Diffusion of CO2 and O2 in Water/Atmosphere System

Consider the following experiment:

• One reservoir of pure water is interfacing the atmosphere at standard condition.
• Both are at rest and at the same temperature.
• Both have no internal gradients (ie: constant concentrations, temperature, ...)
• The interface between them is a perfect flat surface (for sake of simplicity).
• The dissolved $\ce{CO2}$ in the water has a concentration somewhere between 0-50ppm.
• The dissolved $\ce{O2}$ in the water has a concentration somewhere between 0-11ppm.

Questions
in order of importance

1. How to mathematically model the $\ce{CO2}$ and $\ce{O2}$ rate of change per unit area (flux) in the water and in the atmosphere considering only diffusion?
2. Suppose the atmosphere is at constant velocity relative to the water reservoir. How this would change the mathematical model above?

Notes: Analytical methods would be appreciated instead of 'heavy' computer simulations.

• In chemistry the equilibrium conditions are generally what are calculated. The dynamic path to get to equilibrium conditions is much more complex to calculate. Obviously interaction between gas and liquid phases happens at the surface. So there will be some sort of thin layer on gas and liquid side that is not representative of bulk phases which you say are "homogeneously mixed internally." The point is that the "thickness" and profile of the diffusion layers depend on how vigorously the liquid and gas are being mixed. Thus this is a very very complicated problem to handle mathematically. – MaxW Jul 17 '18 at 15:11
• "In chemistry the equilibrium conditions are generally what are calculated." Ok, I see your point. However, at the moment, I prefer not to focus on the equilibrium. – Mark Messa Jul 17 '18 at 15:29
• @MaxW "Thus this is a very very complicated problem to handle mathematically." I imagine someone has already dealt with such problem before (ie: obtain the rate by with gases dissolves into water and vice-versa). Any idea how they might have mathematically model then? – Mark Messa Jul 17 '18 at 15:34
• I'll make this last comment.... // Basically you taking about a diffusion problem. But you haven't even supplied details on the physical dimensions of the isolated system. Is this a one inch tube a mile long half filled with water and half filled with gas, or is it a cylinder a yard in diameter but only one inch high? A dynamic model would go on on on in such details. For an equilibrium calculation they don't matter and you'd just need a few details like temperature, relative volumes, and gas pressure. – MaxW Jul 17 '18 at 15:42
• @MaxW "But you haven't even supplied details on the physical dimensions of the isolated system." For sake of simplicity, we can consider both to be very large reservoirs (ie: the interface between water and air is an infinite flat surface). – Mark Messa Jul 17 '18 at 16:06

Such questions are relatively standard in discussing transport phenomena, so I'll suggest Bird, Stewart, and Lightfoot's Transport Phenomena as a good reference.

We assume a flat plane of contact infinite in extent. We assume that the bulk atmosphere has a constant concentration $c_A$ of some species of interest, and that the bulk liquid has a constant concentration $c_B$ of that species. The species is only allowed to mix between the two bulk phases for times $t > 0$. Then the problem reduces to the one-dimensional diffusion equation (Fick's first and second laws) $$\partial_t c(z,t) = D\,\partial_z^2 c(z,t)$$ subject to the initial condition $$c(z,0) = \frac{c_A+c_B}{2} + \frac{c_B-c_A}{2}\text{sgn}(z),$$ and the boundary conditions $$c(-\infty, t) = c_A, \quad c(+\infty, t) = c_B,$$ where $D$ is the species' diffusion constant and $\text{sgn}(z)$ is the signum function.

For ease of solution, make the substitutions $$\bar{c}(z,t) := c(z,t) - \frac{c_A+c_B}{2}, \quad \Delta c := \frac{c_B-c_A}{2}, \quad \phi(z,t) := \frac{\bar{c}(z,t)}{\Delta c},$$ which yields $$\partial_t\phi(z,t) = D\,\partial_z^2 \phi(z,t), \quad \phi(z,0) = \text{sgn}(z), \quad \phi(\pm\infty, t) = \pm1.$$ By definition $\phi$ is a dimensionless quantity, so it must depend on the variables relevant in the problem in a dimensionless form, for otherwise $\phi$ would have units. The relevant variables in our problem are the diffusion constant $D$, the length $z$, and the time $t$.

