= Solution
At low <temperature>, retain $J=0$ in the even sum and $J=1$ in the odd sum. Their weights are respectively $1$ and $3e^{-2\theta_{\rm rot}/T}$ before the nuclear-spin multiplicity, so
$$
\boxed{K\simeq9e^{-2\theta_{\rm rot}/T}\longrightarrow0\qquad(T\ll\theta_{\rm rot}).}
$$
Equilibrium <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. Hence
$$
\boxed{K\longrightarrow3,\qquad x_o\longrightarrow\frac34,\quad x_p\longrightarrow\frac14
\qquad(T\gg\theta_{\rm rot}).}
$$
This 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.<Compilation of Thermal Properties of Hydrogen in Its Various Isotopic and Ortho-Para Modifications by Harold W. Woolley and others>[Section III.2, printed pp. 394–395 and table 12: ortho–para equilibrium and its experimental checks]{ref}
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