# 13.1: The Gibbs Phase Rule for Multicomponent Systems


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1. Thus, we have two relations involving intensive variables only. Now $$s$$ is 3, $$r$$ is 2, $$P$$ is 1, and the number of degrees of freedom is given by $$F = 2 + s-r-P = 2 \tag{13.1.7}$$ which is the same value of $$F$$ as before.

If we consider water to contain additional cation species (e.g., $$\ce{H5O2+}$$), each such species would add $$1$$ to $$s$$ and $$1$$ to $$r$$, but $$F$$ would remain equal to 2. Thus, no matter how complicated are the equilibria that actually exist in liquid water, the number of degrees of freedom remains $$2$$.

### Example 2: carbon, oxygen, and carbon oxides

Consider a system containing solid carbon (graphite) and a gaseous mixture of O$$_2$$, CO, and CO$$_2$$. There are four species and two phases. If reaction equilibrium is absent, as might be the case at low temperature in the absence of a catalyst, we have $$r = 0$$ and $$C = s - r = 4$$. The four components are the four substances. The phase rule tells us the system has four degrees of freedom. We could, for instance, arbitrarily vary $$T$$, $$p$$, $$y\subs{O\(_2$$}\), and $$y\subs{CO}$$.

Now suppose we raise the temperature or introduce an appropriate catalyst to allow the following reaction equilibria to exist:

1. These equilibria introduce two new independent relations among chemical potentials and among activities. We could also consider the equilibrium $$\ce{2CO}\tx{(g)} + \ce{O2}\tx{(g)} \arrows \ce{2CO2}\tx{(g)}$$, but it does not contribute an additional independent relation because it depends on the other two equilibria: the reaction equation is obtained by subtracting the reaction equation for equilibrium 1 from twice the reaction equation for equilibrium 2. By the species approach, we have $$s = 4$$, $$r = 2$$, and $$P=2$$; the number of degrees of freedom from these values is $$F = 2 + s - r - P = 2 \tag{13.1.8}$$

If we wish to calculate $$F$$ by the components approach, we must decide on the minimum number of substances we could use to prepare each phase separately. (This does not refer to how we actually prepare the two-phase system, but to a hypothetical preparation of each phase with any of the compositions that can actually exist in the equilibrium system.) Assume equilibria 1 and 2 are present. We prepare the solid phase with carbon, and we can prepare any possible equilibrium composition of the gas phase from carbon and O$$_2$$ by using the reactions of both equilibria. Thus, there are two components (C and O$$_2$$) giving the same result of two degrees of freedom.

1. Now to introduce an additional complexity: Suppose we prepare the system by placing a certain amount of O$$_2$$ and twice this amount of carbon in an evacuated container, and wait for the reactions to come to equilibrium. This method of preparation imposes an initial condition on the system, and we must decide whether the number of degrees of freedom is affected. Equating the total amount of carbon atoms to the total amount of oxygen atoms in the equilibrated system gives the relation $$n\subs{C}+n\subs{CO}+n\subs{CO$$_2$$} = 2n\subs{O$$_2$$} + n\subs{CO} + 2n\subs{CO$$_2$$} \qquad \tx{or} \qquad n\subs{C} = 2n\subs{O$$_2$$} + n\subs{CO$$_2$$} \tag{13.1.9}$$ Either equation is a relation among extensive variables of the two phases. From them, we are unable to obtain any relation among intensive variables of the phases. Therefore, this particular initial condition does not change the value of $$r$$, and $$F$$ remains equal to 2.

