Past exam of the chemistry A course of the University of Cambridge / 2023 / Paper 2 / A8 / c

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 ground-state references and the supplied ideal-gas, rigid-rotor and harmonic-oscillator approximations. The two ions have closed electronic shells; take their electronic ground-state degeneracies and the molecular electronic ground-state degeneracy as one, and neglect higher electronic levels. Consistently treated nuclear-spin factors cancel between the species. Since the molecule is heteronuclear, its rotational symmetry number is one.

Let ${\mu }_r=m_{\mathrm{Cs}}{m}_{\mathrm{I}}/m_{\mathrm{CsI}}$ be the reduced mass, neglecting the negligible mass correction from electron transfer. The ratio of translational partition functions per volume reduces toThe molecule has additional rotational and vibrational states, whereas the monatomic ions do not. At $3000\mathrm{K}$,Hence the dimensional equilibrium quotient isThe last form makes the unassigned standard concentration explicit. For the common choice of $1\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$, the corresponding microscopic reference is $c^{\circ }=6.02214\times 10^{26}{\mathrm{m}}^{-3}$, givingEquivalently, the dimensional molar quotient is $8.27\times 10^{-7}\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$. A pressure-based gas standard is a different convention: taking $c^{\circ }=p^{\circ }/(k_{B}T)$ with $p^{\circ }=1\mathrm{bar}$ would give $2.06\times 10^{-4}$, rather than the same numerical constant.

The well-minimum convention gives the identical answer becauseConstants used here are $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, $m_u\simeq 1.66053907\times 10^{-27}\mathrm{kg}$ and $N_A=6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref]