multistage reactions, equilibrium, and detailed balancing: a linear algebra approach

INTERNATIONAL JOURNAL OF CHEMICAL KINETICS, VOL. 111, 491-499 (1971)

Multistage Reactions, Equilibrium, and Detailed Balancing:

A Linear Algebra Approach J . LEE

Department of Chemistry, The Uniuersity of Manchester Institute .f Science and Technology, Manchester M60 140, England

Abstract

Methods of linear algebra are applied to (a) the problem of determining whether detailed balancing for some or all stages of a chemical process is implicit from the existence of steady state for some or all of the chemical species involved, and (b) the formulation of general concentration-type equilibrium relationships solely from kinetic considerations and the assumption of full detailed balancing.

1. Introduction

The second law of thermodynamics necessitates that the equilibrium concentrations of all species involved in a closed chemically reacting system are functions only of temperature, pressure, and initial reactant concentrations. All species attain steady-state concentrations, but it is well known [ l ] that in certain multistage reactions this equilibrium steady state (SS) does not necessarily imply detailed balancing in each component stage (DB). The example most frequently cited [ 11 is the cyclic unirnolecular sequence

' C k2

' B , k ,

A ,

SS requires that

(1)

On the other hand, DB requires that

(2) [Bl/[Al = k l / k ;

and

(3) k,kzki = k;k;ka

49 1 @ 1971 by John Wiley & Sons, Inc.

492 LEE

While eqs. (2) and (3) together necessarily imply eq. ( l ) , the converse is not true, i.e., DB 3 SS but SS Z, DB (Z, meaning here "does not necessarily imply"). This point has been stressed by Landsberg [2, 31, who by using the method of transition probabilities has analyzed in general nonchemical terms the various implication relationships between detailed balancing, steady state, and a number of statistical mechanical principles, including that of microscopic reversibility.

The principal object of the present paper is to generalize in matrix terms the requirements for a closed, uniform, multistage, chemical process to conform (a) to the relationship DB 2 SS and SS Z, DB or (b) to the converse of this, the situation in which both DB 2 SS and SS 2 DB. The treatment is also extended to the nonequilibrium situation where steady state is attained for some of the species present-referred to herein as kinetic steady state (KSS). In certain cases, SS or KSS implies detailed balancing of some but not all of the reaction stages-DBi signifies its occurrence in stage i. Examples have been chosen from both real and hypothetical (generalized) reaction schemes.

2. General Theory

Consider a closed, uniform system in which there occurs a chemical process involving t reaction stages, to each of which the law of mass action applies, and c chemical species, designated A,, A2, * - - -, A,. For the present purpose, let each stage be individually numbered, not necessarily sequentially, and suppose that stage s is represented by the chemical equation

Since all c species will not generally be involved in this particular stage, many of the stoichiometric coefficients ajs, u3f8 will be zero. Furthermore, at least one of each pair ajS and ~~i will usually be zero; this is certainly the case with the examples in section 3. The net rate of reaction in this stage, as defined, in terms of a particular species A,, by (a;, - Q ~ ~ ) - ' d[A, ] /d t is

k , fi - k: [A,]":' 3 = I 3 = I

Thus, letting 6,, = u;, - ups,

C C

d[A,]/dt (by stage s) = 6,,(k, n [AllaJs - k: n [A,]":*) 3 = I 3 = I

and by all t stages of the reaction

t C C

d [A , ] /d t = C 6,,(k, n - k: n [Aj]":.) S = l j = I = I

DETAILED BALANCING AND LINEAR ALGEBRA 493

Some of the 6,, will be zero but this does not affect the present arguments. species A , attains a steady state concentration

When

(4) 1 C C C 6,,(ks r]: [Aj]”” - k ; IT [Aj]”;.a) = 0

8 = 1 j= l j = 1

For a thermodynamic equilibrium, eq. (4) applies for all species, p = 1 to c (SS), but for the usual steady-state assumption of reaction kinetics, eq. (4) applies to a lesser number of species (intermediates), say for p = 1 to d (KSS). Detailed balancing in stage s (DB,) necessitates that d[A,]/dt (by that stage), for any species A,, for which 6,, # 0, is zero, i.e.,

C C

f8 k , n [ A J ~ J ~ - k: IT [A,]“:. = o 3 = 1 ] =1

( 5 )

Clearly the applicability of eq. (5) to all stages of the reaction (DB) would automatically imply the validity of eq. (4)-DB 2 SS-but the converse implications are not automatic.