### Example 3: a solid salt and saturated aqueous solution

Applying the components approach to this system is straightforward. The solid phase is prepared from PbCl$$_2$$ and the aqueous phase could be prepared by dissolving solid PbCl$$_2$$ in H$$_2$$O. Thus, there are two components and two phases: $$F = 2+C-P=2 \tag{13.1.10}$$

For the species approach, we note that there are four species (PbCl$$_2$$, Pb$$^{2+}$$, Cl$$^-$$, and H$$_2$$O) and two independent relations among intensive variables:

1. We have $$s=4$$, $$r=2$$, and $$P=2$$, giving the same result as the components approach: $$F = 2 + s-r-P = 2 \tag{13.1.11}$$

### Example 4: liquid water and water-saturated air

If there is no special relation among the total amounts of N$$_2$$ and O$$_2$$, there are three components and the phase rule gives $$F = 2 + C - P = 3 \tag{13.1.12}$$ Since there are three degrees of freedom, we could, for instance, specify arbitrary values of $$T$$, $$p$$, and $$y\subs{N\(_2$$}\) (arbitrary, that is, within the limits that would allow the two phases to coexist); then the values of other intensive variables such as the mole fractions $$y\subs{H\(_2$$O}\) and $$x\subs{N\(_2$$}\) would have definite values.

Now suppose we impose an initial condition by preparing the system with water and dry air of a fixed composition. The mole ratio of N$$_2$$ and O$$_2$$ in the aqueous solution is not necessarily the same as in the equilibrated gas phase; consequently, the air does not behave like a single substance. The number of components is still three: H$$_2$$O, N$$_2$$, and O$$_2$$ are all required to prepare each phase individually, just as when there was no initial condition, giving $$F = 3$$ as before.

The fact that the compositions of both phases depend on the relative amounts of the phases is illustrated in Prob. 9.5.

We can reach the same conclusion with the species approach. The initial condition can be expressed by an equation such as $$\frac{(n\subs{N$$_2$$}\sups{l} + n\subs{N$$_2$$}\sups{g})} {(n\subs{O$$_2$$}\sups{l} + n\subs{O$$_2$$}\sups{g})} = a \tag{13.1.13}$$ where $$a$$ is a constant equal to the mole ratio of N$$_2$$ and O$$_2$$ in the dry air. This equation cannot be changed to a relation between intensive variables such as $$x\subs{N\(_2$$}\) and $$x\subs{O\(_2$$}\), so that $$r$$ is zero and there are still three degrees of freedom.

Finally, let us assume that we prepare the system with dry air of fixed composition, as before, but consider the solubilities of N$$_2$$ and O$$_2$$ in water to be negligible. Then $$n\subs{N\(_2$$}\sups{l} \) and $$n\subs{O\(_2$$}\sups{l} \) are zero and Eq. 13.1.13 becomes $$n\subs{N\(_2$$}\sups{g} / n\subs{O$$_2$$}\sups{g} = a\), or $$y\subs{N\(_2$$} = ay\subs{O$$_2$$}\), which is a relation between intensive variables. In this case, $$r$$ is 1 and the phase rule becomes $$F = 2 + s - r - P = 2 \tag{13.1.14}$$ The reduction in the value of $$F$$ from 3 to 2 is a consequence of our inability to detect any dissolved N$$_2$$ or O$$_2$$. According to the components approach, we may prepare the liquid phase with H$$_2$$O and the gas phase with H$$_2$$O and air of fixed composition that behaves as a single substance; thus, there are only two components.

### Example 5: equilibrium between two solid phases and a gas phase

Consider the following reaction equilibrium: $\ce{3CuO}\tx{(s)} + \ce{2NH3}\tx{(g)} \arrows \ce{3Cu}\tx{(s)} + \ce{3H2O}\tx{(g)} + \ce{N2}\tx{(g)}$ According to the species approach, there are five species, one relation (for reaction equilibrium), and three phases. The phase rule gives $$F = 2 + s - r - P = 3 \tag{13.1.15}$$

It is more difficult to apply the components approach to this example. As components, we might choose CuO and Cu (from which we could prepare the solid phases) and also NH$$_3$$ and H$$_2$$O. Then to obtain the N$$_2$$ needed to prepare the gas phase, we could use CuO and NH$$_3$$ as reactants in the reaction $$\ce{3CuO} + \ce{2NH3} \arrow \ce{3Cu} + \ce{3H2O} + \ce{N2}$$ and remove the products Cu and H$$_2$$O. In the components approach, we are allowed to remove substances from the system provided they are counted as components.

This page titled 13.1: The Gibbs Phase Rule for Multicomponent Systems is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Howard DeVoe via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.