In the section about fuels, my textbook mentions that the boiling and freezing-points of alkanes (which are non-polar) increase with an increase in molecular mass due to London dispersion forces.

Now, I understand what these are and why they become relevant when talking about non-polar molecules, but I can't figure out how they are connected to the boiling and freezing point of the matter. How do these London dispersion forces influence the boiling and freezing point?

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    $\begingroup$ Short answer: Boiling point and freezing point depends wholly on the inter molecular forces/interaction. The greater the inter molecular forces, the greater the boiling point. $\endgroup$ Dec 28, 2012 at 13:12

1 Answer 1


In general, the larger the molecule, the larger the accessible surface area of the molecule and the more surface polarisable electron density. There are more atoms that can participate in dispersion interactions and the effect stacks up. To melt or boil a substance you need to overcome the intermolecular forces holding the molecules together and dispersion forces - whilst weak even with respect to hydrogen bonding - can add up to large interaction energies. One of the more interesting examples of strong dispersion interactions is the gecko's foot, which permits the reptile to easily stick to sheer surfaces due to the exceptional surface area of the footpad, which is covered in many many tiny hairlike protrusions.

  • $\begingroup$ But how does the molecule's mass contribute to it's intermolecular forces? $\endgroup$ Dec 28, 2012 at 14:05
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    $\begingroup$ @EdwardStumperd - I'm going to assume that the example given in your textbook is a homologous series of alkanes or similar, in which case more massive molecules correspond to later members in the series, with greater chain lengths and correspondingly larger volumes and areas. $\endgroup$ Dec 28, 2012 at 15:00

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