The Phase Rule

The Phase Rule

• It's a physical law governing equilibrium conditions in a system.
• It is used in chemical engineering, including petroleum refining, chemical production, ceramics, glass, cement, and metallurgy.

Terminology

• Key terms are essential for understanding and applying the Phase Rule.

The System

• It is a defined spatial region containing one or more chemical species.
• It has a definite boundary separating it from the surroundings.
• The system's nature depends on the chemical species it contains.
• Example: sand and water in a beaker; the system can either include both or define them separately.
• The contents of the system determines what becomes the surroundings.

Equilibrium

• This indicates a state of no change or no tendency for change.
• For example, in esterification, the reaction rate decreases until constant concentrations are achieved.
• Equilibrium concentrations remain constant if parameters like temperature and dilution are unchanged.
• Adding more alcohol to the mixture can shift the reaction toward more ester formation until a new equilibrium is established.

The Phase

• A phase is a homogeneous portion of a system with a fixed boundary and can be physically separated.
• A sand and sugar mix appears as distinct grains under a microscope, creating two phases.
• These phases are separable by dissolving sugar in water and filtering.
• Sugar in water consists of one phase only since one cannot distinguish sugar particles, even under a microscope.
• A mixture of gases always exists as one phase.

The Component

• It is the minimum number of chemical species needed to define the composition of all phases in the system.
• If system species do not react, the number of components equals the number of chemical species.
• A sugar and sand mixture has two components.
• Sugar solution has two components (water and sugar).
• Air has four components: oxygen, nitrogen, water vapor, and carbon dioxide.
• If species react, the determination of components is different.
• With calcium carbonate decomposition into calcium oxide and carbon dioxide, knowing two species exist implies the third's presence.
• In such a case, the system has 2 components, not 3.
• With 'R' independent reactions among species, the number of components = Number of species - R.

Degrees of Freedom

• It is the minimum number of independent variables (e.g., temperature, pressure, concentration) that must be fixed to define the system's state completely.
• Liquid water has an infinite number of temp and pressure combinations.
• For example, water is in a liquid state at 25°C and 1 atm, or at 60°C and 2 atm.
• Liquid water has 2 degrees of freedom.
• Boiling water occurs at a specific temperature for each pressure; fixing pressure fixes the boiling temperature, therefore, it has 1 degree of freedom.

The Phase Rule Statement

• J.W. Gibbs developed the rule in 1875; it's fundamental for studying equilibrium in complex systems.
• The rule relates degrees of freedom (F) to the number of phases (P) and components (C): F = C - P + 2.
• The rule's proof requires thermodynamics.
• Pressure, temperature, or composition or any combination thereof can be held constant.
• With 'R' restrictions, the rule is: F = C - P + 2 - R.
• In isobaric systems (constant pressure), the rule becomes: F = C - P + 1.

Solved Examples of Finding Degrees of Freedom

• A boiling mixture of alcohol and water: P = 2 (liquid + vapor), C = 2, so F = 2.
• A boiling mixture of oil and water: P = 3 (two liquids + vapor), C = 2, so F = 1.
• Calcium hydroxide decomposition in air: C = 2, P = 3, and R = 1, resulting in F = 0. This is an invariant situation at constant temperature.
• A mixture of NaCl, KCl, and LiCl in equilibrium with their melt in air: C = 3, P = 4, and R = 1, leading to F = 0.
• HI decomposition into iodine and hydrogen: C = 2, P = 1, so F = 3.
• Melting of a cobalt and nickel alloy in air: P = 2, C = 2, R = 1, giving F = 1.

Pig Iron Production Example

• Six solid phases, one gaseous phase, and one liquid phase are present, therefore P=8.
• There are 10 chemical species with 6 independent equations.
• C = 10-6=4
• F=4-8+2 = - 2
• These species cannot co-exist.

One-Component Systems

• For one component, C = 1, so F = 3 – P.
• Minimum phases = 1 making the maximum degrees of freedom = 2, so these are temperature and pressure.
• Equilibria are described using pressure-temperature diagrams.
• Phase transitions classify by volume change: Type I (no change e.g. melting) and Type II (appreciable change e.g. vaporization).

