On the geometry of chemical reaction and phase equilibria

lille

[IWl

"~Uml ELSEVIER

Fluid Phase Equilibria 118 (1996) 77-102

On the geometry of chemical reaction and phase equilibria Y. Jiang, G.R. C h a p m a n , W . R . Smith * Department of Mathematics and Statistics, University of Guelph, Guelph, Ont.. NI G 2W1, Canada Received 22 June 1994; accepted 17 July 1995

Abstract

Kuhn-Tucker optimization theory is employed to obtain new results for the problem of the determination of equilibria in multi-phase multi-reaction systems. The results provide a complete classification of the possible types of behaviour that can occur for such systems. In this classification, there is an essential difference between the cases of systems for which no reactions have a set of stoichiometric coefficients that sum algebraically to zero, and systems for which this is not the case. The results yield a geometric interpretation that can be viewed as an extension of the corresponding interpretation of the geometry of systems undergoing phase equilibria alone. Illustrations are given of all possible cases of binary and ternary reacting systems. Keywords: Theory; Methods of calculation; Vapour-liquid equilibria; Liquid-liquid equilibria; Solid-liquid equilibria; Mixtures; Reactions

1. Introduction

The calculation of the equilibrium composition of complex multi-phase mixtures for which both chemical reaction and phase equilibrium may occur is an important problem in chemical engineering. This problem has been of considerable recent interest, from both the theoretical (Baker et al. (1982), Barbosa and Doherty (1988), Doherty (1990), Michelsen (1982a), Smith et al. (1993)) and the computational (Gupta et al. (1991), Dekker et al. (1993), Michelsen (1982b), Wyczesany (1993)) points of view. The primary difficulty lies in determining the number and compositions of the phases present at equilibrium, a difficulty rooted in the mathematical problem of the non-uniqueness of solutions to the necessary conditions arising from the underlying optimization problem. For systems undergoing phase equilibrium alone, the tangent-plane criterion of Baker et al. (1982), which has a simple geometric interpretation, provides a means to select a globally optimal solution from the set of solutions satisfying the necessary conditions. This criterion has recently been extended to the case of

• Corresponding author, 0378-3812/96/$15.00 (c) 1996 Elsevier Science B.V. All rights reserved SSDI 0~78-3812(95)02828-5

Y. Jiang et al./Fluid Phase Equilibria 118 (1996) 77-102

78

combined reaction and phase equilibrium in the form of the reaction tangent-plane criterion (Smith et al., 1993). It is the purpose of this paper to further extend these results, and to derive a geometric interpretation of the equilibrium conditions for general chemical and phase equilibrium problems. Our results are obtained by employing Kuhn-Tucker optimization theory (Bazaraa et al., 1993) on an appropriate and novel formulation of the problem. This yields a complete classification scheme for the possible types of solutions that can arise. The results are general in the sense that they are independent of any specific assumptions regarding the form of free energy model used to describe the system, and they apply to systems consisting of any number of substances and elements. In what follows, we first describe the problem formulation. We next briefly discuss the phase equilibrium problem in terms of this formulation, and then consider the general chemical and phase equilibrium problem. Our results are valid for any number of species and elements, and we provide concrete illustrations depicting all possible cases for binary and ternary systems.

2. Problem formulation

A key feature of our analysis is the particular formulation of the chemical and phase equilibrium problem. In this section, we discuss this formulation, with special emphasis on the closed system constraints and on comparisons with the simpler phase equilibrium problem. We consider a closed chemical system of N chemical substances, at fixed temperature (T) and pressure (P). The cases of overall thermodynamic constraints other than fixed (T, P ) can be treated by similar arguments. The substances may exist in any of 05 distinct phases, indexed by k = 1, 2 . . . . . 05 (05 is unknown in advance). Let nk.i ___0 be the mass (in moles) of substance i in phase k, and N

Yk = Y'~ nk.i > 0

(1)

i=l

be the total mass of phase k. For convenience (our results are readily extended to other cases), we assume that the composition of each phase is described using mole fractions. Thus, let x'k e~R N denote the composition vector for phase k, so that x'k,~ is the mole-fraction of substance i in phase k, where nk, i = yk X'k.i

(2)

0

(3)

and N

E

= l

(4)

i~l

Conservation of mass implies the elemental-abundance constraints

E ykAX'k = b

(5)

k=l

where A is the M × N system formula matrix, be ~ M is the system elemental abundance vector, and M is the number of elements. We assume that the columns of A satisfy a~ > O, a~ g= O, and that

Y. Jiang et al./Fluid Phase Equilibria 118 (1996) 77-102

79

b > 0, b 4: 0. We note here that the elemental-abundance constraints may be specified either directly as in Eq. (5), or indirectly via the specification of a set of stoichiometric equations describing the system. Both cases may be described by a set of equations of the form (5) (Smith and Missen, 1991, Chapter 2). Equilibrium occurs when the Gibbs free energy of the system is a global minimum subject to (1)-(5). In this paper, we assume that the system Gibbs free energy, F, is given by

F(y, . . . . . yr,x', . . . . . x'6)= E Ykf(X'k)

(6)

k=l

where f ( x ) is a model for the molar free energy. We thus assume that there is only a single phase class in the system (Smith et al., 1993), obtained for example, from an equation-of-state model or a single-phase free energy model. The arguments of this paper can be extended to the case of more than one phase class at the expense of notational complexity. An important general characteristic of f is that, due to ideal-solution terms, its values approach finite limits as any given mole fraction approaches zero, and these limiting values are approached with negatively infinite slope. We thus further assume that f is differentiable for x > 0, and satisfies lim

Of(x')

~',-~o ~

a x'i

-

~

(7)

This condition implies that any phase that is present at equilibrium (i.e. has Yk > 0) has non-zero amount of every substance in the phase (Jiang et al., 1995). In view of (7), we may strengthen the inequalities (1) and (3). (We remark in passing that when f is not differentiable at isolated points, our results also apply provided that such points do not yield equilibrium compositions.) For the elemental-abundance constraints (5), two distinct cases can occur, depending on C = rank(A). These are C = N (phase equilibrium) and C < N (reaction and phase equilibrium). In the former case, 3 an elementary matrix E such that

EA = ( IN

o)

(s)

where I N denotes the N X N identity matrix, and O is an ( M - N ) × N matrix of zeros. If we multiply (5) on the left by E, the elemental abundance constraints may then be written in general as E

y R X k!

=

d t

(9)

k=l

where d' E .~N consists of the first N components of Eb, and gives the abundance of the set of compounds of the system. The phase equilibrium problem can thus be obtained from the chemical equilibrium problem by replacing (5) by (9), i.e. by taking A = I N. The essential feature of both (5) and (9) is their bilinear nature in the variables { yk,X'k}. This will play a crucial role in the following analysis. The goal of the phase equilibrium problem (the case C = N) is to determine the compositions of the phases present and their amounts at equilibrium, i.e. to find values !

