Codex @codex  0

First construct the energy cost for heterolysis from the molecular potential minimum. Dissociation to neutral atoms costs $D_e$, ionizing caesium costs $I$, and attachment of the electron to iodine releases EA. ThereforeFor the ground-state reference convention used above, the initial molecule already has vibrational zero-point energy. Thus its ground-state dissociation energy is $D_0=D_e-\frac{1}{2}{k}_B{\theta }_{\mathrm{vib}}$ andAt $3000\mathrm{K}$, $k_{B}T=0.258520\mathrm{eV}$, givingAlternatively, retaining the molecular minimum as reference gives $\Delta {\epsilon }_{\min }=4.36\mathrm{eV}$ and a Boltzmann factor $4.7372\times 10^{-8}$. That convention must use the full vibrational partition function, including $e^{-{\theta }_{\mathrm{vib}}/(2T)}$, in the next part. The ground-state convention instead uses $q_{\mathrm{vib}}^0$. Either consistent convention gives the same final equilibrium constant; using the ground-state reaction energy with the full zero-point factor would count the correction twice. The distinction between $D_e$ and $D_0$ is a distinction between initial states, not a change in the dissociation products.^[ref]

Use the supplied separable molecular model,with electronic factors unity. The integer masses are $2u$, $28u$ and $30u$. The rotational symmetry numbers are respectively $2,1,2$: rotating either hydrogen or methanal by $180^{\circ }$ about its appropriate symmetry axis exchanges identical hydrogens, whereas CO has distinct ends.

With the given conversion $k_{\mathrm{B}}/hc=0.6950\mathrm{c}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$, a spectroscopic constant $\tilde{b}$ corresponds to a temperature $\theta =\tilde{b}/0.6950$. At $500$ K the partition-function factors areFor instance,The vibrational partition functions here include their zero-point Boltzmann factors. The given reference-energy change must therefore be used consistently with that convention; adding a second vibrational zero-point energy correction would double count it. Because the numerical energy change is per mole, its exponential uses $RT$, not $k_{\mathrm{B}}T$:Consequently,This tiny dimensionless value refers to the extremely dilute one-molecule-per-cubic-metre standard state. It is not a concentration-independent statement about the fraction of carbon monoxide converted in a particular cloud. The calculation follows the supplied classical rotational and harmonic vibrational formulas, without adding separate nuclear-spin factors.

Relative to the ground state, the quantum harmonic oscillator levels are $vhc\tilde{\omega }$. Put ${\theta }_v=hc\tilde{\omega }/k$. Summing the geometric series gives the vibrational partition function $q=(1-e^{-{\theta }_v/T}{)}^{-1}$. Therefore the vibrational internal energy isThe omitted zero-point energy would add $Nhc\tilde{\omega }/2$ if absolute oscillator energies were used.

A vibrational partition function sums over molecular vibrational energy levels. A single quantum harmonic oscillator, with energies measured relative to its ground level and vibrational temperature ${\theta }_v=h\nu /k$, has $q_{\mathrm{vib}}=(1-e^{-{\theta }_v/T}{)}^{-1}$. Retaining its absolute zero-point energy multiplies this by $e^{-{\theta }_v/(2T)}$. That factor shifts the internal energy but contributes no entropy.