6.13: Untitled Page 136

Chapter 6

Here we have followed the same style used in Example 6.2 and placed the pivot species on the right hand side of Eqs. 8. In matrix notation this result can be expressed as (see Sec. 6.4)

R

 

 

R

C

1

1

CO

(9)

R

R

O

1 2

1

2

C





O2 





pivot matrix

column matrix

column matrix

of non‐pivot species

of pivot species

in which the 2  2 matrix is referred to as the pivot matrix since it is the matrix that maps the net rates of production of the pivot species onto the net rates of production of the non‐pivot species. Other possibilities can be constructed by using different pivot species and the development of these has been left as an exercise for the student.

The partial oxidation of carbon is an especially simple example of multiple independent stoichiometric equations, i.e., rank N   N  1

JA

. The partial

oxidation of ethane, illustrated in Eqs. 6‐10 through 6‐12, provides a more challenging problem.

EXAMPLE 6.4. Partial oxidation of ethane

As an example only, we imagine that the process illustrated in Figure 6‐2

is carried out so that ethane is partially oxidized to produce ethylene oxide, carbon dioxide, carbon monoxide and water. Thus the molecular species involved in the process are assumed to be

Molecular species:

C H , O , H O , CO , CO , C H O

(1)

2

6

2

2

2

2

4

and the rates of production for these species are constrained by A  6

Axiom II

N

R

 0 ,

J  C ,

,

(2)

H O

JA

A

A  1

A visual representation of the atomic matrix is given by Molecular Species  C H

O

H O CO CO

C H O

2

6

2

2

2

2

4

carbon

 2

0

0

1

1

2

 (3)

hydrogen

6

0

2

0

0

4

oxygen

 0

2

1

1

2

1

and the matrix representation of Eq. 2 takes the form

Stoichiometry

251

R

C2H6

R

O2

2 0 0 1 1 2 

0

  R H O 

 

Axiom II:

2

6 0 2 0 0 4

 

 

0

(4)

 

R

 

CO

0 2 1 1 2 1

0

 

R CO

2

R

C

2H4O 

By a series of elementary row operations we can transform the atomic matrix to the row reduced echelon form so that Eq. 4 can be expressed as (see Sec. 6.5.2)

R

C2H6

R

O2

1 0 0

1 2

1 2

1 

0

  R H O 

 

Axiom II:

2

0 1 0

5 4

7 4

1

 

 

0

 

(5)

R

  

CO

0 0 1

3 2

3 2

1

0

 

R CO

2

R

C

2H4O 

Here we see that the rank of the atomic matrix is three, r  3 , thus the rank is less than N which is equal to six. Since the rank  N a non‐trivial solution exists consisting of three independent equations given by 1

1

R

 

C

R

R

R

(6a)

2H6

CO

CO

2

2

2

C2H4O

5

7

R

 

O

R

R

R

2

CO

CO

4

4

2

C2H4O

(6b)

3

3

R

R

R

R

(6c)

H2O

CO

CO

2

2

2

C H O

2

4

Here we have chosen CO , CO , and C H O as the pivot species with the 2

2

4

idea that the rates of production for these species will be determined experimentally. Given the rates of production for the pivot species, Eqs. 6

can be used to determine the rates of production for the non‐pivot species, C H , O , and H O .

2

6

2

2

In this section we have illustrated how Axiom II, given by Eqs. 6‐20 or by Eq. 6‐22, is used to constrain the net rates of production. When we have a single independent stoichiometric reaction, such as the complete combustion of ethane, one need only measure a single net rate of production in order to determine all the

252