9.2 Phase Rule, Phase Diagrams of One- and Two-Component Systems, and Clausius-Clapeyron Equation

Phase Rule, Phase Diagrams of One- and Two-Component Systems, and Clausius-Clapeyron Equation

Phase equilibria govern the thermodynamic conditions under which distinct states of matter—solids, liquids, and vapors—coexist in stable thermodynamic equilibrium. On the PRC Chemist Licensure Examination, mastery of phase equilibria requires fluency with the Gibbs Phase Rule, the Clapeyron and Clausius-Clapeyron equations, anomalous single-component behavior ($H_2O$ vs. $CO_2$), and binary phase diagrams involving distillation, azeotropes, and eutectic crystallization.

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1. Thermodynamic Equilibrium Condition and the Gibbs Phase Rule

A phase ($\alpha, \beta, \pi$) is a form of matter that is uniform throughout in chemical composition and physical state, bounded by a distinct interface. When multiple phases of a multicomponent system are in thermal, mechanical, and diffusional equilibrium, three conditions must be satisfied simultaneously:

1. Thermal equilibrium: $T^\alpha = T^\beta = \dots = T^\pi$
2. Mechanical equilibrium: $P^\alpha = P^\beta = \dots = P^\pi$
3. Chemical / Diffusional equilibrium: For every chemical component $i$ present in the system, its chemical potential must be identical in all coexisting phases: $\mu_i^\alpha = \mu_i^\beta = \dots = \mu_i^\pi \quad (i = 1, 2, \dots, C)$

If $\mu_i^\alpha > \mu_i^\beta$, component $i$ spontaneously migrates from phase $\alpha$ to phase $\beta$ until the chemical potentials equalize.

Derivation of the Gibbs Phase Rule

The variance or number of degrees of freedom ($F$) is the number of intensive variables (such as temperature, pressure, and mole fractions) that can be varied independently without altering the number of phases in equilibrium.

• Total intensive variables:

  • System temperature and pressure: $2$ variables ($T, P$).
  • Composition of each phase: In each of the $P$ phases, the composition of $C$ components requires $(C - 1)$ independent mole fractions (since $\sum x_i = 1$).
  • Total variables $= P(C - 1) + 2$.

• Equilibrium constraints:

  • For each component $i$, phase equilibrium imposes the equality $\mu_i^\alpha = \mu_i^\beta = \dots = \mu_i^\pi$, which represents $(P - 1)$ independent algebraic equations.
  • For $C$ components, total constraints $= C(P - 1)$.

Subtracting constraints from variables yields the Gibbs Phase Rule:

$F = [P(C - 1) + 2] - C(P - 1) = PC - P + 2 - CP + C = C - P + 2$

$F = C - P + 2$

For condensed phase systems (solids and liquids) where the vapor phase is absent and experiments are conducted under constant ambient atmospheric pressure ($P = 1\text{ atm}$), pressure is no longer an independent variable, yielding the reduced or condensed phase rule:

$F' = C - P + 1$

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2. One-Component Phase Diagrams ($C = 1$)

For a single-component system, $F = 1 - P + 2 = 3 - P$:

• Single phase ($P = 1$): $F = 2$ (bivariant). Both $T$ and $P$ can be independently varied within a 2D region.
• Two phases in coexistence ($P = 2$): $F = 1$ (univariant). Coexistence occurs along boundary lines; choosing $T$ uniquely fixes $P$.
• Three phases in coexistence ($P = 3$): $F = 0$ (invariant). Coexistence is restricted to a unique, invariant point: the triple point.

Comparative Analysis: Water ($H_2O$) vs. Carbon Dioxide ($CO_2$)

The Water Anomaly: In ice Ih, water molecules form an open, tetrahedral hydrogen-bonded framework containing hollow hexagonal channels. Upon melting to liquid water, this framework partially collapses, allowing molecules to pack more closely together. Consequently, liquid water is denser than ice at $0^\circ\text{C}$ (maximum density occurs at $3.98^\circ\text{C}$), giving $\Delta V_{fus} = V_{liquid} - V_{solid} < 0$. By the Clapeyron equation, this negative volume change produces a negative fusion curve slope.

