# Qualitative Analysis of a Simple Chemical Reaction

We will study a simple chemical reaction described by

$A + A \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} A_2$

where two molecules of $A$ are combined reversibly to form $A_2$ and, $k_1, k_2$ are the reaction rates.

If $x$ is the concentration of $A$, $y$ of $A_2$, then according to the Law of Mass Action,

$\begin{cases}\frac{1}{2}\cdot\frac{dx}{dt}=k_2 y -k_1 x^2 \\ \frac{dy}{dt}=k_1 x^2-k_2 y\\x(0)=x_0, y(0)=y_0\end{cases}$

or equivalently,

$\begin{cases}\frac{dx}{dt}=2 k_2 y - 2 k_1 x^2 \quad\quad(0-1)\\ \frac{dy}{dt}=k_1 x^2-k_2 y\quad\quad\quad(0-2)\\x(0)=x_0, y(0)=y_0\;\quad(0-3)\end{cases}$

We seek first the equilibrium points that represent the steady state of the system. They are the constant solutions where $\frac{dx}{dt}=0$ and $\frac{dy}{dt}=0$, simultaneously.

From $\frac{dx}{dt}=2 k_2 y - 2 k_1 x^2=0$ and $\frac{dy}{dt}=k_1 x^2-k_2 y=0$, it is apparent that

$\forall x \ge 0, (x_*, y_*) = (x, \frac{k_1}{k_2}x^2)\quad\quad\quad(0-4)$

is an equilibrium point.

To find the value of $x_*$, we solve for $y$ from (0-1),

$y = \frac{1}{2k_2}(\frac{dx}{dt}+2 k_1x^2).\quad\quad\quad(0-5)$

Substitute it in (0-2),

$\frac{d}{dt}(\frac{1}{2k_2}(\frac{dx}{dt}+2k_1x^2))=k_1x^2-k_2\frac{1}{2k_2}(\frac{dx}{dt}+2k_1x^2)$

$\implies \frac{d}{dt}(\frac{dx}{dt}+2k_1x^2)=2k_2k_1x^2-k_2(\frac{dx}{dt}+2k_1x^2)$,

i.e.,

$\frac{d^2x}{dt^2} + \frac{dx}{dt}(4k_1x+k_2)=0.\quad\quad\quad(0-6)$

This is a 2nd order nonlinear differential equaion. Since it has no direct dependence on $t$, we can reduce its order by appropriate substitution of its first order derivative.

Let

$p=\frac{dx}{dt}$,

we have

$\frac{d^2x}{dt^2} = \frac{d}{dt}(\frac{dx}{dt})=\frac{dp}{dt}=\frac{dp}{dx}\frac{dx}{dt}=\frac{dx}{dt}\frac{dp}{dx}=p\cdot\frac{dp}{dx}$

so that (0-6) is reduced to

$p\cdot\frac{dp}{dx}+p\cdot(4k_1x+k_2)=0,\quad\quad\quad(0-7)$

a 1st order differential eqution. It follows that either $p=\frac{dx}{dt}=0$ or $\frac{dp}{dx}+(4k_1x+k_2)=0$.

The second case gives

$\frac{dp}{dx}=-4k_1x-k_2.$

Integrate it with respect to $x$,

$p=\frac{dx}{dt}=-2k_1x^2-k_2x + c_0.\quad\quad\quad(0-8)$

Hence, the equilibrium points of (0-1) and (0-2) can be obtained by solving a quadratic equation

$-2k_1x^2-k_2x + c_0=0.$

Notice in order to have $x_* \ge 0$ as a solution, $c_0$ must be non-negative .

Fig. 1

The valid solution is

$x_* = \frac{\sqrt{k_2^2+8c_0k_1}-k_2}{4k_1}.$

Fig. 2

By (0-4),

$y_* =\frac{k_1}{k_2}x_*^2=\frac{(\sqrt{k_2^2+8c_0k_1}-k2)^2}{16k_1 k_2}.$

and so, the equilibrium point is

$(x_*, y_*) = (\frac{\sqrt{k_2^2+8c_0k_1}-k_2}{4k_1}, \frac{(\sqrt{k_2^2+8c_0k_1}-k2)^2}{16k_1 k_2}).$

Next, we turn our attentions to the phase-plan trajectories that describe the paths traced out by the $(x, y)$ pairs over the course of time, depending on the initial values.

For $(x, y) \ne (x_*, y_*), \frac{dx}{dt} \ne 0$. Dividing (0-2) by (0-1) yields

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{k_1 x^2-k_2 y}{2 k_2 y- 2 k_1 x^2}=-\frac{1}{2}$

i.e.,

$\frac{dy}{dx} = -\frac{1}{2}.$

Integrating it with respect to $x$,

$y = -\frac{1}{2} x + c_1.$

By (0-3),

$c_1= y_0 +\frac{1}{2}x_0.$

Therefore,

$y = -\frac{1}{2}x +y_0 + \frac{1}{2}x_0.\quad\quad\quad(1-1)$

Moreover, by (0-5)

$y=\frac{1}{2k_2}(\frac{dx}{dt} + 2 k_1 x^2) \overset{(0-8)}{=} \frac{1}{2k_1}(-2k_1 x^2-k_2 x + x_0 +2 k_1 x^2)=\frac{1}{2k_2}(-k_2x +x_0).$

As a result,

$y_0 = \frac{1}{2k_2}(-k_2 x_0 + x_0).$

Substitute $y_0$ in (1-1), we have

$y = -\frac{1}{2} x + \frac{1}{2k_2}(-k_2x_0+x_0) + \frac{1}{2}x_0.$

This is the trajectory of the system. Clearly, all trajectories are monotonically decreaseing lines.

At last, let us examine how the system behaves in the long run.

If $x < x_*$ then $\frac{dx}{dt}>0$ (see Fig. 2) and $x$ will increase. As a result, $y$ will decrease. Similarly, if $x > x_*, \frac{dx}{dt}<0$ ensures that $x$ will decrease. Consequently, $y$ will increase.

Fig. 3 Trajectories and Equilibriums

It is evident that as time $t$ advances, $(x, y)$ on the trajectory approaches the equilibrium point $(x_*, y_*).$

A phase portrait of the system is illustrated in Fig. 4.

Fig. 4

It shows that the system is asymptotically stable.