There is something I don't understand in the differential of Gibbs free energy for mixtures.

But to explain what I misunderstand, I will make a global recall of what I understood from thermodynamic potential.

A thermodynamic potential is a quantity that is minimum at equilibrium under the good external conditions.

For example, if I fix $N,P,T$, the Gibbs Free energy will be minimum at equilibrium. And all the other variables (chemical potential, volume and entropy) will be function of those $N,P,T$ (they will be imposed by those external conditions).

This is why we say that $G$ is a function of $N,P,T$ (and not a function of $S$ for example). Because we consider its value at equilibrium that will be only function of those $(N,P,T)$.

Thus, we have :

$$dG=-SdT+VdP+\mu dN$$

But in mixtures, we define (first line of "derivation" part : https://en.wikipedia.org/wiki/Gibbs%E2%80%93Duhem_equation )

$$dG=-SdT+VdP+\sum_i \mu_i dN_i$$

If we follow the "logic" of definition of thermodynamic potentials, this would mean that I have to impose a number of particles of each chemical component in my system.

But actually, at equilibrium we have equality of chemical potential and thus $\mu_i=\mu$ and

$$dG=-SdT+VdP+\mu \sum_i dN_i=-SdT+VdP+\mu dN$$

Maybe it is just a question of definition but I find weird that we write the differential with this sum on all the chemical potentials. Indeed, the differential of thermodynamic potentials should only be variables of external constraint we have to put on the system to make it equilibrate.

I am right by saying that this differential of Gibbs free energy with the $\sum_i dN_i$ shouldn't be here ?

I hope my question is clear enough.

  • $\begingroup$ Since you said 'mixtures', that means distinguishable particles. So, no, you need to consider the chemical potential of each type of particle/atom/molecule/whatnot under consideration. $\endgroup$
    – Jon Custer
    Commented Feb 1, 2018 at 23:45
  • $\begingroup$ If you're talking about a chemical reaction, the chemical potentials of the various species can change as a result of the changes in mole fractions, and you must minimize the G at constant T and P over these changes, under the stoichiometric constraints of the reaction. $\endgroup$ Commented Feb 2, 2018 at 0:25

1 Answer 1


But actually, at equilibrium we have equality of chemical potential and thus $\mu_i=\mu$...

There is a little bit of confusion here. If you take two systems $A$ and $B$, each one containing the chemical species which we will label $1,2,\dots,N$, separated by a permeable membrane so that they can exchange particles, you will indeed find that at thermodynamic equilibrium

$$\mu_i^A=\mu_i^B \ \ \ \ \ \ i=1,2,\dots N\label{1}\tag{1}$$

However, what you are usually interested in is one system which can exchange particles with the environment (in statistical mechanics, this is the reason to introduce the grand canonical ensemble).

In this case, if $P$ and $T$ are constant (like in most chemical reactions) and the system is at thermodynamic equilibrium with the environment, the Gibbs free energy is at a minimum: $dG=0$. It follows that

$$\sum_{i=1}^N \mu_i dN_i = 0\label{2}\tag{2}$$

To sum up, eq. \ref{1} is the condition of thermodynamic equilibrium for two open systems at contact which can exchange particles, while eq. \ref{2} is the condition of thermodynamic equilibrium for one system at constant $P,T$ which can exchange particles with the environment.

  • $\begingroup$ Thank you for your answer. I am not sure to understand though. I agree that this is the differential Gibbs for one big system. But if I have different chemical components inside they should be at equilibrium between themself ? If I take one big system with $A$ and $B$ species the chemical potential of those must be equal at equilibrium. The fact I consider in my reasoning the total or the two sub-systems shouldn't change this fact. Indeed the two species will be in contact in my global system. I hope I wrote clearly my misunderstanding. $\endgroup$
    – StarBucK
    Commented Feb 2, 2018 at 18:12
  • $\begingroup$ @StarBucK Indeed, once you have put together your two systems A and B you will obtain a system A+B where at equilibrium every chemical species $i$ has a well defined chemical potential $\mu_i$. But this doesn't mean that the chemical potential of different chemical species is the same. The chemical potentials of different species stays different after equilibrium is reached. $\endgroup$
    – valerio
    Commented Feb 2, 2018 at 18:28
  • $\begingroup$ Ah ok, my confusion was that the chemical potential are equal for different phases of a given chemical specy but they are different even at equilibrium for different chemical species. Was that what you mean ? $\endgroup$
    – StarBucK
    Commented Feb 2, 2018 at 18:37
  • $\begingroup$ @StarBucK Exactly $\endgroup$
    – valerio
    Commented Feb 2, 2018 at 19:57

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