Why don't you use "i" in vapour pressure depression formula?

[sv3]

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I know that you consider the ionizability of a solute (vant hoff factor) when calculating freezing pt, boiling pt, and osmotic pressure effects.
But in both Ek and TPR, they use moles for vapour pressure depression and no look at "i" at all. I dont understand why they don't also consider "i" like with the other three colligative properties? Anyone have any clue?

thanks in advance
steve

 

[Bernoull]

I know that you consider the ionizability of a solute (vant hoff factor) when calculating freezing pt, boiling pt, and osmotic pressure effects.
But in both Ek and TPR, they use moles for vapour pressure depression and no look at "i" at all. I dont understand why they don't also consider "i" like with the other three colligative properties? Anyone have any clue?

thanks in advance
steve

Just taking an educated guess, I'll say it's the difference b/t physical and chemical reactions. In dissolution, the INTRA bonds of the compound are broken and new bonds form Na--Cl ---> Na-(H2O)n + Cl-(H2O)n, since freezing pt, boiling pt, and osmotic pressure effects involve "dissolution" hence the van hoff factor.

Vaporization on the contrary is a physical reaction, wherein the INTER bonds are broken not INTRA therefore the gaseous products are still intact molecules/compounds so van hoff factor is constant and thus not a variable.. Vaporization really breaks dipole interactions..

I hope this helps..

 

[sv3]

Just taking an educated guess, I'll say it's the difference b/t physical and chemical reactions. In dissolution, the INTRA bonds of the compound are broken and new bonds form Na--Cl ---> Na-(H2O)n + Cl-(H2O)n, since freezing pt, boiling pt, and osmotic pressure effects involve "dissolution" hence the van hoff factor.

Vaporization on the contrary is a physical reaction, wherein the INTER bonds are broken not INTRA therefore the gaseous products are still intact molecules/compounds so van hoff factor is constant and thus not a variable.. Vaporization really breaks dipole interactions..

I hope this helps..

Hey thanks,

But I'm seeing some issues here.....
I didn't think this was a chemical vs physical change issue...interesting. B/c for vapourization, i still think the same thing happens - a solute is dissolved in a solvent first, and then you can heat it. But this too would be chemical (although as you'll see below im not sure dissolution is a chemical change....) wouldn't it as it does involve dissolution just like how you mentioned above?

and just for my own sake, melting, freezing, boiling (which i thought was vapourization) - these are all just physical changes aren't they? Just confirming here......

I admit i still always wreck my brain over whether or not dissolution is a physical or chemical change? Like dissolving salt in water - the sodium and chlorine atoms are seperated but what new bonds form? I thought water molecules jsut seperated the sodium from the chorine and therefore its physical. Again I'm not really sure about dissolution so if there's a simple rule i could follow please let me know!! absolutely kills me when the fundamentals still give me headaches.

thanks
steve

 

[Bernoull]

Hey thanks,

But I'm seeing some issues here.....
I didn't think this was a chemical vs physical change issue...interesting. But for vapourization, i still think the same thing happens - a solute is dissolved in a solvent, and then you can heat it. But this too would be chemical wouldn't it as it does involve dissolution just like the others?

Excellent point, I overlooked the solute dissolution aspect of vapor pressure depression/elevation. So my "educated guess" falls flat on it face!!


and just for my own sake, melting, freezing, boiling (which i thought was vapourization) - these are all just physical changes aren't they?

Yes I agree, these are physical reactions H2O(s) <--> H2O(l) <--> H2O(g)... it's still H2O

I also thought dissolution was a physical change too in that molecules of solute are seperated from one another but the molecules stay intact? Where have i got this wrong?

Here's the answer is not necessarily, if u dissolve glucose in water, it's a physical rxn bcos at the molecular level it simply have solvated glucose molecules, BUT ur question concerned "ionizability of a solute" so when I said dissolution is chemical rxn, I meant compounds (ionic) that ionize in H2O hence my example of NaCl, in that case the ionic bonding b/t Na+ n Cl- is broken so water forms hydrogen bonds with each which i think constitutes a chemical rxn.. Having said all this, I know chemical rxns make new compounds, and I don't know that solvated ions are new compounds...😕😕😕
Can someone clarify this???

thanks in advance
steve

You raise excellent points...


Now to get back to ur original question why van hoff factor is not included in the vapor pressure depression Eq, well it is accounted for..

Ideal solution Vapor depression:

dP= Pa - Px (dP=change in vapor Pressure (Pv), Pa=Pv of pure solvent A,
Px = Pv of solution of X in solvent A)

dP = X*Pa where x is the mole fraction, so if u have 100 moles solvent A and 10 moles solute X, when X is glucose, X=0.1, but when X is NaCl, u'll have 20 moles of solute due to ionisation yielding X = 0.2..

So mole fraction of the solute includes van hoff factor. Mole fraction is calculated based on total solutes in solution.

If u have access to
Chemistry 4th Ed, McMurry, Fay this is discussed on page 445.

I hope things are more clear

 

[sv3]

You raise excellent points...


Now to get back to ur original question why van hoff factor is not included in the vapor pressure depression Eq, well it is accounted for..

Ideal solution Vapor depression:

dP= Pa - Px (dP=change in vapor Pressure (Pv), Pa=Pv of pure solvent A,
Px = Pv of solution of X in solvent A)

dP = X*Pa where x is the mole fraction, so if u have 100 moles solvent A and 10 moles solute X, when X is glucose, X=0.1, but when X is NaCl, u'll have 20 moles of solute due to ionisation yielding X = 0.2..

