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The Gibbs phase rule appears to be extremely simple;indeed, that is the source of its elegance: F = C – P + 2, where Cis the number of components in a chemical system, P is thenumber of phases, and F is the number of degrees of freedom.Students of physical chemistry usually have little difficultycounting the number of phases in the systems they encounter.A phase is defined as a state of matter that is uniform through-out, both in chemical composition and in physical properties.In determining P, students need only realize that there canbe at most one gas phase, but that there can be more than onesolid phase (e.g., a mixture of zinc turnings and iron filings)and more than one liquid phase (e.g., a mixture of tolueneand water). However, counting components, although a seem-ingly trivial task, presents unexpected difficulties, especiallyin multiphase systems involving chemical reactions.

In this paper, I discuss the problem of counting compo-nents in multiphase systems involving chemical reactions. Nonew principles are required to solve this problem (1, 2). Itssolution simply requires careful attention to the nature of theconstraints that provide relationships among the species con-stituting the system. As a consequence of this more detailedexamination, we find that the number of components dependson the phase structure of the system. In some situations, thenumber of components can be affected by the temperatureand pressure of the system. In other situations, a systemcontaining arbitrary amounts of two different chemicalcompounds should be regarded as a one-component system.

Counting Components

C, the number of components, is defined as the minimumnumber of independent chemical species necessary to definethe composition of all of the phases present in the system(3). The number of independent species is equal to the numberof constituents (the total number of chemical species used tocharacterize the system) minus the number of constraints onthe concentration of these constituents (4 ). The constraintsinclude both equations involving equilibrium constants, whichprovide relationships among the chemical species at equilibrium,and equations describing stoichiometric relationships amongthe species irrespective of whether the system is in equilibrium.In applying these principles, it is important to remember thatthe phase rule describes the intensive state of the system; therelative amounts of the various phases cannot be determined(5). In other words, the phase rule says nothing about thetotal number of moles in each phase.

The number and even the choice of constituents in asystem is to some extent arbitrary, depending on the problemat hand or on personal choice. However, irrespective of thechoice of constituents, the number of components is an in-variant. For example, in a problem involving the phase equi-librium of a system containing liquid water, ice, and watervapor, there is just one constituent, H2O, and thus one com-ponent. However, in a problem involving the dissociation of

water, there are three constituents, namely, H2O, H3O+, andOH{. There is still only one component because there aretwo constraining relationships among these three constituents:

Kw = [H3O+][OH{] and [H3O+] = [OH{] (1)

In determining the number of components from theconstituents, the textbooks focus on determining the numberof independent chemical reactions without regard to phase.In fact, an article appearing in this Journal discussing theBrinkley method for determining the number of componentsnotes that this method is best suited for one-phase systemssuch as high-temperature mixtures in the gaseous phase orfor complicated aqueous solutions of electrolytes (6 ). TheBrinkley method uses linear algebra to analyze the matrixwhose (i,j)th entry is the number of atoms of element Ej inthe chemical species Si. The authors do apply the Brinkleymethod to a two-phase system, namely a solution of electro-lytes involving an insoluble species (AlCl3 dissolved in waterwith the formation of Al(OH)3(s)). However, they do notdwell on the effect of the second phase in determining thenumber of components.

The complications in determining the number ofcomponents in multiphase systems can be clearly seen in acomparison between two seemingly similar equilibria

NH4Cl(s) = NH3(g) + HCl(g) (2)and

CaCO3(s) = CaO(s) + CO2(g) (3)

Let us first treat the case in which, for each equilibrium,we introduce arbitrary amounts of the three constituents intoa closed container. In the first equation, there are three con-stituents, but only two components. A constraint on the con-stituents is provided by the equilibrium constant expression,which relates the partial pressures of the two gaseous products:

K1 = pNH3 ? pHCl (4)

where K1 is the equilibrium constant and pX denotes the par-tial pressure of the species X.

The same reasoning applies to the second equation. Theequilibrium expression is

K 2 = pCO2 (5)

and again, C = 2.Now consider the case in which, for each equilibrium, we

introduce the reactant only into the container. For the firstequilibrium, there is an additional stoichiometric relationship,n(NH3) = n(HCl), where the notation n(X) means the numberof moles of X. We conclude that the number of componentsis reduced to one. This reasoning is correct. C = 1, P = 2 (onegaseous and one solid phase), and so, and the number ofdegrees of freedom is F = 1. This degree of freedom can betaken to be the temperature. Specifying the temperaturedetermines K1, which together with the stoichiometric rela-tionship determines pNH3 and pHCl.

The Gibbs Phase Rule Revisited:Interrelationships between Components and Phases

Joseph S. AlperDepartment of Chemistry, University of Massachusetts–Boston, Boston, MA 02125; j.alper@umb.edu

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If we apply the same reasoning to the second equilibrium,we note that n(CaO) = n(CO2) and again conclude that C = 1.However, this reasoning is false. If we apply the phase ruleto the decomposition of CaCO3, we have C = 1, P = 3 (twosolid phases and one gaseous phase), and thus F = 0. Thisresult cannot be correct. There is certainly more than onetemperature at which CO2(g) exists in equilibrium withCaCO3(s) and CaO(s).