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3. The Clapeyron and Clausius-Clapeyron Equations

The Clapeyron Equation

Along any phase coexistence line separating phase $\alpha$ and phase $\beta$, the chemical potentials must remain equal during an infinitesimal displacement $(dT, dP)$:

$d\mu^\alpha = d\mu^\beta$

Recalling the fundamental relation $d\mu = -S_m dT + V_m dP$, where $S_m$ and $V_m$ are molar entropy and molar volume:

$-S_m^\alpha dT + V_m^\alpha dP = -S_m^\beta dT + V_m^\beta dP$

Rearranging terms:

$(V_m^\beta - V_m^\alpha) dP = (S_m^\beta - S_m^\alpha) dT \implies \frac{dP}{dT} = \frac{\Delta S_{tr}}{\Delta V_{tr}}$

Because the transition occurs reversibly at temperature $T$, $\Delta S_{tr} = \frac{\Delta H_{tr}}{T}$. Substituting this yields the exact Clapeyron equation:

$\frac{dP}{dT} = \frac{\Delta H_{tr}}{T \Delta V_{tr}}$

The Clausius-Clapeyron Equation

For phase transitions involving a gas phase (vaporization: liquid $\rightarrow$ vapor; sublimation: solid $\rightarrow$ vapor), two approximations are applied:

1. The molar volume of the condensed phase is negligible compared to the vapor phase: $\Delta V_{vap} = V_{m, gas} - V_{m, liq} \approx V_{m, gas}$.
2. The vapor behaves as an ideal gas: $V_{m, gas} = \frac{RT}{P}$.

Substituting these approximations into the Clapeyron equation:

$\frac{dP}{dT} = \frac{\Delta H_{vap}}{T (RT / P)} = \frac{P \Delta H_{vap}}{R T^2} \implies \frac{1}{P}\frac{dP}{dT} = \frac{\Delta H_{vap}}{R T^2}$

$\frac{d\ln P}{dT} = \frac{\Delta H_{vap}}{RT^2} \quad \text{or} \quad \frac{d\ln P}{d(1/T)} = -\frac{\Delta H_{vap}}{R}$

Integrating between $(T_1, P_1)$ and $(T_2, P_2)$, assuming standard enthalpy of vaporization $\Delta H_{vap}$ is independent of temperature:

$\ln\left( \frac{P_2}{P_1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) = \frac{\Delta H_{vap}}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$

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4. Two-Component Systems ($C = 2$): Vapor-Liquid Equilibria

For binary systems ($C = 2$), the full phase rule gives $F = 4 - P$. In boiling and distillation studies, the pressure is held constant at $1\text{ atm}$, resulting in isobaric temperature-composition ($T-x-y$) diagrams where $F' = 3 - P$.

Bubble Point, Dew Point, and Fractional Distillation

• Bubble-point curve (liquidus): Represents the temperature at which the liquid mixture of composition $x$ begins to boil.
• Dew-point curve (vaporus): Represents the temperature at which the vapor mixture of composition $y$ begins to condense.
• Two-phase region: Between the bubble-point and dew-point curves, liquid and vapor phases coexist in dynamic equilibrium along horizontal tie-lines.

