Daily Archives: June 2, 2020

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),


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


\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.



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


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


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}


\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.


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.