000 mol, matching the feed. The vapour’s A recovery is 0.292893/0.50≈58.58%, while its A purity is 58.58 mol%. The numerical equality follows this particular feed and split; recovery and purity are different ratios.
Try the reasoning
Must equilibrium liquid and vapour have equal mole fractions?
Show the answer and reasoning
No. They can partition differently while the component chemical potentials satisfy equilibrium.
Does a point on the equilibrium curve determine the amounts of both phases?
Show the answer and reasoning
No. Feed conditions, overall balances and energy information are also needed.
Why reject the second quadratic root?
Show the answer and reasoning
A mole fraction must lie between 0 and 1; an algebraic root outside that range is not a physical phase composition.
What to carry forward
Use equilibrium to relate phase compositions, then use balances and energy evidence to determine a feasible split.
References
Teaching resources and research supporting this lesson’s explanations and examples.