What Is the Relation of Chemical Potentials in Hydrogen Atom Ionization?

[cozycoz]

Homework Statement

In hydrogen atom ionization $$H→p+e$$ show that $μ_H=μ_p+μ_r$

Homework Equations

G=μN (N is the number of particles)

The Attempt at a Solution

(1) I think the question should say "Find chemical potential relation AT EQUILIBRIUM", don't you think?
(2) My professor said that because $dN_H=-dN_p=-dN_e=dN$, the change of gibbs energy becomes $$dG=μ_HdN_H+μ_pdN_p+μ_edN_e<br /> =(μ_H-μ_p-μ_e)dN$$ And the equilibrium occurs when dG=0, we can derive above relation.
But chemical potential also depends on N, so I think I can't simply write dG as above(cause extra $\frac{∂μ}{∂N}N$ terms should be included). How do you think?

 

[Lord Jestocost]

When there are several species, each has its own chemical potential so the total Gibbs free energy has to be written as:

G = μ_H N_H + μ_p N_p + μ_e N_e

 

[Chestermiller]

I think that, from the Gibbs-Duhem equation, we know that, at equilibrium, $$N_Hd\mu_H+N_pd\mu_p+N_ed\mu_e=0$$. This is pretty much the same thing that @Lord Jestocost said.

 

[TeethWhitener]

Maybe I'm reading too much into this, but I think where @cozycoz is getting hung up is the notion that
$$dG \neq \mu dN + Nd\mu$$
but rather
$$dG = \mu dN$$
The best way I can think to explain this is to look at the definition of $G$: $G = U-TS+pV$. Here, the dependence on $\mu$ and $N$ is entirely contained within the definition of internal energy $U$. But $U$ is defined as a function of extensive variables only: $U = U(S,V,N)$. So taking the total differential of $U$ gives;
$$dU = \frac{\partial U}{\partial S}dS + \frac{\partial U}{\partial V}dV +\frac{\partial U}{\partial N}dN$$
and we define $\frac{\partial U}{\partial S} \equiv T$, $\frac{\partial U}{\partial V} \equiv p$, $\frac{\partial U}{\partial N} \equiv \mu$.

 

[Chestermiller]

Maybe I'm reading too much into this, but I think where @cozycoz is getting hung up is the notion that
$$dG \neq \mu dN + Nd\mu$$

For a single chemical species at equilibrium, this should be an equality, since, from the Clausius-Duhem equation, $$-SdT+VdP+Nd\mu=0$$