The concentration change of gas in reversible reactions

Introduction

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 $$\left[\ce{H2O}\right]=20+ \frac{5.53\left(1-1.0046^t\right)}{1.0046^t-0.1929} \quad\left(\text{SI}\right).$$