5
$\begingroup$

I'm currently studying relative lowering of vapor pressure with the addition of a solid non volatile solute.

The book explains it with the notion that some of the non volatile solute also appears at the surface between the liquid solution and the vapor. However, i don't understand why don't the molecules of the solute at the surface just escape? Obviously, a solid can't vaporize (Unless you raise the temperature, (And maybe reducing the pressure?)) however the molecules of the solute that are at the surface between the solution and the vapor and should not be held by the strong intermolecular bonds of the solid.

Hence, if i understand this correctly, they, just like the water (Or other solvent) molecules should be constantly moving around in the solution and sometimes 'bump' into other molecules to gain enough velocity to escape the solution. The few molecules that are are the surface between the solution and the air should be getting frequently hit with high energy molecules from below, which should constantly knock such molecules into the vapor. This would mean two things:

  1. The solute also has a small partial vapor pressure
  2. The vapor pressure should still lower, as once solute particles are knocked out into vapor, a few of the solvent particles should escape through the 'opening' before another solute can take it's place. Rinse and repeat, like a valve being opened and closed. Hence, while there would still be a lowering of vapor pressure, there shouldn't be any permanent layer being formed between the solution and the vapor.

But both of these are in contradiction to what's in the book. So where did i go wrong in my reasoning?

$\endgroup$
  • 1
    $\begingroup$ 1. Many solids have vapor pressures. Imagine any solid you can smell. 2. I don't think that explanation is giving the hard physics behind vapor pressure reduction due to solute (it's apparently entropy-based). It's a massively oversimplified representation. $\endgroup$ – SendersReagent Mar 27 '16 at 13:23
4
$\begingroup$

It is true that "non-volatile" is a relative term and that virtually all compounds have some, possibly infinitesimal, vapor pressure, but the idea is that this is negligible. And you are also correct that "...the molecules of the solute that are at the surface between the solution and the vapor and should not be held by the strong intermolecular bonds of the solid."

Concerning that latter statement however, there are many reasons that a compound might maintain very low volatility as a dilute solute, even though the intermolecular forces within the bulk solid are not at play. For example, compounds having a large molecular weight tend to be non-volatile, and these will still require much more kinetic energy to be "bumped out" of the solvent, meaning this will be a statistically rare event compared to the transfer of solvent molecules to the vapor phase.

Ionic solids are another type of solute that are typically quite non-volatile. In order for an ionic species occupying a solute surface site to escape into the gas phase, in the case of an aqueous solvent for example, a rather large amount of energy would be required to dissociate the ion from it's solvation sphere and kick it out into the gas phase as a free ion.

I hope this answers your question "where did I go wrong in my reasoning". Don't hesitate to ask for clarification in the comments if I missed something.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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