8. Solute (1) / Solvent (2) Systems 12.7 SVNA Until now, all the components we have considered in VLE calculations have been below their critical temperature. Their pure component liquid fugacity is calculated using: Our VLE equation that describes the distribution of each component between liquid and vapour has the form: How do we deal with components that, at the temperature of interest, are above Tc and no longer have a Pisat? CHEE 311 J.S. Parent

VLE Above the Critical Point of Pure Components CHEE 311 J.S. Parent

Pure Species Fugacity of a Solute The difficulty in handling a component that is above its critical temperature or simply unstable as a pure liquid is to define a pure component fugacity for the purpose of VLE calculations. While this component must have a liquid solution fugacity, f1l, it does not have a pure liquid fugacity, f1l at x1 = 1. The tangent line at x1=0 is the Henry’s constant, k1. It is useful for predicting the mixture fugacity of a dilute component, but it cannot be extrapolated to x1=1 with any degree of accuracy. CHEE 311 J.S. Parent

Pure Component Fugacity of a Solute The pure component fugacity of a solute is calculated from a combination of Henry’s Law and an activity coefficient model. Recall that Henry’s Law may be used to represent the mixture fugacity of a minor (xi<0.02) component in a liquid. defines the Henry’s constant and is accurate as long as x1 < 0.02 Unfortunately, we cannot extrapolate the above equation to x1 = 1 to give us the pure component f1. An activity coefficient model can refine this approach CHEE 311 J.S. Parent

Pure Component Fugacity of a Solute Recall that the activity coefficient is the ratio of the mixture fugacity of a component to its ideal solution fugacity: At infinite dilution (x10), the activity coefficient becomes: Since the pure component fugacity is a constant at a given T, we can write this expression as: Using the definition of the Henry’s Constant, ki, we have: or 12.34 CHEE 311 J.S. Parent

Pure Component Fugacity of a Solute Equation 12.34 is a rigorous thermodynamic equation, 12.34 for the fugacity of a “pure” solute. However, it is evaluated at P2sat (where x1 = 0) and its use requires us to assume that pressure has an insignificant influence on the solute’s fugacity. To apply 12.34, we require a Henry’s constant for the system at the temperature of interest, ki(T), and an excess Gibbs energy model for the system, also at the T of interest. CHEE 311 J.S. Parent

VLE Relationship for a Supercritical Component Consider a system where one component is above Tc (species 1) and the other component is below Tc (species 2). The equilibrium relationship for component 2 is unchanged: or However, component 1 is handled differently, using a Henry’s constant (k1) and the infinite dilution activity coefficient (1). Both are properties specific to this mixture. 12.36 CHEE 311 J.S. Parent

Solute (1) / Solvent (2) Systems: Example CHEE 311 J.S. Parent

9. Phase Stability and Liquid-Liquid Equilibria Throughout the course we have developed methods of calculating the thermodynamic properties of different systems: Gibbs energy of pure vapours and liquids Gibbs energy of ideal and real mixtures Definition of vapour liquid equilibrium conditions As we apply these methods, we assume that the phases are stable. Recall our calculation of the Gibbs energy of a hypothetical liquid while developing Raoult’s law. In our flash calculations that we calculated Pdew and Pbubble before assuming that two phases exist A slight extension of the thermodynamic theory covered in CHEE 311 provides us with a means of assessing the stability of a phase. Answers the question: “Will the system actually exist in the state I have chosen?” CHEE 311 J.S. Parent

Phase Stability A system at equilibrium has minimized the total Gibbs energy. Under some conditions (relatively low P, high T) it assumes a vapour state Under others (relatively high P, low T) the system exists as a liquid Mixtures at specific temperatures and pressures exist as a liquid and vapour in equilibrium Consider the mixing of two, pure liquids. We can observe two behaviours: complete miscibility which creates a single liquid phase partial miscibility which creates two liquid phases in the extreme case, these phases may be considered completely immiscible. CHEE 311 J.S. Parent

