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- Howard DeVoe
- University of Maryland

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In Sec. 8.1.7, the Gibbs phase rule for a pure substance was written \(F = 3 - P\). We now consider a system of more than one substance and more than one phase in an equilibrium state. The phase rule assumes the system is at thermal and mechanical equilibrium. We shall assume furthermore that in addition to the temperature and pressure, the only other state functions needed to describe the state are the amounts of the species in each phase; this means for instance that surface effects are ignored.

The derivations to follow will show that the phase rule may be written either in the form \begin{equation} F = 2 + C - P \tag{13.1.1} \end{equation} or \begin{equation} F = 2 + s - r - P \tag{13.1.2} \end{equation} where the symbols have the following meanings:

\(F=\) the number of degrees of freedom (or variance)

\(\hspace2ex =\) the maximum number of intensive variables that can be varied independently while the system remains in an equilibrium state;

\(C=\) the number of components

\(\hspace2ex =\) the minimum number of substances (or fixed-composition mixtures of substances)

\(\hspace 3ex\) that could be used to prepare each phase individually;

\(P=\) the number of different phases;

\(s=\) the number of different species;

\(r=\) the number of independent relations among intensive variables of individual phases other than relations

\(\hspace 3ex\) needed for thermal, mechanical, and transfer equilibrium.

If we subdivide a phase, that does not change the number of phases \(P\). That is, we treat noncontiguous regions of the system that have identical intensive properties as parts of the same phase.

## 13.1.1 Degrees of freedom

Consider a system in an equilibrium state. In this state, the system has one or more phases; each phase contains one or more species; and intensive properties such as \(T\), \(p\), and the mole fraction of a species in a phase have definite values. Starting with the system in this state, we can make changes that place the system in a new equilibrium state having the same kinds of phases and the same species, but different values of some of the intensive properties. The number of different independent intensive variables that we may change in this way is the **number of degrees of freedom** or **variance**, \(F\), of the system.

Clearly, the system remains in equilibrium if we change the *amount* of a phase without changing its temperature, pressure, or composition. This, however, is the change of an extensive variable and is not counted as a degree of freedom.

The phase rule, in the form to be derived, applies to a system that continues to have complete thermal, mechanical, and transfer equilibrium as intensive variables change. This means different phases are not separated by adiabatic or rigid partitions, or by semipermeable or impermeable membranes. Furthermore, every conceivable reaction among the species is either at reaction equilibrium or else is frozen at a fixed advancement during the time period we observe the system.

The number of degrees of freedom is the maximum number of intensive properties of the equilibrium system we may independently vary, or fix at arbitrary values, without causing a change in the number and kinds of phases and species. We cannot, of course, change one of these properties to just any value whatever. We are able to vary the value only within a certain finite (sometimes quite narrow) range before a phase disappears or a new one appears.

The number of degrees of freedom is also the number of independent intensive variables needed to specify the equilibrium state in all necessary completeness, aside from the amount of each phase. In other words, when we specify values of \(F\) different independent intensive variables, then the values of all other intensive variables of the equilibrium state have definite values determined by the physical nature of the system.

Just as for a one-component system, we can use the terms *bivariant*, *univariant*, and *invariant* depending on the value of \(F\) (Sec. 8.1.7).

## 13.1.2 Species approach to the phase rule

This section derives an expression for the number of degrees of freedom, \(F\), based on *species*. Section 13.1.3 derives an expression based on *components*. Both approaches yield equivalent versions of the phase rule.

Recall that a *species* is an entity, uncharged or charged, distinguished from other species by its chemical formula (Sec. 9.1.1). Thus, CO\(_2\) and CO\(_3\)\(^{2-}\) are different species, but CO\(_2\)(aq) and CO\(_2\)(g) is the same species in different phases.

Consider an equilibrium system of \(P\) phases, each of which contains the same set of species. Let the number of different species be \(s\). If we could make changes while the system remains in thermal and mechanical equilibrium, but not necessarily in transfer equilibrium, we could independently vary the temperature and pressure of the system as a whole and the amount of each species in each phase; there would then be \(2 + Ps\) independent variables.

The equilibrium system is, however, in transfer equilibrium, which requires each species to have the same chemical potential in each phase: \(\mu_i\bph = \mu_i\aph\), \(\mu_i\gph = \mu_i\aph\), and so on. There are \(P - 1\) independent relations like this for each species, and a total of \(s(P - 1)\) independent relations for all species. Each such independent relation introduces a constraint and reduces the number of independent variables by one. Accordingly, taking transfer equilibrium into account, the number of independent variables is \(2 + Ps - s(P - 1) = 2 + s\).

