2
$\begingroup$

I have learnt that in order for dissolution to occur, the solute-solvent intermolecular forces must be strong enough to overcome the solute-solute and solvent-solvent forces. However if this is the case, why are non-polar substances such as methane able to partially dissolve in polar solvents such as water? Aren't the dispersion forces always too weak to break the pre-existing hydrogen bonds present in water?

$\endgroup$

1 Answer 1

1
$\begingroup$

In solids, reorientational and especially translational molecular motion are highly hindered. The motion of individual molecules is constrained about mean fixed positions within a regular lattice. Such regularity is absent in liquids, where significant thermal energy is associated with rearrangements in molecular position and orientation. This is of course one defining property of a liquid (and more generally of fluids). The corollary of this structural irregularity is that it leads to a greater number of potential "defects" in the network of interactions within the substance.

Consider water. The wikipedia explains that in crystalline ice, water forms hydrogen bonds with 4 other water molecules. Meanwhile,

[from] TIP4P liquid water simulations at 25 °C, it was estimated that each water molecule participates in an average of 3.59 hydrogen bonds. At 100 °C, this number decreases to 3.24 due to the increased molecular motion and decreased density, while at 0 °C, the average number of hydrogen bonds increases to 3.69.[35]

The net result of this structural irregularity is that even molecules that interact weakly through dispersion interactions (or, in principle, not at all) with water have a significant if small chance of persisting within the liquid. Therefore, it is less a matter of dispersion interactions "breaking" hydrogen bonds, and more of apolar solutes occupying "defects" in the liquid.

References in wikipedia article

[35] Jorgensen, W. L.; Madura, J. D. (1985). "Temperature and size dependence for Monte Carlo simulations of TIP4P water". Mol. Phys. 56 (6): 1381.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.