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

  1. Dec 16, 2017 #1
    1. The problem statement, all variables and given/known data
    In hydrogen atom ionization [tex]H→p+e[/tex] 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 [tex]dG=μ_HdN_H+μ_pdN_p+μ_edN_e
    =(μ_H-μ_p-μ_e)dN[/tex] 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. jcsd
  3. Dec 18, 2017 #2
    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
     
  4. Dec 18, 2017 #3
    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.
     
  5. Dec 18, 2017 #4

    TeethWhitener

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    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##.
     
  6. Dec 18, 2017 #5
    For a single chemical species at equilibrium, this should be an equality, since, from the Clausius-Duhem equation, $$-SdT+VdP+Nd\mu=0$$
     
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