# 8.24: Untitled Page 213

## Chapter 9

Figure 9‐1. Production of hydrogen bromide

to completion since both hydrogen and bromine appear in the product stream.

Here it is important to note that the products of a chemical reaction are determined by experiment, and in this case experimental data indicate that hydrogen bromide can be produced by reacting hydrogen and bromine. For the process illustrated in Figure 9‐1, the visual representation of the atomic matrix takes the form

Molecular species  H

Br

HBr

2

2

hydrogen

 2

0

1 

(9‐1)

bromine

 0

2

1 

and the elements of this matrix can be expressed explicitly as

2 0 1

2 0 1

N  

or

A

(9‐2)

JA

0 2 1

0 2 1

The components of N

are used with Axiom II

JA

A N

Axiom II

N R

 0 ,

J  1 , 2 ,...,T

(9‐3)

JA A

A  1

to develop the stoichiometric relations between the three net rates of production represented by R

, R

, and R

. For the atomic matrix given by Eq. 9‐2 we

H2

2

Br

HBr

see that Axiom II provides

Reaction Kinetics

397

R

H

2

2 0 1

0

Axiom II:

R

 

 

(9‐4)

2

Br

0 2 1 

0

R

 HBr 

and the use of the row reduced echelon form of the atomic matrix leads to

R

H

2

1 0

1 2

0

R

 

 

(9‐5)

2

Br

0 1 1 2 

0

R

 HBr 

If hydrogen bromide ( HBr ) is chosen to be the single pivot species (see Sec. 6.4) we can express Axiom II in the form

R

H

1 2

Pivot Theorem:

2

 

R

(9‐6)

 HBr

R

1 2

2

Br 



pivot matrix

This matrix equation provides the following representations for the net rates of production of hydrogen and bromine

Local Stoichiometry:

1

1

R

  R

,

R

  R

(9‐7)

H2

HBr

2

2

Br

HBr

2

At this point we wish to apply Axioms I and II to the control volume illustrated in Figure 9‐1. The axioms are given by Eqs. 7‐4 and 7‐5 and repeated here as d

Axiom I:

c dV

c

dA  R ,

A  1 , 2 ,..., N

v n

(9‐8)

A

A

A

A

dt V

A

A N

Axiom II

N R

 0 , J  1 , 2 ,...,T

(9‐9)

JA A

A  1

For the particular case under consideration, Eq. 9‐8 leads to

Q c

Qc

 R ,

A  H , Br , HBr

(9‐10)

A entrance

A exit

A

2

2

while Eq. 9‐9 takes the special form given by

Global Stoichiometry:

R

  1 R

,

R

  1

H

R

2

HBr

Br

2

2

2

HBr

(9‐11)   398