Statistical physics of molecular dissociation of a diatomic gas Say there are $N$ atoms of type $A$ in a box of volume $V$ and they are undergoing a reversible association-dissociation reaction $A + A = A_2$. 
Let an $A$ atom have mass $m$, and hence the molecule $A_2$ has mass $2m$. 
Assume that the temperature $T$ is high-enough that Boltzmann statistics applies. 
Say $N_1$ be the number of $A$ atoms and say $N_2$ be the number of $A_2$ molecules at equilibrium. Then $N_1 + 2N_2 = N$. 
If there had to be $N$ $A$-type atoms in the the box then I guess the partition function $Z$ would be given by,
$$Z = \frac{1}{N!} [ \int e^{-\frac{\beta p^2}{2m}} \frac{d^3p d^3q}{h^3} ] ^N = \frac{1}{N!} [ \frac{V}{h^3} (2\pi m kT)^{\frac{3}{2}} ] ^N .$$
Say that the dissociation energy of the reaction is $E_0$.  Say $N_1$ be the number of $A$ atoms and say $N_2$ be the number of $A_2$ molecules at equilibrium. Then $N_1 + 2N_2 = N$. 
I want to write the partition function for the above mixed system in equilibrium. 
Using the expression for $Z$ as stated above, I am guessing that this partition function (say $Z_N$)  would look like, 
$$Z_N = \sum _ {N_1 + 2N_2 = N} e^{-\beta N_2 E_0} \frac{1}{N_1 !}  [ \frac{V}{h^3} (2\pi m kT)^{\frac{3}{2}} ] ^{N_1}  \frac{1}{N_2 !}  [ \frac{V}{h^3} (4\pi m kT)^{\frac{3}{2}} ] ^{N_2}.    $$
(In the above I am using the idea that for $N_2$ molecules to have formed $N_2$ $E_0$ (binding) energy would have been released and hence that much of extra energy has to be accounted for which is not captured in the partition functions over the $N_1$ atoms and $N_2$ molecules...) 


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*Is the above expression for the partition function correct ? 


One further introduces the grand canonical partition function (say $Z_{\mu}$ )  for the above system with a chemical potential $\mu$ to "take care of the constraint $N_1 + 2N_2 = N$",
$$Z_{\mu} = \sum _ {N=0} ^\infty e^{\beta \mu N} Z_N.$$


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*Firstly I don't understand the meaning of the $\mu$ in the above. (I can't relate it to what I would usually think as chemical potential) I mean..isn't the above expression for $Z_{\mu}$ independent of $N_1$ and $N_2$? 

*How does the $Z_{\mu}$ help find the expectation value for $N_1$ and $N_2$?

*If my expressions for $Z_N$ is correct then does $Z_{\mu}$ have a closed form expression? 
All this is to understand the claim that as expectation values the following relationship apparently holds, 
$$\frac{N_1 ^2} {N_2} \sim e ^{-\frac{|E_0|}{kT}} .$$ 
I would be glad to know of other insights along the above. 
 A: Yes, the Maxwell-Boltzmann partition functions of the two subsystems are
$$Z_1(N_1) ~=~ \frac{1}{N_1!}\left[\frac{V}{h^3} \left(\frac{2\pi m}{\beta}\right)^{\frac{3}{2}}\right] ^{N_1},   $$
$$Z_2(N_2) ~=~ \frac{1}{N_2!}\left[\frac{V}{h^3} \left(\frac{4\pi m}{\beta}\right)^{\frac{3}{2}} e^{-\beta E_0} \right] ^{N_2},   $$
where we ignore the rotational and vibrational modes of the molecule $A_2$. Here $E_0<0$ is the binding energy of the chemical reaction $2A \leftrightarrows A_2$. The full partition function, for a fixed number $N=N_1+2N_2$, is
$$ Z(N) ~=~ \sum_{N_2=0}^{[\frac{N}{2}]}Z_1(N-2N_2)Z_2(N_2). $$
The corresponding grand partition functions for the two subsystems are
$$Z_1(\mu)~:=~\sum_{N_1=0}^{\infty}Z_1(N_1)e^{\beta\mu N_1} ~=~\exp\left[\frac{V}{h^3} \left(\frac{2\pi m}{\beta}\right)^{\frac{3}{2}}e^{\beta\mu } \right] ,   $$
$$Z_2(\mu)~:=~\sum_{N_2=0}^{\infty}Z_2(N_2)e^{2\beta\mu N_2} ~=~\exp\left[\frac{V}{h^3} \left(\frac{4\pi m}{\beta}\right)^{\frac{3}{2}}e^{\beta(2\mu-E_0) } \right] ,   $$
where we have inserted a factor of $2$ to prepare for the implementation of the constraint $N_1+2N_2=N$ of the chemical reaction $2A \leftrightarrows A_2$. The full grand partition function, that is relevant for the chemical reaction $2A \leftrightarrows A_2$, is
$$Z(\mu)~:=~\sum_{N=0}^{\infty}Z(N)e^{\beta\mu N}~=~Z_1(\mu)Z_2(\mu).$$
The average occupation numbers are
$$\left< N_1\right>_{\mu} ~=~ \frac{1}{\beta}\frac{\partial\ln Z_1(\mu)}{\partial\mu} = \ln Z_1(\mu) , $$
$$\left< N_2\right>_{\mu} ~=~ \frac{1}{2\beta}\frac{\partial\ln Z_2(\mu)}{\partial\mu} = \ln Z_2(\mu) . $$
The following fraction is independent of the chemical potential $\mu$:
$$ \frac{\left< N_1\right>^2_{\mu=0}}{\left< N_2\right>_{\mu=0}}~=~\frac{\left< N_1\right>^2_{\mu}}{\left< N_2\right>_{\mu}} ~=~\frac{V}{h^3} \left(\frac{\pi m}{2\beta}\right)^{\frac{3}{2}}e^{\beta E_0} ~\sim~ e^{\beta E_0}~=~e^{-\beta |E_0|} ,$$
where the "$\sim$" sign means "is proportional to".