(Yl,Y2 .....

t

YTr,XI,X2 . . . . . Xt~r)

80

Y. Jiang et al. / Fluid Phase Equilibria 118 (1996) 77-102

that minimize (6) subject to (1)-(4) and (9), with (1) and (3) strengthened as noted. 7r is the number of phases actually present in non-zero amounts at equilibrium (i.e. Yl,Y2 . . . . . Y,~> 0), whereas 4' is the potential number of phases that may exist. The case C < N corresponds to reacting systems, for which 3 a complete set of R = ( N - C) linearly independent vectors, {vl .... t,R}, satisfying a ~ ' j = 0(1 < j < R )

(10)

which determine the stoichiometry of the system (Smith and Missen, 199 I, Chapter 2). The general chemical reaction and phase equilibrium problem, which for brevity we refer to as the chemical equilibrium problem, is to minimize (6) subject to (1)-(5), with (1) and (3) strengthened as noted. A feature common to both the phase equilibrium and chemical equilibrium problems is that each can be reformulated by reducing the dimension of the composition space, i.e. by rewriting (4) as N-1 Xk,N = 1-- E Xk,i (1 < k < i=l

7r)

(11)

and then eliminating Xk.N. To this end, let Xk = (X,k. ' 'Xk,2 . . . . . X,k.N_, ) r

(12)

The strengthened forms of (1) and (3) are

xL>0

(13)

Yk > 0

(14)

Eq. (13), together with (4), is then equivalent to (15)

xk~S

where S is the standard open simplex {x • .cRN- t. EN__-jtxi < 1,xi > 0}. There is thus a reduction of one dimension in passing from the original composition space to the simplex S, which we call the reduced composition space. For a binary system (N = 2), S is the 1-dimensional composition interval (0, 1). For a ternary system ( N = 3), S is the 2-dimensional triangular composition space defined by x t + x 2 < 1, x~,x 2 > 0. Finally, we define the free energy as a function of the reduced composition variables using (16)

g(x)=f(x') 77

G ( y I . . . . . y,~,x, . . . . . x,~)= Y'~ y k g ( X k ) k=l

(17)

This reduction in dimension allows the familiar geometric interpretation of the solution to the phase equilibrium problem in terms of common tangent planes to g in the reduced composition space (Baker et al., 1982; Michelsen, 1982a). The main results arising from the following analysis will be seen to be the fact that, for a reacting system, the optimality of G is determined by two separate sets of conditions, the intersection of which determines the ultimate solution. The first set requires identifying a certain set of tangent planes to g, as in the case of the phase equilibrium problem. However, in reacting systems the tangent planes must

Y. Jiang et al./Fluid Phase Equilibria 118 (1996) 77-102

8l

satisfy additional conditions, the details of which depend on the two possible cases for the rank of the matrix AS, defined by

where 1 denotes a 1 × N matrix of l's. By construction, the rank of AS is either C or (C + 1). It will be seen that the case rank(if) = C corresponds to the condition that every stoichiometric vector for A has coordinates that sum to zero, and rank(A") = (C + 1) corresponds to all other cases. When r a n k ( A ' ) = ( C + 1), we will show that the tangent plane is constrained to intersect the reduced composition space in a prescribed subspace determined solely by A, which we call the stoichiometric constraint space. When rank(A") = C we will show that the tangent plane is constrained to have a certain inclination, which is also determined solely by A. The second set of conditions is the feasibility conditions, which require the satisfaction of the elemental-abundance constraints. These conditions select a tangent plane from those identified as described above. These are more complicated than their counterparts in the phase equilibrium problem, and we are led to determine the feasible region in two steps. The first step is to find the one-phase feasible region. This is the set of x e S such that ( y , x ) is feasible, for some scalar y. All other feasible points can then be obtained from the one-phase feasible region.

3. The phase equilibrium problem In this section, we briefly discuss the phase equilibrium problem in terms of our formulation and the tangent-plane criterion of Baker et al. (1982), in order to introduce the terminology that will subsequently be generalized to the case of the chemical equilibrium problem. The phase equilibrium problem is to minimize (6) subject to (4), (9), (13) and (14). We eliminate x],,N via (4) to reduce the dimension of the problem, and sum Eqs. (9) from i = 1 to N and use (4) and (14), Thus yk = ] ~ d i = Y k=l

(19)

i~l

where Y > 0 is a constant giving the total amount of substances in the system. Substituting (11) in the final equation of (9) gives Yk 1 -k=l

Xk,i = d~

(20)

i=l

or equivalently, N-I

Y-

E i=l

w

N-I

EYkX'k,i = Y k=l

E

d'i=dN

(21)

i=l

However, this is a trivial consequence of the first N member of (9) is redundant. Finally, let t

1 members of Eq. (9), and hence the last

d = (d'~ ,d'2 . . . . . d N_ l)v/Y

(22)

z = y/Y

(23)

82

Y. Jiang et al. / Fluid Phase Equilibria 118 (1996) 77-102

Then, using (12) and (16), the phase equilibrium problem in reduced dimension is to minimize

..... z

E zkg(xk)

,x, . . . . .

k=l

subject to the constraints 77"

E zkx

(24)

=d

k=l 77"

E

= 1

(25)

k=l

Zk > 0, X k eS

(26)

where S is the reduced composition space. We note that (24)-(26) state that d is a convex combination of x k. This is an essential feature of the geometric interpretation of the phase equilibrium problem, and is a higher dimensional generalization of the concept of a tie-line in a binary system. The solution to the phase equilibrium problem is determined in 2 sequential steps, which involve the satisfaction of two respective sets of conditions. The first set of conditions is concerned with the properties of g itself, for x • S. The second set of conditions concerns the satisfaction of the feasibility constraints of (24)-(26). The initial step requires the identification of a certain set of tangent planes to g, and is carried out in 2 stages. The first stage is to identify the common tangent sets (CT-sets) of g. These are sets of points {X 1. . . . . XS} in the reduced composition space S, for which { ( x t , g ( x l ) ) . . . . . (x~,g(x~))} are points of tangency to g for a single plane Eg. Note that, in general, for any x • S, the singleton set {x} is trivially a CT-set. We call O the associated tangent plane to the CT-set. It is an ( N - 1)-dimensional plane, so that for a binary system it is a line. Examples are depicted in Figs. 1 and 2 for N = 2 and N = 3, respectively. The projections in the composition space of the points of tangency of the planes yield the indicated CT-sets and their associated tangent planes.