Clearly, the source of the problem lies in the additionalsolid phase in the second equilibrium. After the equilibriumhas been established, we can add arbitrary amounts of CaOwithout affecting the equilibrium. Thus C = 2. Note that inthe equilibrium expression K 2, CaO(s) does not appear. Sinceit is in a phase by itself, the composition (concentration) ofthe solid CaO phase is independent of the amount of thatphase. For the usual choice of standard states, this con-centration can be set equal to 1. Thus the existence of thestoichiometric relationship n(CaO) = n(CO2), relating thenumber of moles of the two species, is irrelevant in deter-mining the number of components. In the NH4Cl example,on the other hand, addition of either HCl or NH3 drivesthe equilibrium to the left. Both HCl and NH3 appear inthe equilibrium expression K1, and the stoichiometric re-lationship between the two species does affect the numberof components.

The general principle becomes clear from these examples.The only stoichiometric relationships that reduce the num-ber of components are those in which every species in therelationship appears in an equation involving an equilibriumconstant. Stoichiometric relationships involving at least onespecies in a solid phase or at least one pure liquid species ina phase by itself are irrelevant in determining the number ofcomponents.

This approach to the determination of the number ofcomponents clarifies the example mentioned above involvingan aqueous solution of AlCl3. Let us take as constituents thespecies H2O, Al3+, Cl{, H+, OH{, and Al(OH)3(s). There aretwo relevant equilibrium constants: Kw and Ksp, the solubilityproduct constant for Al(OH)3. There is only one stoichio-metric constraint involving species all of which appear in theequilibrium equations involving Kw and Ksp. This constraintis the charge balance equation

3[Al3+] + [H+] = [Cl{] + [OH{] (6)

which comprises all the charged species in solution. Thus thenumber of components is 3 (6 constituents, 2 equilibriumconstants, 1 stoichiometric constraint). These componentscan be chosen to be H2O, AlCl3, and Al(OH)3. This resultcan also be obtained by noting that the component Al(OH)3is required to specify the composition of the solid phase.

Suppose now, contrary to fact, that Al(OH)3 were solubleand appeared in the aqueous phase. In this case, the equationinvolving Ksp would be replaced by

K = [Al3+][OH{]/[Al(OH)3] (7)

The species Al(OH)3 now appears in an equation involvingan equilibrium constant, so that a second stoichiometric con-straint becomes relevant. This constraint is the mass balanceequation

3[Al(OH)3] + 3[Al3+] = [Cl{] (8)

reflecting the fact that no matter what aqueous species form,the ratio of the total (analytical) concentration of Al to thatof Cl must remain 1:3. Thus, assuming Al(OH)3 is soluble,there would be just two components, H2O and AlCl3.

C as a Function of Temperature and Pressure

It is obvious that the number of phases present in a sys-tem depends on the temperature T and pressure p. However,it is not often appreciated that the number of components isnot an invariant of a system, but that it too is a function of pand T. As a first example, let us return to the decompositionof NH4Cl. At a pressure of 1 atm and between temperatures191 and 239 K, NH3 exists in the liquid phase while HClremains a gas. Now only HCl appears in the equilibriumequation because NH3 is a pure liquid in a phase by itself.Consequently, there are no longer any stoichiometric constraintsso that the number of components is increased from 1 to 2. Itseems strange that changing the temperature (and by extension,the pressure) of a system can result in a change in the numberof components as well as a change in the number of phases.

Another situation in which the number of componentscan vary depending upon the temperature and pressure occursunder conditions of metastable equilibrium. An arbitrarymixture of hydrogen, oxygen, and water vapor at equilibriumhas two components because of the equilibrium constantrelating the three species. In this case P = 1 and there arethree degrees of freedom: for example, p, T, and the molefraction of one of the three constituents.

A mixture of these three gases might also be preparedby introducing arbitrary amounts of each of the gases into acontainer being careful to avoid an explosion. The rate ofreaction is so slow at normal temperatures and pressures thatwe can consider the three unreacted gases to be in equilibrium.In Richard Feynman’s words, a system is in equilibrium “ifall the ‘fast’ things have happened and all the ‘slow’ things not”(7 ). Under these conditions, C = 3 (there is no equilibriumconstant expression) and there are four degrees of freedom:p, T, and the mole fractions of two of the three gases.

Distinguishing Components from Phases

In most cases, that is, exercises in physical chemistrytexts, there is never any question about whether two speciesrepresent two components or should be considered to be twophases of the same component. Liquid water and ice are twophases of the same component. Similarly, C(graphite) andC(diamond) are also regarded as two phases of the samecomponent, namely carbon. But are things always so simple?

