# Diffusion down chemical potential gradients

• Hypatio
In summary, the equation predicts that a flux of a component will continue to remove it from a mixture if its concentration approaches zero, even if mass conservation is "preserved." This behavior can be corrected by enforcing non-negative concentrations, but this is not always necessary.
Hypatio
Ignoring cross-diffusion, diffusive mass fluxes down chemical potential gradients can be described by the equation (I am working from de Groot and Mazur's 1984 text on non-equilibrium thermodynamics):

$\frac{\partial C_k}{\partial t} = L_{kk}\frac{\partial (\mu_k-\mu_n)}{\partial x}$

where $C_k$ is the concentration of species $k$, $L_{kk}$ is a phenomenological coefficient, $\mu_k$ is the chemical potential of species k, $\mu_n$ is the chemical potential of a different species, and $x$ is distance.

However, if the chemical potential of a species is high (relative to others) when its concentration approaches zero, this equation will predict that a flux will continue to remove the component, resulting in negative concentrations. Mass conservation is "preserved" because the concentration of a different component will then be greater than 1, but this is obviously still incorrect.

How can this behavior be corrected? Must you simply enforce non-negative concentrations ad hoc, or is there a more obvious method.

This equation appears strange to me. Why should the diffusion current of species k depend on mu_n, if cross currents are neglected?
Furthermore, if the concentration of a species goes to zero, it will behave as an ideal solution, i.e. ##\mu_k=\mu_{k0}+RT \ln(c_k/c_{k0})##. That is, if c_k becomes small, the chemical potential of this component gets very negative.

The more complete equation is

$\frac{\partial C_k}{\partial t} =\nabla\cdot\sum_{k=1}^{n-1} L_{ik}\nabla (\mu_k-\mu_n)$

which I reduced to one dimension (sorry I forgot the first d/dx before the coefficient L in the OP equation) and ignored cross coefficients (i =/= k) which removes the summation. The reason for the -\mu_n term is for mass conservation in an n-component system. If we write all chemical potentials relative to the chemical potential of species n, then we only need to write n-1 equations and mass is conserved. Otherwise you need to write n+1 equations (flux equations plus a mass conservation equation).

Regarding your second point, it depends on the what the mechanical gibbs free energy function looks like. If there is a large difference in free energy between two end members, then the ideal part will be small. So, if you draw a plot of gibbs free energy as a function of binary composition, you could write two phases as a big X as: |X|

In this case, energy could continue to be minimized if you allowed one phase to take on negative concentrations and the other phase to take on >1 mass fraction concentrations.

Last edited:
Here is a better visualization. If alpha phase was all component A, and beta phase was all component B, the system would be minimized as much as possible. However, the chemical potentials allow further diffusion, with alpha phase accepting more A and beta phase accepting more B. The dashed part of the curve shows that because of the chemical potentials, the system thinks that this can happen, but the components are obviously not available.

Your picture is wrong. A pure component can never be the equilibrium state and I already explained you why.

DrDu said:
Your picture is wrong. A pure component can never be the equilibrium state and I already explained you why.
I don't think my figure is exactly wrong, but I think I am seeing what you mean. It looks like the problem is that the large changes in chemical potential only occur very close to the limit of pure concentrations. I am solving non-equilibrium transport with finite difference methods, and if a phase has a concentration close to pure (e.g., within 0.1-1.0 percent mass fraction) then the system will behave exactly like in the picture. A huge overstep will occur in the flux and compositions will go negative. However, it looks like if I only allow concentrations to change by a mass fraction less than 10^-12 , then the overstep will not occur because this effect is only significant for these very low concentrations.

Maybe you could solve the transport equations using ln c as basic variable, not c itself?

I'm not sure how that equation would look. How then would you scale the chemical potentials and time?

DrDu said:
Your picture is wrong. A pure component can never be the equilibrium state and I already explained you why.
"a pure component"

What does that even mean? That "activity coefficient" can theoretically never be zero? Isn't that size-dependent?

Assumptions of a statistical ensemble is the required qualification, I'm naively guessing, but this still seems to raise issues with that unqualified assertion in a phase-separated heterogeneous system.

Last edited:

## 1. What is diffusion and how does it occur?

Diffusion is the movement of particles from an area of high concentration to an area of low concentration. This occurs due to the random motion of particles and does not require any external force.

## 2. What is a chemical potential gradient?

A chemical potential gradient refers to a difference in the concentration of a particular chemical or substance between two areas. This gradient is the driving force for diffusion to occur.

## 3. How does diffusion down chemical potential gradients affect biological processes?

Diffusion down chemical potential gradients plays a crucial role in biological processes such as the exchange of gases in respiration and the movement of nutrients and waste products across cell membranes.

## 4. What factors affect the rate of diffusion down chemical potential gradients?

The rate of diffusion is affected by factors such as the concentration gradient, temperature, molecular weight of the diffusing substance, and the surface area and permeability of the membrane through which diffusion occurs.

## 5. Can diffusion down chemical potential gradients be actively controlled?

While diffusion itself does not require any external energy, it can be influenced and controlled by various factors such as the presence of transport proteins and active transport mechanisms in biological systems.

• Chemistry
Replies
13
Views
1K
• Chemistry
Replies
9
Views
1K
• Other Physics Topics
Replies
1
Views
778
• Materials and Chemical Engineering
Replies
2
Views
3K
• Materials and Chemical Engineering
Replies
1
Views
2K
• Other Physics Topics
Replies
1
Views
2K
• Biology and Medical
Replies
4
Views
7K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Materials and Chemical Engineering
Replies
10
Views
3K