Past exam of the chemistry A course of the University of Cambridge / 2025 / Paper 2 / A7

For identical, non-interacting molecules, the translational canonical partition function is $Q_N=q_{\mathrm{trans}}^N/N!$. Stirling's approximation gives the Helmholtz free energyAt fixed $V$, $q_{\mathrm{trans}}\propto T^{3/2}$, so differentiation gives $S=Nk[\log (q_{\mathrm{trans}}/N)+5/2]$. For one mole at the standard-state molar volume,Here the quoted molar translational quantity is per mole, and its ratio to $N_A$ is dimensionless. The derivative defining entropy is taken at fixed volume before inserting the standard-state volume; differentiating along a constant-pressure path would give the wrong additive constant.

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 $q_{\mathrm{rot}}$.

Under the stated vibrational-ground-state approximation and a nondegenerate electronic ground state, only translation and rotation contribute. ThereforeA temperature-independent ground-state energy, including zero-point energy, contributes no entropy. This estimate deliberately omits thermal vibrational excitation; it need not equal a tabulated value that includes all modes.

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.

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.

As a first estimate, neglect the small mass change and changes in geometry and vibrational populations. Removing an electron creates an unpaired spin, giving at least the spin-only electronic entropy increaseFor a more specific estimate, the actual cation ground term is $X{}^2{\Pi }_g$, rather than an orbitally nondegenerate doublet. Ignoring spin-orbit coupling, its two orbital components and two spin components give electronic degeneracy four andThe ground-state assignment is confirmed by state-resolved photoionization measurements.^[ref] Spin-orbit splitting and changed vibrational/rotational constants refine this estimate; the question's supplied partition functions alone do not determine those corrections. The $R\log 2$ answer is a spin-only estimate, while $R\log 4$ additionally retains the orbital degeneracy.