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mtod92 committed Dec 19, 2023
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Expand Up @@ -53,29 +53,25 @@ In the first section of this work, we define the general problem, in the second

# Problem Definition
Let’s start with a simple example of interacting species:
\begin{equation}\label{eq:1}
\frac{[AB_2]}{[A][B]^2}=K_1
\end{equation}
\begin{equation}\label{eq:2}
\frac{[AB_2C]}{[AB_2][C]}=K_2
\end{equation}
$$A + 2B ↔ AB_2$$
$$AB_2 + C ↔ AB_2C$$

And their associated equilibrium constants, defined as the ratio between forward and backward reaction rates:
\begin{equation}\label{eq:3}
\frac{[AB_2]}{([A][B]^2)} = K_1
\begin{equation}\label{eq:1}
\frac{[AB_2]}{[A][B]^2} = K_1
\end{equation}
\begin{equation}\label{eq:4}
\frac{[AB_2C]}{([AB_2][C])} = K_2
\begin{equation}\label{eq:2}
\frac{[AB_2C]}{[AB_2][C]} = K_2
\end{equation}

And the associated mass conservations:
\begin{equation}\label{eq:5}
\begin{equation}\label{eq:3}
[A]_{tot} = [A] + [AB_2] + [AB_2C]
\end{equation}
\begin{equation}\label{eq:6}
\begin{equation}\label{eq:4}
[B]_{tot} = [B] + 2[AB_2] + 2[AB_2C]
\end{equation}
\begin{equation}\label{eq:7}
\begin{equation}\label{eq:5}
[C]_{tot} = [C] + [AB_2C]
\end{equation}

Expand All @@ -88,21 +84,41 @@ The algorithm presented here the advantage of operating on a user-friendly set o

# Mathematical Treatment
In a system with n different species $X_{1…n}$, the mass conservation relationship for the $i^{-th}$ species can be stated as the sum over all the species contributions with their relative stoichiometries (a). We can define the conservation of mass for species $X_i$ as:
\begin{equation}\label{eq:8}
\begin{equation}\label{eq:6}
a_1[X_1] + a_2[X_2] + ... + a_n[X_n] = [X_i]_{tot}
\end{equation}

Or equivalently:
\begin{equation}\label{eq:9}
\begin{equation}\label{eq:7}
\sum_{j=1}^n a_j[X_j] = [X_i]_{tot}
\end{equation}

In order to express such conservation of mass as a linear function of the logarithm of concentrations of the reactants, following the approach by Wall we must first transform the summations to products using the theory of the arithmetic-geometric mean inequality from Passy 3 as applied by Baker 4. We reorganize Eq. 7 so that the summation over all strictly positive terms *a* and *X* is rewritten as the following:

\begin{equation}\label{eq:9}
\begin{equation}\label{eq:8}
\frac{[X_i]_{tot}}{\sum_{j=1}^n a_j[X_j]} = 1
\end{equation}

Then we “condense” the sum in the denominator of Eq. 8 into a product:
\begin{equation}\label{eq:9}
\sum_{j=1}^n a_j[X_j] = \prod_{j=1}^n \bigg(\frac{a_j[X_j]}{W_j}\bigg)^{W_j}
\end{equation}

With *W* for a given species *j* part of a mass conservation relationship being equal to:
\begin{equation}\label{eq:10}
W_j = \frac{a_j[X_j]}{\sum_{p=1}^n a_p[X_p]}
\end{equation}

So that Eq. 8 becomes:
\begin{equation}\label{eq:11}
\frac{[X_i]_{tot}}{\bigg(\frac{a_1[X_1]}{W_1}\bigg)^{W_1} * \bigg(\frac{a_2[X_2]}{W_2}\bigg)^{W_2} * ... * \bigg(\frac{a_n[X_n]}{W_n}\bigg)^{W_n}} = 1
\end{equation}

We can then reorganize the fraction:
\begin{equation}\label{eq:12}
\left\{ [X_1]^{-1}\*[X_2]^{-2}\*...\*[X_n]^{-n} \right\}\*[X_i]_{tot}
\end{equation}

# Old Stuff

Single dollars ($) are required for inline mathematics e.g. $f(x) = e^{\pi/x}$
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