120100 80-

g ( x ) 60 40 20 E1

012

014

X

0~6 X~2~X3 0:8

Fig. 1. Phase equilibrium in a binary system.

x4

83

Y. Jiang et al. / Fluid Phase Equilibria 118 (1996) 77-102

g(x)

(a)

g(x)

x ,.:.,.~::

x~

x~

:(x)

(c)

X

(b)

======================================: :.: :...

~

~ g(x)

Xz (d)

.............~ ~:~}~!}!!ii:~ili~"~. " '.~ • ~,'%~2!~::~i" a:~:-<~................

===================== ......

..............

x2 1 .............

Fig. 2. Phase equilibrium in a ternary system.

The second stage of the first step of the solution procedure is to select those CT-sets which satisfy the tangent-plane criterion (Baker et al., 1982; Michelsen, 1982a). These are CT-sets for which the associated plane O is a supporting hyperplane, i.e. O lies nowhere above the graph of g, and which we call supporting CT-sets (SCT-sets). They correspond to potential (depending on the particular value of d in Eq. (24)) global minima of the objective function G, while the CT-sets discarded in this selection process correspond to local maxima and saddle (thermodynamically unstable) points. In Fig. 1, {xl,x2}, {x3,x 4} are SCT-sets whereas the points of tangency of the lines 13 and 14 are CT-sets but not SCT-sets. In Fig. 2, all planes shown except that of Fig. 2(d) are planes associated to SCT-sets. By retaining only SCT-sets and their associated tangent planes, we identify the (convex) envelope of g. This is the body obtained by replacing portions of the graph of g with the planes associated with the SCT-sets. In general, the cardinality of the SCT-set is the number of phases represented by it. Thus, {xj,x 2} and {x3,x 4} in Fig. 1 represent potential two-phase solutions to the binary phase equilibrium problem. As noted, the initial step of the solution procedure outlined above is concerned only with the identification of SCT-sets of g. The final step requires that the feasibility conditions (24)-(26) be satisfied, and this requires that we consider a specific choice of the vector d that stipulates the relative abundance of the substances within the system. This choice selects a particular SCT-set from those identified in the initial step of the solution procedure. In order for the problem to have a feasible solution (i.e. for (24)-(26) to be satisfied) d must be a convex combination of vectors in S, and so must itself lie in S. Note also that, for any d • S, the point z = 1, xl = d is trivially feasible. A SCT-set which determines the solution to the phase equilibrium problem for the given value d is determined as follows. If, at d, the envelope of g coincides with g, then the corresponding SCT-set is the singleton {d}. The point given by z = 1, x = d is feasible and yields the (one-phase) solution to the phase equilibrium problem. On the other hand, at some values of d the envelope of g lies strictly below the graph of g. In such a case, d lies below a plane (for a binary system, a line) O associated with an SCT-set {x~ . . . . . x~}. This SCT-set then provides the solution for the prescribed value of d,

Y. Jiang et al./Fluid Phase Equilibria 118 (1996) 77-102

84

since d can then be expressed as a convex combination of {x~ . . . . . x,}, i.e. 3 Zl . . . . . z= > 0, such that ~=k=l Zk = 1, Zk > 0 and Eq. (24) holds. Thus the point ( z l . . . . . z ~ , x i , . . ., X,r) satisfies the feasibility conditions and yields a 7r-phase solution to the phase equilibrium problem for the given value d. In Fig. i, when d lies in either of the intervals ( x ~ , x 2) or ( x 3, x4), a two-phase solution is indicated, whose compositions are given by the respective SCT-sets; for other values of d, a one-phase solution results, with composition x = d. Fig. 2 displays the possible results for a ternary system. It is seen that generically, for a given (T, P), the possible number of phases in a binary system ranges from 1 to 2 and for a temary system from 1 to 3, depending on the value of d. This is so because 2 points in general determine a line, and 3 points in general determine a plane. In such a case, the solution to the equilibrium problem is unique.

4. The chemical equilibrium problem The chemical equilibrium problem is to minimize the total system Gibbs free energy 7/" ¢

t

F(y, ..... y~,x, ..... x~)=

Y'. y k f ( X ' k )

(27)

k=l

subject to the constraints (28)

Y'. YkAX'k = b k=l N E

Xk,i =

1

(29)

i=l

y k > 0 , X ~ > 0 (1 < k < T r )

(30)

where the M × N matrix A has rank < N. We note that there is no loss in generality in assuming that C, i.e. the rank of A, is equal to M, the number of rows of A. As in the case of the phase equilibrium problem, a reduction in dimension is effected by eliminating the variables X~,.N,via Eq. (29). The chemical equilibrium problem, in reduced dimension, is then to minimize 7T

G(y, ..... y=,x I..... x=)=

~_~ y k g ( X k )

(31)

k=l

subject to 7]"

yk(eXk +aN)=b

(32)

k=l

(33)

Yk > O, X k e S

where x k = ( 'X k , I ,Xk.2 ' ••. g(Xk)=f(x'k) P =

(a~

-

. ,

x'k N - I ) r (I < k < 7r)

a N , a 2 -- a N . . . . . aN_ . -- aN)

( 34) (35) (36)

Y. Jiang et al./Fluid Phase Equilibria 118 (1996) 77-102

85

and S is the reduced composition space, as defined previously in the case of the phase equilibrium problem. Similarly to the case of the phase equilibrium problem, the solution to the chemical equilibrium problem is determined in a sequential 2-step process, each step of which ensures the satisfaction of a particular set of conditions. The first set again requires the identification of a certain set of tangent planes to the graph of g, which we call the reaction tangent-plane conditions, and the second set requires the satisfaction of the feasibility constraints. Each set of conditions is similar to, but more complicated than, the corresponding case for the phase equilibrium problem.

4.1. The reaction tangent-plane conditions There is a dichotomy in the nature of the solution to the chemical equilibrium problem, depending on rank(A"), where A is defined in Eq. (18). By its construction, rank(A") is either (C + 1) or C. The possible sets of stoichiometric vectors take different forms, depending on rank(A"), as shown by the following two lemmas, whose proofs are given in Appendix A.

4.1.1.1. Lemma 1. Let {v 1. . . . . v R} be a complete set of stoichiometric vectors for A (see Eq. (10)). Then precisely one of the following two cases applies: 1. {u~ . . . . . u R} can be chosen so that N

Y'. uj.i:g0V j, (1 < j < R )

(37)

i=l

2. For all choices of u~ . . . . . VR, N

E

/'~j.i ~--0' (1 < j < _ R )

(38)

i=l

4.1.1.2. Lemma 2, Case 2 applies if and only if rank(A') = C. It is mathematically convenient to relate cases 1 and 2 to rank(P), where P is defined in Eq. (36), rather than to rank(A"). The following two lemmas achieve this, the proofs of which are also given in Appendix A. 4.1.I.3. Lemma 3. Rank(P) = C if and only if Px = - a N has a solution. 4.1.1.4. Lemma 4. With reference to Lemma 1, Case 1 applies if and only if r a n k ( P ) = C or, equivalently, Case 2 applies if and only if rank(P) = (C - 1). The chemical interpretation of these results will be seen to be reflected in an intrinsic difference in the geometry of the solution to the chemical equilibrium problem for the two cases of system stoichiometry. The simplest illustration of the two cases occurs for a binary system, for which case 1 corresponds to the reaction A = mB, (m-4= 1); and case 2 corresponds to the isomerization reaction A=B.

