Question

Does CCl4 have a higher boiling point than CH2Cl2? I though London forces were weaker than dipole-dipole forces?

Answer 1

Apparently yes, but London dispersion forces ARE weaker than dipole-dipole forces.

It looks like the reason for the exception here in boiling point trends is that there is a greater increase in entropy due to boiling `"CH"_2"Cl"_2` than `"CCl"_4`, and it requires less thermal energy to boil `"CH"_2"Cl"_2` than `"CCl"_4`. (These are not competing data.)

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`"CCl"_4` is completely symmetrical, and `"CH"_2"Cl"_2` is not. Therefore, the latter has dipole-dipole interactions, and the former has only London dispersion.

[I went into this in more detail here.](https://socratic.org/questions/when-are-tetrahedral-molecules-polar-and-when-are-they-nonpolar?source=search)

As a result, `"CH"_2"Cl"_2` is anticipated to have the HIGHER boiling point; it is EXPECTED to interact with itself more strongly than `"CCl"_4` interacts with itself, and is EXPECTED to be harder to boil.

[However, `"CH"_2"Cl"_2` does indeed have a lower boiling point than `"CCl"_4`. Look these up on NIST.]

`T_b("CH"_2"Cl"_2) = ul"313 K"`, `DeltaH_"vap" ~~ ul"28.06 kJ/mol"`
`T_b("CCl"_4) = ul"349.8 K"`, `DeltaH_"vap" ~~ ul"29.82 kJ/mol"`

which is opposite to the trend we expected.

It may have to do with the idea that `"Cl"` is so much heavier than `"H"` when comparing `"CH"_2"Cl"_2` with `"CCl"_4`. I would suspect that the sheer size of `"CCl"_4` compared with `"CH"_2"Cl"_2`, as well as its RMS speed, is enough to push its boiling point higher from a borderline position.

Both the changes in enthalpy and entropy agree and do not compete (they are not conflicting trends).

Higher `DeltaH_"vap"` `->` harder to boil

For any phase change at constant temperature and pressure,

`DeltaH_"tr" = TDeltaS_"tr"`

Therefore,

`DeltaH_"vap" = T_bDeltaS_"vap"`

And so, the changes in entropy due to boiling are:

`DeltaS_"vap"("CH"_2"Cl"_2) = ("28.06 kJ/mol")/("313 K") xx "1000 J"/"kJ" = ul("89.6 J/mol"cdot"K")`

`DeltaS_"vap"("CCl"_4) = ("29.82 kJ/mol")/("349.8 K") xx "1000 J"/"kJ" = ul("85.2 J/mol"cdot"K")`

Lower `DeltaS_"vap"` `->` harder to boil

As it turns out, `"CH"_2"Cl"_2` boils at a lower temperature because its increase in entropy is larger, and `"CCl"_4` boils at a higher temperature because its increase in entropy is lower. But also, it requires less thermal energy to boil `"CH"_2"Cl"_2`.

[So, both enthalpy and entropy conspire to boil `"CH"_2"Cl"_2` at a lower temperature.]