# 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 $$\PageIndex{1}$$: Production of butadiene from ethanol8

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 ($$\ce{C4} \ce{H6}$$) from ethanol ($$\ce{C2H5OH}$$):

$\ce{C2}\ce{H5} \ce{OH} , \quad \ce{C2} \ce{H4} , \quad \ce{H2O} , \quad \ce{CH3} \ce{CHO} , \quad \ce{H2} , \quad \ce{C4} \ce{H6} \nonumber$

Since both ethanol and ethylene are reactants, it is reasonable to arrange the chemical composition matrix in the form

$\text{ Molecular Species} \to \ce{C2H5OH} \quad \ce{C2} \ce{H4} \quad \ce{H2O} \quad \ce{CH3} \ce{CHO} \quad \ce{H2} \quad \ce{C2} \ce{H6} \\ \begin{matrix} {carbon} \\ { hydrogen} \\ {oxygen} \end{matrix} \begin{bmatrix} {2} & {2} & { 0} & { 2} & { 0} & { 4 } \\ {6} & {4} & { 2} & { 4} & { 2} & { 6 } \\ {1} & {0} & { 1} & { 1} & { 0} & { 0 } \end{bmatrix} \label{1a}\tag{1}$

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 $$\ce{C2H5OH}$$, $$\ce{C2} \ce{H4}$$ and $$\ce{H2O}$$. The atomic matrix can be expressed explicitly by

$\mathbf{A}=\begin{bmatrix} {2} & {2} & {0} & {2} & {0} & {4} \\ {6} & {4} & {2} & {4} & {2} & {6} \\ {1} & {0} & {1} & {1} & {0} & {0} \end{bmatrix} \label{2a}\tag{2}$

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

$\mathbf{A}^{{\prime} {\prime} {\prime} } =\begin{bmatrix} {1} & { 1} & { 0} & { 1} & { 0} & { 2} \\ {0} & { 1} & {-1} & { 1} & {-1} & { 3} \\ {0} & { 0} & { 0} & { 1} & {-1} & {1} \end{bmatrix} \label{3a}\tag{3}$

This indicates that the rank of the matrix is three ($$rank=3$$) and we have three independent equations to determine the six rates of production. The calculation represented by Equation $$(6.2.8)$$ is given by

$\begin{bmatrix} {1} & { 1} & { 0} & { 1} & { 0} & { 2} \\ {0} & { 1} & {-1} & { 1} & {-1} & { 3} \\ {0} & { 0} & { 0} & { 1} & {-1} & {1} \end{bmatrix} \begin{bmatrix} R_{\ce{C2H5OH} } \\ R_{\ce{C2} \ce{H4} } \\ R_{ \ce{H2O}} \\ R_{\ce{CH3} \ce{CHO} } \\ R_{\ce{H2} } \\ R_{\ce{C4} \ce{H6} } \end{bmatrix}=\begin{bmatrix} {0} \\ {0} \\ {0} \end{bmatrix} \label{4a}\tag{4}$

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. $$(6.2.8)$$ we express a judicious choice of a column/row interchange as

$N_{J C} R_{C} \rightleftarrows N_{J D} R_{D}, \quad B, D \Rightarrow \ce{H2O},\ce{CH3CHO}, \quad J \Rightarrow \ce{C},\ce{H},\ce{O} \label{5a}\tag{5}$

in order to express Equation \ref{4a} as

$\begin{bmatrix} {1} & { 1} & { 1} & { 0} & { 0} & { 2} \\ {0} & { 1} & { 1} & {-1} & { -1} & { 3} \\ {0} & { 0} & { 1} & { 0} & {-1} & {1} \end{bmatrix} \begin{bmatrix} R_{\ce{C2H5OH} } \\ R_{\ce{C2} \ce{H4} } \\ R_{ \ce{CH3} \ce{CHO} } \\ R_{\ce{H2O} } \\ R_{\ce{H2}} \\ R_{\ce{C4} \ce{H6} } \end{bmatrix}=\begin{bmatrix} {0} \\ {0} \\ {0} \end{bmatrix} \label{6a}\tag{6}$

Here we note that our original choice of non-pivot species, $$\ce{C2H5OH}$$, $$\ce{C2H4}$$ and $$\ce{H2O}$$, has been changed by the application of Equation \ref{5a} that leads to the non-pivot species represented by $$\ce{C2H5OH}$$, $$\ce{C2H4}$$ and $$\ce{CH3CHO}$$

At this point we make use of some routine elementary row operations to obtain the desired row reduced echelon form

$\begin{bmatrix} {1} & { 0} & { 0} & { 1} & { 1} & {- 1} \\ {0} & { 1} & { 0} & {-1} & { 0} & { 2} \\ {0} & { 0} & { 1} & { 0} & {-1} & {1} \end{bmatrix} \begin{bmatrix} R_{\ce{C2H5OH} } \\ R_{\ce{C2} \ce{H4} } \\ R_{ \ce{CH3} \ce{CHO} } \\ R_{\ce{H2O} } \\ R_{\ce{H2} } \\ R_{\ce{C4} \ce{H6} } \end{bmatrix}=\begin{bmatrix} {0} \\ {0} \\ {0} \end{bmatrix} \label{7a}\tag{7}$

