Molecules having higher enthalpy of vaporization and boiling point than another but with lower entropy of vaporization?

Question

I came across the Trouton's rule that predicts the entropy of vaporization of most molecules to be around 85~88kJ/(K mol). It is said to fail when there is hydrogen bond between molecules.

When I checked the entropy of vaporization for linear alkanes, it increases with molecular mass which looks reasonable to me because a liquid with greater volume should expand more once it transits completely into gas than those with lesser volume. Eventually, the increase does slow down as the values get close to ~90kJ/(K mol).

For linear monohydric alcohols, their entropies of vaporization were higher than that predicted by Trouton's rule which I have found to be explained by their lower entropies in liquid phase due to hydrogen bond. However, I found that the entropy of vaporization of alcohols decreases instead with molecular mass. But since the boiling point and enthalpy of vaporization both increase with molecular mass. Shouldn't their entropy of vaporization increase with stability like alkanes?

So my question is: Is there any way to explain the decreasing ratio between the enthalpy of vaporization and the boiling point with increasing molecular mass for linear monohydric alcohols?

Answer 1

The enthalpy of vaporization is greater for hydrogen-bonding molecules than for plain alkanes. For low-molecular weight alcohols, this effect is pronounced. The longer the alkane chain becomes, the more the compound behaves like a pure alkane. For example, using data obtained from the NIST Webbook:

Table 1. Enthalpy of vaporization in kj/mol:
$\ce{Methanol\ \ \ \ \ 38}$
$\ce{Ethanol\ \ \ \ \ \ \ 42}$
$\ce{n-propanol\ \ 47}$
$\ce{n-butanol\ \ \ \ 52}$
$\ce{n-pentanol\ \ 57}$
$\ce{n-hexanol\ \ \ \ 61}$
$\ce{n-heptanol\ \ 67}$
$\ce{n-octanol\ \ \ \ 71}$
$\ce{n-nonanol\ \ \ 77}$
$\ce{n-decanol\ \ \ \ 82}$

Keeping in mind the relative molecular weights of the compounds, you can see there is a decreasing effect of the hydrogen bonding (and other) effects on the n-alcohol series as we move to larger chains and become less alcohol-like and more alkane-like. This is much more obvious in the next table, which is simply Table 1 normalized to a per-carbon basis:

Table 2. Enthalpy of vaporization on a per-carbon basis in kj/mol:
$\ce{Methanol\ \ \ \ \ 38}$
$\ce{Ethanol\ \ \ \ \ \ \ 21}$
$\ce{n-propanol\ \ 16}$
$\ce{n-butanol\ \ \ \ 13}$
$\ce{n-pentanol\ \ 11}$
$\ce{n-hexanol\ \ \ \ 10}$
$\ce{n-heptanol\ \ 10}$
$\ce{n-octanol\ \ \ \ 9}$
$\ce{n-nonanol\ \ \ 9}$
$\ce{n-decanol\ \ \ \ 8}$

Table 2 even more drastically illustrates the deviation in $\mathrm{\Delta}H$ changes as we move from alcohol-like compounds to alkane-like compounds. So, given that the enthalpy of vaporization is greater for hydrogen-bonding molecules than for plain alkanes, this should explain your findings.