86

Y. Jiang et al. / Fluid Phase Equilibria 118 (1996) 77-102

4.2. Case 1: rank(A) = C + 1 In the case rank(A') = C + 1, we have rank(P) = C. Under these conditions, Jiang et al. (1995) obtained a solution to the chemical equilibrium problem that has a geometric interpretation. This interpretation follows closely, and extends, the geometric solution to the phase equilibrium problem. In this section we review this solution and examine closely those aspects of it that relate to the geometry of the function g in the reduced composition space. A necessary condition for (Yl . . . . . y,~ ,x I . . . . . x~ ) to be a local minimum for (31) subject to (32) and (33) is that it be a Kuhn-Tucker point (Bazaraa et al., 1993), i.e. that 3 o~ * e ~ c such that

g ( x ~ ) - - o t * V ( p x ~ + a N ) = 0 (1 < k < v g(x;)-eT

*

= 0 (1 _
7r)

(39)

(40)

7/"

E y;(Px; +aN) =b

(41)

k=l

y;>O,x;eS

(1


(42)

The Kuhn-Tucker conditions separate naturally into the reaction tangent-plane conditions (involving only {xl*,..., x~}) of Eqs. (39) and (40), and the feasibility conditions of Eqs. (41) and (42). Similarly as for the phase equilibrium problem, the solution is determined using a 2-step procedure that focuses on each of these sets of conditions in turn. The initial step of the solution procedure focuses on the identification of compositions satisfying the reaction tangent-plane conditions. These are equivalent to the existence of a single plane 6 ) ( x ) = a *T(px + aN)

(43)

that is simultaneously tangent to g at the points {(x~,g(x~)); k = 1,2 . . . . . s}. Of particular importance are the sets of points {x~*, . . . . x s } in the reduced composition space for which (39) and (40) hold. We call such a set a CT-set relative to P (CTe-set). The identification of CTp sets is the first stage of the initial step of the solution procedure. The key feature that distinguishes the geometry of the chemical equilibrium problem from that of the phase equilibrium problem arises from the special structure of Eq. (43), which links the behaviour of the tangent planes to the matrix P, and hence to the system formula matrix, A. It follows from Eq. (43) that the tangent plane 6) associated to a CTp-set intersects the x-plane (!~ N- J), in a set of points satisfying

Px +

a N =

0

(44)

The solution to this equation defines an (R - 1)-dimensional linear space in .~ N- J, which we call the stoichiometric constraint space and denote by ~ . By Lemma 3, ~ =/= ~b. We call Eq. (44) the formula matrix constraint on the tangent planes, since it depends only on the original system formula matrix A. The stoichiometric constraint space ~ can be described in terms of the stoichiometric vectors as follows. Let {~,t . . . . . v R} be a complete set of stoichiometric vectors such that N Sj ~-~" E Pj,i ::~0 i=l

87

Y. J iang et al. / Fluid Phase Equilibria 118 (1996) 77-102

120

120

(a)

100 80

(b)

100.

14

80

g(X)6o

g(x) 60

40

40 _

l

20 X

Xl

i

2

0:2

0:4

X

0:6 xt 0:8

Fig. 3. Chemical equilibrium in a binary system: (a) case 1; (b) case 2.

It is easy to show that if Vj is normalized so that its coordinates sum to unity, i.e. if we put vj=vj/sj

(I_
then v j ' = (vl. t . . . . . v[N_ t,0) v lies in ~. Consequently, ~ is the linear space determined by the points {v~' . . . . . v~}. The fact that the tangent plane O associated to a SCTp-set meets the x-plane in ~ represents a geometric constraint, derived from the formula matrix and thus the stoichiometry, that 6) must satisfy. Each vector vj in a complete set of stoichiometric vectors gives rise to a point v I' in the x-plane that must lie on t9. Thus 19 is constrained to pass through the points { v[ . . . . . v~}, and hence through the linear space determined by these points. This constraint has a significant effect on the geometry of the solution to the chemical equilibrium problem, and differentiates it from the solution to the phase equilibrium problem. An important feature of the stoichiometric constraint space is that it is disjoint from S; i.e., when a i ~ 0 , a i ¢ 0 , (1 _
(45)

This fact is proven in Appendix B. Similarly as for the phase equilibrium problem, the existence of a CTp-set is only one of the necessary conditions for a local minimum of G. Many such sets may exist. A further necessary condition is that the associated tangent plane 19 lies nowhere above the graph of g,

19(x) < g( x) V x e S

(46)

Eq. (46) is the reaction tangent-plane criterion, and we call a CTp set whose associated tangent plane satisfies this criterion a supporting CT-set relative to P (SCTp set). The second stage of the first step of the solution procedure involves identifying all supporting SCTp sets. Fig. 3(a) illustrates the general situation in Case 1 for a binary system of I element (C = 1) in which a single reaction occurs. A typical example is the dimerization reaction 2A = B

(47)

Y. Jiang et aL / Fluid Phase Equilibria 118 (1996) 77-102

88

g(x)

(a)

g(x)

(b)

x2

x2 Xl 1

Pl

Fig. 4. Chemical equilibrium in a ternary system, C = 1: (a) case 1; (b) case 2.

corresponding to a=(1

2)

(48)

A=(11 ~)

(49)

In this case, rank(A~) = C + 1 = 2. A stoichiometric vector is ~'l = ( - 4,2)T" Hence v'~ = ( 2 , - 1) T, v'~' = (2,0) T, and ~.~ = {(2,0)T}. Thus the reaction tangent plane (line) is required to intersect the mole fraction axis at x = 2. Four different tangent lines are drawn in Fig. 3 satisfying this stoichiometric constraint, whose points of tangency to g yield CTp-sets. However, only l~ satisfies the tangent-plane criterion of Eq. (46). The (single-phase, unique) solution to the problem is given by the SCTp-set, {xl}, to which I l is associated. Fig. 4(a) illustrates the general situation in Case 1 for a ternary system consisting of 1 element ( C - - 1), in which there are 2 reactions. This is typified by the example A=(1

10

5)

(50)

giving

~=(111015)1

(51)