The Clausius-Clapeyron Equation

• This equation describes vapor pressure dependence on temperature during phase transition: dp/dT = ΔHv / (T.ΔV)
• T = temperature (K), ΔV = molar volume change (m3.mol-1), ΔHv = heat effect (J.mol-1), Δp = vapor pressure increase (Pa).
• The slope of the p-T diagram reflects the equation's RHS.

Class I Transitions

• ΔV ≈ 0, slope = ∞.
• This means the p-T line is vertical, and transition temperature does not change with pressure.
• Most solids increase in volume on melting. Solid iron's density solid iron is 7.87 g.cm-³, with molten irons density at 6.98 g.cm-³.
• This means that their specific volumes are: For Fe(s): 0.127 cm³.g-¹ and for Fe(l) = 0.143 cm/-¹g. Since the atomic mass of Fe = 55.84, their molar volumes are: V₁ = 7.09 cm³.mol-¹ and V₂ = 7.98 cm³.mol-¹.
• Where ∆V = 0.89×10-6 m³.mol-1 > 0 resulting in a slightly positive slope of the Pº/T melting line.
• Water and bismuth contract upon melting implying a negative delta V.

Class II Transitions

• This is where volume change is significant, as with vaporization.
• Assuming ideal gas behavior, molar vapor volume relates to temperature, where V = RT/p
• With negligible liquid volume, the Clapeyron equation simplifies to: dp/dT = p.ΔHv / RT2.
• Assuming constant ∆Ηv, integrating yields: ln p = -ΔHv / RT + A.
• This is the Clausius-Clapeyron equation.
• Vapor pressure increases exponentially with temperature.
• A plot of ln p against 1/T yields a straight line with a slope of -ΔHv/R.

Example Calculation: Latent Heat of Vaporization of Ethanol

• The data was obtained by following the variation of vapor pressure of ethanol with temperature over the range 10 – 60°C.
• It is used to measure average latent heat.
• From the properties of logarithms any pressure units may be implemented.

Prediction of Boiling Point Change With Pressure

• Integrating the equation between two temperature limits T1 and T2 gives: ln(P2/P1) = (ΔHv/R) * (1/T1 - 1/T2).
• ΔHv must be constant, or its temp relation must be included in the integral.

Example: Water Boiling Point Estimation

• Estimate the boiling point of water at a pressure of 10 atm if the latent heat of vaporization = 540 cal.g-¹.
• First, convert the latent heat to proper units:
• ΔΗ, = 540×18×4.18 = 40629.6 J.mol-1
• Hence ∆HR = 4886.9 J.mol-¹K-1
• The boiling temperature at 1 atm is 373 K.
• Applying equation (1.4):
• In10/1 = 4886.9×(1/373-1/T2)
• From which: T2 = 452.5 K = 179.5°C

Empirical Equations to Predict Vapor Pressure Dependence

• Antoine equation: ln p° = A - B / (T + C), where A, B, and C are constants.
• Differentiating gives ΔHv/R²T = B / (T + C)².

The Water Diagram

• Water exists in solid, liquid, and gaseous forms.
• The BC curve shows the vapor pressure and freezing point relation, called the freezing curve.
• If going from solid to liquid, water's slope is negative due to water's volume decrease.
• The BD curve shows the relation between vapor pressure and boiling temperature called the vaporization curve.
• Point D is water's critical point at 374°C.
• Under atmospheric pressure (101.3 kPa), a line crosses the freezing curve at 0°C and the vaporization line at 100°C.
• The curve AB shows the transition from solid to vapor at low pressures called the sublimation curve.
• Inside one-phase regions, pressure and temperature can vary independently.
• Along any curve, two phases exist, and one variable fixes the other.
• The three curves meet at the triple point, where all three phases are at equilibrium: 0.01°C and 611.73 Pa.
• Slight temp or pressure changes shift equilibrium.
• Water can sustain its liquid state below freezing as a metastable state which requires pure water because any solid impurity will cause rapid crystallization of ice.

