Take the 2-minute tour ×
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required.

This is the boiling-point elevation equation:

$$ \Delta\mathrm{T}_{b}=ib\mathrm{K}_{eb}. $$

If we do the limit of the above, for $b\rightarrow\infty$ we find that the boiling-point elevation is infinite. How this could be explained under a physical view?

$$ \lim_{b~\rightarrow~\infty}ib\mathrm{K}_{eb}=\infty,~\text{if}~i\wedge b > 0. $$

For example, if we use a solute molality that is so big that the solution would be something like the pure solute, the final boiling-point of the solution, in the case of water as solvent for example:

$$ \mathrm{T}_{b}=ib\mathrm{K}_{eb} + 100\mathrm{°C}. $$

wouldn't be like the boiling-point of the pure solute? I mean, something like this:

$$ \lim_{b~\rightarrow~\infty}\mathrm{T}_{solution}=\mathrm{T}_{solute}. $$

Now, please forget what I've said above about $b\rightarrow\infty$. And take a look at this:

$$ T(b) = 0.86b + 64.7. $$

enter image description here

As you can see is, the boiling-point of a solution of $\mathrm{CH_3OH}$ with $\mathrm{H_2O}$ as solute would be $322.7\mathrm{°C}$ for $b = 300$. This it's weird to me 'cause that temperature would be about three times higher then the boiling-point of the pure water!

So the following:

$$ \mathrm{T}_{b}=ib\mathrm{K}_{eb} + \mathrm{T_{solvent}}. $$

holds true if

$$ b = \frac{x~\mathrm{g}_{solute}}{1000~\mathrm{g}_{solvent}} \le 1 $$

right? If yes why (in the above case):

$$ 0.86\frac{1000~\mathrm{g}}{18.01528~\mathrm{g}/\mathrm{mol}}\frac{1~\mathrm{mol}^2}{1~\mathrm{kg}} + 64.7~\mathrm{°C} = 112.4~\mathrm{°C} $$

?

share|improve this question
    
I hope my answer actually answers your question. A little side note: Please add descriptions of the symbols you use to your future questions. That way prospective answerers have an easier time to help you with your problem. –  Philipp Mar 2 at 1:44
    
After your edit... To use that equation you're imposing the constraint that the solute won't boil. You're looking at the boiling point of the solution, under the assumption that the methanol is the only component that will boil. –  Brian Mar 2 at 3:07
    
Could you please add units to $b$ and your $T(b)$ equation? If $b = 300$ means something like $b=300 \, \text{mol}_{\ce{H2O}}/\text{kg}_{\ce{CH3OH}}$ then $300$ is not a realistic value as per the arguments I provided in my answer. If you considered realistic molalities ($< 50 \, \text{mol}_{\ce{H2O}}/\text{kg}_{\ce{CH3OH}}$) you would be well in the reasonable temperature regime below 100 °C in your plot. –  Philipp Mar 2 at 3:22
    
@Brian take a look at my updates –  FormlessCloud Mar 2 at 4:05
    
You have inserted the wrong values for the molality in your equation. The molality is $b = n(\text{solute})/m(\text{solvent})$. So, say, you want to know the boiling point of a methanol/water mixture where you have $1 \, \text{mol}$ of water in $1 \, \text{kg}$ of methanol. Then you'd get $T(b) = 0.86 \frac{\text{K} \, \text{kg}}{\text{mol}} \cdot \frac{1 \, \text{mol}}{1 \, \text{kg}} + 64.7~\mathrm{°C} = 65.56~\mathrm{°C}$. –  Philipp Mar 2 at 4:29

2 Answers 2

In reality, when you're saying that the molality is infinite, you are indeed basically asking for solute with infinitesimal amounts of solvent, and it will be true that $T_\text{solution} \approx T_\text{solute}$.

This equation describes the boiling point of the solvent under the effects of the non-volatile solute. Ie. you're making the assumption that the solute does not enter the gas phase at the temperatures you're considering.

A substance boils when the chemical potential of the liquid phase equals the chemical potential of the gas phase: $\mu_\text{liquid} = \mu_\text{gas}$, or when the vapor pressure equals the surrounding pressure.

So to elevate the boiling point, we're making the claim that for a given temperature, the chemical potential of the liquid phase is lowered (it's made more stable), or equivalently, that the vapor pressure is lowered. The effect is entropy based.

