For a chemist, expressions of the type:
$$\displaystyle CH_4 + 2\ O_2 \longrightarrow CO_2 + 2\ H_{2}O$$
They are intuitive and describe the global behavior of the substances that appear in them. This way of representing chemical reactions is simple and derives naturally from the use of chemical formulas as a written representation of the substances involved in any type of reaction. A chemist quickly identifies the name of the substance methane with its formula $CH_4$ and with everything that both representations, name and formula, mean: stoichiometric composition of the substance, molecular mass, type of chemical compound, physical characteristics... etc. . Years of individual study and the previous work of many generations of researchers hide all this behind a simple formula and a name. But on many occasions, it is necessary to convert this symbolism into something much more general. Mathematical calculations require the use of general expressions that can be treated numerically and that have a solid and rigorous notation system, a numerical formulation.
Let us consider a set made up of $S$ chemical compounds (it is understood that we are talking about pure substances in general, that is, both compounds in the strict sense, for example $CH_4$, and elements, $O_2$). We will designate each compound with a name: $A_1$, $A_2$, ..., $A_i$, ..., where $i$ $=$ $1$, $2$, ..., $S$. According to this form of representation, the reaction that we have mentioned before could be written:
$$\displaystyle CH_4 + 2\ O_2 \longrightarrow CO_2 + 2\ H_{2}O \quad\equiv\quad A_1 + 2\ A_2 = A_3 + 2\ A_4$$
where, in this case:
$$\displaystyle \begin{array}{c} S = 4\\ CH_4 \equiv A_1\\ O_2 \equiv A_2\\ CO_2 \equiv A_3\\ H_{2}O \equiv A_4 \end{array} $$
In this way, through this notation, we have converted a chemical reaction into a mathematical equation that we can handle as such. It is convenient to write all the chemical compounds on the same side of the equation and give positive signs to the coefficients of the products of the equation. In our case, therefore, the reaction will be symbolized by the equation:
$$\displaystyle - A_1 - 2\ A_2 + A_3 + 2\ A_4 = 0$$
The coefficients that appear in this equation are the stoichiometric coefficients , which we will designate as $\alpha_1$, $\alpha_2$, ..., $\alpha_i$, ..., $\alpha_S$, such that any reaction can be symbolized as the mathematical equation given by:
$$\displaystyle \sum_{i = 1}^{S} \alpha_i \cdot A_i = \alpha_1 \cdot A_1 + \alpha_2 \cdot A_2 + ... + \alpha_S \cdot A_S = 0$$
According to the convention that we have adopted, the chemical compounds whose stoichiometric coefficient is positive will be called the products of the reaction and those with negative coefficients will be the reactants of the reaction . Products are formed from reactants by direct reaction . If there is an inverse mathematical equation of ours, that is, a reaction that uses our products as reactants and our reactants as products, we will call it an inverse reaction . In the vast majority of real systems, the two reactions, direct and inverse, take place simultaneously, reaching a stationary situation when both reactions reach equal and opposite speeds. We will call this stationary state chemical equilibrium .
Let's go back to the chemical reaction of the example... An expression of this type can be interpreted in several ways. We can assume that it describes the microscopic behavior of the compounds that participate in it. That is, one could imagine that two oxygen molecules approach a methane molecule and from that collision one carbon dioxide molecule and two water molecules are obtained. In this particular case, we know that this is not so. On the other hand, the notation of the reaction could be a macroscopic description of the proportions in which different chemicals participate in the process. In other words, what we see written means that methane and oxygen react with each other in a total ratio of 1 to 2, obtaining carbon dioxide and water in ratios 1 and 2 with respect to methane. From the above, it is evident that if a reaction expression is true in a kinetic sense, it will also be true in the stoichiometric sense, but on the contrary it does not have to be true. A reaction like the one in the example is what we will call an integer reaction . An entire reaction does not provide a description of what happens at the microscopic level. For that we need to know what we will call the reaction mechanism , that is, the set of elementary stages into which we can decompose the entire reaction. Each of these elementary stages is characterized in that the stoichiometric equation coincides with the kinetics.
The stoichiometric coefficients of a reaction can all be multiplied by a nonzero constant without changing the reaction. Indeed, if we have a constant $\lambda$, such that $\lambda \not= 0$, then the equation:
$$\displaystyle \sum_{i=1}^{S} \lambda\cdot\alpha_{i}\cdot A_i = 0$$
has exactly the same meaning as:
$$\displaystyle \sum_{i=1}^{S}\alpha_{i}\cdot A_i = 0$$
Exercises
Exercise 1: We will define a chemical compound $A_j$ as a linear combination of the elements of the Periodic Table $\lbrace E_1, E_2, ..., E_T\rbrace$ given by:
$$\displaystyle A_j = \sum_{k=1}^{T} \epsilon_{j,k}\ \cdot E_k$$
Show that:
$$\displaystyle \forall\ k \in \lbrace 1, 2, ..., T\rbrace \quad \mbox{ we have that } \sum_{j=1}^{S} \alpha_{j}\cdot \epsilon_{jk} = 0$$
Exercise 2: If $M_j$ is the molecular mass of $A_j$, prove that:
$$\displaystyle \sum_{j=1}^{S} \alpha_j \cdot M_j = 0$$
On what physical law do we base ourselves to affirm this?