We obtain the same result if a species present in one phase is totally excluded from another. For example, solvent molecules of a solution are not found in a pure perfectly-ordered crystal of the solute, undissociated molecules of a volatile strong acid such as HCl can exist in a gas phase but not in aqueous solution, and ions of an electrolyte solute are usually not found in a gas phase. For each such species absent from a phase, there is one fewer amount variable and also one fewer relation for transfer equilibrium; on balance, the number of independent variables is still \(2 + s\).

Next, we consider the possibility that further independent relations exist among intensive variables in addition to the relations needed for thermal, mechanical, and transfer equilibrium. (Relations such as \(\sum_i p_i=p\) for a gas phase or \(\sum_i x_i=1\) for a phase in general have already been accounted for in the derivation by the specification of \(p\) and the amount of each species.) If there are \(r\) of these additional relations, the total number of independent variables is reduced to \(2 + s - r\). These relations may come from

- In the case of a reaction equilibrium, the relation is \(\Delsub{r}G=\sum_i\!\nu_i \mu_i = 0\), or the equivalent relation \(K=\prod_i(a_i)^{\nu_i}\) for the thermodynamic equilibrium constant. Thus, \(r\) is the sum of the number of independent reaction equilibria, the number of phases containing ions, and the number of independent initial conditions. Several examples will be given in Sec. 13.1.4.
There is an infinite variety of possible choices of the independent variables (both extensive and intensive) for the equilibrium system, but the total

*number*of independent variables is fixed at \(2+s-r\). Keeping intensive properties fixed, we can always vary how much of each phase is present (e.g., its volume, mass, or amount) without destroying the equilibrium. Thus, at least \(P\) of the independent variables, one for each phase, must be extensive. It follows that the maximum number of independent*intensive*variables is the difference \((2 + s - r) - P\).Since the maximum number of independent intensive variables is the number of degrees of freedom, our expression for \(F\) based on species is \begin{equation} F = 2 + s - r - P \tag{13.1.3} \end{equation}

## 13.1.3 Components approach to the phase rule

The derivation of the phase rule in this section uses the concept of

**components**. The number of components, \(C\), is the minimum number of substances or mixtures of fixed composition from which we could in principle prepare each individual phase of an equilibrium state of the system, using methods that may be hypothetical. These methods include the addition or removal of one or more of the substances or fixed-composition mixtures, and the conversion of some of the substances into others by means of a reaction that is at equilibrium in the actual system.It is not always easy to decide on the number of components of an equilibrium system. The number of components may be less than the number of substances present, on account of the existence of reaction equilibria that produce some substances from others. When we use a reaction to prepare a phase, nothing must remain unused. For instance, consider a system consisting of solid phases of CaCO\(_3\) and CaO and a gas phase of CO\(_2\). Assume the reaction CaCO\(_3\)(s) \(\ra\) CaO(s) + CO\(_2\)(g) is at equilibrium. We could prepare the CaCO\(_3\) phase from CaO and CO\(_2\) by the reverse of this reaction, but we can only prepare the CaO and CO\(_2\) phases from the individual substances. We could not use CaCO\(_3\) to prepare either the CaO phase or the CO\(_2\) phase, because CO\(_2\) or CaO would be left over. Thus this system has three substances but only two components, namely CaO and CO\(_2\).

In deriving the phase rule by the components approach, it is convenient to consider only intensive variables. Suppose we have a system of \(P\) phases in which each substance present is a component (i.e., there are no reactions) and each of the \(C\) components is present in each phase. If we make changes to the system while it remains in thermal and mechanical equilibrium, but not necessarily in transfer equilibrium, we can independently vary the temperature and pressure of the whole system, and for each phase we can independently vary the mole fraction of all but one of the substances (the value of the omitted mole fraction comes from the relation \(\sum_i x_i = 1\)). This is a total of \(2 + P(C - 1)\) independent intensive variables.

(Video) # Phase Rule # Gibb's Phase Rule # System # Component # Phase #When there also exist transfer and reaction equilibria, not all of these variables are independent. Each substance in the system is either a component, or else can be formed from components by a reaction that is in reaction equilibrium in the system. Transfer equilibria establish \(P - 1\) independent relations for each component (\(\mu_i\bph = \mu_i\aph\), \(\mu_i\gph = \mu_i\aph\), etc.) and a total of \(C(P - 1)\) relations for all components. Since these are relations among chemical potentials, which are intensive properties, each relation reduces the number of independent intensive variables by one. The resulting number of independent intensive variables is \begin{equation} F = [2 + P(C - 1)] - C(P - 1) = 2 + C - P \tag{13.1.4} \end{equation}

If the equilibrium system lacks a particular component in one phase, there is one fewer mole fraction variable and one fewer relation for transfer equilibrium. These changes cancel in the calculation of \(F\), which is still equal to \(2 + C - P\). If a phase contains a substance that is formed from components by a reaction, there is an additional mole fraction variable and also the additional relation \(\sum_i\!\nu_i \mu_i = 0\) for the reaction; again the changes cancel.

