To disambiguate, “element” refers to element in math, while “chemical element” refers to element in chemistry.

For convenience, if there is no ambiguity, denote the zero element in all kinds of algebraic structures as $0$.

From the perspective of algebra, $m$ different chemical elements forms a free $\mathbb Z$-module $\mathscr M$, whose rank (a.k.a. dimension) is $m$. Chemicals (whether pures or mixtures) consisting of these chemical elements can be regarded as elements in $\mathscr M$.

Suppose the basis of $\mathscr M$ is $B\in\mathscr M^m$, whose components are the respective chemical elements. Then, any $c\in\mathscr M$ can be represented as an inner product $c=\left<p,B\right>$, where $p\in\mathbb Z^m$.

With the isomorphism $f:p\mapsto\left<p,B\right>$, one can find that $\mathscr M\simeq\mathbb Z^m$.

Suppose a chemical reaction equation involves $n$ chemicals $C\in\mathscr M^n$. Then, balancing the equation is in fact to find $x\in\mathbb Z^n$ such that $\left<x,C\right>=0$. In other words, it is to solve the homogeneous linear equation $\left<x,C\right>=0$. The solution $x$ is the “signed stoichiometries”. Chemicals with positive stoichiometries and those with negative ones stay at different side of the chemical reaction equation.

One can define the concept of “rank” in $\mathscr M^n$. Let $r\coloneqq\operatorname{rank}C$.

1. $r\ge n$. In this case, $x$ has the only solution $x=0$. A mistake made when writing out the reaction equation may lead to this case.
2. $r=n-1$. In this case, $x$ has infinite solutions, all with the form $x=ts$, where $s\in\mathbb Z^n$ is the solution basis, and $t\in\mathbb Z$ is an arbitrary parameter. Then $s$ is the result of balancing. Usually, it is required that components of $s$ be mutually prime. Such chemical reactions can be called “simple reactions”.
3. $r<n-1$. In this case, $x$ has infinite solutions, with multiple (to be clear, $n-r$) solution basis. Such chemical reactions can be called “complicated reactions”. This means the reaction equation can be decomposed into $n-r$ linearly independent simple reaction equations, where the stoichiometries are the solution basis we previously solved out.

Note that the solution basis of a complicated reaction are not unique, so you may derive conclusions of completely different chemical meanings.

A reversible elementary reaction takes place inside a closed, highly thermally conductive container of constant volume, whose reactants are all gases, and the reaction equation is $$\sum_ka_kX_k\rightleftharpoons\sum_kb_kY_k,$$ where $X_k$ and $Y_k$ are reactants, and $a_k$ and $b_k$ are stoichiometries.

Use square brackets to denote concentrations. Our goal is to find $\left[X_k\right]$ and $\left[Y_k\right]$ as functions with respect to time $t$.

The approach

It is easy to write out the rate equations $$\begin{split} \frac{\mathrm d\left[X_j\right]}{\mathrm dt}= a_j\left(\mu_Y\prod_k\left[Y_k\right]^{b_k}- \mu_X\prod_k\left[X_k\right]^{a_k}\right),\\ \frac{\mathrm d\left[Y_j\right]}{\mathrm dt}= b_j\left(\mu_X\prod_k\left[X_k\right]^{a_k}- \mu_Y\prod_k\left[Y_k\right]^{b_k}\right), \end{split}$$ $$(1)$$ where $\mu_X$ and $\mu_Y$ are rate constants derived by experimenting.

Apply a substitution $$\begin{split} x_j\coloneqq\frac{\left[X_j\right]}{a_j},\quad y_j\coloneqq\frac{\left[Y_j\right]}{b_j},\\ \mu_x\coloneqq\mu_X\prod_ka_k^{a_k},\quad \mu_y\coloneqq\mu_Y\prod_kb_k^{b_k} \end{split}$$ $$(2)$$ to Formula 1, and then it becomes $$\begin{split} \frac{\mathrm dx_j}{\mathrm dt}= \mu_y\prod_ky_k^{b_k}-\mu_x\prod_kx_k^{a_k},\\ \frac{\mathrm dy_j}{\mathrm dt}= \mu_x\prod_kx_k^{a_k}-\mu_y\prod_ky_k^{b_k}, \end{split}$$ $$(3)$$ which means the changes of $x_j$ are all equal, the changes of $y_j$ are all equal, and the changes of $x_j$ are opposite to the changes of $y_j$.

Denote the changes of $x_j$ are equal to $s$, the initial value of $x_j$ is $A_j$, the initial value of $y_j$ is $B_j$, which means $$\begin{split} x_j=A_j+s,\\ y_j=B_j-s. \end{split}$$ $$(4)$$ Substitute Formula 4 into Formula 3, and it can be derived that $$\frac{\mathrm ds}{\mathrm dt}=F\!\left(s\right),$$ by which we can reduce the problem to an integral problem $$t=\int_0^s\frac{\mathrm ds}{F\!\left(s\right)},$$ $$(5)$$ where $$F\!\left(s\right)\coloneqq\mu_y\prod_k\left(B_k-s\right)^{b_k}- \mu_x\prod_k\left(A_k+s\right)^{a_k}$$ $$(6)$$ is a polynomial of $n$th degree, where $$n\coloneqq\max\!\left(\sum_ka_k,\sum_kb_k\right)$$ is the larger of the orders of the forward and reverse reactions. The degree of $F$ may be lower if the high-order term is offset, but only mathematicians believe in such coincidences.