The following questions arise: (a) When does SS 3 DB? (b) When does SS or KSS 2 DB,?

The set of equation (4) may be written in matrix form:

(6) 6f = 0

where

6 = (! 2 ::: ’:), = (I) 6bt f i b l . . . . . . . . .

and h = c (SS) or d (KSS, d < c).

A. b >, t (Type I )

square matrix 6 satisfying

Two possibilities then arise:

When b > t , a set of b - t rows may be omitted from the matrix 6 to form a

6,f = 0

When b = t , 6 ,, = 6. If, for any one square matrix so obtained, 16 I # 0, then f = 0 and SS (or KSS) 2 DB; note that if KSS 3 DB, then, since DB 3 SS, automatically KSS 2 SS. All 16 I will be zero when one of the columns of 6 (a) comprises all zero elements--this is a KSS possibility which arises when one stage (at least) does not involve steady-state species, or (b) is a linear combination of the other columns. This will occur when there is a cycle in the reaction scheme (as in the A, B, C sequence referred to above). In the event of all 6 I3 being singular, it is only possible to relate the elements of f to one another. For example, if

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there are no zero columns and if there is a single cycle in the reaction scheme, the rank of the 6 matrices will be t - 1. Suppose that column t is a linear combination of other rows and that omission of this column and the last b - t + 1 rows of 6 produces a nonsingular matrix, then transference offt terms to the right side gives

( 7) ( f t )

ft -1 6, -1,t

If there are two cycles and the rank of the 6 is t - 2, it will be necessary to trans- fer two elements of f to the right side of the equation, etc.

B. b < t (Type IZ)

This situation is most commonly found for KSS ( b = d < c). Equation (6) in this case can be written

t 8)

Here, the last t - b columns of 6 and the corresponding elements of f have been transferred to the right side of the equation, but of course any set of b columns of 6 and the corresponding f elements may be so transferred. The resultant generali- zation may be written

If 6, is nonsingular,

f = - a - a 's& (9)

If row s of 6,'6, comprises all zero elements, fa8 = 0 and SS (or KSS) 3 DB,, (as being the stage corresponding to the zero element of fa). In the largely hypothetical case where 6 , comprises all zero elements,fl,f2, * . . . , fb = 0. The elements of f, have arbitrary values, and one can make no conclusion about microscopic reversibility in the B stages, i.e., SS (or KSS) 2j DB. These considerations also apply to eq. (7) and its generalizations. I t will always be possible to find at least one nonsingular 6, if the number of stages not involving steady-state species or the number of cycles does not exceed t - b. If all 6, are singular, it is necessary to adopt a procedure analogous to that for type I when all 6, are


singular, last row and the last column,

If, for example, in eq. (8), 6, is made nonsingular by omission of the

3. Examples

A. Generalized k ,

(a) A B C; A C (as h u e ) . Number of species, c = 3; number

of stages, t = 3; number of cycles = 1 (6x3 = 6Xl + 6x2 for X = A, B, or C). For b , the number of steady state species = c, this example is of type I, and the existence of a cycle then necessitates that SS Z, DB (as is well known). The rank of 6 is 2, and eq. (7) becomes

Insertion of numerical values for the 6 elements then gives fl = f 2 = -f3 at SS. kr

(b) A + B F C. Number of species, c = 3; t = 2; number of cycles = 0. When b = c, this is a type-I example for which all 6 , are nonsingular and for which therefore SS 3 DB. This can in fact be readily confirmed by kinetic analysis. The assumption of steady state in any two of the species (6 = d = 2) still conforms to type I with 16 I # 0 and KSS 3 DB 3 SS. The common steady- state assumption in reaction kinetics, however, is for species B only (6 = l) , and now we have type I1 in which both 6, are nonsingular. One of the two eqs. (7) is