In this case, rank(A')= C + 1 = 2. A complete set of stoichiometric vectors is {(-5,0,1) v, ( - 10,1,0) v} which, when normalized, becomes the set { ( 5 / 4 , 0 , - 1/4) T, ( 1 0 / 9 , - 1/9,0) T} Thus the reaction tangent plane must pass through the points {(5/4,0,0) v, ( 1 0 / 9 , - 1/9,0)V}, and hence must intersect the x-plane in the line determined by these two points. The (unique, single-phase) SCTp-set solving the problem is determined by the intersection of the indicated tangent plane with the g surface, as depicted in Fig. 4(a). Fig. 5(a) and 5(b) illustrate two cases of the general situation in Case 1 for a ternary 2-element system (C = 2), for which there is 1 reaction. An example is 9A = B + C

(52)

Y. Jiang et al. / Fluid Phase Equilibria 118 (1996) 77-102

89

[] 1-phase

[] 1-phase [] 2-pha~e

g(x)

(a)

[] 2-ph e

g(x)

Co)

lX2 .... .---

[

Xl

Xl

Fig. 5. Chemicalequilibrium in a ternary system, C = 2, case 1. corresponding to (154) 1 6

A=

3

so that

if=

16 1

(54) 1

and rank(A") = C + 1 = 3. A normalized stoichiometric vector is ( 9 / 7 , - 1 / 7 , - 1 / 7 ) T, which gives rise to the point ( 9 / 7 , - 1/7,0) T in the x-plane through which O must pass. This point is indicated in the figures. There are many possible reaction tangent planes satisfying the stoichiometric constraints, one of which is shown in each figure. Since the reaction tangent plane is only constrained by the single point in space, it has 2 rotational degrees of freedom. The SCTp-set in Fig. 5(a) is a one-phase set, and that in Fig. 5(b) is a two-phase set. The feasibility conditions, discussed in Section 4.2, determine which of the two cases yields the solution to the problem. 4.3. Case 2: rank(A) = C

In the case rank(A~) = C, we have r a n k ( P ) = C - 1. Under these conditions, Eqs. (32) are not of full rank, and the results of Jiang et al. (1995) do not apply. However, in this case we may reformulate the description of the feasible region as follows. Eq. (32) may be rewritten P

YuXk k=l

=b-a

N Y',yk k=l

(55)

90

Y. J iang et al. / Fluid Phase Equilibria 118 (1996) 77-102

Since rank(P) = C - 1, P may be row-reduced using a matrix E so that

and then (55) becomes YkXk

0

0

k=l

= E

b

-

aN

Yk

k=l

(57)

The final equation of (57) gives a linear equation for the total number of moles in the system, Y, which is a constant in this case in terms of the entries of b, a N. Thus for fixed b, this is given by rr

C

C Y = ~ Yk = ~] Ec.ibi/ E Ec,iaN,i k=l i=l i=l

Now let beg~ c-1 denote the first ( C - 1) entries of E ( b - a N Y ) , Then the first ( C - I) equations of (57) become

ykQXk = b

(58) and denote ( I c - i Z) by Q.

(59)

k=l Eq. (32) has thus been replaced by a similar equation, Eq. (59), together with an additional equation, Eq. (58). Moreover, Q is now of full rank, and the Kuhn-Tucker conditions may hence be applied, giving

g ( X k ) -- a *TQxk + / 3 " = 0 , (1 _
(60)

V g ( x ; ) - QTa~ * = 0 , (1 < k N 7r)

(61)

together with the feasibility conditions. The Lagrange multiplier/3 * that appears in (60) (but not in (39)) arises from equation (58), which has no counterpart in Case 1. Eqs. (60) and (61) are the reaction tangent-plane conditions for Case 2, in the sense that they are equivalent to the existence of a single tangent plane

O ( x ) = ~ *TQx + /3"

(62)

that is simultaneously tangent to g at the points {(x k , g ( x k )); k = 1,2 . . . . . s} A set of points {x k } for which (60) and (61) hold is called a CT-set relative to Q (CTo-set). To elucidate the geometric properties of CTQ-Sets, and the tangent planes associated with them, consider (x, g(x))-space (~N). Note that (0 T, /3 *) lies on O (i.e. O intercepts the g(x)-axis at 13 * ). Furthermore, the point ( oT, /3 * ) in ( x, g(x))-space lies on /9 if and only if /3* ~" lfl¢*TQ p ..~ /3*

(63)

i.e. if and only if Qp = 0

(64)

By definition of Q, (64) is equivalent to

Pp = 0

(65)

Y. Jiang et al./Fluid Phase Equilibria 118 (1996) 77-102

91

The set of vectors {( RT,0)T; PR = 0} forms an R-dimensional subspace of ,~ N, contained in x-space, which we call the truncated stoichiometric space, and denote by ..,~'. Since (0P) = ( ~ P . ) - (/30,)

(66)

and since both vectors on the right-hand side of Eq. (66) lie on O, it follows that ..~ is parallel to the plane O. Thus the feature that distinguishes CTe-sets, and characterizes the geometry of the case rank(A") = C is that the inclination of the tangent plane O is constrained to be parallel to the truncated stoichiometric space .~. N

Vj.ia~ = 0. Since

The truncated stoichiometric space is so named because if A v i = 0 then i=l

N

v i,i = 0, we may write i=l N

i=l

l.e° N-I

l~j.i( ai -- am ) = 0 i=l

This means that P-#j = 0 where ~jT = (vj. 1. . . . . vj.N-t)- Thus _£a is generated by the set of vectors

(/ vj. l . . . . . Vj.N_I,O; ; I < j < R } obtained from a complete set of stoichiometric vectors by replacing the last coordinate of each vector by 0. Fig. 3(b) illustrates the general situation in Case 2 for a binary system of 1 element (C = 1), for which the only reaction is the isomerization A= B

(67)

corresponding to A = (1

1)

(68)

giving /f=

1

1

so that rank(A") = C = 1. A stoichiometric vector is ( 1 , - 1)T. and so the truncated stoichiometric space .Y is the vector space generated by {(1, 0)'r}. Thus, the reaction tangent plane must be parallel to the x-axis. Four tangent lines are drawn in Fig. 3(b) satisfying this constraint. However, only 1~ satisfies the tangent-plane criterion of Eq. (46). The (single-phase, unique) solution to the problem is thus given by the SCTe-set {xl}, to which l~ is associated. (SCTQ-Sets are defined analogously to SCTp-sets in Case 1.)

