I know that both molecular size and complexity increase overall entropy, but between the two I'm not sure which increases it more. Would 1 mole of hydrogen have the greater entropy because it is two atoms instead of one, or would 1 mole of neon have the greater entropy because it is a larger atom?
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$\begingroup$ Entropy inherent to atoms itself is treated as zero, because it can't be lowered with cooling. $\endgroup$– MithoronCommented Feb 7, 2019 at 2:36
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2 Answers
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Assuming you have equal moles of two different perfect gases, according to the Sackur–Tetrode equation the difference in their molar translational entropy is:
$$\frac{\Delta S_\mathrm{trans}}{R} = \frac{3}{2}\ln\left(\frac{M_2}{M_1}\right)$$
where the $M_i$ are the molecular masses. For instance, since $\ce{D_2}$ and $\ce{He}$ have identical masses this term would vanish, i.e. $ \Delta S_\mathrm{trans} = 0$.
If the atoms are in their electronic ground states then the electronic state makes no contribution to the entropy. On the other hand, a diatomic molecule has rotational and vibrational degrees of freedom. Assuming it's a rigid rotor, then in the high temperature limit (in which many rotational quantum levels are significantly occupied),
$$\frac{S_\mathrm{rot}}{R} = 1 + \ln\left(\frac{8\pi^2IkT}{h^2\sigma}\right)$$
where $I$ is the moment of inertia, $\sigma$ is a symmetry number.
Assuming the diatomic molecule is a harmonic oscillator, then
$$\frac{S_\mathrm{vib}}{R} = x(e^x - 1)^{-1} - \ln(1 - e^{-x})$$
where $x = hν/kT$ and $T$ is around room temperature. However, the contribution of the vibrational entropy is tiny compared to the rotational component, so it can be ignored.
That would wrap up the story, in any case, if nuclear spin didn't have something to say about it. Nuclear spin contributes additionally to the entropy of $\ce{H2}$ and $\ce{D2}$ as explained in this post, resulting in an altered rotational entropy term. This rotational/spin contribution lifts the entropy of these molecules above those of Ne and He.
Therefore at room temperature and $\pu{1 atm}$ pressure, the molar entropy of $\ce{D2}$ is expected to be larger than that of helium due to the rotational contribution of $\pu{31.3 J mol-1 K-1}$.
Similarly, the translational entropy of neon is $\pu{3.5 J mol-1 K-1}$ higher than that of hydrogen gas, but $\ce{H2}$ has a rotational entropy of $\pu{18.7 J mol-1 K-1}$ so the entropy of neon is expected to be smaller by about $\pu{15.2 J mol-1 K-1}$.
answered Feb 7, 2019 at 9:17
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$\begingroup$ The entropy of deuterium is a little larger than you quote (about 35 rather than 22 ) due to nuclear spin statistics, see the answer chemistry.stackexchange.com/questions/70620/… $\endgroup$ Commented Feb 8, 2019 at 17:42
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$\begingroup$ @porphyrin well done catching that- I'll have to edit... $\endgroup$– Buck Thorn ♦Commented Feb 8, 2019 at 18:01
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Assuming the same conditions, there is a difference in translational entropy proportional to log of the mass, see the Sakur-Tetrode eqn. A diatomic molecule has the ability to vibrate and rotate and this adds to the entropy but an atom has neither of these motions.
