Find the relation of chemical potentials in chemical reaction In hydrogen atom ionization

1. Dec 16, 2017

cozycoz

1. The problem statement, all variables and given/known data
In hydrogen atom ionization $$H→p+e$$ show that $μ_H=μ_p+μ_r$

2. Relevant equations
G=μN (N is the number of particles)

3. 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 $$dG=μ_HdN_H+μ_pdN_p+μ_edN_e =(μ_H-μ_p-μ_e)dN$$ 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?

2. Dec 18, 2017

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 = μH NH + μp Np + μe Ne

3. Dec 18, 2017

Staff: Mentor

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.

4. Dec 18, 2017

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 $\frac{\partial U}{\partial S} \equiv T$, $\frac{\partial U}{\partial V} \equiv p$, $\frac{\partial U}{\partial N} \equiv \mu$.

5. Dec 18, 2017

Staff: Mentor

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