The Sulfur Diagram

• Sulfur exists in two allotropic forms: rhombic (low temperature) and monoclinic (high temperature).
• At normal pressure, rhombic sulfur transforms to monoclinic at 95.4°C, which melts at 119.2°C and transitions to vapor at 444.6°C.
• At pressures above 154°C and 103 atm, rhombic sulfur directly melts, skipping the monoclinic state.

Triple Points

• At very low pressures, sublimation happens where allotropic forms occur directly without passing through the liquid state.
• Below 10-5 atm, the rhombic form is sublimated directly.
• Pressures between 10-5 and 10-4 atm see the rhombic form transform first to monoclinic, then sublimate to vapor.
• At 10-4 to 103 atm transitions are the same as atmospheric pressures.
• The triple point E exists because the volume change from rhombic to monoclinic is less than the melting of the monoclinic form.
• Extremely slow heating of rhombic sulfur forms a triple point where rhombic sulfur, liquid melt, and vapor are in equilibrium (Point H).

Two-Component (Binary) Systems

• Components, C = 2, so F = 4 – P
• The minimum number of phases = 1, then degrees of freedom = 3
• This entails having three coordinates to fully represent a two-component system: Composition of one of the two components, temperature and pressure.
• In solid-liquid systems, pressure is usually kept constant at one atmosphere, with F = 3 – P. Therefore, use composition and temperature.
• In liquid-vapor systems, pressure is important. You can fix pressure or temperature and study equilibrium on temperature-composition or pressure-composition diagrams.

Liquid-Vapor Systems - Isothermal Diagrams

• Temperature is constant, and equilibrium is a relation between pressure and composition.
• Phase rule is F = 3 – P.
• An Isothermal diagram plots total pressure against molar composition of the more volatile component.
• Draw the relation between T and xĀ.
• If pĀ and pв are partial pressures of A and B, then the total pressure is: P = pĀ + pв.
• The relation between component partial pressure and liquid state molar composition requires special thermodynamic knowledge.

Ideal Solutions

• These solutions obey Raoult's law, with direct proportionality between partial pressure and mole fraction in the liquid phase: pĀ = pĀ.*xA and pв = pв.*xB
• Resulting in: P = pĀ.xA + pв. хв
• For ideal solutions to form, conditions must be fulfilled: complete solubility, similar chemical composition, and absence of chemical reaction.
• Ideal solutions do not have a heat of mixing. Also, there is no change volume following mixing.
• Since xB = 1 – xA, the equation rearranges to: P = (pĀ - pв). xA + pв
• This equation represents a linear relation between P and xA.
• This linear relation between pressure and composition of A in liquid phase is called the liquidus curve.
• Gradual lowering of pressure will cause boiling to start if considering any composition xA in the liquid phase.
• Let yA represent the molar vapor composition that will be at equilibrium with the liquid (xA). Usually, yA > xA. This yields the equilibrium relation.

Relating to Raoult's and Dalton's Laws

• According to Raoult's law: P = (pĀ - pв).xA + p
• Dalton's law states: yA = pA/P
• The result: yA = pĀxA / ((pĀ - pв) xA + pв)
• By defining of relative volatility of A with respect to B as α = (yA/xA) / (yB/xB)

Simplification

• For the ideal solution simplifies to: α = pĀ / pв , resulting in: yA = α.xA / ((1-α)xA +1)
• As eq is nonlinear, the relation between total pressure P and yA is nonlinear as well referred to as the vaporous curve, whereas the example illustrates drawing both liquidus and vaporous curves.

Prediction and Application

• The liquidus curve is easily obtained by a simple joining of the two vapor pressures with a straight line, with the custom to deduce it from the equilibrium curve. The value of α = 750 / 400 = 1.875.
• The values are the data points for total pressure against values of xA (A = methanol) calculated from equation (3.4) and for yA calculated from equation (3.9).

Positive Deviation

• In other words positive deviation is, partial pressure of A is always higher than that calculated for an ideal solution.
• With this is called called a positive deviation from Raoult's law.
• In that figure the dotted straight line represents ideal behavior, and the actual curve is above that line.
• Also, the sloe of the initial portion of the curve represents Henry's law constant. In that case: HA > pĀ.