We may need to

*remove*a component from a phase to achieve the final composition. Note that it is not necessary to consider additional relations for electroneutrality or initial conditions; they are implicit in the definitions of the components. For instance, since each component is a substance of zero electric charge, the electrical neutrality of the phase is assured.We conclude that, regardless of the kind of system, the expression for \(F\) based on components is given by \(F=2+C-P\). By comparing this expression and \(F=2+s-r-P\), we see that the number of components is related to the number of species by \begin{equation} C = s - r \tag{13.1.5} \end{equation}

## 13.1.4 Examples

The five examples below illustrate various aspects of using the phase rule.

### Example 1: liquid water

For a single phase of pure water, \(P\) equals \(1\). If we treat the water as the single species H\(_2\)O, \(s\) is 1 and \(r\) is 0. The phase rule then predicts two degrees of freedom: \begin{equation} \begin{split} F & = 2 + s - r - P\cr & = 2 + 1 - 0 - 1 = 2 \end{split} \tag{13.1.6} \end{equation} Since \(F\) is the number of intensive variables that can be varied independently, we could for instance vary \(T\) and \(p\) independently, or \(T\) and \(\rho \), or any other pair of independent intensive variables.

Next let us take into account the proton transfer equilibrium \[ \ce{2H2O}\tx{(l)} \arrows \ce{H3O+}\tx{(aq)} + \ce{OH-}\tx{(aq)} \] and consider the system to contain the three species H\(_2\)O, H\(_3\)O\(^+\), and OH\(^-\). Then for the species approach to the phase rule, we have \(s = 3\). We can write two independent relations:

- 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 \begin{equation} F = 2 + s-r-P = 2 \tag{13.1.7} \end{equation} 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 \(y\subs{CO}\).

(Video) Introduction to Multicomponent Systems (Sept. 20, 2017)Now suppose we raise the temperature or introduce an appropriate catalyst to allow the following reaction equilibria to exist:

- 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 \begin{equation} F = 2 + s - r - P = 2 \tag{13.1.8} \end{equation}
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.

- 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 \begin{equation} 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} \end{equation} 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: \begin{equation} F = 2+C-P=2 \tag{13.1.10} \end{equation}

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:

- We have \(s=4\), \(r=2\), and \(P=2\), giving the same result as the components approach: \begin{equation} F = 2 + s-r-P = 2 \tag{13.1.11} \end{equation}
### 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 \begin{equation} F = 2 + C - P = 3 \tag{13.1.12} \end{equation} 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 \begin{equation} \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} \end{equation} 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 \begin{equation} F = 2 + s - r - P = 2 \tag{13.1.14} \end{equation} 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 \begin{equation} F = 2 + s - r - P = 3 \tag{13.1.15} \end{equation}

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.

(Video) 3.1. Phase Equilibrium

- We have \(s=4\), \(r=2\), and \(P=2\), giving the same result as the components approach: \begin{equation} F = 2 + s-r-P = 2 \tag{13.1.11} \end{equation}

- 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 \begin{equation} 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} \end{equation} Either equation is a relation among extensive variables of the two phases. From them, we are unable to obtain any relation among

- 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 \begin{equation} F = 2 + s - r - P = 2 \tag{13.1.8} \end{equation}

- 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 \begin{equation} F = 2 + s-r-P = 2 \tag{13.1.7} \end{equation} which is the same value of \(F\) as before.

## FAQs

### What is Gibbs phase rule for multi component system? ›

The Gibbs phase rule **p+n=c+1** gives the relationship between the number of phases p and components c in a given alloy under equilibrium conditions at constant pressure, where n is the number of thermodynamic degrees of freedom in the system.

**What is the Gibbs phase rule explanation? ›**

Gibbs' Phase Rule provides the theoretical foundation, based in thermodynamics, for characterizing the chemical state of a (geologic) system, and predicting the equilibrium relations of the phases (minerals, melts, liquids, vapors) present as a function of physical conditions such as pressure and temperature.

**What is the example of Gibbs phase rule? ›**

An example showing that for a single phase of a pure substance, F=2: For a glass of liquid water, specify one of the independent intensive variables to be pressure. Choose this pressure to be 1 atm. If liquid is in the glass, the temperature can take any value between 0 'C and 100 'C.

