29 Multicomponent Equilibrium
Learning Objectives
By the end of this section, you should be able to:
Apply properties of ideal mixtures to describe vapour-liquid separations using Raoult’s law and Henry’s law
An ideal mixture contains substances with similar molecular interactions (which usually means similar structures or functional groups). e.g. benzene and toluene, with both being non-polar and of a similar molecular size
Image from Darkness3560 via Wikimedia Commons/CC0
A combination of two substances is called a binary mixture. For a binary mixture, there is no longer a set temperature and pressure for a boiling point. Instead, there is a range of temperature or pressure values where the mixture is in vapour-liquid equilibrium. These are bounded by two points called the bubble point and dew point. We’ll explore these concepts more thoroughly when looking at diagrams for these binary mixture systems.
Vapour-Liquid Equilibrium
In a closed container, vapour-liquid equilibrium can be achieved by the same number of molecules vapourizing and condensing at any given period of time (meaning no net change in the system). The temperature and pressure have to be kept constant if we are truly at equilibrium (again, no net change in any intensive variable).
We usually denote:
[latex]T[/latex]: overall temperature
[latex]P[/latex]: overall pressure
[latex]x_{i}[/latex]: the mole fraction of substance [latex]i[/latex] in the liquid phase
[latex]y_{i}[/latex]: the mole fraction of substance [latex]i[/latex] in the vapour phase
Estimating the Molar Fraction of a Component at Vapour-Liquid Equilibrium
Raoult’s law
Raoult’s law is used when all components are in relatively significant quantities or they are chemically very similar (generally in terms of intermolecular interactions).
[latex]p_{i}=y_{i}×P=x_{i}×p^*_{i}(T)[/latex]
[latex]pi^*(T)[/latex] means that the vapour pressure is a function of temperature, which can be calculated using the Antoine Equation (there are other equations, but in this class we will stick to the Antoine equation).
Exercise: Estimating Vapour Pressure for Ethanol
At 298 K, the vapour pressure of water is 3,138 Pa and the vapour pressure of ethanol is 7,817 Pa. Suppose we have a mixture of only water and ethanol in a closed container, at 298 K, what is the mole fraction of ethanol in the vapour phase ([latex]y_{enthaol}[/latex]) at equilibrium if the liquid is 50 mol% water and 50 mol% ethanol?
Solution
Step 1: First note here ethanol and water both have very polar OH groups, so we will consider their intermolecular interactions similar and apply Raoult’s law. Rearrange the Raoult’s law to isolate [latex]y_{i}[/latex]:
\begin{align*}
y_{i}×P & =x_{i}×p^*_{i}(T)\\
y_{i} & = \frac{x_{i}×p^*_{i}(T)}{P}
\end{align*}
Here we know both [latex]x_i[/latex], which was given to be 0.5 (50 mol%) for ethanol, and [latex]p^*_{i}[/latex] for ethanol, again given in the question prompt. But now we need to find pressure ([latex]P[/latex]).
Step 2: