# 6.15: Untitled Page 138

## Chapter 6

a b

a b

a b

a b

a b

c

11 1

12 2

13 3

14 4

15 5

1

 

a b

a b

a b

a b

a b

c

(6‐62)

 21 1

22 2

23 3

24 4

25 5 

 2 

a b a b

a b

a b

a b

c

31 1

32 2

33 3

34 4

35 5 

 3

This is an alternate representation of Eq. 6‐52 that immediately leads to Eqs. 6‐53.

A more detailed discussion of matrix multiplication and partitioning is given in Appendix C1; however, the results in this section are sufficient for our treatment of Axiom II. In the following paragraphs we learn that partitioning of the atomic matrix leads to especially useful forms of Axiom II.

6.3 Pivots and Non‐Pivots

In the previous section we illustrated that the number of constraining equations associated with Axiom II is equal to the rank  r of the atomic matrix which is less than or equal to the number of atomic species, T . Because of this, the number of pivot species must be equal to N  r and the number of the non-pivot species must be equal to the rank of the atomic matrix, r . The choice of pivot species and non‐pivot species is not completely arbitrary since it is a necessary condition that all the atomic species be present in at least one non‐pivot species. In this section we consider the issue of pivots and non‐pivots in terms of an example and some analysis using matrix partitioning that was discussed in Sec. 6.2.6.

EXAMPLE 6.5. Production of butadiene from ethanol (Kvisle et al., 1988) Ethanol produced by fermentation of natural sugars from grain can be used in the production of butadiene which is a basic feedstock for the production of synthetic rubber. The following molecular species are involved in the production of butadiene ( C H ) from ethanol ( C H OH ): 4

6

2

5

C H OH ,

C H ,

H O ,

CH CHO ,

H ,

C H

2

5

2

4

2

3

2

4

6

Since ethanol is the reactant, it is reasonable to arrange the atomic matrix in the form

Molecular Species  C H OH C H

H O CH CHO H

C H

2

5

2

4

2

3

2

4

6

carbon

2

2

0

2

0

4 

hydrogen

6

4

2

4

2

6 

oxygen

1

0

1

1

0

0 

(1)

Stoichiometry

255

If we assume that the rank of the matrix is three ( r  rank  3 ) we are confronted with six unknowns and three equations, thus the non‐pivot species are represented by C H OH , C H and H O . The atomic 2

5

2

4

2

matrix can be expressed explicitly by

2 2 0 2 0 4

A

6 4 2 4 2 6

(2)

1 0 1 1 0 0

and a series of elementary row operations leads to the row echelon form given by

 1

1

0

1

0

2 



A

0

1 1

1 1

3

(3)

 0

0

0

1 1

1 

This indicates that the rank of the atomic matrix is three ( rank  3 ) and we have three independent equations to determine the six rates of production. The calculation indicated in Eq. 6‐22 is given by

R

 C2H5OH 

R

C H

1

1

0

1

0

2 

2

4

0

  R H O 

 

0

1 1

1 

2

1

3 

 

0

(4)

 

R

0

0

0

1 1

1  CH CHO 

 

3

0

 

R H2

R C4 H6 

At this point we see that the atomic matrix is not in the row reduced echelon form; however, we can obtain this form by means of a column / row interchange. In terms of Eq. 6‐20 we express a judicious choice of a column/row interchange as

N

R N

R ,

B, D  H O, CH CHO ,

J  C, H, O

(5)

JC

C

JD

D

2

3

in order to express Eq. 4 in the form

256