Why does the combustion of graphite have a positive Delta S?


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> The following reaction is exothermic and entropically favorable:
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> $$\ce{C (s) + O2 (g) -> CO2 (g)}$$
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> My question: Why is this reaction entropically favorable?
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Clearly there are more ways to arrange $\ce{C (s)}$ and $\ce{O2 (g)}$ in a container than $\ce{CO2}$.


Sure, the $\ce{CO2}$ has degrees of freedom in it's vibrations, but these shouldn't play a role in determining the entropy of the reaction since the reaction is completely unaffected by the type of vibration $\ce{CO2}$ is doing.




There are a number of possible explanations for the entropy difference, which can be computed based on [standard free energies](https://en.wikipedia.org/wiki/Standard_Gibbs_free_energy_of_formation) and [enthalpies](https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation) of formation as $\pu{3.0 J/molK}$. First, it could be that the entropy change is so small ($TΔS^∘<\pu{1 kJ}$, compare this to $ΔH^∘=\pu{−393.5kJ/mol}$) and measurement too imprecise for this difference to be significant. 


However, an alternate conclusion can be drawn from basic expressions for the translational and rotational entropy of a linear molecule in a gas, as derived from statistical mechanics:


$$\begin{align} S\_{\text{rot}}/R &\propto \log{\left( I\right)} \\ S\_{\text{trans}}/R &\propto 1.5\log{\left( m\right)}\end{align}$$


where m and I are respectively the mass and moment of intertia of the molecule. The moment of intertia of $\ce{CO2}$ is roughly 4 times that of $\ce{O2}$, therefore the above two contributions suggest that the molar entropy of $\ce{CO2}$ should be about $\pu{15.5 J/molK}$ greater than that of $\ce{O2}$. 


Finally, note that vibrational contributions to the entropy of the gases are expected to be negligible based on basic statistical mechanical expressions. Vibrations in graphite contain softer contributions and therefore vibrations would be expected to contribute more to the entropy of the solid than to that of either gas. [NIST](https://webbook.nist.gov/cgi/cbook.cgi?ID=C7782425&Type=SS) provides a value of $\approx \pu{5.5 J/molK}$ for the standard entropy of graphite. This doesn't entirely account for the small value of $\Delta\_r S^\circ =\pu{3.0 J/molK}$ based on enthalpies and free energies of formation, but brings the estimate a little closer, to $\Delta\_r S^\circ \approx \pu{10 J/molK}$.


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