Stability and the Gibbs Energy of Mixing We have already discussed the property changes of mixing, in particular the Gibbs energy of mixing. Before After GA GB G nA moles nB moles nA + nB moles liquid A liquid B of mixture The Gibbs energy of mixing is defined as: which in terms of mole fractions becomes: + CHEE 311 J.S. Parent

Stability and the Gibbs Energy of Mixing The mixing of liquids changes the Gibbs energy of the system by: Clearly, this quantity must be negative if mixing is to occur, meaning that the mixed state is lower in Gibbs energy than the unmixed state. CHEE 311 J.S. Parent

Stability Criterion Based on Gmix If the system can lower its Gibbs energy by splitting a single liquid phase into two liquids, it will proceed towards this multiphase state. A criterion for single phase stability can be derived from a knowledge of the composition dependence of Gmix. For a single phase to be stable at a given temperature, pressure and composition: Gmix and its first and second derivatives must be continuous functions of x1 The second derivative of Gmix must satisfy: 14.5 CHEE 311 J.S. Parent

Phase Stability Example: Phenol-Water CHEE 311 J.S. Parent

Phase Stability Example: Phenol-Water (25C) CHEE 311 J.S. Parent

Liquid-Liquid Equilibrium: Phenol-Water CHEE 311 J.S. Parent

9. Liquid Stability 14.1 SVNA Whether a multi-component liquid system exists as a single liquid or two liquid phases is determined by the stability criterion: 14.5 If this condition holds, the liquid is stable. If not, it will split into two (or more) phases. Substituting Gmix for a binary system, (A) we derive an alternate stability criteria based on component 1: (A) into 14.5 This quantity must be positive for a liquid to be stable. CHEE 311 J.S. Parent

Wilson’s Equation and Liquid Stability Given our phase stability criterion: what we require to gauge liquid stability is an activity coefficient model. Wilson’s equation for component 1 of a binary system gives: 11.17 Applying our stability criterion to 11.17 yields: Given that all Wilson’s coefficients Lij are positive, all quantities on the right hand side are greater than zero for all compositions Wilson’s equation cannot predict liquid instability, and cannot be used for LLE modeling. CHEE 311 J.S. Parent

Liquid-Liquid Equilibrium (LLE) SVNA 14.2 CHEE 311 J.S. Parent

Liquid-Liquid Equilibrium: Phenol-Water CHEE 311 J.S. Parent

Liquid-Liquid Equilibrium Relationships Two liquid phases (a,b) at equilibrium must have equivalent component mixture fugacities: In terms of activity coefficients: If each component can exist as a liquid at the given T,P, the pure component fugacities cancel, leaving us with: 14.10 Note that the same activity coefficient expression applies to each phase. CHEE 311 J.S. Parent

Liquid-Liquid Equilibrium-NRTL Consider a binary liquid-liquid system described by the NRTL excess Gibbs energy model. For phase a, we have: 11.20 11.21 For phase b, we have: The activity coefficients of the two liquids are distinguished solely by the mole fractions of the phases to which they apply. CHEE 311 J.S. Parent

Liquid-Liquid Equilibrium Calculations In CHEE 311, we will consider only binary liquid-liquid systems at conditions where the excess Gibbs energy is not influenced by pressure. The phase rule tells us F=2-p+C =2-2+2 = 2 degrees of freedom If T and P are specified, all intensive variables are fixed For this two-component system we can write the following equilibrium relationships: and 14.10 The latter can be stated in terms of component 1, to yield: and 14.12 The activity coefficients gia and gib are functions of xia and xib. CHEE 311 J.S. Parent

Liquid-Liquid Equilibrium Calculations Once we establish that a LLE condition exists, we are interested to know the composition of the two phases. Given T (and P), find the composition of the two liquids. Start with our equilibrium relationships. Component 1: Component 2: The natural logarithm is usually simpler to solve: We have two equations, and two unknowns xia and xib. CHEE 311 J.S. Parent