Since Formula 5 is to integrate a rational function, it is easy.

After deriving $s$ as a function of $t$, substitute it into Formula 4 and then Formula 2. We can derive $$\begin{split} \left[X_j\right]=a_j\left(A_j+s\right),\\ \left[Y_j\right]=b_j\left(B_j-s\right) \end{split}$$ $$(7)$$ as the answer.

Properties of $F$

As we all know, here exists a state where the system is in chemical equilibrium. Denote the value of $s$ in this case as $q$. It is easy to figure out that $q$ is a zero of $F\!\left(s\right)$ on the interval $$I\coloneqq\left(-\min_kA_k,\min_kB_k\right),$$ which is the range of $s$ such that the concentration of all reactants are positive.

It is obvious that the value of $q$ is unique. It is because $F$ is monotonic over $I$ and the signs of its value at ends of interval $I$ are different.

Note that $q$ is a flaw of $\frac1{F\left(s\right)}$ and that the improper integral $\int_0^q\frac{\mathrm ds}{F\left(s\right)}$ diverges, so we can imagine how $s$ changes with respect to $t$. $s=0$ when $t=0$, and then $s$ changes monotonically, and finally $s\rightarrow q$ when $t\rightarrow+\infty$. Thus, the range of $s$ over $t\in\left[0,+\infty\right)$ is $\left[0,q\right)$ for $q>0$ or $\left(q,0\right]$ for $q<0$. $q=0$ is not considered because only mathematicians believe in such coincidences.

Suppose $F$ has $n$ different complex zeros $r_\alpha$, one of which is $q$. The possible existence of multiple roots is ignored because only mathematicians believe in such coincidences. Decompose the rational function $\frac1{F\left(s\right)}$ into several partial fractions, and it can be derived that $$\frac1{F\!\left(s\right)}=\sum_\alpha\frac{C_\alpha}{r_\alpha-s},$$ $$(8)$$ where $C_\alpha$ are undetermined coefficients.

Integrate Formula 8, and then it can be derived that $$t=-\sum_\alpha C_\alpha\ln\!\left(1-\frac s{r_\alpha}\right)$$ $$(9)$$ In most cases, Formula 9 cannot be solved analytically and can only be solved numerically.

Note that if the coefficients $C_\alpha$ are in general commensurable, Formula 9 can be reduced into a rational equation. However, only mathematicians believe in such coincidences. However, if $n=2$, it can be proved that the equation can be reduced into a rational one.

Example

The closed container that is highly thermally conductive is in a certain temperature environment, and the water-gas shift reaction $$\ce{CO +H2O\rightleftharpoons CO2 +H2}$$ occurs under the catalysis of a certain catalyst, where the forward rate constant $$\mu_1=2.07\times10^{-4}\quad\left(\text{SI}\right),$$ and the reverse rate constant $$\mu_2=8.29\times10^{-6}\quad\left(\text{SI}\right).$$ Initial concentrations are $$\begin{split} \left[\ce{CO}\right]_0=10.00\quad\left(\text{SI}\right),\\ \left[\ce{H2O}\right]_0=20.00\quad\left(\text{SI}\right),\\ \left[\ce{CO2}\right]_0=30.00\quad\left(\text{SI}\right),\\ \left[\ce{H2}\right]_0=40.00\quad\left(\text{SI}\right). \end{split}$$ Find $\left[\ce{H2O}\right]$ as a function of time.

Formula 6 becomes $$F\!\left(s\right)\coloneqq8.29\times10^{-6}\left(30-s\right)\left(40-s\right) -2.07\times10^{-4}\left(10+s\right)\left(20+s\right) \quad\left(\text{SI}\right).$$ It is a polynomial of $2$nd degree. Its two roots are $$\begin{split} r_1=-28.65\quad\left(\text{SI}\right),\\ r_2=-5.53\quad\left(\text{SI}\right). \end{split}$$ Decomposing $\frac1{F\left(s\right)}$ into partial fractions, we can derive that $$\begin{split} C_1=-217.654\quad\left(\text{SI}\right),\\ C_2=217.654\quad\left(\text{SI}\right). \end{split}$$ Thus, $$t=217.654\ln\!\left(1-\frac s{-28.65}\right)- 217.654\ln\!\left(1-\frac s{-5.53}\right) \quad\left(\text{SI}\right).$$ Since $C_1$ and $C_2$ are in general commensurable, we can solve the equation analytically into $$s=\frac{5.53\left(1-1.0046^t\right)}{1.0046^t-0.1929} \quad\left(\text{SI}\right).$$ Use Formula 7, and then we can find the answer $$\left[\ce{H2O}\right]=20+ \frac{5.53\left(1-1.0046^t\right)}{1.0046^t-0.1929} \quad\left(\text{SI}\right).$$