When particles are interspersed between the solvent molecules, the solvent molecules become less ordered, and are thus at a state of higher entropy than the pure solvent. This translates into a lower chemical potential. The gas phase will have only the solvent molecules, because we made the assumption that the solute is non-volatile. Therefore, the gas-phase is unaffected by this effect, and the chemical potential of the gas phase remains the same.

It's easy to visualise from the vapour pressure perspective. Imagine the liquid-gas interface. At a given temperature, a certain portion of the liquid solvent molecules at the liquid-gas-interface will enter the gas phase. The gaseous solvent molecules at the liquid-gas-interface will likewise enter the liquid phase when they come in contact with it.

The amount of solvent molecules going from gas->liquid depends on the atmospheric pressure. The amount of solvent molecules going from liquid->gas depends on the kinetic energy of the molecules (the temperature). A molecule may have enough energy to break free of the solvent-solvent interactions, or it may not. The higher the temperature, the more likely.

Now imagine that we add solute molecules. There will be less liquid solvent-molecules present at the liquid-gas interface, and thus a lower rate of liquid -> gas (meaning a lower vapour pressure), simply because the probability of a liquid molecule hitting the phase interface is lower... So the temperature has to be higher to compensate, if we want to counter the unchanged rate solvent molecules going from gas -> liquid.

Why boiling point goes infinite: At the limit where molality goes towards infinity, there will be an infinitesimal amount of liquid solvent molecules at the phase interface... And if we want that infinitesimal amount of molecules to sustain a vapour pressure that equals the atmospheric pressure, we need a temperature that approaches infinity as well.

share|improve this answer
1  
I've heard the explanation about solute molecules "pushing" the liquid molecules out of the liquid-gas phase boundary, but something doesn't feel quite right to me. Would this explanation still be valid if I added a surfactant? Even at low concentrations, surfactants quickly cover a large part of the liquid-gas phase boundary. Does that mean surfactant solutions have a higher effect of the boiling point of a liquid compared to a normal solute, on a per mole basis? –  Nicolau Saker Neto Mar 2 at 2:18
    
That's an interesting thought. I don't think I can come up with anything right now. THe surfactant will form its own extremely thin liquid phase, that the liquid solvent will interact with. And I suppose it impedes both entry from liquid into the gaseous solvent phase but from the gas into the liquid phase. Does that make any sense? –  Brian Mar 2 at 3:24
    
I'm not sure. Honestly I feel like the whole idea of blocking the surface is sort of an over-explanation. The correct answer would simply be the change in entropy of the solution (as you mention), and the idea of covering the surface would be trying to go too far in order to provide further physical insight. I'm searching for some articles on water boiling point elevation by surfactants, but I haven't found anything enlightening yet. –  Nicolau Saker Neto Mar 2 at 11:41

The flaw in your reasoning is that the molality $b = n(\text{solute})/m(\text{solvent})$ cannot be infinite even if you use a solute molality that is so big that the solution would be something like the pure solute. Take water as an example: How concentrated is water? Very concentrated you may say - but the concentration is limited. We know that one mole of pure water has a mass of 18 g and occupies 18 cm³. So, in one dm³, there are 1000/18 = 55.56 mol which gives you a water concentration of 55.56 mol/dm³. You cannot get more concentrated water than this (unless you did something drastic like taking it into a black hole!). For the molality it is quite similar: If you dissolve something in water the solute concentration cannot exceed the concentration of water, so $n(\text{solute})$ (and therefore $b$) has an upper bound.

share|improve this answer
    
This is not true. Molality is n(solute)/m(solvent)... That is not m(solution). Unless I'm entirely mistaken. –  Brian Mar 2 at 1:44
    
@Brian Ah, yes your right about the definition of molality. Sorry about that. But my point remains valid I think: If you dissolve something in water the solute concentration cannot exceed the concentration of water, so n(solute) has an upper bound, right? And the situation is analogous in other solvents. –  Philipp Mar 2 at 1:49
    
@Brian I've edited my answer in order to treat molality the right way. –  Philipp Mar 2 at 1:58
1  
What if I took 1 kg of my solvent and decided to add as many moles of solute as I like. Then I can make b as large as I like, since I would have as many moles of solute in my kg of solvent as I like. :-) –  Brian Mar 2 at 2:05
    
@Brian That is the nice part: At some point before reaching the 55.56 mol limit the solute would either stop dissolving and precipitate or there will be some kind of phase inverion such that water will then be dissolved in the solute. In that case the maximum concentration the pure solute could attain (analogous to the little calculation I did for water) would be the upper bound for $n(\text{solute})$. –  Philipp Mar 2 at 2:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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