Given this representation of Axiom II we can apply a column/row partition illustrated by

$\begin{bmatrix} {1} & { 0} & { 0}& \vdots & { 1} & { 1} & {- 1} \\ {0} & { 1} & { 0} & \vdots & {-1} & { 0} & { 2} \\ {0} & { 0} & { 1} & \vdots & { 0} & {-1} & {1} \end{bmatrix} \begin{bmatrix} R_{\ce{C2H5OH} } \\ R_{\ce{C2} \ce{H4} } \\ R_{ \ce{CH3} \ce{CHO} } \\ \hdashline R_{\ce{H2O} } \\ R_{\ce{H2} } \\ R_{\ce{C4} \ce{H6} } \end{bmatrix}=\begin{bmatrix} {0} \\ {0} \\ {0} \end{bmatrix} \label{8a}\tag{8}$

$\underbrace{\begin{bmatrix} {1} & { 0} & { 0} \\ {0} & { 1} & { 0} \\ {0} & { 0} & { 1} \end{bmatrix}}_{\text{non-pivot submatrix}} \begin{bmatrix} R_{\ce{C2H5OH} } \\ R_{\ce{C2} \ce{H4} } \\ R_{ \ce{CH3} \ce{CHO} } \end{bmatrix} + \underbrace{\begin{bmatrix} { 1} & { 1} & {- 1} \\ {-1} & { 0} & { 2} \\ { 0} & {-1} & {1} \end{bmatrix}}_{\text{pivot submatrix}} \begin{bmatrix} R_{\ce{H2O} } \\ R_{\ce{H2} } \\ R_{\ce{C4} \ce{H6} } \end{bmatrix} =\begin{bmatrix} {0} \\ {0} \\ {0} \end{bmatrix} \label{9a}\tag{9}$

Here the non-pivot submatrix is the unit matrix that maps a column matrix onto itself as indicated by

$\underbrace{\begin{bmatrix} {1} & { 0} & { 0} \\ {0} & { 1} & { 0} \\ {0} & { 0} & { 1} \end{bmatrix}}_{\text{non-pivot submatrix}} \begin{bmatrix} R_{\ce{C2H5OH} } \\ R_{\ce{C2} \ce{H4} } \\ R_{ \ce{CH3} \ce{CHO} } \end{bmatrix} = \begin{bmatrix} R_{\ce{C2H5OH} } \\ R_{\ce{C2} \ce{H4} } \\ R_{ \ce{CH3} \ce{CHO} } \end{bmatrix} \label{10a}\tag{10}$

Substitution of this result in Equation \ref{9a} provides the simple form given by

$\begin{bmatrix} R_{\ce{C2H5OH} } \\ R_{\ce{C2} \ce{H4} } \\ R_{ \ce{CH3} \ce{CHO} } \end{bmatrix} = -\underbrace{\begin{bmatrix} { 1} & { 1} & {- 1} \\ {-1} & { 0} & { 2} \\ { 0} & {-1} & {1} \end{bmatrix}}_{\text{pivot submatrix}} \begin{bmatrix} R_{\ce{H2O} } \\ R_{\ce{H2} } \\ R_{\ce{C4} \ce{H6} } \end{bmatrix} \label{11a}\tag{11}$

From this we extract a representation for the column matrix of non-pivot species in terms of the pivot matrix of stoichiometric coefficients and the column matrix of pivot species. This representation is given by

$\underbrace{\begin{bmatrix} R_{\ce{C2H5OH} } \\ R_{\ce{C2} \ce{H4} } \\ R_{ \ce{CH3} \ce{CHO} } \end{bmatrix}}_{\text{column matrix of non-pivot species}} = \underbrace{\begin{bmatrix} { -1} & { -1} & {1} \\ {1} & { 0} & { -2} \\ { 0} & {1} & {-1} \end{bmatrix}}_{\text{pivot submatrix}} \underbrace{\begin{bmatrix} R_{\ce{H2O} } \\ R_{\ce{H2} } \\ R_{\ce{C4} \ce{H6} } \end{bmatrix}}_{\text{column matrix of pivot species}} \label{12a}\tag{12}$

This is a special case of the pivot theorem in which we see that the net rates of production of the pivot species are mapped onto the net rates of production of the non-pivot species by the pivot matrix. The matrix multiplication indicated in Equation \ref{12a} can be carried out to obtain

$R_{\ce{C2H5OH}} =-R_{\ce{H2O}} - R_{\ce{H2}} + R_{\ce{C4} \ce{H6} } \label{13aa}\tag{13a}$

$R_{\ce{C2} \ce{H4} } =R_{ \ce{H2O}}+0 - 2R_{\ce{C4} \ce{H6} } \label{13ba}\tag{13b}$

$R_{\ce{CH3} \ce{CHO}} =0+R_{\ce{H2}} - R_{\ce{C4} \ce{H6} } \label{13ca}\tag{13c}$

In this example we have provided a template for the solution of Eq. $$(6.2.10)$$ in which the mathematical steps are always the same, i.e.,

1. develop the row reduced echelon form of the atomic matrix,
2. partition Axiom II as indicated by Eqs. \ref{8a} and \ref{9a}, and
3. carry out the matrix multiplication to obtain the net rates of production for the non-pivot species.

