Question about Gibbs free energy for mixture 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. 
 A: 
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. 
