Question

A typical chemical formula for aerobic growth of a microorganisms is

C2H5OH + a(O2) + (b)NH3 —> (c)CH1.7H0.15O0.4 + (d)H2O + (e)CO2

where term CH_1.7H_0.15O_0.4 represents metabolism of the microorganism. The ratio of moles of CO₂ produced per mole of O₂ consumed is called the respiratory quotient, RQ, which can be determined experimentally. Given this ratio we have 4 constants, a-d that are unknown. We can perform a mass balance on each of the four key elements

Carbon: 2 = c + (RQ)a

Hydrogen: 6 + 3b = 1.7c + 2d

Oxygen: 1 + 2a = 0.4c + d + 2(RQ)a

Nitrogen: b = 0.15c 

Let RQ = 0.8 and then find the vector [a b c d] using Gauss Elimination Method.

Answer 1

The equations may be written as

$\begin{matrix} \text{Carbon: }& (RQ)a + c = 2 \\ \text{Hydrogen: }& -3b + 1.7c + 2d = 6\\ \text{Oxygen: }& (2(RQ)-2)a + 0.4c + d = 1\\ \text{Nitrogen: }& b - 0.15c = 0 \end{matrix}$

Or substituting RQ = 0.8 in matrix form

$\begin{bmatrix} 0.8 & 0 & 1 & 0 \\ 0 & -3 & 1.7 & 2 \\ -0.4 & 0 & 0.4 & 1 \\ 0 & 1 & -0.15 & 0 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 2 \\ 6 \\ 1 \\ 0 \end{bmatrix}$

Adding 1/2 of the first row to the third one:

$\begin{bmatrix} 0.8 & 0 & 1 & 0 \\ 0 & -3 & 1.7 & 2 \\ 0 & 0 & 0.9 & 1 \\ 0 & 1 & -0.15 & 0 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 2 \\ 6 \\ 2 \\ 0 \end{bmatrix}$

Add 1/3 of the second row to the fourth:

$\begin{bmatrix} 0.8 & 0 & 1 & 0 \\ 0 & -3 & 1.7 & 2\\ 0 & 0 & 0.9 & 1 \\ 0 & 0 & 0.417 & 0.667 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 2 \\ 6 \\ 2 \\ 2 \end{bmatrix}$

Subtract 4.17/9 of the third row from the fourth:

$\begin{bmatrix} 0.8 & 0 & 1 & 0 \\ 0 & -3 & 1.7 & 2\\ 0 & 0 & 0.9 & 1 \\ 0 & 0 & 0 & 0.204 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 2 \\ 6 \\ 2 \\ 1.07 \end{bmatrix}$

Therefor

$d = 1.07 / 0.204 \\ c = (2 - d) / 0.9 \\ b = (6 - 1.7 c - 2 d) / (-3) \\ a = (2 - c) / 0.8$

$\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 7.0 \\ -0.55 \\ -3.6 \\ 5.24 \end{bmatrix}$