 Solution

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]