**What is the maximum number of phases can remain is equilibrium for two components system? ›**

Thus , the maximum number of phases that can be in mutual equilibrium in 2 component system is **4** .

**How will you define the Gibbs phase rule for a multicomponent system which is at thermal and mechanical equilibrium? ›**

The phase rule, in the form to be derived, **applies to a system that continues to have complete thermal, mechanical, and transfer equilibrium as intensive variables change**. This means different phases are not separated by adiabatic or rigid partitions, or by semipermeable or impermeable membranes.

**What is the phase rule for two-component system? ›**

Two-component systems

For binary mixtures of two chemically independent components, C = 2 so that F = 4 − P. In addition to temperature and pressure, the other degree of freedom is the composition of each phase, often expressed as mole fraction or mass fraction of one component.

**What does it mean if the Gibbs energy change for a process is positive? ›**

It is important to recognize that a positive value of ΔG° for a reaction does not mean that no products will form if the reactants in their standard states are mixed; it means only that **at equilibrium the concentrations of the products will be less than the concentrations of the reactants**.

**What happens when Gibbs is positive? ›**

If ΔG is positive, then **the reaction is nonspontaneous** (i.e., an the input of external energy is necessary for the reaction to occur) and if it is negative, then it is spontaneous (occurs without external energy input).

**How do you find the number of components in a system? ›**

The number of components is **equal to the number of distinct chemical species (constituents), minus the number of chemical reactions between them, minus the number of any constraints** (like charge neutrality or balance of molar quantities).

**How do you write a Gibbs Reflective Cycle example? ›**

- Description – what happened?
- Feelings – what were you thinking and feeling?
- Evaluation – what was good and bad about the experience? What went well and what went badly?
- Analysis – what sense can you make of the situation?
- Conclusion – what else could you have done?
- Action plan – if it arose again, what would you do?

### How do you calculate phase rule? ›

The phase rule states that **F = C − P + 2**. Thus, for a one-component system with one phase, the number of degrees of freedom is two, and any temperature and pressure, within limits, can be attained.

**What is the maximum number of phases which can coexist in a one component system? ›**

Abstract. For different phases to coexist in equilibrium at constant temperature T and pressure P, the condition of equal chemical potential μ must be satisfied. This condition dictates that, for a single-component system, the maximum number of phases that can coexist is **three**.

**What is the maximum of how many phases can coexist at equilibrium in a one component system and why? ›**

In general, **no more than three phases** of a substance made from a single component can coexist at thermal equilibrium, a state governed by what is widely called the Gibbs phase rule. That is, the maximum number of phases of a single-component substance that can coexist is widely accepted as three.

**What is the maximum number of phases in equilibrium in 3 component system? ›**

If there are 3 components, using the formula, F=M+2-P and solving for P under equilibrium conditions there would be a total of **5 phases**.

**Which of the following is Gibbs phase rule for metallurgical system? ›**

In other words: assuming that the pressure is constant reduces the maximum degrees of freedom by one. Gibbs phase rule reduces to P + F = C + 1, and we end up with the following table: Pressure p = const.

**Which of the following is represents the correct equation for Gibbs's phase rule? ›**

1. Which of the following equation represents the Gibbs phase rule? Explanation: The Gibbs phase rule is represented as: **F = C – P + 2**, where F = number of intensive degrees of freedom, P = number of phases and C = minimum number of independent constituents.

**How do you calculate heat capacity from Gibbs free energy? ›**

The standard Gibbs free energy change, ΔG°, indicates the thermodynamic favorability of a physical or chemical process. When ΔG° < 0, the process is thermodynamically favored. For a given process, the value of ΔG° can be calculated directly from the values of ΔH° and ΔS° using the following equation: **ΔG° = ΔH° - TΔS°**.

**What is the phase rule for one and two-component system? ›**

For these systems, C = 1 and the phase rule is stated **Φ = 1 − P + 2 = 3 − P**. Logically, the variance cannot be greater than 2, and the presentation of phase equilibria in a two-dimensional temperature (T) versus pressure (P) plot is possible.

**How many phases are there in a two-component system? ›**

At the eutectic point in this two component system, all **three phases**, that is Liquid, crystals of A and crystals of B, all exist in equilibrium.

**What is an example for a two-component system? ›**

While many two-component systems control basic cellular processes such as motility, nutrient uptake and cell division, bacterial pathogens use two-component systems to control virulence. A prominent example is the **gram-negative pathogen Salmonella Typhimurium**, which is a frequent cause of bacterial gastroenteritis.

