# 6.17: Untitled Page 140

## Chapter 6

R

R

 0  2 R

(13b)

C2H4

H2O

C4H6

R

0

R

R

(13c)

CH3CHO

H2

C H

4

6

In this example we have provided a template for the solution of Eq. 6‐22 in which the mathematical steps are always the same, i.e., (I) begin with the atomic matrix given by Eq. 2 and develop the row reduced echelon form given by Eq. 3, (II) partition Axiom II as indicated by Eqs. 8 and 9, and (III) carry out the matrix multiplication to obtain the net rates of production for the non‐pivot species indicated by Eqs. 13.

In the design of a butadiene production unit, the representation for the net rate of production of the non‐pivot species is a crucial part of the analysis. In Chapter 7 we will apply this type of analysis to several rocesses.

p

6.3.1 Rank of the atomic matrix

In the previous paragraphs we have seen several examples of the atomic matrix when the rank of that matrix was less than the number of molecular species, i.e., r  N . Here we wish to illustrate two special cases in which r  N

and r  N . First we consider a reactor containing only methyl chloride ( CH Cl ), ethyl chloride ( C H Cl ), and chlorine ( Cl ). We illustrate the atomic 3

2

5

2

matrix by

Molecular Species  CH Cl C H Cl

Cl

3

2

5

2

carbon

 1

2

0 

(6‐63)

hydrogen

3

5

0

chlorine

 1

1

2 

and use elementary row operations to obtain the row reduced echelon form given by

 1 0 0 

A

0 1 0

(6‐64)

 0 0 1 

Use of this result in Axiom II provides

Stoichiometry

259

 1 0 0  R CH

0

3Cl 

 

0 1 0  R

 

0

(6‐65)

C

 

2H5Cl

 0 0 1 

0

 

R

 

Cl

2

and we immediately obtain the trivial solution expressed as R

 0

(6‐66a)

CH3Cl

R

 0

(6‐66b)

C2H5Cl

R

 0

(6‐66c)

Cl2

This indicates that no net rate of production can occur in a reactor containing only methyl chloride, ethyl chloride and chlorine. If we allow molecular hydrogen to be present in the reactor, our system is described by

Molecular Species  CH Cl C H Cl

Cl

H

3

2

5

2

2

carbon

 1

2

0

0 

(6‐67)

hydrogen

3

5

0

2

chlorine

 1

1

2

0 

and the atomic matrix takes the form

 1 2 0 0 

A

3 5 0 2

(6‐68)

 1 1 2 0 

The row reduced echelon form of this atomic matrix can be expressed as

 1 0 0

4 

A

0 1

0 2

(6‐69)

 0 0 1 1 

and from Axiom II we obtain

R

CH3Cl

 1 0 0

4 

0

R

C H Cl

 

2

5

0 1

0 2 

 

0

(6‐70)

 

 

R Cl

 

2

0 0

1

1

0

 

 

R

H

2

260