The Thermodynamics of Irreversible Processes
Diffusion and Fick’s First Law
You’ve no doubt learned about the process of diffusion and its description in the form of
Fick’s First Law:
dc
Equation 1
J i   Di i
dx
where dci is the difference in concentration of solute i between two points separated by distance dx,
and Ji is the flux of i between the two points, usually measured in moless-1:
dx
ci ,1
J i12
ci , 2
The diffusion coefficient, Di, reflects the ease with which i can move through the medium separating
the two points. In keeping with our lecture theme of driving forces causing things to happen, we
ci ,1  ci , 2
dc
can think of the concentration gradient, i 
, as the driving force for diffusion, and the
dx
dx
minus sign simply ensures that matter diffuses out of regions of high concentration towards regions
of low concentration.
Fick’s First Law is adequate for describing diffusional flux of many solutes, but it can fail
when ionic solutes are involved. Recall from lecture that all spontaneous processes result in a
negative value for G. Well, this includes the process of diffusion, which always proceeds in the
direction that results in G < 0. When an uncharged solute such as glucose diffuses from a region
of high concentration towards a low-concentration region, the resulting G is negative, and c is an
adequate descriptor of the driving force for diffusion of uncharged solutes. However, when dealing
with diffusion of charged solutes such as ions, the Gibbs Free Energy change that results from
diffusion depends not only on concentration differences, but also on any electric potential gradients
() that may be present.
Fick’s First Law clearly lacks a  (Equation 1), so we need an alternative model when we’re
trying to understand diffusion of solutes such as Na+, K+, Ca++, Cl-, etc. Let’s see if we can use the
body of theory known as irreversible thermodynamics to develop a model for diffusion of ions
through a membrane.
The Thermodynamics of Diffusion
Consider the following thermodynamic system, which is divided into Side 1 and Side 2 by a
membrane that is perforated by pores through which molecules of substance A can freely pass.
Side 1
Side 2
n A1
n A2
 A1
 A2
This system would thus be analogous to a cell separated from the extracellular space by its plasma
membrane, with A representing a charged solute such as Na+ , K+, or Ca++, and the pores
representing ion channels. Start our thought experiment by putting nA1 moles of A in Side 1 and nA2
moles in Side 2: From irreversible thermodynamics theory, we know that the contribution of ni
moles of substance i to the chemical potential in any system is given by:
 i   *  RT ln ni  ,
Equation 2
where * is a standard state chemical potential that can’t be calculated or measured. (not to
worry…..since we’re interested in difference in the chemical potentials at two points in the system,
* will ‘drop out’ when we do the subtraction.)
For our system, then, the chemical potential of A on Side 1 due to the nA1 moles of A will be:
 A1   *  RT ln n A1 
Equation 3a
 A2   *  RT ln n A2 
Equation 3b
Similarly, for Side 2 we have:
The difference in chemical potential between Side 1 and Side 2 is then given by:
 A2   A1   *  RT ln n A2    *  RT ln n A1 
 RT ln n A2  RT ln n A1
 RT ln n A2  ln n A1 
 RT ln
n A2
n A1
Equation 4
Notice that as promised earlier  * doesn’t appear in the final expression. Also, pay attention to the
form of Equation 4. We’re going to make use of it, or something similar to it, several times during
future lectures.
Now, let’s allow dnA moles of A move from Side 1 to Side 2, and see if we can learn anything
of interest.
Side 1
Side 2
n A1
n A2
 A1
dnA
 A2
The movement of dnA moles of A out of Side 1 will reduce the amount of A in Side 1 will by dnA
moles (so dnA < 0 in Side 1), while the amount of A in Side 2 will be increased by dnA moles (so dnA
> 0 in Side 2). Recall from our lecture discussion of Gibbs Free Energy that:
 the Gibbs Free Energy change associated with the movement of matter is dni, where
dni is the number of moles of matter i moving, and
 the total Gibbs Free Energy change must be < 0 for all spontaneous processes.
In the present case, nothing has changed except the amount of A in Side 1 and in Side 2; everything
else is constant (i.e., dP = dT = dV = …. = 0). This lets us write the expression for the total Gibbs
Free Energy change accompanying the movement of matter from Side 1 to Side 2 as the sum of two
terms:
dG    Ai dn Ai   A1dn A1   A2 dn A2  0
Equation 5
i
Put in words, Equation 5 tells us that the change in nA1 and the change in nA2 are together
responsible for the change in G that results from the movement of dnA1 moles of solute A from Side
1 to Side 2 of our system.
Let’s now work with only the right-hand part of Equation 5, i.e.:
A1 dnA1 + A2 dnA2 < 0
Equation 6
We first note that dnA2 = – dnA1 (because whatever enters Side 2 had to come from Side 1). By
substituting – dnA1 for dnA2 in Equation 4, it’s easy to show that:
dG = (A1 - A2 )dnA1 < 0
Equation 7
This is an insightful result. To see why I say this, consider the situation in which A1 > A2,
meaning that the quantity (A1 - A2 ) will be positive (> 0). When this is the case, in order for
Equation 7 to be true, dnA1 must be negative (i.e., < 0). As you’ll recall from your calculus class,
a negative value for dnA1 means that the amount of A in Side 1 is decreasing. In other words, if A1
> A2, molecules of A will leave Side 1 and enter Side 2.
What if A2 > A1? Well, you should be able to show that this would require dnA1 > 0,
meaning that A would leave Side 2 and enter Side 1. By extension, if A1 = A2 there will be no
net flux of A. Let’s put these results in tabular form:
Condition
Equation 6
Requires:
Direction of Flux
A1 > A2
dnA1 < 0
Side 1  Side 2
A1 < A2
dnA1 > 0
Side 2  Side 1
A1 = A2
n/a
No net flux; equilibrium
These results are extremely important and are the main reason we spend time talking about
thermodynamics at the start of the semester. The significance of those results, which cannot be
overstated, can be summed up as follows:
Matter tends to move from regions of high chemical
potential toward regions of low chemical potential, and it is
gradients in chemical potential that serve as the driving
force for flux of matter from one region to another.
The General Transport Equation
The conclusion we just arrived at is formalized in the General Transport Equation, which is
the charged-solute analog of Fick’s First Law:
d
J  L
Equation 8
dx
d  A2   A1
Here, the chemical potential gradient,
, serves as the driving force for diffusion,

dx
dx
and the Conductivity Coefficient, L, measures basically the same thing as the diffusion coefficient,
D, in Equation 1. In the case of non-ionic solutes, Equation 8 reduces to Equation 1, because the
d
contribution of  to
disappears (because  doesn’t affect the chemical potential of an
dx
uncharged solute), and d then depends only on concentration of the solute.
The reason that our conclusion is so very important to a budding physiologist such as yourself
derives from the dependence of  on a number of parameters, not just on the amount of matter in a
system. I.e.,
 = f(n, T, P, V,  , etc. ).
That  term means that gradients in electric potential (i.e., potential differences between two points
=   ) can contribute to chemical potential gradients, which means that electric potential gradients
can cause flux of ionic solutes. In fact, if   is big enough, ionic solutes will actuall y
move against their concentration gradient s, resulting in large concentration
gradients being maintained at equilibrium.