### Do you want Gibbs free energy to be positive or negative? ›

Correct answer:

**In order for a reaction to be spontaneous, Gibb's free energy must be negative**. Looking at the equation, we can see that the value for will ALWAYS be negative if enthalpy is negative and entropy is positive.

**What does it mean when Gibbs is negative? ›**

If the Gibbs Free Energy is negative, **the reaction is random**; if it is positive, the reaction is non-spontaneous. When G: is negative, the process is random and will proceed forward according to the written instructions.

**How do you know if Gibbs free energy is positive or negative? ›**

...

Gibbs Free Energy.

**What does it mean when Gibbs free energy is 0? ›**

Gibbs free energy can be greater than 0, less than 0, and equal to 0. If Gibbs free energy is greater than 0 the reaction is a non-spontaneous reaction. If Gibbs free energy is less than 0 the reaction is a spontaneous reaction. If Gibbs free energy is equal to 0 **the reaction is in equilibrium**.

**What does a higher Gibbs free energy mean? ›**

Gibbs free energy (G) is a value that defines how spontaneous a reaction is, with a negative value meaning the reaction is spontaneous, and a positive value meaning **the reaction is nonspontaneous**.

**Does Gibbs have to be negative? ›**

Explanation: **Gibb's free energy does not have to be negative all of the time**. Gibb's free energy is negative when the reaction is spontaneous at a particular temperature, and positive when the reaction is non-spontaneous.

**What are the 3 types of components system? ›**

**Every computer system has the following three basic components:**

- Input unit.
- Central processing unit.
- Output unit.

**What are the 5 main components of a system? ›**

An information system is essentially made up of five components **hardware, software, database, network and people**. These five components integrate to perform input, process, output, feedback and control. Hardware consists of input/output device, processor, operating system and media devices.

**What are the 3 components of a system? ›**

Every System has an IPO: **Input, Process, Output**. When you look at any system, in its simplest form, it has 3 components. It's what I'll call IPO: Input, Process, Output.

**Is Gibbs phase rule applicable to heterogeneous system? ›**

5.1 The Gibbs Phase Rule

**The phase rule relates the observed state of the heterogeneous system with number of thermodynamic variables required for the energetic description of its state as equilibrium**.

### What is Gibbs phase rule for metallurgical system? ›

2. What is Gibbs phase rule for metallurgical system? Explanation: For metallurgical system pressure has no appreciable effect on phase equilibrium and hence, **F = C – P + 1**.

**What is the phase rule for three component system? ›**

A ternary phase diagram has the shape of a triangular prism with an equi- lateral triangle as a foundation which is used to describe a three component system.

**What is the phase rule for one and two component system? ›**

For these systems, C = 1 and the phase rule is stated **Φ = 1 − P + 2 = 3 − P**. Logically, the variance cannot be greater than 2, and the presentation of phase equilibria in a two-dimensional temperature (T) versus pressure (P) plot is possible.

**What are the limitations of Gibbs phase rule? ›**

Limitations of Phase rule:

**It can be applied only for system in equilibrium**. Consequently, it is of little value in case of very slow equilibrium state attaining system. 2. It applies only to a single equilibrium system; and provide no information regarding any other possible equilibria in the system.

**What is the maximum number of co existing phases in a c component system? ›**

This condition dictates that, for a single-component system, the maximum number of phases that can coexist is **three**. Historically this is known as the Gibbs phase rule, and is one of the oldest and venerable rules of thermodynamics.

**What is Gibbs phase rule for one component system? ›**

The phase rule states that **F = C − P + 2**. Thus, for a one-component system with one phase, the number of degrees of freedom is two, and any temperature and pressure, within limits, can be attained.

**What is a 3 component system? ›**

A ternary system is one with three components. We can independently vary the temperature, the pressure, and two independent composition variables for the system as a whole. A two-dimensional phase diagram (triangular coordinate) for a ternary system is usually drawn for conditions of constant temperature and pressure.

**How many phases are there in a two component system? ›**

At the eutectic point in this two component system, all **three phases**, that is Liquid, crystals of A and crystals of B, all exist in equilibrium.

**What is a phase give an example of a two-phase system? ›**

1 Behavior of Two-Phase Systems. The definition of a phase, as given by SB&VW, is ``**a quantity of matter that is homogeneous throughout**. '' Common examples of systems that contain more than one phase are a liquid and its vapor and a glass of ice water.

**What is the difference between one component and two-component system? ›**

Answer: An example of one-component system is a system involving one pure chemical, while two-component systems, such as mixtures of water and ethanol, have two chemically independent components, and so on. Typical phases are solids, liquids and gases.