Page 319 - Thermodynamics of Biochemical Reactions

Matrices in Chemical and Biochemical Thermodynamics 319

1
TableForm[Transpose [NullSpace [a] 1
-2 -1
-2 -3
1 1
0 1

These are the same two reactions as in equation 5.1-20.

(b) Equations 5.1-22 to 5.1-26.
Now we start with the transpose of the stoichiometric number matrix.

nutr=Transpose[nul
{{-I, -3, 1, 1, O}, I-1, 1, 0, -1, 111

0 0 0
0 0 0

Now calculate a basis for the transpose of the conservation matrix from the transposed stoichiometric number matrix.

TableForm[NullSgace [nutrl I

3 - 1 0 0 4
-11 0 2 0
1 1 4 0 0

This looks different from equation 5.1-8, but yields 5.1-15 when RowReduction is used.

TableFonn[RowReduce~NullSpace[nutrll1
1 0 0 1 2
0 1 0 3 2
0 0 1 -1 -1

The rank of the conservation matrix is the number of components, which is 3.

Dimensions [a]
(3, 5)

The rank of the stoichiometric number matrix is equal to the number of independent reactions, which is 2.

Dimensions [nu]
(5, 2)

(c) Calculate the amounts of the components C, H, and 0 for a system containing one mole of each of the five species.
Equation 5.1-27 yields

a. {l,l,l, 1,l)