# Example of chemical reaction with a given form for the kinetic equations

I’m an applied mathematician looking for examples from various applications fields (right now, chemistry, obviously) to illustrate the following reaction-diffusion system $$\begin{cases} \frac{\partial u}{\partial t} - d_u \Delta u = k_1 v + (k_2-k_3) u - k_4 u^{p_1} v^{p_2} - k_5 u^{p_3+p_4} \\ \frac{\partial v}{\partial t} - d_v \Delta v = k_6 u + (k_7-k_8) v - k_9 u^{q_1} v^{q_2} - k_{10} v^{q_3+q_4} \end{cases}$$

Above, all constants $$d_u$$, $$d_v$$, $$k_i$$, $$p_i$$, $$q_i$$ are positive, and $$p_i,q_i\geq 1$$. $$k_2-k_3$$ and $$k_7-k_8$$ might be of any sign. The system need not be self-contained — if you need to add a third equation in order to make these two exist, feel free to do so.

If the diffusion part is bothering you, then I’m also fine with $$\begin{cases} \frac{\text{d} u}{\text{d} t} = k_1 v + (k_2-k_3) u - k_4 u^{p_1} v^{p_2} - k_5 u^{p_3+p_4} \\ \frac{\text{d} v}{\text{d} t} = k_6 u + (k_7-k_8) v - k_9 u^{q_1} v^{q_2} - k_{10} v^{q_3+q_4} \end{cases}$$

And if the $$p_i$$ and $$q_i$$ are bothering you, then one example where they are all equal to $$1$$ is $$\begin{cases} \frac{\text{d} u}{\text{d} t} = k_1 v + (k_2-k_3) u - k_4 u v - k_5 u^2 \\ \frac{\text{d} v}{\text{d} t} = k_6 u + (k_7-k_8) v - k_9 u v - k_{10} v^2 \end{cases}$$

One more precision: if $$k_2-k_3\leq 0$$, then $$k_5=0$$ is allowed, and similarly, if $$k_7-k_8\leq 0$$, then $$k_{10}=0$$ is allowed. Therefore the following system would be a satisfying answer: $$\begin{cases} \frac{\partial u}{\partial t} - d_u \Delta u = \tilde{k}_1 v - \tilde{k}_2 u - \tilde{k}_3 uv \\ \frac{\partial v}{\partial t} - d_v \Delta v = \tilde{k}_2 u - \tilde{k}_1 v - \tilde{k}_4 uv \end{cases}$$ (where the constants $$\tilde{k}_i$$ differ a priori from the previous constants $$k_i$$).

I looked for actual examples of first-order reversible reactions. I found $$\ce{Cu+ <-> Cu^{2+}}$$ there but I do not know if having an irreversible reaction “$$\ce{Cu^+ + Cu^{2+} ->}$$ some product” in this configuration is possible.

Can you think of any “real” chemical system that would satisfy such kinetic equations?

• This one seems to be a homework problem. Please show your efforts before you expect an answer here. – Mitradip Das May 28 at 4:56
• You need to have rate constants that are common between the two reactions. – porphyrin May 28 at 6:27
• @MitradipDas: I tried but I have to admit that I am no chemist. I think that the $k_1$ and $k_6$ part corresponds to a reversible monomolecular reaction. The $k_4$ and $k_9$ part could produce a third quantity, let’s say $U+V\to W$, and similarly we could have $2U\to X$ and $2V\to Y$. Is that more convincing? – Elvith May 28 at 6:41
• @porphyrin: yes, some rates can be equal if needed, it’s fine by me. You can also have $k_2-k_3=k_7-k_8=0$ and forget about this part of the equations, if needed. – Elvith May 28 at 6:44
• @MitradipDas: I added more details in the opening post. Hope this is satisfying. – Elvith May 28 at 13:21

A chemist friend found an answer for the simplest case: $$\begin{cases} \frac{\partial u}{\partial t} - d_u \Delta u = \tilde{k}_1 v - \tilde{k}_{-1} u - \tilde{k}_2 uv \\ \frac{\partial v}{\partial t} - d_v \Delta v = \tilde{k}_{-1} u - \tilde{k}_1 v - \tilde{k}_2 uv \end{cases}$$
• $$u$$ is the concentration of ethenol;
• $$v$$ is the concentration of ethanal;
Nevertheless I’m still interested in other answers, especially with autocatalysis of either $$u$$ or $$v$$ (that would decrease the first-order consumption of it) or with autoreactions (that would give the term $$-k_i u^2$$ or $$-k_i v^2$$). As a matter of fact, the French Wikipedia claims that an aldol reaction of ethanal with itself in presence of a base can also produce 3-hydroxybutanal, so that the same example would also give a term $$-k v^2$$? But maybe these two aldol reactions correspond in fact to the same elementary mechanism?