92

Y. Jiang et al. / Fluid Phase Equilibria 118 (1996) 77-102

1-phase

1-phase

g(x)

2-phase

Xl

(a)

', L

~

xl

2-phase

1 ~

g(x)

(b)

x2 ", L

Fig. 6. Chemicalequilibrium in a ternary system, C = 2, case 2. Fig. 4(b) illustrates the general situation in Case 2 for a ternary system consisting of I element (C = 1) in which there are 2 reactions, as typified by the isomerization reactions (70)

A=B B=C

corresponding to A=(1

1

l)

(71)

giving

,~=(11 11 11)

(72)

so that again, rank(A')= C = 1. A complete set of stoichiometric vectors is {(1,- 1,0) T, ( 0 , 1 , - 1)T} and so the truncated stoichiometric space ..~ is the vector space generated by { P~,P2} = {(1,- 1,0)T,(0,1,0)T}, which is the entire x-plane. The (unique, single-phase) SCTQ-Set solving the problem is determined by the intersection of the indicated tangent plane with the g surface. Fig. 6(a) and 6(b) illustrate two cases of the general situation in Case 2 for a ternary 2-element system (C = 2), for which there is 1 reaction. An example is Z + B = 2C

(73)

corresponding to A=

(12 4l 31)

(74)

so that .4=

2

4

3

1 1 1

(75)

Y. Jiang et al./Fluid Phase Equilibria 118 (1996) 77-102

93

and rank(A') = C = 2. A stoichiometric vector is ( 1 , 1 , - 2) T and the truncated stoichiometric space S a is the line determined by the vector 01 = (1,1, 0) w in the x-plane. The reaction tangent plane must be parallel to this direction. Many reaction tangent planes satisfy this constraint, and one is shown in each figure. Since the reaction tangent plane is constrained to be parallel to only one direction, it has one rotational and one translational degree of freedom. The SCTQ-Set in Fig. 6(a) is a two-phase set and that in Fig. 6(b) is a one-phase set. The feasibility conditions, discussed in the next section determine which of the two cases yields the solution to the problem.

4.4. The feasibility conditions For a given system formula matrix, A (and hence P and aN), only a certain set, 5~', of elemental-abundance vectors b is consistent with the feasibility conditions (41) and (42). We say that vectors b E ~ are A-consistent, or allowable for the system. A given particular b e ~ gives rise to a particular feasible region in the reduced composition space, which we denote by ~r b In this section, we describe the geometry of ~ and of ~qrb. A geometric interpretation of ~ ' may be obtained by referring to Eq. (5). This implies that, to be allowable, b must be expressible as a linear combination of the columns of the matrix A, with strictly positive coefficients, i.e. ~ is the positive, open cone

Y'~wiai;wi>O

(76)

i=l

Fig. 7 depicts ,~' for the system with the formula matrix of Eq. (53). Given a particular allowable b, a difficulty in describing the geometry of ,,,atb arises because the number of phases is unknown a priori, and feasible points can take the form ( y~, x ~), ( y I, Y2, x ~, x 2), etc. With this in mind, we define the one-phase feasible region, ~rb, 1, to be the set of points x • S for which 3 y e !~ such that ( y , x) is feasible. Thus, ~-b,l is given by "~i'b,l =

~ b 0

(77)

S

where

(78)

"~b = { x : y ( P x + aN) = b,y > 0}

132

(5,6)

5.0 4.0 3.0

(4,3)

2.0 1.O

1.O

2.0

3.0

4.0

5.0

6.0

bl

Fig. 7. Illustration o f the allowable set ~ ' for a ternary s y s t e m , C = 2.

94

Y. Jiang et a l . / F l u i d Phase Equilibria 118 (1996) 7 7 - 1 0 2

The significance of ~'-b,l stems from the following lemma, the proof of which is simple and is omitted. The lemma is a consequence of the fact that (41) is bilinear in y and x.

4.4.1.1. L e m m a 5.~rLet x I. . . . . x= E S, yj . . . . . y~ > 0. Then (yj . . . . . y ~ , x I . . . . . x~) is feasible if and only if (Y,Y - l ~ YkXk) is feasible, where Y = ~ Ykk=l k=l Thus to any feasible point in ~ r u, there corresponds a one-phase feasible point in ~b,~" The nature of the correspondence is illustrated by reformulating the statement of Lemma 5.

4.4.1.2. L e m m a 6. Let x~ . . . . . x . E S and suppose 7]" X0= E AkXk, E Ak= 1,Ak > 0 k=l k=l

(79)

Then (Y, x o) is feasible if and only if (YA~. . . . . Y A ~ , x I . . . . . x~) is feasible. This means that all feasible points in J b can be obtained by expressing a one-phase feasible point Xo as a convex combination of the vectors x~ . . . . . x . in S. The corresponding y~ . . . . . y~ values are obtained from Y, where (Y, x 0) is feasible, via multiplication by the convex coefficients A~. . . . . , ~ . This is the generalization of the lever rule for binary systems. Thus, to determine 5rb we first determine ~rb, ~, which is characterized by 3 b" From Eq. (78), ~b is the set of x o satisfying P x 0 =- Y -

lb -

(80)

aN

where Y > 0. When C > 1, the nature of the solution to (80) depends on the rank of P, so we consider the two cases separately.

4.5. Feasibility f o r Case 1, rank (A~) = C + 1" In this case r a n k ( P ) = C. Consequently the matrix P elementary matrix E such that ee

may be row-reduced to produce an

(Sl)

= (/c:Z)

where I c denotes the C × C identity matrix. Let

(Pl,P2 ..... OR-l) =

IR_I

The general solution to (80) can then be written Xo =

~

t~ p~ + Y - '

_

N

(82)

i=l Thus ~,~b is a linear space of dimension R. From Eq. (77), 5rb, ~ is obtained by finding t z. . . . . ,Y > 0 such that the above expression for x o lies in S. We note that, as Y ~ ~, (80) becomes

Px o = -a N

t R_ ~ e

(83)

Y. Jiang et al. / Fluid Phase Equilibria 118 (1996) 77-102

95

which is independent of b, and is the definition of the (R - 1)-dimensional stoichiometric constraint space ,~ of Eq. (44). Thus ~-~ G -~ b V be...~

(84)

Although our results apply to all values of N and C, the analysis may be carried further for the special case C = 1 (a one-element system) and C = N - 1 (a one-reaction system). Furthermore, these two special cases exhaust all possibilities for binary and ternary systems, and this enables us to present concrete geometric realizations of such systems in two- and three-dimensional space. When C = 1, we have A = ( a I . . . . . a N ) , P = ( c l . . . . . CN 1), where c i = a i - - a N ( 1 < i < N 1). Eq. (80) may be re-written as b Y=

