Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 2 / A6

Treat each species as a dilute classical ideal gas: molecules of the same species are indistinguishable, while different species are distinguishable. Let $N_i$ be the number of molecules of species $i$, $q_i(T,V)$ its molecular partition function with energies measured from a chosen internal reference, and ${\epsilon }^{0,i}$ that reference energy per molecule relative to a common chemical energy convention. Then the mixture's canonical partition function factorizes:The $N_i!$ corrects the overcounting of indistinguishable particles. The exponential restores the species-dependent reference energies that matter for reactions. Thus the Helmholtz free energy isUsing Stirling's approximation, $\ln (N_i!)\simeq N_i\ln N_i-N_i$, and differentiating at fixed $T,V,N_{j\ne i}$ yieldsThe ideal-gas $q_i$ does not depend on the number of molecules, so no additional $\partial q_i/\partial N_i$ term appears. These are chemical potentials per molecule; multiply by the Avogadro constant for a molar chemical potential. The result assumes the thermodynamic, nondegenerate-gas limit.

Let ${\rho }_i=N_i/V$ be a molecular number density and define the volume-independent molecular partition function density $f_i=q_i/V$. Then $q_i/N_i=f_i/{\rho }_i$. Also define the reaction reference-energy differenceSubstitution into the equilibrium chemical-potential relation givesAn equilibrium constant expressed using concentration activities $a_i={\rho }_i/c^{\circ }$ is dimensionless:Here every $f_i$ has units ${\mathrm{m}}^{-3}$, and $c^{\circ }$ is the chosen standard concentration, initially one molecule per cubic metre. The factor $c^{\circ }$ is essential: without it the association quotient has units of volume.

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.

Convert the new standard concentration to the number-density convention:For this association, $K_c={\rho }_{{\mathrm{H}}_2\mathrm{CO}}{c}^{\circ }/({\rho }_{{\mathrm{H}}_2}{\rho }_{\mathrm{CO}})$, so it is proportional to $c^{\circ }$. ThereforeBoth dimensionless constants describe the same physical equilibrium; changing the reference changes the numerical activity quotient because two reactant molecules form one product molecule.