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?

  • $\begingroup$ Watch your units! $\endgroup$
    – mcocdawc
    Commented Jan 15, 2017 at 11:03
  • $\begingroup$ Are you talking about alcohols with one OH group where you are just extending the chain length or is the number of OH groups increasing as well. $\endgroup$
    – mcocdawc
    Commented Jan 15, 2017 at 11:06
  • $\begingroup$ just extending the chain length for monohydric alcohols. $\endgroup$
    – Jfrey
    Commented Jan 15, 2017 at 13:26

1 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.

  • $\begingroup$ I realised the increase in boiling point for alcohols decreases as well. Both boiling point and enthalpy of vaporisation should reflect the strength of intermolecular forces, but since they refer to the vapour pressure needed and energy needed for vaporisation(which might not be used only to overcome the intermolecular forces) respectively, they probably don't change by the same percentage, what exactly is behind the difference? This is what still puzzles me. $\endgroup$
    – Jfrey
    Commented Feb 6, 2017 at 10:16
  • $\begingroup$ Well, I may not have done a good job of answering your question then. The main point of my answer lies in table 2, which shows that 2 different processes control the enthalpy of vaporization, and similarly the saturation vapor concentration (also known as vapor pressure) or boiling point. At the low-molecular weight end, hydrogen bonding dominates, so we see the behavior common to polar, hydrogen-bonding compounds. At the high-MW end, we see the pattern observed for alkanes. I'll try to think of a better way to explain, but no guarantees ;) $\endgroup$
    – airhuff
    Commented Feb 6, 2017 at 10:41

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