Under ordinary conditions, C(gr) and C(d) are not inchemical equilibrium; C(gr) is the more stable form. How-ever, as we know from ordinary experience, the two forms ofcarbon are in metastable equilibrium and can exist indepen-dently of each other. For all practical purposes, the graphite–diamond system can be regarded as being in equilibrium andsubject to the phase rule, just as is the case for the hydrogen–oxygen–water vapor system discussed above. Consequently,under these conditions for the graphite–carbon system, C = 2.It is clear that P = 2. Applying the phase rule, we find F = 2.The two degrees of freedom are p and T. (Each of the twosolid phases consists of only one species, so the compositionsof both are determined.)

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If however, C(gr) and C(d) are at true chemical equilib-rium, the chemical potentials of the two species become equal.As a result of this constraint, C = 1. Since P remains equalto two, there is just one degree of freedom, p or T. Fromthe constraint on the chemical potentials, we can derive theClausius equation

dp/dT = ∆trsS/∆ trsV (9)

where ∆trsS and ∆ trsV are the changes in entropy and volumeacross the phase boundary, which provides the functionalrelationship between p and T.

In view of this discussion, consider a hypotheticalisomerization reaction, A = B, that has reached equilibrium.(We assume that the initial amount of each species is arbi-trary.) If one or both of the two isomers are gases, C = 1because there are two constituent species, one equilibriumexpression, but no stoichiometric relation between A and B.In the case of one gaseous constituent P = 2, and so F = 1.If both constituents are gases, P = 1 and F = 2, the molefraction of A in the gas phase providing the additional degreeof freedom.

If both A and B are solids, the system is totally analogousto the graphite–diamond example. If the rate of interconver-sion of the two solids is very slow, A and B are in metastableequilibrium, and we can take C = 2. There are two differentcompounds A and B with no equilibrium constant expressionrelating them. However, there may be some regions of thep–T plane in which p and T take on values for which theisomerization proceeds and the reaction reaches equilibrium. Forthese values of p and T, A and B are in chemical equilibrium.However, unlike a typical chemical reaction involving gases orsolutions, there is no equilibrium constant for this reactionrelating the concentrations of A and B. Each of the chemicalspecies A and B is in a phase by itself so that the mole fraction(or concentration) of each species is unity. Even at equilibrium,the ratio of the number of moles of A and B is arbitrary.

Although there is no equilibrium constant, we can analyzethis equilibrium. Instead of treating the system consisting ofA(s) and B(s) as a two-component system, we treat it as aone-component system with two phases that are in equilibrium.The reduction in the number of components from 2 to 1results from the fact that at equilibrium, the chemical poten-tials of the two species are equal, just as is the case in thegraphite–diamond example. Since C = 1 and P = 2, the phaserule tells us that F = 1. Consequently, there must be a relation-ship between p and T. This relationship is given by a Clausiusequation which is derived by equating the chemical potentialsof A and B. Starting from dµA( p,T ) = dµB( p,T ) we obtainthe Clausius-like relation

dp/dT = ∆S/∆V (10)

where ∆S and ∆V are the changes in entropy and volumeupon isomerization.

But is it legitimate to call a system containing A(s) andB(s) a one-component system? In the derivation of the usualClausius equation, as in the determination of the number ofconstraints in the derivation of the phase rule itself, we equatechemical potentials of the same substance in two differentphases. I have purposely introduced the new word “substance”to leave open the question of how the relative similaritybetween graphite and diamond compares with that betweenthe two isomers A and B. Graphite and diamond differ in thearrangement of their carbon atoms; in graphite the carbonatoms lie in sheets, in diamond they form a tetrahedralstructure. Supposing for specificity that the two isomers arehydrocarbons, then the isomers differ only in the arrangementof their carbon and hydrogen atoms. Thermodynamics istotally independent of the details or even the existence of themolecular structure. Consequently, if chemical equilibriumcan be established between any two isomers (they need notbe hydrocarbons), these two species are thermodynamicallyjust as similar to each other as are graphite and diamond. Ifthe latter pair can be considered to be a one-componentsystem, so can the former.

This isomerization example shows that there are surprisingsubtleties associated with the application of the Gibbs phaserule. For certain chemical equilibria, such as one involvingthe solid phases of two isomers, a seemingly two-componentsystem should be regarded as a one-component system. Moregenerally, in determining the number of components, athorough understanding of the phase structure of the systemis critical. C is not independent of P.

Acknowledgments

I would like to thank R. I. Gelb for his critical reading ofthe manuscript and for several helpful discussions. I would alsolike to thank the reviewers for several valuable suggestions.

Literature Cited

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3. Atkins, P. Physical Chemistry, 6th ed.; Freeman: New York, 1998;pp 192–193.

4. Noggle, J. H. Physical Chemistry, 3rd ed.; Harper Collins: NewYork, 1996; pp 332–333.

5. Alberty, R. A.; Silbey, R. J. Physical Chemistry, 2nd ed.; Wiley:New York, 1997; pp 141–143.

6. Zhao, M.; Wang, Z.; Xiao, L. J. Chem. Educ. 1992, 69, 539–542.7. Feynman, R. Statistical Mechanics: A Set of Lectures; Benjamin:

Reading, MA, 1972; p 1.