# Gibbs phase rule for azeotrope and Solid-Liquid phase diagram

## Azeotrope

An azeotrope is a mixture of two or more components in fluidic states whose proportions cannot be altered or changed by simple distillation

Applying the phase rule to azeotrope. There are 2 phases: liquid and vapor, there are 2 components: X and Y. As the pressure is held constant, so F = 2 - 2 + 1 = 1.

Question:

1. What is the "1" degree of freedom represents? I cannot seem to find any proper explanation.

Thought: Is it because when we change the composition fraction, the mixture which includes both liquid and vapor still maintain?

1. Is there any exception to Gibbs phase rule?

## Solid-liquid phase diagram

I wonder if the mixture is on the red line, what is the degree of freedom?

Thought: Since this is still a binary system, so F = 2 - 3 + 1 = 0. So, there is no intensive variable should be changed here. It is a bit strange here since the mixture is unchanged when increasing temperature (in the red line), so it should be 1?

• Be aware that additional relations like reactions or the same phase co position for azeotropes decrease the number of degrees for freedom. So for the given pressure , there is no degree left and all is defined by the value of pressure. Jan 10 at 7:17
• So, you are saying that for azeotrope, the composition fraction is well defined, so another constraint like x(liquid) = ky(vapor) will decrease the degree of freedom? Jan 10 at 7:55
• The system has 1 degree of freedom,spent of choosing the pressure. It could be temperature too. But the 2nd parameter is determined by the chosen one. (It is not exactly correct as composition of azeotrope is slightly dependent on pressure. // see also chem.libretexts.org - The Gibbs Phase Rule for Multicomponent Systems Jan 10 at 8:34
• What do you mean by choosing the pressure? The liquid-vapor phase diagram above has the pressure held constant Jan 10 at 11:36
• For ethanol/water azeotrope composition, the pressure is (pre)chosen the only degree of freedom. By setting the pressure, you cannot choose the temperature. By setting the temperature, you cannot choose the pressure. Jan 10 at 12:20

What is the "1" degree of freedom represents? I cannot seem to find any proper explanation.

The degrees of freedom of a fluid is the number of intensive variables that you must fix in order to have a complete thermodynamic description of that fluid. If you have $$F = 2$$, then you need two intensive variables (such as $$p$$ and $$T$$), so that $$G$$, $$S$$, $$A$$, $$U$$, $$V$$, etc. have a defined value.

For azeotropy, the degrees of freedom can be inferred by writing down the equilibrium problem. However, since we have an azeotrope, an additional mathematical restriction must be included (I will name the components just $$\ce{A}$$ and $$\ce{B}$$) \begin{align} \mu_A^\mathrm{vapor} (p,T,y_A) &= \mu_A^\mathrm{liquid} (p,T,x_A) \tag{1} \\ \mu_B^\mathrm{vapor} (p,T,y_B) &= \mu_B^\mathrm{liquid} (p,T,x_B) \tag{2} \\ y_A + y_B &= 1 \tag{3} \\ x_A + x_B &= 1 \tag{4} \\ x_A &= y_A \hspace{3 cm} \color{blue}{\text{(Azeotropy condition)}} \tag{5} \\ \end{align} I count six unknowns ($$p,T,y_A,y_B,x_A,x_B$$) and five equations. Hence, $$F = 1$$. All the equations are well known to us, the $$\color{blue}{\text{blue}}$$ one is new. This agrees with your diagram. Once you set the pressure $$p$$, you can point at the azeotrope directly with your finger without specifying anything more.

You may think that $$x_B = y_B$$ is another piece of information that we must add, but it is not. You can check that this equation can be derived combining Eqs. (3-5), so it is linearly dependent.

Is there any exception to Gibbs phase rule?

I think that one exception is the critical point, but I have not obtained a definite answer although I have asked many people. The critical point is zero-variant ($$F = 0$$). If you change the critical pressure $$p_\mathrm{c}$$ or the critical temperature $$T_\mathrm{c}$$, then the critical state is gone. If we apply Gibbs' phase rule, this means that \begin{align} F &= 2 - \text{Num. of phases} + \text{Num. of components} \\ \text{Num. of phases} & = 2 + \text{Num. of components} - F \\ \text{Num. of phases} & = 2 + 1 - 0 \rightarrow \boxed{\text{Num. of phases = 3}} \tag{6} \\ \end{align} This is indeed what happens at the triple point, there are three phases in equilibrium, and thus $$F = 0$$.

However, when you grab any Physical Chemistry book, the description of the critical point contradicts $$P = 3$$ ($$P$$ from now on is number of phases). Generally, it is spoken as an 'impossibility to distinguish the gas and liquid phases', in which both merge. You can go to YouTube for some videos on this, and if you are asked to point with your finger where the liquid and vapour phase are, you won't be able to do so.

If the liquid and vapour merge together, then we may just say that $$P = 1$$, but $$P$$ should be three. However, we cannot say that $$P = 3$$ either, because we cannot distinguish three phases.

The question for me is open, but some tend to state that the Gibbs's phase rule breaks down in the critical point.

I wonder if the mixture is on the red line, what is the degree of freedom?

For solid-liquid phase diagrams, that red line has the same interpretation as the simple ones in liquid/vapour equilibrium. It has $$F = 1$$.

• Is the F = 1 in this case different from my understanding? In the liquid-vapor diagram that I browsed above, saying that we have a point in the liquid region, then F = 2 implies that I could change both T and composition fraction in two dimensions without breaking the phase. While the F = 1 in the azeotrope meaning that we should choose the pressure, and cannot be interpreted by the diagram? Jan 10 at 16:25
• @Shira Hi! I am very sorry, but I couldn't understand what you meant. Could you rephrase? Jan 10 at 16:33
• Lets say, I heard that we can calculate the degree of freedom of azeotrope like this: F = C - P + 1 (P = constant). There are 2 components X and Y, 2 phases (liquid and vapor) and there is another constraint x(A) = y(A) which decrease further by 1. Therefore F = 2 - 2 + 1 - 1 = 0. What do you think about this idea? Jan 10 at 16:42
• @Shira I understood. The conclusion is the same, but you must have in mind that "pressure is fixed". That is fine, so your formula now becomes $F = C - P + \color{red}{1}$, instead of the most common one which is $F = C - P + \color{red}{2}$. I would suggest you to use this last one. Your conclusion in words would be all right ("all I need to fix the state is pressure"), but the statement "the degrees of freddom are zero" is wrong. Just use $F = C - P + 2$. Jan 10 at 16:48
• In case F = C - P + 2 is used, how could we come to the result F = 1? I am still a bit unclear Jan 10 at 16:56