The Lever Rule

To determine the relative molar amounts of coexisting liquid ($n_L$) and vapor ($n_V$) from a total mixture composition $z_0$ lying on a tie-line connecting liquid composition $x_L$ and vapor composition $y_V$, we apply the conservation of mass (the mechanical lever principle):

$n_L (z_0 - x_L) = n_V (y_V - z_0)$

$\text{Fraction of Liquid: } f_L = \frac{n_L}{n_{tot}} = \frac{y_V - z_0}{y_V - x_L}$

$\text{Fraction of Vapor: } f_V = \frac{n_V}{n_{tot}} = \frac{z_0 - x_L}{y_V - x_L}$

Azeotropic Systems

An azeotrope is a liquid mixture whose vapor has the exact same composition as the boiling liquid ($x_i = y_i$). At the azeotropic point, $F' = C - P + 1 - 1 = 0$ (due to the composition constraint), meaning the mixture boils at a constant temperature without fractional composition change.

• Minimum-Boiling Azeotropes (Positive Deviations from Raoult's Law): Arise when $A-B$ intermolecular attractions are significantly weaker than $A-A$ and $B-B$ interactions. Total vapor pressure exhibits a maximum; boiling point exhibits a minimum.
  • Classic Example: Ethanol-Water mixture forms a minimum-boiling azeotrope at $95.63\text{ wt}\%$ ($89.4\text{ mol}\%$) ethanol boiling at $78.15^\circ\text{C}$ (below pure ethanol at $78.37^\circ\text{C}$ and pure water at $100.00^\circ\text{C}$).
• Maximum-Boiling Azeotropes (Negative Deviations from Raoult's Law): Arise when $A-B$ intermolecular attractions (such as hydrogen bonding) are stronger than like-molecule interactions. Total vapor pressure exhibits a minimum; boiling point exhibits a maximum.
  • Classic Example: Hydrochloric acid-Water forms a maximum-boiling azeotrope at $20.22\text{ wt}\%$ $HCl$ boiling at $108.6^\circ\text{C}$ (above water at $100^\circ\text{C}$ and pure anhydrous $HCl$ at $-85^\circ\text{C}$); Chloroform-Acetone ($bp = 64.7^\circ\text{C}$).

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5. Solid-Liquid Binary Phase Diagrams: Simple Eutectic Systems

When two components $A$ and $B$ are completely miscible in the liquid state but mutually insoluble in the solid state (crystallizing as pure solid $A$ and pure solid $B$), the system forms a simple eutectic phase diagram.

The Eutectic Point ($E$)

The eutectic point represents the lowest melting point of any mixture composition. At this point, the liquid phase coexists simultaneously with pure solid $A$ and pure solid $B$:

$\text{Liquid}(x_E) \underset{\text{cooling}}{\overset{\text{heating}}{\rightleftharpoons}} \text{Solid } A + \text{Solid } B$

Applying the condensed phase rule ($F' = C - P + 1$) at the eutectic point ($C = 2$, $P = 3$: Liquid, Solid $A$, Solid $B$):

$F' = 2 - 3 + 1 = 0 \quad (\text{Invariant})$

The eutectic mixture behaves like a pure compound during solidification, freezing at a sharp, constant temperature ($T_E$).

Cooling Curves and Thermal Analysis

Cooling curves plot temperature versus time as a molten melt cools:

1. Pure Component: Shows an initial steady cooling slope, a completely horizontal thermal plateau (thermal arrest, $F'=0$) during solidification, followed by continued solid cooling.
2. Off-Eutectic Mixture ($x \neq x_E$): As temperature drops to the liquidus line, the primary component begins to crystallize, causing a noticeable decrease in cooling rate (thermal break, $F'=1$). As primary solid continues to freeze out, the remaining liquid composition tracks down the liquidus line until it reaches $x_E$ at $T_E$. At $T_E$, the entire remaining liquid solidifies isothermally (eutectic arrest, $F'=0$).
3. Eutectic Composition ($x = x_E$): Cools uniformly until reaching $T_E$, where it undergoes a single, prolonged horizontal thermal arrest with no preliminary thermal break.

Incongruent Melting and Peritectic Reactions

In systems where components react to form an intermediate compound $A_m B_n$ that is thermally unstable at elevated temperatures, the compound undergoes incongruent melting (melting with decomposition):

$\text{Compound } AB_2(s) \xrightarrow{\text{heating}} \text{Solid } B(s) + \text{Liquid}(x_p)$

This reversible transformation is known as a peritectic reaction ($Solid_1 + Liquid \rightleftharpoons Solid_2$).

