Is there a known relation between the diffusion of a molecule and the diffusion of the reactants?

I have a plasma consisting of elements A and B (which have a similar mass). These elements can react to form molecules AB.

$$\ce{A + B -> AB}$$

Now if my plasma or gas is expanding based on pressure gradients, I will have some sort of diffusion of these two element concentrations. In fact, this diffusion is the dominant process affecting the concentrations A and B. Based on my spectroscopic analysis, the rate of decay of both elements is similar/the same, so I model them as $$[\ce{A}](t) = A_0 \exp(-k_0t) \,, \\ [\ce{B}](t) = B_0 \exp(-k_0t) \,.$$ Then the concentration of AB changes with $$\frac{d[\ce{AB}]}{dt} = k_1 A_0 B_0 e^{-2k_0t}\ - k_2 [\ce{AB}] \,,$$ where $$k_2$$ is the rate constant for diffusion. The obtained formula for [AB] is then $$[\ce{AB}](t) = \frac{k_1A_0B_0}{2k_0-k_2} \left(e^{-k_2t} - e^{-2k_0t} \right) \,.$$ This actually fits my data quite nicely, but one thing confuses me: When fitting my data for $$[\ce{AB}]$$ with this model, the parameter $$k_2$$ is exactly $$2k_0$$.

Or rather, they are very close, since the formula has a singularity at $$k_2 = 2k_0$$. But if the formula is derived with this assumption directly, one obtains $$[\ce{AB}](t) = (k_1A_0B_0)\, t\, e^{-2k_0t} \,,$$ which also fits my data very nicely.

I don't understand why $$k_2 = 2k_0$$ and I am worried that I might say something very stupid in my paper when I try to explain this observation. But this is what I am thinking: $$k_2$$ is the rate of diffusion for AB, which is twice as heavy as A and B. It will have twice the momentum, which could lead to a doubling of the diffusion rate as well. Does this seem like a probable explanation? If not, which other effects will play a role?

Note that the plasma I am looking at is short-lived and changes constantly, which makes it difficult to apply textbook chemistry to it. Basically all signals I am looking at can be modeled as exponential functions or sums of exponential functions, and sometimes multiple processes might be involved. It's a lot of guesswork, and I am worried that I am actually on the wrong track here. But at least intuitively it makes sense to me that the diffusion of a heavier molecule should happen faster.

• It might help if you posted a sample of data to play around with. Obviosuly the cancellation of terms makes one wonder how you managed to fit anything with your first model. Jul 29 '19 at 8:03
• @BuckThorn Well, my question is not really about my data, it's about my interpretation of the fit. I am asking if it makes sense that a reaction product has a diffusion rate that is the sum of the diffusion rates of the reactants due to how momentum is conserved. My equations are shown here to demonstrate how I came to this idea. And: there is no cancellation of terms. That is a misunderstanding of my description. If you do the math, you will see that it works out fine. Jul 29 '19 at 16:20
• Sure, but you get some odd results when performing the fits with your standard model, which led you to posit what @porphyrin regards as a flawed or odd model, so you may want to give it a try. You have nothing to loose afai can see. It's called providing a minimal working problem. Jul 29 '19 at 16:53
• What I am suggesting is also that numerical issues may be what's determining whether you can fit the data with a particular model. Cancellation of terms is usually a sign of an overparameterized model. Jul 31 '19 at 8:13
• @BuckThorn This is pretty vague. How can the model be overparameterized? It is rather simple and only has three free parameters (of which two are found to be equal). Again, there is no cancellation of terms; if you could point out what you mean by that phrase, that would be helpful. The only part of my model that porphyrin thought of as flawed was the part that was directly derived from my experimental data. I don't think his argument was very solid, but I also had the feeling we were talking past each other. Jul 31 '19 at 8:31

I don't understand why you make A and B vary with time irrespective of your reaction scheme.

You should really analyse using Fick's diffusion equations with the reaction scheme added and do so for each species. This will have to be a numerical calculation in space and time. (There are well established methods to do this. e.g. Crank-Nicholson). By the symmetry of your experiment you may be able to reduce this to one dimension and time.

You should be able to 'guestimate' the diffusion coefficient from kinetic theory at least their ratio which should be quite accurate.

Your second equation ('new formula') is that for a reaction $$\ce{A + B ->[k_1] AB ->[k_1] }$$ where both rate constants are the same. In your case there seems to be no reason why the two rate constants are the same, only you know enough about your experiment to explain this :)

• You say there is only one rate constant in the second case, but there are two: $k_1$ and $k_2 = 2k_0$, or am I missing something? Isn't it $\ce{A + B ->[k1] AB ->[k2] }$? As for why I didn't use a more complex model - because that would be very complicated, and it's not really the focus of my work. And I was hoping the simple model would be sufficient, since it fits the data and I don't want to do unnecessary assumptions about the plasma. Jul 21 '19 at 12:42
• When $k_2=2k_0$ then both are the same and your first solution fails as you point out hence you have the scheme I suggest in my answer. Jul 21 '19 at 15:24
• In your answer you imply that $k_1 = k_2$, and I don't see a good reason for this, since they will also have different units. I guess it was just a small mistake, but I was confused. Jul 23 '19 at 16:23
• They just have to be numerically the same, which was my point. The scheme you propose seems to have no real physical basis unfortunately. You do really need to model this with combined diffusion and reaction. Jul 24 '19 at 16:07
• They do not have to be numerically the same. I don't know how you come to this conclusion, but as you see in the equations, $k_1$ is just a proportionality factor in the end. As for the physical basis: my model is based on diffusion and reaction, but I make the assumption that diffusion is more dominant than reaction, so that the small effect of the reaction on the concentrations of A and B can be neglected. This assumption is justified by the fact that equilibrium calculations estimate the AB concentration as very low. Jul 24 '19 at 17:27