Why is the spectroscopic entropy of carbon monooxide higher than that of nitrogen?

Question

While solving a question set, I found a question with this data table (at $\pu{298K}$):

$$ \begin{array}{lrrr} \hline \text{Substance} & T_\mathrm b/\pu{K} & S^⦵_\mathrm{cal}/R & S^⦵_\mathrm{spec}/R \\ \hline \text{Nitrogen} & 77.35 & 18.3 & 18.30 \\ \text{Carbon monoxide} & 81.61 & 18.6 & 19.22 \\ \hline \end{array} $$

The first part of the question asked why $\ce{CO}$ has higher $S_\mathrm{spec}$ (spectroscopic entropy) than $S_\mathrm{cal}$ (calorimetric entropy). I know that this is due to the residual entropy of $\ce{CO}$.$^*$

However, the next part of the question asked why the value of spectroscopic entropy of $\ce{CO}$ is larger than that of $\ce{N2}$. The question supplied the information that the rotational reduced temperature ($\theta_T$) was same for both molecules (I think that means the rotational partition function $q_\mathrm{rot}$ will be same for both); and that the contribution to entropy from the vibrational modes to entropy is negligible in both cases.

Now the spectroscopic entropy can be written in terms of the contribution of each degree of freedom: $$S_\mathrm{spec}=S_\mathrm{trs}+S_\mathrm{rot}+S_\mathrm{vib}+S_\mathrm{elec}$$ The translational entropy contribution depends on the translational partition function $q_\mathrm{trs}$ , which depends on the mass of the molecule. $\ce{CO}$ and $\ce{N2}$ has almost the same mass (28.01 u), so I don't think there would be any difference in $S_\mathrm{trs}$. The rotational and vibrational terms are assumed to be same in the question. The electronic term should also be zero in both cases, as the electronic partition function would be 1 for both. (I think the excited states are all inaccessible at room temperature).

So where does the difference in the calculated entropy come from?

I feel like the answer is something simple that I have missed.

$^*$ For the concept to discern calorimetric entropy from spectroscopic entropy, see for example this Chemistry LibreText; an literature example e.g., in 1958JAmChemSoc1054.

Answer 1

One contribution is the number of identical conformations of a molecule that can be generated by symmetry operations, which is described by a symmetry number $\sigma$. In the high temperature limit

$$S_\mathrm{rot} /R = 1 + \ln\left( \frac{8\pi^2IkT}{h^2\sigma} \right) $$

A larger symmetry number decreases the entropy. This number is 2 for nitrogen and 1 for carbon monoxide. That alone can account for $\log(2)= 0.69 \times R$. The larger symmetry number for nitrogen decreases the entropy by this amount.

In addition the moments of inertia (inversely proportional to the rotational temperature) of the two molecules are not really identical (it is ~5% larger for $\ce{CO}$), but the difference accounts for only ~$0.05 \times R$.