So mole fraction of the solute includes van hoff factor. Mole fraction is calculated based on total solutes in solution.

If u have access to
Chemistry 4th Ed, McMurry, Fay this is discussed on page 445.

I hope things are more clear

Hi, I think i get the vapour pressure thing now as i didnt realize the mole fraction was actually a "particle" fraction - meaning it takes into account the ionizability. So if you were given just moles, you would have to convert them according to their "i" factors (i x moles)? (neither Ek or TPR mention that i plays any role and always talk about mole fraction.........so thats why i never saw the i factor playing a role....). I would have just used the formulas they have given and used mole fractions without considering the i factor!

thanks for that explanation though, it certainly helped!

Yup the dissolution thing is a pain. I'm not sure dissolving NaCl in H2O is chemical either. I didnt know the water hydrogen bonded with the Na and Cl ions - how does that work anyhow? I just knew the ions get seperated but since nothing new forms......it wasnt chemical. Hopefully someone will clear it up

cheers
steve

 

[bkz]

I didnt know the water hydrogen bonded with the Na and Cl ions - how does that work anyhow? I just knew the ions get seperated but since nothing new forms.....

lone pairs on water molecules...Na+ are attracted to the negative oxygen pairs, and Cl- is attracted to H

i didnt realize the mole fraction was actually a "particle" fraction - meaning it takes into account the ionizability. So if you were given just moles, you would have to convert them according to their "i" factors (i x moles)?

van't hoff factor is moles solute in solution...complete ionization is complete dissociation of particles in solution...I don't think mole fraction has anything to do with ability to ionize. Maybe # of particles and mole fraction are just being used interchangably..cos they both quantify particles?
hopefully bernoulli or someone else can clear this up, now you have me a little confused too 😛

 

[bkz]

ok i think I figured out what you are asking...ALWAYS ALWAYS look at the equations! See http://en.wikipedia.org/wiki/Van_'t_Hoff_equation

this goes back to your question about vapor pressure...so what is vapor pressure?..the #of molecules escaping the surface...usually evaporation...and as you can see in the wiki link the derived equation for Van't hoff involves enthalpy...vapor pressure varies with temperature change...make sense? 🙂

 

[sv3]

ok i think I figured out what you are asking...ALWAYS ALWAYS look at the equations! See http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation

this goes back to your question about vapor pressure...so what is vapor pressure?..the #of molecules escaping the surface...usually evaporation...and as you can see in the wiki link the derived equation for Van't hoff involves enthalpy...vapor pressure varies with temperature change...make sense? 🙂

first, thanks for the H2O bonding to NA+ and H- explanation, I think I get it now - so they actually do bond to the sodium and chlorine atoms? .....thus constituting a chemical change? Everywhere I read it says dissolution is a physical change! I also noticed you said attracted to and not bonded so wondering if the only thing that happens is solvation and in that case it would be physical wouldn't it?

As for the wiki equation, ever so slowly it is coming to me. I knew vapour pressure increased with Temperature since the molecules of a liquid get more energy and therefore can break free and go into the gaseous phase.
It was just that in every equation i saw for vapour pressure depression, the equations stated that you take the mole fraction of solute and multiply it by the vapour pressure the pure solvent would have to see how much the vapour pressure declines (since the presence of solute makes it impossible to have 100% solvent). The vant hoff factor never got mentioned like it does explicity in the equations for the other three colligative properties (freezing pt, boiling pt, osmotic pressure). For example, freezing pt depression: constant x i x molality. The "i" is nowhere to be seen in the formulas for vapour pressure depression. Im guessing somehow its taken into account but cant figure it out. I officially freaking hate vapour pressure now.

steve

 

[bkz]

lol me too. I found this which uses partial derivatives to go from mole fraction to vapor pressure, but it's giving me a headache can't wait for p-chem.

 

[bkz]

ok here it is...

Again, this is not in the same form you saw in freshman chemistry. To get there we have to convert from mole fraction to molality as before. We get,

(37)​

The part in brackets is the freezing point depression constant or the "cryoscopic" constant.

x2 is concentration, or solute in solution.

 

[sv3]

lol me too. I found this which uses partial derivatives to go from mole fraction to vapor pressure, but it's giving me a headache can't wait for p-chem.

thanks

i took a look and noticed this:

"We see that the change in the vapor pressure is negative which means that the vapor pressure decreased and it decreased by an amount that is proportional to the vapor pressure of the pure liquid and proportional to the fraction of solute molecules in the solution (that is, the ratio of the number of solute molecules to the total number of molecules).

So it does mention "Fraction of solute molecules". Perhaps when I'm using the equation i've got to make sure the question explicity states the moles of the solute. If it doesn't, then i have to take a look at the substance the solute came from and check out the ratio of dissolution (i.e if it was MgOH2, and i was looking at OH, i would have to realize there's twice as much in the solution than there was MgOH2 to begin with - i know OH isnt a good example but work with me.....)

ah well...........perhaps i'll wait till i get screwed on a question...haven't yet.....but haven't written the real MCAT yet and do not want to have that be the time where this comes up!

thanks alot
steve

 

[bkz]

no prob! thinking this through and trying to explain it helped me too.