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Latent heat of vaporization and boiling point are both chemical properties related to the bond strength of the molecule. In general, the higher the boiling point is, the higher the heat of vaporization will be. However, there are some molecules that break this trend. For example, ammonia has a higher heat of vaporization than you’d expect based on its boiling point relative to other molecules with similar boiling points. See table here: https://en.m.wikipedia.org/wiki/Enthalpy_of_vaporization. Why does ammonia (and other substances) break this trend?

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  • $\begingroup$ Im guessing this has something to do with the change in heat capacity between liquid and gas. Different rotational/vibrational degrees of freedom. ? $\endgroup$ – Karl Dec 11 '19 at 21:35
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Hm, the proportionality is not so very good. (data from the wikipedia link above) enter image description here The line is a linear regression through all ten data points.

The answer to your question is given on the same wp article. The quotient of enthalpy of vaporisation and temperature is the difference in entropy between the liquid and gaseous phase.

According to Trouton´s rule, it is about 85 to 88 J/molK (ca. $10.5 R$) for most liquids. https://en.wikipedia.org/wiki/Trouton%27s_rule explains the reasons for deviations, like hydrogen bonding, small rotational excitation, dimerisation in the gas phase etc. enter image description here

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  • $\begingroup$ In the plot graph above, why does ammonia have a higher latent heat of vaporization than butane but butane has a higher boiling point? $\endgroup$ – Ryan Dec 11 '19 at 21:22
  • $\begingroup$ Th logscale plot looks a lot more convincing ... ;) $\endgroup$ – Karl Dec 11 '19 at 22:48
  • $\begingroup$ Some of those values from the wp article seem to be wrong, btw. Definitely butane and propane. $\endgroup$ – Karl Dec 12 '19 at 19:06
  • $\begingroup$ Also isoprop and ethanol are wrong. $\endgroup$ – Karl Dec 12 '19 at 22:36
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Latent heat of vaporization is a physical property of a substance, which is defined as the heat required to change one mole of liquid at its boiling point under standard atmospheric pressure. Hence, it is usually expressed as $\pu{kJ mol-1}$.

The physical properties of a substance (e.g., its boiling point, melting point, latent heat of vaporization, etc.) will be effected by the types of intermolecular forces (e.g., H-bonding, dipole-dipole interactions, van der Waals forces, etc.) that occur in a substance. For instance, substances with weak intermolecular forces will have low melting and boiling points since only less thermal energy is needed to overcome these forces. On the other hand, those with strong intermolecular forces (e.g., water and alcohols) will have high melting and boiling points as more thermal energy is required to overcome these forces. Similarly, latent heat of vaporization is very much depends on the strength of these intermolecular forces.

For example, let's look at first ten linear alkanes, which have the least intermolecular forces (strong intermolecular forces such as H-bonding or dipole-dipole interactions are absent).

$$ \begin{array}{lrr} \hline n\text{-Alkane} & T_\mathrm{b}/\pu{K} & \Delta_\mathrm{vap}H(T_\mathrm{b})/\pu{J mol-1}\\\hline \text{Methane} & 112 & 8176\\ \text{Ethane} & 184 & 14640\\ \text{Propane} & 231 & 18832\\ n\text{-Butane} & 273 & 22390\\ n\text{-Pentane} & 309 & 26352\\ n\text{-Hexane} & 342 & 28850\\ n\text{-Heptane} & 372 & 31800\\ n\text{-Octane} & 399 & 33972\\ n\text{-Nonane} & 424 & 36910\\ n\text{-Decane} & 447 & 38750\\ \hline \end{array}$$

Data sources: NIST Chemistry WebBook; The Engineering Tool Box.

When you plot these values, you'd get nice correlation:

Relationship between latent heat of vaporization and boiling point of alkanes

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    $\begingroup$ A nice linear fit, but not a proportionality. ;) $\endgroup$ – Karl Dec 12 '19 at 18:23
  • $\begingroup$ @Karl: I'm not physical chemist, so I didn't understand why there is a negative intercept here. I thought to beat van der Waals forces (and/or London forces), we need positive input. Do you understand why? I'd love to hear your opinion. $\endgroup$ – Mathew Mahindaratne Dec 12 '19 at 18:43
  • $\begingroup$ If you add a line through zero and decane, you will find it has a slope of 86 or so, like Trouton predicts. The smaller alkanes deviate, because kT at their boiling point is too small to allow (reaching equillibrium between translational and) rotational excitation. $\endgroup$ – Karl Dec 12 '19 at 18:59
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    $\begingroup$ Thank you both for correct explanation in this regard. :-) $\endgroup$ – Mathew Mahindaratne Dec 12 '19 at 20:53
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    $\begingroup$ @Karl Boiling temperature :-) Not tuberculosis $\endgroup$ – Buck Thorn Dec 12 '19 at 21:30

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