Sometimes it pays to be persistent in thinking about things you don’t understand (if you have the time as I do). The chemical potential is of enormous interest to chemists, and yet is defined thermodynamically in 5 distinct ways. This made me wonder if the definitions were actually describing essentially the same thing (not to worry they are).

First, a few thousand definitions

Chemical potential of species i — mu(i)

Internal energy — U

Entropy — S

Enthalpy — H

Helmholtz free energy — F or A (but, maddeningly, never H)

Gibbs free energy — G

Ni — number of elements of chemical species i

Pressure — p

Volume — V

Temperature — T

Just 5 more

mu(i) == ∂H/∂Ni constant S, p

mu(i) == ∂S/∂Ni constant U, V

mu(i) == ∂U/∂Ni constant S, V

mu(i) == ∂F/∂Ni constant T, V

mu(i) == ∂G/∂Ni constant T, p

Clearly, at a given constellation of S, U, F, G the mu(i)’s won’t all be the same number, but they’re essentially the same thing. Here’s why.

Start with a simple mathematical problem. Assume you have a simple function (f) of two variables (x,y), and that f is continuous in x and y and that its partial derivatives u = ∂f/∂x and w = ∂f/∂y are continuous as well so you have

df = u dx + w dy

u and dx are conjugate variables, as are w and dy

Suppose you want to change df = u dx + w dy to

another function g such that

dg = u dx – y dw

which is basically flipping a pair of conjugate variables around

Patience, the reason for wanting to do this will become apparent in a moment.

The answer is to use what is called the Legendre transform of f which is simply

g = f – y w

dg = df – y dw – w dy

plug in df

dg = u dx + w dw – y dw – w dy == df – y dw – w dy Done.

Where does the thermodynamics come in?

Well, you have to start somewhere, so why not with the fundamental thermodynamic equation for internal energy U

dU = ∂U/∂S dS + ∂U/∂V dV + ∑ ∂U/∂Ni dNi

We already know that ∂U/Ni = mu(i)

Because of the partial derivative notation (∂) it is assumed that all the other variables say in the expression for dU e.g. V and Ni are held constant in ∂U/∂S. This will reduce the clutter in notation which is already cluttered enough.

We already know that ∂U/∂Ni is mu(i). One definition of temperature T, is as ∂U/∂S, and another for p is -∂U/∂V (which makes sense if you think about it — decreasing volume relative to U should increase pressure).

Suddenly dU looks like what we were talking about with the Legendre transformation.

dU = T dS – p dV + ∑ mu(i) dNi

Apply the Legendre transformation to U to switch conjugate variables p and V

H = U + pV ; looks suspiciously like enthalpy (H) because it is

dH = dU + p dV + V dp + ∑ mu(i) dNi

= T dS – p dV + ∑ mu(i) dNi + p dV + V dp

= T dS + V dp + ∑ mu(i) dNi

Notice how mu(i) here comes out to ∂H/dNi at constant S and P

Start with the fundamental thermodynamic equation for internal energy

dU = T dS – p dV + ∑ mu(i) dNi

Now apply the Legendre transformation to T and S and you get

F = U – TS ; looks like the Helmholtz free energy (sometimes written A, but never as H) because it is.

You get

dF = – S dT – p dV + ∑ mu(i) dNi

Who cares? Chemists do because, although it is difficult to hold U constant or S constant (and it is impossible to measure them directly) it is very easy to keep temperature and volume constant in a reaction, meaning that changes in Helmholtz free energy under those conditions is just

∑ mu(i) dNi. So here mu(i) = ∂F/∂Ni at constant T and p

If you start with enthalpy

dH = T dS + V dp + ∑ mu(i) dNi

and do the Legendre transform you get the Gibbs free energy G = H – TS

I won’t bore you with it but this gives you the chemical potential mu(i) at constant T and p, conditions chemists easily arrange all the time.

To summarize

Enthalpy (H) is one Legendre transform of internal energy (U)

Helmholtz free energy (F) is another Legendre transform of U

Gibbs free energy (G) is the Legendre transform of Enthalpy (H)

It should be clear that Legendre transforms are all reversible

For example if H = U + PV then U = H – PV

If you think a bit about the 5 definitions of chemical potential, you’ll see that it can depend on 5 things (U, S, p, V and T). Ultimately all thermodynamic variables (U, S, H, G, F, p, V, T, mu(i) ) often have relations to each other.

Examples include H = U + pV, F = U – TS, G = H -TS

Helping keep things clear are equations of state from the things you can easily measure (p,V, T). The most famous is the ideal gas law p V = nRT.