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6. Worked Examination Calculations

Problem A: Depression of Ice Melting Point Under Pressure (Clapeyron)

Task: Calculate the pressure required to lower the melting point of ice from $0.00^\circ\text{C}$ to $-1.00^\circ\text{C}$ ($272.15\text{ K}$).

• Data: $\Delta H_{fus} = 6010\text{ J/mol}$; density of ice $\rho_{ice} = 0.9167\text{ g/cm}^3$; density of water $\rho_{water} = 0.9998\text{ g/cm}^3$; molar mass $M = 18.015\text{ g/mol}$.

Solution: $V_m(ice) = \frac{18.015\text{ g/mol}}{0.9167\text{ g/cm}^3} = 19.652\text{ cm}^3/\text{mol} = 19.652 \times 10^{-6}\text{ m}^3/\text{mol}$ $V_m(water) = \frac{18.015\text{ g/mol}}{0.9998\text{ g/cm}^3} = 18.019\text{ cm}^3/\text{mol} = 18.019 \times 10^{-6}\text{ m}^3/\text{mol}$ $\Delta V_{fus} = V_m(water) - V_m(ice) = (18.019 - 19.652) \times 10^{-6} = -1.633 \times 10^{-6}\text{ m}^3/\text{mol}$

Using the Clapeyron equation: $\frac{dP}{dT} = \frac{\Delta H_{fus}}{T \Delta V_{fus}} = \frac{6010\text{ J/mol}}{(273.15\text{ K})(-1.633 \times 10^{-6}\text{ m}^3/\text{mol})} = -1.347 \times 10^7\text{ Pa/K} = -134.7\text{ bar/K}$

To achieve $\Delta T = -1.00\text{ K}$: $\Delta P = \left( \frac{dP}{dT} \right) \Delta T = (-1.347 \times 10^7\text{ Pa/K})(-1.00\text{ K}) = +1.35 \times 10^7\text{ Pa} = 135\text{ bar} \approx 133\text{ atm}$

Problem B: Atmospheric Boiling Point in Baguio City (Clausius-Clapeyron)

Task: Baguio City is situated at an elevation of approximately $1500\text{ m}$, where we take the atmospheric pressure as $620.0\text{ mmHg}$ ($0.8158\text{ atm}$). Determine the boiling temperature of pure water in Baguio City.

• Data: Normal boiling point of water $T_1 = 373.15\text{ K}$ at $P_1 = 760.0\text{ mmHg}$; $\Delta H_{vap} = 40.66\text{ kJ/mol} = 40660\text{ J/mol}$; $R = 8.3145\text{ J}/(\text{mol}\cdot\text{K})$.

Solution: $\ln\left( \frac{P_2}{P_1} \right) = \ln\left( \frac{620.0}{760.0} \right) = \ln(0.81579) = -0.20356$ Using the Clausius-Clapeyron equation: $-0.20356 = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) = -\frac{40660}{8.3145} \left( \frac{1}{T_2} - \frac{1}{373.15} \right)$ $-0.20356 = -4890.25 \left( \frac{1}{T_2} - 0.0026799 \right)$ $\frac{1}{T_2} - 0.0026799 = \frac{-0.20356}{-4890.25} = 4.1626 \times 10^{-5}\text{ K}^{-1}$ $\frac{1}{T_2} = 0.0026799 + 0.0000416 = 0.0027215\text{ K}^{-1} \implies T_2 = \frac{1}{0.0027215} = 367.44\text{ K}$ $T_2 = 367.44 - 273.15 = 94.29^\circ\text{C}$

Pure water boils at approximately $94.3^\circ\text{C}$ in Baguio City due to the reduced barometric pressure.