(1 1) (fl) = - (6Bl)-1(~E?>(f2>

the matrices are of one element and the example is rather trivial. aB1 = 1, 6,, = - 1, and so fl = fi (a fact easily established by normal kinetic analysis), but neither fl nor f2 are necessarily zero.

k i kr k h - I B; A X(2) R; . . . .; A 7 X(n) A B.

m -1 T (c) A X(') k; Number of species, c = n + 2 ; t = 2n; number of cycles = n - 1 (Sj1 + Sj2 =

identical to example (b). When n = 2, c = t , and there is a single cycle, a type-I situation for which SS Z, DB and for which application of eq. (7) gives fl = f2 =

-f3 = -f4. When n > 2, c < t and the number of cycles (= n - 1) > t - c (= n - 2). This is a type-11 case for which all 6, are singular and for which therefore eq. (9) is inapplicable; the maximum rank of 6, matrices is n + 1, and this is therefore a situation where an analog of eq. (10) applies. The result of the

63, + 6, = . . . . - - 6j,2n-1 + 6j,2,, for all speciesj). For n = 1, the example is

496 LEE

applicationisthatfi = f i , f 3 = f 4 , . - - , f i n - - l =finandfi + f 3 + + f i n - 1 = 0. If it is assumed that only the intermediates (X"' to X'"') attain steady-state concentrations, d = n and conformation to type I1 exists for all n. Furthermore, eq. (9) is always applicable-number of cycles < t - d (= n) - and results in

f 1 = f i , f 3 = f4, ..., f i n - l = fin, only as one obviously would expect.

B. T h e hydrogen-bromine reaction

This reaction will be treated in two ways (a) as a five-stage process: kz

Br, 2Br; Br + Hi H + HBr; H + Br, HBr + Br;

Hi 2H; H + Br HBr

'k, k, kr

and (b) as a three-stage process comprising the first three of the above stages. For b = c, this is a type-I example, and in fact 16 I = 0.

This arises because of the existence of two cycles in the reaction scheme: chemical eq. (2) is the sum of eqs. (4) and ( 5 ) , and eq. (3) is the sum of eqs. (1) and (5) i.e., 6j2 = Sj4 + ~$5, and Sj3 = 6 j l + 4 5 for all speciesj. The existence of such cycles necessitates that SS D DB. The rank of 6 is 3, and the analog of eq. (7) is

(a) c = 5 , t = 5 .

f i = - 6Brl :::2 :::3 1' f::4 ?5 ) (k) (t) ( I L r l 6HBr2 "HBr3 6HBr4 8HBr5

Insertion of values for the elements of 6 then gives f i = -f4 + f 5 , , f 2 = - f4 and f 3 = f 4 - j5. If only H and Br are assumed to attain steady-state concentrations, b = d = 2, and since (2) d < 1, (ii) the number of cycles (= 2) > I - b (= 3), (iii) H and/or Br are involved in all stages, eq. (9) may be applied:

H3 8H4 "H5) (i) (2) = - (?I :::)-' c B r 3 8Br4 6Br5

whence f i = -f4 + f 5 , and f 2 = f 3 - 2f4 + f5. Neither for SS nor KSS are any of the f elements necessarily zero and SS (or KSS) D DB, (s = 1, 2, * * * , or 5 ) .

(b) c = 5 , t = 3, number of cycles = 0. This type-I situation is thus charac- terized by SS 2 DB; this is in contrast to the fuller reaction scheme of (a). (The implication can be established also by kinetic analysis.) The assumption of steady state for H and Br only and the satisfaction of criteria (i), (ii), and (iii) as in (a) above permits the application of eq. (9), which gives

Insertion of numerical values then gives


ix . , f l = 0, f 2 = f3 (in agreement with the result of kinetic analysis). KSS (for H and Br) 2 DB1.

