An azeotrope is defined as the common composition of vapor and liquid when they have the same composition.
Azeotropes can be either maximum boiling or minimum boiling, as show in Figure \(\PageIndex{2; left}\). Regardless, distillation cannot purify past the azeotrope point, since the vapor and the liquid phases have the same composition. If a system forms a minimum boiling azeotrope and has a range of compositions and temperatures at which two liquid phases exist, the phase diagram might look like Figure \(\PageIndex{2; right}\):
Another possibility that is common is for two substances to form a two-phase liquid, form a minimum boiling azeotrope, but for the azeotrope to boil at a temperature below which the two liquid phases become miscible. In this case, the phase diagram will look like Figure \(\PageIndex{3}\).
Example \(\PageIndex{1}\):
In the diagram, make up of a system in each region is summarized below the diagram. The point e indicates the azeotrope composition and boiling temperature.
- Single phase liquid (mostly compound A)
- Single phase liquid (mostly compound B)
- Single phase liquid (mostly A) and vapor
- Single phase liquid (mostly B) and vapor
- Vapor (miscible at all mole fractions since it is a gas)
Solution
Within each two-phase region (III, IV, and the two-phase liquid region, the lever rule will apply to describe the composition of each phase present. So, for example, the system with the composition and temperature represented by point b (a single-phase liquid which is mostly compound A, designated by the composition at point a, and vapor with a composition designated by that at point c), will be described by the lever rule using the lengths of tie lines lA and lB.
Gibbs-Duhem and Henry's law
What happens when Raoult does not hold over the whole range? Recall that in a gas:
\[μ_j = μ_j^o + RT \ln \dfrac{P_j}{P^o} \label{B} \]
or
\[μ_j = μ_j^o + RT \ln P_j \nonumber \]
after dropping \(P^o=1\; bar\) out of the notation. Note that numerically this does not matter, since \(P_j\) is now assumed to be dimensionless.
Let's consider \(dμ_1\) at constant temperature:
\[dμ_1 = RT\left(\dfrac{\partial \ln P_1}{ \partial x_1}\right)dx_1 \nonumber \]
likewise:
\[dμ_2 = RT\left(\dfrac{\partial \ln P_2}{ \partial x_2}\right)dx_2 \nonumber \]
If we substitute into the Gibbs-Duhem expression we get:
\[x_1 \left(\dfrac{∂\ln P_1}{ ∂x_1}\right) dx_1+x_2 \left(\dfrac{∂\ln P_2}{∂x_2} \right) dx_2=0 \nonumber \]
Because \(dx_1= -dx_2\):
\[x_ 1 \left( \dfrac{∂\ln P_1}{ ∂x_1} \right) =x_2 \left( \dfrac{∂\ln P_2}{∂x_2} \right) \nonumber \]
(This is an alternative way of writing Gibbs-Duhem).
If in the limit for \(x_1 \rightarrow 1\) Raoult Law holds then
\[P_1 \rightarrow x_1P^*_1 \nonumber \]
Thus:
\[ \dfrac{∂ \ln P_1}{∂x_1} = \dfrac{1}{x_1} \nonumber \]
and
\[\dfrac{x_1}{x_1}=x_2 \dfrac{∂ \ln P_2}{∂x_2} \nonumber \]
\[1=x_2 \dfrac{∂ \ln P_2}{ ∂x_2} \nonumber \]
\[\dfrac{1}{x_2}= \dfrac{∂ \ln P_2}{∂x_2} \label{EqA12} \]
We can integrate Equation \(\ref{EqA12}\) to form a logarithmic impression, but it will have an integration constant:
\[\ln P_2 =\ln x_2 + constant \nonumber \]
This constant of integration can be folded into the logarithm as a multiplicative constant, \(K\)
\[\ln P_2 = \ln \left(K x_2 \right) \nonumber \]
So for \(x_1 \rightarrow 1\) (i.e., \(x_2 \rightarrow 0\)), we get that
\[P_2=K x_2 \nonumber \]
where \(K\) is some constant, but not necessarily \(P^*\). What this shows is that when one component follows Raoult the other must follow Henry and vice versa. (Note that the ideal case is a subset of this case, in that the value of \(K\) then becomes \(P^*\) and the linearity must hold over the whole range.)