The only dimensionless variable that arises from this set of quantities is $\eta := z/\sqrt{Dt}$, so we know $\phi(z,t) = \phi(\eta)$. Substituting this result into our diffusion equation and applying the chain rule yields the ordinary differential equation $$-\frac{\eta}{2}\phi'(\eta) = \phi''(\eta).$$ Integrating twice yields the result $$\phi(\eta) = \alpha + \beta\int_0^\eta\text{d}\eta'\,\exp\left(-\frac{\eta'^2}{4}\right),$$ and it remains to determine the constants $\alpha$ and $\beta$ from our boundary conditions. Actually, for standardization, we will first take $\eta' \to 2\eta'$ and $2\eta \to \eta$, so $$\phi(\eta) = \alpha + 2\beta\int_0^\eta\text{d}\eta'\,\exp(-\eta'^2).$$ By symmetry $\alpha = 0$, and by explicit evaluation $\beta = 1/\sqrt{\pi}$, and hence we have $$\phi(\eta) = \text{erf}(\eta), \quad \text{or} \quad \phi(z,t) = \text{erf}\left(\frac{z}{\sqrt{4Dt}}\right).$$ Going back and untangling all our substitutions, we end up with $$c(z,t) = \frac{c_A+c_B}{2}+\frac{c_B-c_A}{2}\text{erf}\left(\frac{z}{\sqrt{4Dt}}\right).$$ The flux then follows as $$J(z,t) = -D\,\partial_zc(z,t) = -D\,\frac{c_B-c_A}{\sqrt{4\pi Dt}}\exp\left(-\frac{z^2}{4Dt}\right),$$ and we are done.

Regarding your second question, I don't believe the movement of one phase relative to the other will have any effect. As we've seen above, the xy-plane essentially drops out of the picture, and you won't be introducing any further gradients in chemical potential by translating one phase relative to the other, since the concentration of each phase is a function of $z$ only. Of course, if you had a finite area of contact, then things would be different. Perhaps see BSL 18.5, in this case.

In addition, keep in mind that there's important chemistry that's being neglected in this simplistic model. For example, the dissolution of carbon dioxide in water follows not just from diffusion but also from reaction with water, forming bicarbonate ion and related species in solution. This model doesn't account for that.

• You've correctly understood where I am trying to go, ie, obtain the concentration development with time. However, the focus of my original post was only a small part of your question. Consider the one-dimensional diffusion equation you've written. The diffusion coefficient is constant for z < 0 (inside the water) and constant for z > 0 (inside the atmosphere). But what will be its value for z = 0 (at the interface)? I don't think you can model the interface this way. Fick's equation, in this simplest form, assumes an homogeneous medium, which is not the case at the interface. I guess ... – Mark Messa Jul 18 '18 at 2:28
• Math details: your boundary condition c(z,0) is not equal to ca for negatives z values. – Mark Messa Jul 18 '18 at 2:40
• "keep in mind that there's important chemistry that's being neglected in this simplistic model." I need to consider this. At least the first order terms. This is another reason why I prefer to focus only on the instantaneous rate of change at t=0 instead of the time development. BTW, this is another reason why I don't feel much confidence on Fick's equations at the interface. There is more stuff going on ... – Mark Messa Jul 18 '18 at 2:50
• "not just from diffusion but also from reaction with water" Yeap, I guess you summarize it well. Your one-dimensional diffusion equation is perfectly correct (including at the interface) if we consider only diffusion. However, there is also some chemical reactions going on. I'm not sure whether such chemical reactions would play a major role or not. Any idea how can we estimate the influence of such reactions? – Mark Messa Jul 18 '18 at 3:04
• "subject to the initial condition" I guess the correct would be: c(z,0) = ca.(1-H) + cb.H . Agree? – Mark Messa Jul 18 '18 at 3:09