In the design of a butadiene production unit, the representations for the net rate of production of the non-pivot species represent a crucial part of the analysis. In Chapter 7 we will apply this type of analysis to several processes.

## 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 ($$\ce{CH3CL}$$), ethyl chloride ($$\ce{C2H5Cl}$$), and chlorine ($$\ce{Cl2}$$). We illustrate the atomic matrix by

$\text{ Molecular Species} \to \ce{CH3Cl} \quad \ce{C2H5Cl} \quad \ce{Cl2} \\ \begin{matrix} {carbon} \\ { hydrogen} \\ {chlorine} \end{matrix} \begin{bmatrix} {1} & { 2} & { 0} \\ {3} & { 5} & {0 } \\ {1} & { 1} & {2 } \end{bmatrix} \label{63}$

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

$\mathbf{A}^* = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Use of this result in Axiom II provides

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} R_{\ce{CH3} \ce{Cl}} \\ R_{\ce{C2} \ce{H5} \ce{Cl}} \\ R_{\ce{Cl2}} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

and we immediately obtain the trivial solution expressed as

$R_{\ce{CH3} \ce{Cl}} = 0$

$R_{\ce{C2} \ce{H5} \ce{Cl}} = 0$

$R_{\ce{Cl2}} = 0$

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

$\text{ Molecular Species} \to \ce{CH3Cl} \quad \ce{C2H5Cl} \quad \ce{Cl2} \quad \ce{H2} \\ \begin{matrix} {carbon} \\ { hydrogen} \\ {chlorine} \end{matrix} \begin{bmatrix} {1} & { 2} & { 0} & {0 }\\ {3} & { 5} & {0 } & {2 }\\ {1} & { 1} & {2 } & {0 } \end{bmatrix} \label{67}$

and the atomic matrix takes the form

$\mathbf{A} = \begin{bmatrix} 1 & 2 & 0 & 0 \\ 3 & 5 & 0 & 2 \\ 1 & 1 & 2 & 0 \end{bmatrix}$

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

$\mathbf{A}^* = \begin{bmatrix} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & -1 \end{bmatrix}$

and from Axiom II we obtain

$\begin{bmatrix} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & -1 \end{bmatrix} \begin{bmatrix} R_{\ce{CH3Cl} } \\ R_{\ce{C2H5Cl} } \\ R_{ \ce{Cl2}} \\ R_{\ce{H2} } \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \label{70}$

In this case we have $$r = rank =3$$ and $$N = 4$$, and the net rates of production for methyl chloride, ethyl chloride and chlorine can be represented in terms of the net rate of production of hydrogen. The column/row partition of Equation \ref{70} is illustrated by

$\begin{bmatrix} 1 & 0 & 0 & \vdots & 4 \\ 0 & 1 & 0 &\vdots & -2 \\ 0 & 0 & 1 &\vdots & -1 \end{bmatrix} \begin{bmatrix} R_{\ce{CH3Cl} } \\ R_{\ce{C2H5Cl} } \\ R_{ \ce{Cl2}} \\ \hdashline R_{\ce{H2} } \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \label{71}$

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} R_{\ce{CH3Cl} } \\ R_{\ce{C2H5Cl} } \\ R_{ \ce{Cl2}} \end{bmatrix} + \begin{bmatrix} 4 \\ -2 \\ -1 \end{bmatrix} \begin{bmatrix} R_{\ce{H2}} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \label{72}$

This can be solved for the column matrix of non-pivot species according to

$\underbrace{ \begin{bmatrix} R_{\ce{CH3Cl} } \\ R_{\ce{C2H5Cl} } \\ R_{ \ce{Cl2}} \end{bmatrix}}_{\text{column matrix of non-pivot species}} = \underbrace{\begin{bmatrix} -4 \\ 2 \\ 1 \end{bmatrix}}_{\text{pivot matrix}} \underbrace{ \begin{bmatrix} R_{\ce{H2}} \end{bmatrix}}_{\text{column matrix of pivot species}} \label{73}$

Here we have a non-trivial solution in which the three rates of production of the non-pivot species are specified in terms of the rate of production of the pivot species, $$R_{\ce{H2}}$$. One should always keep in mind that the null solution is still possible for this case, i.e., all four net rates of production may be zero depending on the conditions in our hypothetical reactor.