1

N-i i=l

(85)

i=l

Provided the chemical system under consideration does not consist only of ionic charge, we may assume b > 0, a~ > 0 (1 _< i < N). The term in the denominator in Eq. (85) is then positive for all x o -,- b = ~ N - I , the one-phase feasible region is • S, and hence (85) yields a positive value of Y. Thus ,s all of S (for all b > 0), and any configuration of points in S is feasible. Thus the points depicted in Fig. 3(a) and Fig. 4(a) yield solutions to their respective problems. If C = N - 1, then P is C X C. Since we are assuming P has full rank, it is invertible and the solution to (80) is x o = Y-IP-lb-

(86)

P-la N

Thus, ;~b is a line in the x-plane (i.e. in ~RN-I), which has direction P - l b and passes through ~'~ = - P - laN" ,s Consider, for example, the formula matrix of the ternary system of Eq. (53). In this case, Eq. (80) becomes, for b = ( b j , b 2 ) "r,

- 2

3 } ~ Xo.2

b2

Following Eq. (86) the solution is Xo,~ = y-~ Xo.2

+ (3b 2 - 2b,)/7

]

- 1/7

and so .~, the stoichiometric constraint space, is the point { ( 9 / 7 , can be eliminated to yield

1 / 7 ) v} in the x-plane. Further, Y

(b 2 - 3b,)x0. 2 = (3b 2 - 2 b , ) x 0 . , + (3b, - 4b2) Thus 3 ~ is a line. Different values of b lead to different lines, and the one-phase feasible region "~b.l is that segment of the line that passes through S. Fig. 8 shows in detail the triangular regions in the ( x 1, x 2) plane of Fig. 6. Fig. 8(a) shows the one-phase feasible regions for b = ( 1 , 1 ) T and b ' = (8,7) w. Observe that .~b,.~ b' meet at .~ = ( 9 / 7 , - 1/7), which is not in S. Multi-phase feasible points can be found from Lemma 6. In the

96

Y. Jiang et al. / Fluid Phase Equilibria 118 (1996) 77-102

~2

X02

l.O[

(a)

Co)

1.0

//I

"-4 7

b'=(8,7~r

"-....

X2o iJ J

Xt.

I ~XOl " ~ ~

S

•'

1.0

i Lx01

(9/7,-1/7)

Fig. 8. The one-phase feasible region ,grb, ~ for a ternary system, C = 2; (a) case 1, (b) case 2.

figure, the set {x l, X2}is feasible, in the sense that 3 Yl, Y2 such that (Yl, Y2, xj, x 2) satisfies (41), (42) when b' = (8,7) T. However, ( x l, x a) is not feasible for b = (1,1) v. The feasibility conditions allow us to determine, for the formula matrix under consideration, which of the SCTr,-Sets indicated in Fig. 5(a), 5(b) yields the solution to the chemical equilibrium problem. In Fig. 5(a), b = (1,1) w and ~b is denoted by L. The reaction tangent plane has a single point of tangency to g(x) which, when projected onto the x-plane, lies on the one-phase feasible line ~qrb.I. This single point is therefore feasible, and the problem has a one-phase solution. This situation applies for any value of b that leads to a one-phase feasible line lying in the region denoted by 1-phase in the figure. Fig. 5(b) is for b = (8,7) v. There is no reaction tangent plane that is tangent to g(x) in a single point which, when projected onto the x-plane, lies on L. Rather, 3 two points of tangency x l, x 2 which, by construction, must lie on different sides of L. Thus the pair (x~, x 2) is feasible, and there is a two-phase solution. This situation arises whenever b gives rise to a one-phase feasible line ,~rb~, that lies in the region denoted by 2-phase in the figure.

4.6. Feasibility for Case 2, rank (A") = C In case 2 rank(P) = C - 1, and when P is reduced Eq. (56) results. Thus, provided C > 1, Eq. (80) becomes

Ic_ 0

Z) 0

x°

=E(Y_,b_aN)

(87)

Y. J iang et al. / Fluid Phase Equilibria 118 (1996) 77-102

97

The last equation of (87) determines the value of Y in terms of the entries of b, a N. Let

(01,02 . . . . . OR)=

(z) IR

Then the solution to (80) can be written

X0= E ti tai + i=l Thus ,~ b, the solution set of (80), is a linear space of dimension R. The one-phase feasible region is the intersection of ,%0 with S, i.e. it is obtained by finding t~ . . . . . t R e ~R, such that the expression for x o lies in S. As b varies, only the quantity

in Eq. (88) varies. This means that for different values of b, the linear spaces -~ b are parallel to one another. It is clear from the construction of ta~ . . . . . ta R that etai = 0 (1 < i < R )

(89)

and that tat . . . . . tar generate the space of P-null directions _~. In fact, if b is chosen so that Y - l b - - a N = O,

i.e.

b = Ya N

then ,~ b = - ~ , and the spaces ~ u corresponding to other values of b are obtained by translating . ~ in the x-plane. When C = 1, the solution to Eq. (78) is independent of rank(P), and so the result of case 1 will apply here. For any elemental abundance vector b, any set of points in S is feasible. Thus the points depicted in Fig. 3(b), Fig. 4(b) yield solutions to their respective problems. When C = N - 1,R = 1 and the set of x 0 given by (88), i.e. ~ b , is a line in the direction talDifferent values of b give different lines, but all are parallel. Consider the ternary system with formula matrix given by Eq. (74). Recall that the truncated stoichiometric space _~ is generated by the single vector (1,1,0) T. Eq. (80) is, for b = ( b 1, b2) T,

(0 0)(xo -

1

1

Xo,2

Following (87), we obtain

(, ,)(. 0

0

x0,2

b) (:1)

-- b 2

Thus Y = b 3, and "~b is the line Xo,2 = x0,1 -t-(b2/b I - 3). As b varies we obtain a family of parallel lines, with slope 1. When b = a N = (1,3) v, -~b is the line Xo.2 = Xo,~ in the x-plane, which is the line generated by the truncated stoichiometric vector (1,1,0) f. The portion of these lines that lies in S is O~rb.l, the one-phase feasible region, Fig. 8(b) shows the one-phase feasible regions for b = (10,29) 7 and b ' = (10,31) v. Observe that ."~b,.~b, are

Y. Jiang et a l . / Fluid Phase Equilibria 118 (1996) 77-102

98

parallel. Multi-phase feasible points can be found using Lemma 6. In the figure, the set {xl,x 2} is feasible, in the sense that B YJ,Y2 such that (y~,y2,xl,x2) satisfies (41), (42) when b = (10,29) "r. However, (xl,x 2) is not feasible for b' = (10,31) "r. The feasibility constraints allow us to determine, for the formula matrix under consideration, which of the SCTQ-Sets indicated in Fig. 6(a), 6(b) yields the solution to the chemical equilibrium problem. In Fig. 6(b), b = (10,31) v and ~b is denoted by L. The reaction tangent plane has a single point of tangency to g ( x ) which, when projected onto the x-plane, lies on the one-phase feasible line ~rb. I. This single point is therefore feasible, and the problem has a one-phase solution. This situation applies for any value of b that leads to a one-phase feasible line lying in the region denoted by 1-phase in the figure. Fig. 6(a) corresponds to b = (10,29) T. There is no reaction tangent plane that is tangent to g(x) at a single point which, when projected onto the x-plane, lies on L. Rather, B two points of tangency x~, x 2 which, by construction, must lie on different sides of L. Thus the pair (x~, x 2) is feasible, and there is a two-phase solution. This situation arises whenever b gives rise to a one-phase feasible line 9-b. ~ that lies in the region denoted by 2-phase in the figure.

