2. We typically think of dipole-dipole interactions as being stronger than Londo

Question

2. We typically think of dipole-dipole interactions as being stronger than London Dispersion forces for molecules that are similar in size. Think about candle wax (a long chain hydrocarbon) which is a solid as compared to a polar molecule like acetone, which is a liquid. Comment on the relative strength of the intermolecular forces, and how that translates to the observed states of matter. 3. Two substances, A, and B, have very similar boiling points. Substance A is much smaller in size just a few atoms bound together) than Substance B (many atoms bound together). How can this be the case? Explain in terms of the intermolecular forces that must be present, and the types of atoms present in each substance.

Explanation / Answer

For candelwx, assume this is mostly nonpolar, therefore, the dispersion forces are the only forces present which are representative

Note that these are long linear chains, therefore, expect to have a higher melting/boiling point.

That is why we see "solid wax" which can be melted at relatively high temperautre ( 100°C or so ) therefore, the forces are stronger than many other nonpolar species due to size.

Acetone is slightly polar, the C=O group is polar, but there are many carbon/hydrogne between, therefore, this is expected to be liquid.

If we compare both species, they have overall high intermoelcular values compared to many nonpolars epcies.

Acetone will melt at realtively higher melting point than many other samples.

The direct comparison of a polar vs. nonopolar sample, let us compare how intermoelcular forces differ.

If the candel wax was to be a polar molecule, this will be most likely a liquid as well...

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