7.3 Vapor-Liquid Equilibrium (VLE) and Phase Equilibrium

Thermodynamics of Phase Equilibrium

Phase equilibrium is of paramount importance in chemical engineering separation processes such as distillation, absorption, and extraction. For a system containing multiple species in multiple phases at equilibrium, the temperature and pressure must be uniform throughout all phases, and the chemical potential ($\mu_i$) of each species $i$ must be equal in every phase. Because chemical potential approaches negative infinity as concentration approaches zero, Gilbert Lewis introduced fugacity ($f_i$), which acts as a 'corrected pressure' and has units of pressure. The criterion for phase equilibrium in a vapor-liquid system is: $\hat{f}_i^v = \hat{f}_i^l$ where $\hat{f}_i^v$ and $\hat{f}_i^l$ are the fugacities of species $i$ in the vapor and liquid mixtures, respectively.

The vapor-phase fugacity of species $i$ is related to its mole fraction $y_i$ and the system pressure $P$ by: $\hat{f}_i^v = y_i \hat{\phi}_i P$ where $\hat{\phi}_i$ is the mixture fugacity coefficient, which accounts for vapor-phase non-idealities. For an ideal gas, $\hat{\phi}_i = 1$, and $\hat{f}_i^v = y_i P$ (the partial pressure).

The liquid-phase fugacity of species $i$ is related to its liquid mole fraction $x_i$ and its pure liquid fugacity by: $\hat{f}_i^l = x_i \gamma_i f_i^{\text{sat}} \exp\left[\frac{V_i^l(P - P_i^{\text{sat}})}{RT}\right]$ where $\gamma_i$ is the activity coefficient (which accounts for liquid-phase non-idealities due to intermolecular interactions), $f_i^{\text{sat}}$ is the fugacity of pure species $i$ at its vapor pressure $P_i^{\text{sat}}$, and the exponential term is the Poynting correction (accounting for pressure effects on liquid volume $V_i^l$). At low to moderate pressures, the Poynting correction is very close to $1.0$, and the pure saturated fugacity is approximately equal to $P_i^{\text{sat}}$. This simplifies the liquid fugacity to: $\hat{f}_i^l = x_i \gamma_i P_i^{\text{sat}}$

Equating vapor and liquid fugacities under these conditions yields Modified Raoult's Law: $y_i P = x_i \gamma_i P_i^{\text{sat}}$

Ideal Systems: Raoult's Law and Henry's Law

For an ideal liquid solution, the activity coefficients are $\gamma_i = 1.0$ for all components. Modified Raoult's Law simplifies to Raoult's Law: $y_i P = x_i P_i^{\text{sat}}$ Raoult's Law is applicable when the components are chemically similar (e.g., benzene and toluene). The vapor pressure $P_i^{\text{sat}}$ is typically calculated using the Antoine Equation: $\ln P_i^{\text{sat}} = A_i - \frac{B_i}{T + C_i}$ where $A_i, B_i, C_i$ are species-specific constants. Candidates must check the units of $P$ and $T$ specified in the handbook tables for these constants.

For highly dilute solutions (as $x_i \to 0$ for a solute), the solute molecules are surrounded entirely by solvent molecules, leading to behavior that deviates from Raoult's Law. In this region, Henry's Law is applied: $y_i P = x_i \mathcal{H}_i$ where $\mathcal{H}_i$ is the Henry's law constant for solute $i$ in the solvent.

Activity Coefficient Models

In non-ideal liquid mixtures, activity coefficients describe how the mixture behaves relative to an ideal solution. They are mathematically related to the excess Gibbs free energy ($G^E$) of the mixture: $\ln \gamma_i = \left[\frac{\partial (n G^E / RT)}{\partial n_i}\right]_{T,P,n_{j \ne i}}$ Several semi-empirical equations are used to model $G^E$ and predict activity coefficients as a function of liquid composition:

• Margules Equation (Symmetric, Binary): Assuming a symmetric behavior: $\ln \gamma_1 = A x_2^2, \quad \ln \gamma_2 = A x_1^2$ where $A$ is a parameter related to the molecular interactions. For asymmetric systems, the two-parameter Margules equations are: $\ln \gamma_1 = x_2^2 [A_{12} + 2(A_{21} - A_{12})x_1]$ $\ln \gamma_2 = x_1^2 [A_{21} + 2(A_{12} - A_{21})x_2]$
• van Laar Equation (Binary): $\ln \gamma_1 = \frac{A_{12}}{\left(1 + \frac{A_{12} x_1}{A_{21} x_2}\right)^2}, \quad \ln \gamma_2 = \frac{A_{21}}{\left(1 + \frac{A_{21} x_2}{A_{12} x_1}\right)^2}$

Bubble and Dew Point Calculations

VLE calculations generally fall into bubble point or dew point categories. The equilibrium ratio is defined as $K_i = y_i / x_i$. Under Raoult's Law, $K_i = P_i^{\text{sat}}/P$.

• Bubble Point Pressure: Given liquid compositions $x_i$ and temperature $T$ (so $P_i^{\text{sat}}$ are known). Since $\sum y_i = 1$, we have: $P = \sum x_i \gamma_i P_i^{\text{sat}}$
• Dew Point Pressure: Given vapor compositions $y_i$ and temperature $T$. Since $\sum x_i = 1$, we have: $P = \frac{1}{\sum \left(\frac{y_i}{\gamma_i P_i^{\text{sat}}}\right)}$
• Bubble/Dew Point Temperature: Require iterative solutions because $P_i^{\text{sat}}$ is a non-linear function of $T$ via the Antoine equation. We solve $\sum x_i K_i = 1$ for bubble $T$, or $\sum (y_i/K_i) = 1$ for dew $T$.

VLE Diagrams and Azeotropes

Vapour-liquid equilibrium behavior is visualized using phase diagrams:

• $T-x-y$ Diagram (Constant $P$): Temperature is plotted on the y-axis, and mole fraction is on the x-axis. The lower curve is the bubble-point line (saturated liquid) and the upper curve is the dew-point line (saturated vapor).
• $P-x-y$ Diagram (Constant $T$): Pressure is plotted on the y-axis. The upper curve is the bubble-point line and the lower curve is the dew-point line.
• $x-y$ Diagram: Vapor fraction $y_1$ is plotted against liquid fraction $x_1$.

An azeotrope is a mixture that boils at a constant temperature and has the same composition in both the liquid and vapor phases ($y_i = x_i$). At an azeotropic point: $\gamma_i P_i^{\text{sat}} = P$ Azeotropes cannot be separated by simple distillation because the vapor has the same composition as the liquid.

Worked Example: Binary Bubble Point Pressure

A liquid mixture contains 40 mol% benzene (1) and 60 mol% toluene (2). At $80^\circ\text{C}$, the vapor pressures of pure benzene and toluene are $P_1^{\text{sat}} = 101.3\text{ kPa}$ and $P_2^{\text{sat}} = 39.9\text{ kPa}$, respectively. Assuming the mixture forms an ideal solution, calculate the bubble point pressure of the mixture and the composition of the vapor phase in equilibrium with the liquid.

Step 1: Identify the given data. $x_1 = 0.40, \quad x_2 = 0.60$ $P_1^{\text{sat}} = 101.3 \text{ kPa}, \quad P_2^{\text{sat}} = 39.9 \text{ kPa}$ Since the solution is ideal, $\gamma_1 = \gamma_2 = 1.0$.

Step 2: Calculate the bubble point pressure ($P$). $P = x_1 P_1^{\text{sat}} + x_2 P_2^{\text{sat}}$ $P = (0.40 \times 101.3 \text{ kPa}) + (0.60 \times 39.9 \text{ kPa})$ $P = 40.52 \text{ kPa} + 23.94 \text{ kPa} = 64.46 \text{ kPa}$ The bubble point pressure is $64.46\text{ kPa}$.

Step 3: Calculate the equilibrium vapor composition ($y_1$ and $y_2$). Using Raoult's Law ($y_i = x_i P_i^{\text{sat}}/P$): $y_1 = \frac{x_1 P_1^{\text{sat}}}{P} = \frac{40.52 \text{ kPa}}{64.46 \text{ kPa}} \approx 0.629$ $y_2 = 1 - y_1 = 1 - 0.629 = 0.371$ The vapor phase is enriched in benzene ($62.9\text{ mol}%$), which is the more volatile component (having a higher vapor pressure). This concentration difference is the basis for separation in distillation columns.