For example a reaction is given as ($\pu{T = 1000 K}$),

$$\ce{2Na(g) <--> Na2 (g)}$$

The equilibrium constant,

$$\ce{K_{c} = \frac{\frac{q_{Na_{2}}}{V}}{(\frac{q_{Na}}{V})^2}}$$

But when considering the partition function for $\ce{Na2(g)}$. Teacher multiplied by another term which I don't understand.

The partition function for $\ce{2Na(g)}$,

$$\ce{q_{Na} = q^{t}.q^{e}}$$

The partition function for $\ce{Na2(g)}$,

$$\ce{q_{Na_{2}} = q^{t}.q^{r}.q^{v}.q^{e}.e^{β(D_{o} - \frac{hν}{2})}}$$

  1. Where dose this $\ce{e^{β(D_{o} - \frac{hν}{2})}}$ term come from why it has been used in the partition function from my understanding its coming from Morse Oscillator. Looks like it has some thing to do with bond dissociation for a diatomic molecule but I don't understand how did it get resemble to the partition function.

  2. Teacher took $\ce{q^{e}}$ for $\ce{Na(g)}$ as $\pu{2}$ and for $\ce{Na_{2}(g)}$ $\pu{1}$ how did that work according to my understanding,

The electronic partition function is given as,

$$\pu{q^{e} = g_{o}e^{-βE_{o}} + g_{1}e^{-βE_{1}} + g_{2}e^{-βE_{2}} ....}$$

Teacher only considered the ground state energies how did the other terms got neglected if ground state energy is $\pu{0}$ then only degeneracy matter.Further more I did not find any doubts about the other partition functions they are ok for me.

  • 3
    $\begingroup$ The Morse potential term arises from the nuclear degree of freedom term, thus for your diatomic $q = q_{\rm trans}q_{\mathrm rot}q_{\mathrm vib}q_{\mathrm elec}q_{\mathrm nucl}$. Your expression for $q_{\mathrm elec}$ is missing the first term - $\omega_{e1}$ - which is what the term converges to in a common approximation. The factor of 2 is simply because there are 2 $\ce{Na}$. Finally, your teacher copied this example (exactly) from p. 144 (section 9-2 A) in McQuarrie ${\it Statistical\, Mechanics}$ (1976), so you can look there to clear up the rest of your concerns. $\endgroup$ Commented Dec 10, 2021 at 17:42


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