Rotational partition functions of hydrogen ($q_{\mathrm{eq}}=q_{\mathrm{p}}+3q_{\mathrm{o}}$)

Write the linear rigid rotor energy eigenvalues as $E_J=k_{\mathrm{B}}{\theta }_{r}J(J+1)$, where ${\theta }_r={\hslash }^2/(2I{k}_{\mathrm{B}})=hB/k_{\mathrm{B}}$ is the rotational temperature scale, $I$ the moment of inertia, $B$ the frequency-valued rotational constant, and $k_{\mathrm{B}}$ the Boltzmann constant. Put $t=T/{\theta }_r$ and ${ϵ}_J=J(J+1)$. Define sums that include rotational multiplicity but exclude nuclear-spin multiplicity:The canonical partition function for one equilibrating internal rotor is $q_{\mathrm{eq}}=q_{\mathrm{p}}+3q_{\mathrm{o}}$, so its equilibrium orthohydrogen fraction isFor $t\ll 1$, the lowest allowed levels give $q_{\mathrm{p}}\simeq 1$ and $q_{\mathrm{o}}\simeq 3e^{-2/t}$, hence $x_{\mathrm{o}}^{\mathrm{eq}}\simeq 9e^{-2/t}/(1+9e^{-2/t})\to 0$. For $t\gg 1$, the densely populated even and odd ladders have equal leading sums, $q_{\mathrm{p}}\sim q_{\mathrm{o}}\sim t/2$, hence $x_{\mathrm{o}}^{\mathrm{eq}}\to 3/4$. The factor nine belongs to the low-temperature two-level approximation, and the factor three to the high-temperature spin multiplicities.

A frozen mixture requires two separately normalized probability distributions,Multiply these by the prescribed fractions $1-x$ and $x$, respectively. The nuclear-spin factor three cancels from the conditional orthohydrogen distribution. For fixed counts $N_{\mathrm{o}}$ and $N_{\mathrm{p}}$, the internal part of the many-molecule partition function is proportional to $q_{\mathrm{p}}^{N_{\mathrm{p}}}(3q_{\mathrm{o}}{)}^{N_{\mathrm{o}}}$, with combinatorial factors independent of temperature. Its logarithm yields the two conditional means weighted by the fixed counts. Using $q_{\mathrm{p}}+3q_{\mathrm{o}}$ instead would silently permit conversion and change the physical constraint.

Let $u_s=⟨ϵ{⟩}_s$ and $v_s={\mathrm{Var}}_s(ϵ)$ for the conditional distributions of parahydrogen and orthohydrogen, with $s=\mathrm{p},\mathrm{o}$. Differentiating the normalized Boltzmann factors gives $d{u}_s/dt=v_s/t^2$. Thus the rotational heat capacity at constant volume per molecule of a composition frozen at $x$ isAt conversion equilibrium, the law of total variance adds a nonnegative contribution from the changing species populations:The last term is the contribution from conversion. Compare the first two terms with a frozen sample at the same instantaneous composition; a frozen $3:1$ sample generally has different weights too. This formula follows directly from the variance of a mixture: subtract the overall mean inside each species, expand the square, and the cross terms vanish because each conditional deviation has zero mean. It also proves nonnegative rotational heat capacity in either model.

For an original worked example at $t=1/2$, summing the allowed ladders gives $q_{\mathrm{p}}\approx 1.00003072$, $q_{\mathrm{o}}\approx 0.05494692$, and $x_{\mathrm{o}}^{\mathrm{eq}}\approx 0.14150982$. The calculated rotational heat capacities are $c_{\mathrm{eq}}/k_{\mathrm{B}}\approx 1.94719633$ and $c_{\mathrm{fr}}/k_{\mathrm{B}}\approx 0.00110733$ for frozen normal hydrogen. The contrast is large because changing species accesses the $J=0$ to $J=1$ separation, whereas fixed species must use the larger $J=0$ to $2$ or $J=1$ to $3$ separations. These are model numbers at a chosen dimensionless temperature, not measured hydrogen data.

Both rotational heat capacities approach zero as $t\to 0$, although frozen normal hydrogen retains nonzero rotational internal energy. Both approach $k_{\mathrm{B}}$ as $t\to \infty$ within the rotor model. The high-temperature limit is a formal limit with all other excitations suppressed; actual molecular hydrogen eventually needs molecular vibration, nonrigidity, and dissociation. Multiplying the per-molecule result by the Avogadro constant gives rotational molar heat capacity in units of the molar gas constant $R$. A dilute classical ideal gas additionally contributes $3R/2$ from translation to its heat capacity at constant volume.