Thus, The zero element of -6a-16, is in fact

(8H38Br2 - 6H26Br3)/(6H26Brl - 6H18Br2)

for which the numerator is zero, since clearly 8H2/6H3 = 6&2/1&

4. Operational Analysis of Reaction Schemes

The procedures described in section 2 and exemplified in section 3 are sum- marized as an operational flowsheet in Figure 1.

5. Detailed Balancing and Equilibrium Constants

General equilibrium relationships are normally deduced from the second law equilibrium requirement that, at constant temperature and pressure, the affinities of each reaction stage and of all linear combinations of these are zero. The em- ployment of kinetic arguments to deduce equilibrium relationships has often been criticized [4], and certainly such criticism is justified (a) when no account is taken of the multistage nature of a process or (b) when the effects of nonideality are to be contemplated. Nevertheless, general equilibrium relationships (based upon concentration) can be properly deduced by kinetic considerations of a multistage process on the basis of DB, established as a result of SS or assumed additionally to SS; this assumption is often taken as a starting point for the Onsager reciprocal relationships [ 11. The treatment of equilibrium may be generalized in linear algebraic terms, akin to those used above and to those used by Copley [5] in the consideration of chemical equations.

As a result of DB, and the applicability of eq. ( 5 ) ,

where m, is an arbitary power factor, appropriate to stage s. eq. (13) will apply to all stages, i.e., s = 1 to t . and right sides of eq. (13),

If DB is assumed, Thus, taking products of left sides

This represents the most general form of an equilibrium relationship; but suppose that we wish to be more restrictive and make

1 C rns6 j s = o

for some of the chemical species present in the reaction scheme, say, for j = 1 to d. In this case,

6m = 0

s =1

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e a c t i o n e t a

xceed nurbe

pres..ed as cornbin ions or o t h e r s , on of o t h e r s , o r (10) nnd t h e i r g:cnerol-

b of t h e f i can be expressed i n t e r m of remain ing t-b --see equations (8) and ( 9 )

Figure 1. Operational analysis of reaction schemes.

where

The equation is analogous to eq. (6) and it may be subjected to a similar kind Equations analogous to eqs. (7) to (lo), with m replacing f elements,

The case of m = 0 of analysis. may be applied, and types I and I1 may be distinguished.


would, of course, be completely trivial and of no interest. Certainly, the common situation here is type I1 (d < t ) and will be illustrated here by two of the kinetic schemes given above, viz., (Ab) and (Bb).

(Ab) Suppose that species B is to be eliminated from eq. (13): d = 1, t = 2, number of cycles = 0, number of stages not involving B = 0. The analog of eq. (9) is then equivalent to eq. (1 l ) , and thus ml = m2. Combination of this result with eq. (13) and the insertion of values for aB1 and SB2 then gives

[CI/[AI = k i k ~ ’ k ; G

(Bb) Suppose that the two atomic species are to be eliminated from eq. (13). Again the analog of eq. (9) can be applied, and this is equivalent to eq. (12), with the consequent result that m, = 0 and m2 = m3. Combination of these results with numerical values of 6 j , ( j = H, Br; s = 1,2,3) and eq. (13) yields

[HBrI2/ [H21 [BrA = k&,/kSG

The examples here are perhaps rather trivial and the results may certainly They do, however, serve to illustrate the principles be derived in simpler ways.

and the analogy to detailed balancing considerations.

Bibliography

[ l ] K. G. Denbigh, “The Thermodynamics of the Steady State,” Methuen, London, 1951,

[2] P. T. Landsberg, “Thermodynamics,” Interscience, New York, 1961, Chap. V, p. 356. [3] P. T. Landsberg, Phys. Rev., 96, 1420 (1954). [4] P. G. Ashmore, Educ. Chem., 2, 160 (1965). [5] G. N. Copley, Chemistry, 41, 22 (1968).

Chap. 111, p. 31.

Received October 2, 1970. Revised April 8, 197 1.

multistage reactions, equilibrium, and detailed balancing: a linear algebra approach

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