5. Summary The chemical equilibrium problem has been formulated as a generalisation of the phase equilibrium problem. The solution has been characterized algebraically and interpreted geometrically in a way that generalises the familiar solution to the phase equilibrium problem. As for phase equilibrium, the solutions are characterized in terms of tangent planes to the free energy surface, but these tangent planes are subject to additional constraints, that can be interpreted in terms of the stoichiometry of the system. There are two distinct cases, depending on whether all stoichiometric vectors have coordinates summing to zero, which can be distinguished by a simple algebraic criterion concerning the formula matrix. The tangent planes must also satisfy the reaction tangent-plane criterion (Smith et al., 1993) which is a generalization of the tangent-plane criterion of Baker et al. (1982) for the phase equilibrium problem. The feasibility constraints determine precisely which tangent plane yields the solution. By considering the feasible points that correspond to a single-phase state of the system, a connection is established between the tangent plane and the feasibility conditions that allows a complete geometric description of the solution to the chemical equilibrium problem. Although our analysis applies to any number of species and elements, we have provided concrete illustrations for all possible cases that may arise for binary and ternary systems. Finally, we believe that our methods and approaches are applicable to other aspects of chemical and phase equilibria. These include theoretical questions such as the phase rule, as well as the computational implications of our results.

6. List of symbols a i

A b

formula vector of ith substance system formula matrix elemental-abundance vector

Y. Jiang et al. / Fluid Phase Equilibria 118 (19961 77-102

C d d' E

f F g G I

M fl k, i

N P P

Q R sj S T X k

Yk Y

matrix A, augmented with row of unity elements rank of A t ! ( d I. . . . . d N_ l)'r / Y first N components of Eb elementary matrix molar Gibbs free energy function Gibbs free energy function molar Gibbs free energy function, in reduced dimension Gibbs free energy function, in reduced dimension identity matrix number of elements in system number of moles of species i in phase k number of species in system pressure, bar (a~

-

a N .....

a N - i-aN)

row-reduced form of P number of reactions sum of coordinates of a stoichiometric vector vj standard open simplex temperature composition vector for phase k, in reduced dimension composition vector for phase k total mass (moles) in phase k 7"/"

Y'. Yk, total number of moles in the system k~-I

Y/Y matrix arising in row reduction Z Greek letters" o~ :~ Lagrange multiplier /3 ~ Lagrange multiplier vj stoichiometric vector vl vJsj u".i point in the stoichiometric constraint space, ,~ ~number of phases present in non-zero amount at equilibrium p vector in the truncated stoichiometric space, ..~ O the reaction tangent plane & potential number of phases implied by the given G function Script letters set of allowable elemental abundance vectors ~b feasible region for given b one-phase feasible region for given b stoichiometric constraint space solution set of P x + a N 0 truncated stoichiometric space Z

=

99

100

Y. Jiang et a l . / F l u i d Phase Equilibria 118 (1996) 7 7 - 1 0 2

Subscripts i,j,k

d u m m y indices

Superscripts T

transposed vector or matrix

Acknowledgements Financial support has been received from the Natural Sciences and Engineering Research Council of Canada.

Appendix A A.1. Proof of Lemma 1 N

N

If 3j0 such that Y'~ ltjO, i ~ O, then any ~,j such that ~ i~l

~'j,i = 0 can be replaced by ( Uj,o + v j) and

i=l

a complete set of stoichiometric vectors is maintained. This proves the lemma.

A.2. Proof of Lemma 2 N

Now ~

uj, i = 0 V j if and only if

i=l

•

°

=0

,,;) i This is equivalent to (1 . . . . . 1) v being a linear combination of the columns of A T, i.e. a linear combination of the rows of A. This in turn is equivalent to rank(A') = rank(A) = C which proves the lemma.

A.3. Proof of Lemma 3 Now

C = rank(a I . . . . . aN) =

rank(aj

-- a N .....

aN-I

-- aN,--

aN)

= r a n k ( P , - aN) Thus the condition r a n k ( P ) = C is equivalent to r a n k ( P ) = r a n k ( P , - a N) which is necessary and sufficient for Px = - a N to have a solution.

Y. Jiang et al. / Fluid Phase Equilibria 118 (1996) 77-102

101

A.4. Proof of Lemma 4 N

Suppose =! uj such that ~

uj. i ~ 0. Then

i=l Pj,ial + ...+

Vj,Na N = 0

can be rewritten

~j.,(a,-an)

+ ...+ ~j.N-,(aN-,--a~)

= -a~

where vj. i = uj,i/(ujA + ...+ r'j,N). Hence Pv~ = - a N and so Px = - a N has a solution. By the last lemma, r a n k ( P ) = C. N

For the converse, suppose all stoichiometric vectors satisfy ~

uj,~ = 0. This means that

i=l

•

= 0

Thus ( 1 . . . . . 1)T is a linear combination of the columns of A T, or equivalently a linear combination of the rows of A. Hence ::! scalars 11 . . . . . I c not all 0, such that c Eaj.ihj j=l

-- 1 ( l < i < N )

Subtracting the Nth equation from the first ( N -

1) equations gives

C

Y'. ( a j . i - a j . N ) A j = 0 (1 < i < N -

1)

j= 1

If (rp)j is the jth row of P, the last equation is C

E Aj(rp)j = 0 j=l

Thus there is a dependency between the rows of P, and so rank(P) L e m m a 4.

< C. This completes the proof of

Appendix B Proof that ~ (3 S = 05 . Suppose x0e ,"~. Then Eq. (44) can be written N-

I

(90)

x 0 , i ( a i - aN) = --a N i=l

i.e.

E

i=l

Xo.iai q-- a N

x0, i i=l

~ 0

(91)

102

Y. Jiang et al. / Fluid Phase Equilibria 118 (1996) 77-102

If, in addition, x o E S then N-I

Y'~ x0, i < 1,x0. i > 0

(1 < i N N -

1)

(92)

i=l

Eq. (92), together with the assumption that a~ > 0, a i :g 0, contradict (91). Thus no such x o exists, and ~.~ A S = q~.

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