Exercise 3: If the following two reactions:
$$\displaystyle A_1 + A_2 - A_3 - A_4 = 0$$
$$\displaystyle - A_2 + 2\ A_3 - A_4 = 0$$
take place between the four chemical species $A_1$, $A_2$, $A_3$ and $A_4$, prove that:
$$\displaystyle \frac{1}{2} M_4
Exercise 4: Show that the following reaction scheme:
$$\displaystyle A_1 + A_2 - A_3 + A_4 = 0$$
$$\displaystyle A_1 - A_2 + A_3 - A_4 = 0$$
it is impossible.
Solutions
Exercise 1:
Let us consider a chemical reaction in which the compound $A_j$ participates:
$$\displaystyle \sum_{j=1}^{S} \alpha_j \cdot A_j = 0$$
Since each of the compounds involved in the reaction can be written as a linear combination of the $T$ elements of the Periodic Table, we will have to:
$$\displaystyle \sum_{j=1}^{S} \left ( {\alpha_j \cdot \sum_{k=1}^{T} \epsilon_{jk} \cdot E_k} \right ) = 0$$
That is, we will have:
$$\displaystyle \sum_{j=1}^{S} \left [ {\alpha_j \cdot \left ( {\epsilon_{j1} \cdot E_1 + \epsilon_{j2} \cdot E_2 + ... + \ epsilon_{jT} \cdot E_T} \right ) } \right ] = 0$$
That is to say:
$$\displaystyle \alpha_1 \cdot \epsilon_{11} \cdot E_1 + \alpha_1 \cdot \epsilon_{12} \cdot E_2 + ... + \alpha_1 \cdot \epsilon_{1T} \cdot E_T +{}$ $
$$\displaystyle {}+ \alpha_2 \cdot \epsilon_{21} \cdot E_1 + \alpha_2 \cdot \epsilon_{22} \cdot E_2 + ... + \alpha_2 \cdot \epsilon_{2T} \cdot E_T + {}$$
$$\displaystyle \vdots$$
$$\displaystyle {}+ \alpha_S \cdot \epsilon_{S1} \cdot E_1 + \alpha_S \cdot \epsilon_{S2} \cdot E_2 + ... + \alpha_S \cdot \epsilon_{ST} \cdot E_T = 0$$
Now we group the coefficients of each element $E_k$, removing it as a common factor, and we are left with:
$$\displaystyle (\alpha_1 \cdot \epsilon_{11} + \alpha_2 \cdot \epsilon_{21} + ... + \alpha_S \cdot \epsilon_{S1}) \cdot E_1 +{}$$
$$\displaystyle {}+ (\alpha_1 \cdot \epsilon_{12} + \alpha_2 \cdot \epsilon_{22} + ... + \alpha_S \cdot \epsilon_{S2}) \cdot E_2 +{}$$
$$\displaystyle \vdots$$
$$\displaystyle {}+ (\alpha_1 \cdot \epsilon_{1T} + \alpha_2 \cdot \epsilon_{2T} + ... + \alpha_S \cdot \epsilon_{ST}) \cdot E_T = {}$$
$$\displaystyle {}= \sum_{k=1}^{T} {\left ( {\sum_{j=1}^{S} {\alpha_j \cdot \epsilon_{jk}}} \right ) \ cdot E_k} = 0$$
Since the chemical elements are linearly independent of each other (there are no chemical reactions that transform one element into another, as the alchemists were able to verify), the only possibility that a linear combination of these is zero is that all the coefficients are zero (trivial solution ), with which $\forall\ k = 1, 2, ..., T$ we have:
$$\displaystyle \sum_{j=1}^{S} \alpha_j \cdot \epsilon_{jk} = 0$$
Note:
It must be said that this demonstration has a very important utility. The fact that the implication holds:
$$\displaystyle \sum_{j=1}^{S} \alpha_j \cdot A_j = 0 \quad \Longleftrightarrow \quad \sum_{j=1}^{S} \alpha_j \cdot \epsilon_{jk} = 0 $$
It will allow us to solve the calculation of the stoichiometric coefficients ($\alpha_j$) of the reactions by means of linear equation systems. When we have a chemical reaction it is common to know the species that participate in it as well as their empirical formulas. But instead, stoichiometric reaction coefficients often have to be calculated. Such a procedure is known as equalizing (or balancing) the reaction equation. And the system of linear equations that we must solve to find the values of the coefficients $\alpha_j$ is, precisely, the one formed by the equations:
$$\displaystyle \sum_{j=1}^{S} \alpha_j \cdot \epsilon_{jk} = 0 \mbox{ , where } k = 1, 2, ..., T$$
Exercise 2:
Suppose we have any reaction involving $S$ chemical compounds:
$$\displaystyle \sum_{j=1}^{S} \alpha_j \cdot A_j = 0$$
whose molecular masses are $\{ M_j \}, j = 1, 2, ..., S$. To calculate the value of the expression:
$$\displaystyle \sum_{j=1}^{S} \alpha_j \cdot M_j$$
Let's break it down a bit... Of the $S$ compounds that participate in the reaction, a part will be reactants and another part will be products. If we have $p$ products, for example, then we can rewrite the previous expression as follows:
$$\displaystyle \sum_{j=1}^{S} \alpha_j \cdot M_j = \sum_{j=1}^{p} \alpha_j \cdot M_j + \sum_{j=p+1}^{S } \alpha_j \cdot M_j$$
Remembering the convention adopted for the writing of the equations of the reactions, we will have to:
$$\displaystyle \left. \begin{matrix} \mbox{ If } & {1 \leq j \leq p} & \Rightarrow & A_j = product & \Rightarrow & \alpha_j> 0 \\ \mbox{ If } & {p
so that the coefficients $\alpha_j$ that appear in the first summation are positive, while those of the second summation are all negative. This is best visualized by writing it as follows:
$$\displaystyle \sum_{j=1}^{S} \alpha_j \cdot M_j = \sum_{j=1}^{p} \alpha_j \cdot M_j + \sum_{j=p+1}^{S } \alpha_j \cdot M_j = \sum_{j=1}^{p} \alpha_j \cdot M_j - \sum_{j=p+1}^{S} \left | \alpha_j \right | \cdot M_j$$
The product $\alpha_i \cdot M_i$ represents the mass $m_i$ (in grams) of substance $A_i$ that is reacting, since:
$$\mbox{# of moles (mol)} \cdot \mbox{ Molecular mass (g)} \cdot mol^{-1} = \mbox{mass (g)}$$
Therefore, we have to:
$$\begin{eqnarray*} \displaystyle \sum_{j=1}^{S} \alpha_j \cdot M_j & = & \sum_{j=1}^{p} \alpha_j \cdot M_j - \sum_{j =p+1}^{S} \left | \alpha_j \right | \cdot M_j ={}\\ & = & \sum_{j=1}^{p} m_j \mbox{(products)} - \sum_{j=p+1}^{S} m_j \mbox{reactants} ={}\\ & = & m_{total} (products) - m_{total} (reactants) \end{eqnarray*}$$
Now, by the principle of conservation of mass (Lavoisier's Law), we know that, in the course of any chemical process, it is true that:
$${mass}_{reactants} = {mass}_{products}$$
so it must hold that:
$$\displaystyle \sum_{j=1}^{S} \alpha_j \cdot M_j = m_{total} (products) - m_{total} (reactants) = 0$$
Exercise 3:
Thanks to the demonstration of the previous exercise, we can establish the following relationship:
$$\displaystyle \left\begin{matrix} A_1 & + & A_2 & - & A_3 & - & A_4 & = & 0 \\ {} & - & A_2 & + & 2\ A_3 & - & A_4 & = & 0 \end{matrix} \right \}\quad \Longrightarrow \quad \left{\begin{matrix} M_1 & + & M_2 & - & M_3 & - & M_4 & = & 0 \\ {} & - & M_2 & + & 2\ M_3 & - & M_4 & = & 0 \end{matrix} \right $$
The first inequality is proved immediately from the second equation:
$$\displaystyle - M_2 + 2\ M_3 - M_4 = 0 \Longrightarrow M_3 = \frac{1}{2}\ M_2 + \frac{1}{2}\ M_4> \frac{1}{2}\ M_4 $$
since $M_2 \neq 0$. On the other hand, if we isolate $M_2$ from the second equation:
$$\displaystyle - M_2 + 2\ M_3 - M_4 = 0 \Longrightarrow M_2 = 2\ M_3 - M_4$$
and we substitute the result in the first equation, we have that:
$$\displaystyle M_1 + (\ 2\ M_3 - M_4\ ) - M_3 - M_4 = 0 \Longrightarrow M_1 + M_3 - 2\ M_4 = 0$$
Therefore:
$$\displaystyle 2\ M_4 = M_1 + M_3> M_3$$
since it is evident that $M_1 \neq 0$.
Exercise 4:
If we add the two equations, we have:
$$\displaystyle \frac{\left. \begin{matrix} A_1 & + & A_2 & - & A_3 & + & A_4 & = & 0 \\ A_1 & - & A_2 & + & A_3 & - & A_4 & = & 0 \end{matrix} \right \} }{\begin{matrix} 2\ A_1 & {\quad} & / & {\quad } & / & {\quad } & / & \ = & 0 \end{matrix}}$$
which is impossible: matter cannot disappear!