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The six-operation molecular point group with a threefold proper rotation axis and three vertical mirror planes. Its rotational symmetry number is three because mirror operations do not count as proper spatial rotations.

Define ${\theta }_{\mathrm{rot}}={\hslash }^2/(2I{k}_{\mathrm{B}})$. Summing the spatial states counted in part (c),When $T\gg {\theta }_{\mathrm{rot}}$ the closely spaced terms can be replaced by a Gaussian integral:Here labelled means that the two ends of the molecular axis were distinguished in the angular count. In the usual high-temperature molecular convention that removes equivalent orientations and factors out nuclear-spin multiplicity, divide by the rotational symmetry number $s=2$ for homonuclear hydrogen:The full exact spin isomers of hydrogen count is instead the even-$m$ spatial sum plus three times the odd-$m$ sum; its high-temperature limit is $2\sqrt{\pi T/{\theta }_{\mathrm{rot}}}$, including the four nuclear-spin states. These conventions differ only by temperature-independent factors in this limit. All give $U_{\mathrm{rot},\mathrm{m}}=RT/2$ and $C_{V,m}^{\mathrm{rot}}=R/2$, consistent with one quadratic rotational degree of freedom. By comparison, the free linear rigid rotor has $q_{\mathrm{rot}}\propto T$ and $C_{V,m}^{\mathrm{rot}}=R$.

For a high-temperature linear rigid rotor, $q_{\mathrm{rot}}=T/{\theta }_{\mathrm{rot}}$ up to temperature-independent symmetry/spin factors. Therefore $q^{\prime }/q=1/T$ and $q^{\prime \prime }=0$, givingThis agrees with the equipartition theorem for two quadratic rotational coordinates. It is the rotational contribution; adding free three-dimensional translation gives $C_{V,m}=5R/2$ when vibrations remain unexcited. A constant rotational symmetry number does not affect this derivative.

For independent internal rotational states, the rotational factor in the many-molecule partition function is $q_{\mathrm{rot}}^N$. The indistinguishability factor has already been included in the translational part and is not applied a second time to the internal factor. ThusA nonlinear rigid rotor has three rotational quadratic degrees of freedom; in the high-temperature limit, $q_{\mathrm{rot}}\propto T^{3/2}$. ThereforeThis is the rotational entropy contribution, with the appropriate rotational symmetry number already inside $q_{\mathrm{rot}}$. It has no ideal-gas pressure dependence.

Chloroform has the C3v molecular point group. Its three identical ${}^{35}\mathrm{Cl}$ atoms are interchanged by three proper rotations, so $\sigma =3$. The three mirror operations do not increase the rotational symmetry number to six. Isotopic identity matters for this count.

Using the supplied wavenumber form of $k_{\mathrm{B}}$,All are far below $298\mathrm{K}$, justifying the high-temperature nonlinear rigid rotor expression. ThenandThe calculation treats the molecule as rigid and separates electronic, vibrational and nuclear-spin contributions from the requested rotational term.

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]

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.

For a linear rigid rotor in the high-temperature regime, $q_{\mathrm{rot}}\propto T$. Its molar rotational entropy isThe rotational symmetry number must already be included in $q_{\mathrm{rot}}$.

The substituted molecule has the larger entropy. Increased mass and moment of inertia reduce translational and rotational spacings and make more states thermally accessible. The largest contribution here is the loss of the twofold rotational indistinguishability of the oxygen ends, contributing $R\log 2$. It is a rotational symmetry number correction for one isotope-defined species, not an entropy of mixing between different molecules.

Use the same standard pressure and temperature, the same bond length, and the same electronic and vibrational-ground-state approximations. The translational partition function scales as mass to the power $3/2$, giving the factor $(47/44{)}^{3/2}$. For the original isotopologue, $I=32m_u{R}^2$ and the rotational symmetry number is two. For the isotope-substituted molecule, the two oxygen ends are distinguishable, so the symmetry number is one. HenceThe change in standard molar entropy issoThis uses a convention omitting a separate nuclear-spin entropy. If nuclear-spin degeneracies are instead included explicitly, the spin-$1/2$ carbon-13 nucleus adds a further $R\log 2$ relative to the spin-zero original isotopologue; that convention must be stated rather than silently mixed with the quoted partition functions.

For identical atoms, use $\mu =m_A/2$ and the homonuclear rotational symmetry number $\sigma =2$. With a rate defined as reaction events per volume and time, this halves the geometric prefactor relative to distinguishable labelled pairs. Do not independently count both ordered pairs and then add another unrelated factor of two. If the rate is instead defined by disappearance of A, each event consumes two atoms, so $-d{n}_A/dt=2r$; that rate coefficient is twice the event-rate coefficient.

The entropy contribution from populated molecular rotational states. A nonlinear rigid rotor has $q_{\mathrm{rot}}\propto T^{3/2}$ at high temperature, giving $S_{\mathrm{rot},\mathrm{m}}=R(\ln q_{\mathrm{rot}}+3/2)$. Its rotational symmetry number removes redundant orientations.

The rotational partition function sums Boltzmann factors over rotational energy levels, including their degeneracies and the appropriate indistinguishability convention. For a linear rigid rotor at high temperature, $q_{\mathrm{rot}}=8{\pi }^{2}IkT/(\sigma h^2)$, with moment of inertia $I$ and rotational symmetry number $\sigma$.

The rotational symmetry number corrects a rotational phase-space count for indistinguishable molecular orientations. A linear molecule with identical ends, such as carbon dioxide with identical oxygen isotopes, has $\sigma =2$; making its ends isotopically distinct gives $\sigma =1$. In the high-temperature linear rigid rotor approximation, $q_{\mathrm{rot}}=8{\pi }^{2}IkT/(\sigma h^2)$. Holding the spectrum otherwise fixed, doubling $\sigma$ lowers molar rotational entropy by $R\log 2$.