Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 2 / A8 / a

The two protons are identical spin-one-half fermions, so the total state is antisymmetric under exchange. In the electronic and vibrational ground state, an even rotational $J$ has a symmetric spatial wavefunction and therefore uses the antisymmetric nuclear-spin singlet state; odd $J$ uses the symmetric triplet state. Thus parahydrogen has one nuclear-spin state, and orthohydrogen has three. Each rotational level also has spatial state degeneracy $2J+1$.

The translational partition functions and all common electronic/vibrational factors cancel in the ratio of equilibrium populations. Using the same zero of energy for the two spin species givesSubtracting the $J=1$ energy separately from all odd levels without restoring that offset in the equilibrium ratio would incorrectly remove the low-temperature preference for parahydrogen. The nuclear spin statistics account for the factor three, independently of the spatial factor $2J+1$.

At low temperature, retain $J=0$ in the even sum and $J=1$ in the odd sum. Their weights are respectively $1$ and $3e^{-2{\theta }_{\mathrm{rot}}/T}$ before the nuclear-spin multiplicity, soEquilibrium hydrogen then becomes almost entirely parahydrogen. At high temperature, many even and odd rotational levels contribute, and their spatial sums have the same leading value. HenceThis is normal hydrogen. Its familiar high-temperature ratio is a nuclear-spin multiplicity ratio, not the $9:1$ ratio of the two lowest levels. Maintaining conversion is essential: a sample cooled without an effective conversion mechanism can retain its earlier composition instead of following $K(T)$. NBS thermodynamic data distinguish equilibrium composition from normal hydrogen and compare the state-sum prediction with spectroscopic and thermal evidence.^[ref]