Negative Deviation

• This means that if the partial pressure of A is always lower than that calculated for an ideal solution, then this is called a negative deviation from Raoult's law.
• In that case: HA < p. This deviation's presence results in a curved liquidus rather than a straight line.
• Both liquidus and vaporous lines in Fig (3.1) will be curved.
• Significant ideality leads to maximum or minimum pressure values, referred to as azeotropes illustrated in Fig (3.6) and (3.67)

Application

• Elevation in boiling point: Assuming the boiling point of pure solvent A is To at some pressure value po.
• If a quantity of a non-volatile material is added where the pressure will drop to pxA = pĀ (1 - xB) accompanied by an elevation = ∆T.
• (1- xB) / P = ΔHv / R * (1/To - 1/To + ΔT) or ∆T / To (To + ∆T)
• Since ∆T << To, we can write approximately: To (To + ∆T) ≈ T², which means that the equation = xB = ΔHv∆T / RT

Isobaric Diagrams

• If pressure is constant, then equilibrium shows the mole fraction effect of volatile component (xA) on boiling temperature.
• Binary solutions generally do not boil at constant temperature since F = 1.
• A typical isobaric temperature-composition diagram to show and better visually realize the various factors at play.
• If a vapor of composition corresponding to point C (za) is cooled under constant pressure, as the temperature reaches a value Ta, (point D), with the first drop of liquid begins to show. This temperature is the dew point.
• On further cooling more liquid will form at the expense of vapor, as the temperature reaches a value (Te /point E) these there phases are at equilibrium, as the liquid phase has a composition that equals something, and the vapor phase is now something else too.
• As the temperature (Tb) is hit at point F, where the vapor has now disappeared, and only liquid has remained. This is the bubble point, and shows that the original composition is bubble and has remained static.

The Lever Arm Principle

• This principle, helps to find the ratio of liquid and vapor mass coexisting at equilibrium.
• It is applicable if written not as mol fraction, but mass fraction.
• With an F mass and composition of point zA , you have phase L = L , and phase mass of V = V
• Overall and A component mass balances are the F = L+V liquid mass of F with composition of A = LxA + VyA
• Eliminating, you get two components: L/V =(ZA-XA)/(YA-ZA)
• L/V = EV/EL
• There are some useful principles like finding out what the liquid % is = (EV/(EL+EV)) *100

Aetrope Formation

• Systems of positive Raoult's law form have strong isothermal diagrams. While temperature diagrams will have a minimum boiling point. It has water and ethenol as the baseline.
• Negavtive Ruoults law have a maximum boiling point.
• If over it bubble the mix of the point (mix bubble) formed vapor contains (A) more than liquid or y > x
• If greater: the mix over azero: this contain less A than liquid ( y<x).
• Liquid retains comp it it transform into vapor over degree freedom etc
• Point where boiling is at constant temp.

Miscibility Gap Formation

• Components can be not soluble ( totally).
• For mix ( one is the phenol - wather mixture (temperately lowe).
• A layer of water is rich and one is phenol.
• Mass ratio (temp calcul lever arm ) close the composition over the temperature when point coincides
• Consolete temperature one phase only.
• Minimal boiling often assoc for azeoto formed
• comp in A that it is the 2 layers and represents X1 and x2.

Two-component systems – Isobaric Diagrams

• At two-component system under constant pressure shows mole fraction on volatility component on temp boiling/
• The general binary solution: isn't constant with phase rule eqaul constant.
• Diagram of comp isobar ( liquid + vapor).
• If it coold the composition C at Za below
• First point of liquid with Tb is called point Dew/
• Then the A the follow will made it to the exp:

Solid Liquid System

• This system under atmosph, under atmosph pressure show the formulation:F=3-P+2-1=4-P
• In a system that is at mini then has degree that is almost those for sys .
• Eventually ( the ternar form shows.

Systems with complete solid solubitiy

This is this under the conditions of comp solubility and the diagram has a ternat form (Fig 4.

Systems with comp

• Total insoubliity three com don tr react the result has three type comp , melt a at of resp 1080 1160 type comp

Allotropic transitions in two component systems

• Transition allop ( A ) tran then occ isotherm A) .3 .9) if con dis is situ diff n is form