oxygen gas has a higher boiling point than Nitrogen gas - Why?

[TalinAagam]

is there any pattern involved with this?
answer says that this is the case because the molar mass of O2 is higher than that of N2, but I dont see what that has to do with boiling point?

is it implying that more moler mass = more van der waals forces?

---

Members don't see this ad.

 

[deh2ase]

A higher molar mass means the molecule is larger, thus a higher surface area. Yes, you are right since molecules with more surface area have more vander waal forces.

 

[gettheleadout]

A higher molar mass means the molecule is larger, thus a higher surface area. Yes, you are right since molecules with more surface area have more vander waal forces.

This is incorrect. Higher molar mass in itself simply means more literally massive molecules, but says nothing of their size. N2 actually has a greater molecular volume than O2 (see here: http://www2.ucdsb.on.ca/tiss/stretton/database/van_der_waals_constants.html)

Always keep periodic trends in mind; oxygen atoms are more massive but smaller than nitrogen atoms. This can be extrapolated to reason this case out without knowing the van der Waals constants.

The reason the boiling point of O2 is higher is not because of increased van der Waals interactions, but simple physics. The mass of a molecule of O2 is greater than that of a molecule of N2, so the molecule of O2 traveling at a speed sufficient to break out of the liquid phase has a greater kinetic energy than an analogous N2 molecule.

The net effect is that more energy must be distributed throughout a sample of O2 to achieve a given vapor pressure (in this case equal to atmospheric pressure) than for a sample of N2. More energy means greater temperature.

*I'm making some simplifying assumptions here about van der Waals interactions and heat capacities, but this isn't terribly important for this problem given the gases in question.

 

[QrtrLifeCrisis]

This is incorrect. Higher molar mass in itself simply means more literally massive molecules, but says nothing of their size. N2 actually has a greater molecular volume than O2 (see here: http://www2.ucdsb.on.ca/tiss/stretton/database/van_der_waals_constants.html)

Always keep periodic trends in mind; oxygen atoms are more massive but smaller than nitrogen atoms. This can be extrapolated to reason this case out without knowing the van der Waals constants.

The reason the boiling point of O2 is higher is not because of increased van der Waals interactions, but simple physics. The mass of a molecule of O2 is greater than that of a molecule of N2, so the molecule of O2 traveling at a speed sufficient to break out of the liquid phase has a greater kinetic energy than an analogous N2 molecule.

The net effect is that more energy must be distributed throughout a sample of O2 to achieve a given vapor pressure (in this case equal to atmospheric pressure) than for a sample of N2. More energy means greater temperature.

*I'm making some simplifying assumptions here about van der Waals interactions and heat capacities, but this isn't terribly important for this problem given the gases in question.

I would tend to agree with this explanation because the difference between O and N is only one atomic number. However, in general...and especially in organic chemistry when you are comparing between larger molecules, such as alkanes, dispersion forces do matter. Large molecules have more van der Waals interactions and therefore, have higher boiling points (all else being equal).

 

[ToldYouSo]

gettheleadout explained it well; more mass = more energy needed to enter the gas phase

 

[anegron519]

.

 

[gettheleadout]

I would tend to agree with this explanation because the difference between O and N is only one atomic number. However, in general...and especially in organic chemistry when you are comparing between larger molecules, such as alkanes, dispersion forces do matter. Large molecules have more van der Waals interactions and therefore, have higher boiling points (all else being equal).

I agree, but I wasn't implying that the difference in surface area and thus difference in intermolecular interaction is never significant. Also we must be clear that "larger" refers to surface area. Two skeletal isomers of an alkane have identical molecular volume and mass, but one is only "larger" in any meaningful context because it has greater surface area for van der Waals interactions, and thus alkane branching reduces boiling point.

If you're comparing compounds of different masses, both mass and surface area would contribute to the differences in phase change points.

The question isn't referring to long hydro-carbon chains. It's referring to N2 vs O2.

O2 needs more energy to break apart. Follow the period trend as stated before. Left to right means higher ionization energy is needed. O2needs more energy to go into a gaseous phase than N2 does. It's as simple as that without having to be over-complicated.

Let's not confuse ionization energies of individual atoms with those for compounds, and let's definitely not confuse ionization with vaporization (my point being that ionization energy of component atoms is not reflective of enthalpy of vaporization for a compound). Just saying.

 

[anegron519]

.

 

[QrtrLifeCrisis]

The question isn't referring to long hydro-carbon chains. It's referring to N2 vs O2.

O2 needs more energy to break apart. Follow the period trend as stated before. Left to right means higher ionization energy is needed. O2needs more energy to go into a gaseous phase than N2 does. It's as simple as that without having to be over-complicated.

gettheleadout put it correctly. This has nothing to do with "breaking apart" as it pertains to ionization. Phase change is a physical transformation, not a chemical one.

And in regards to long hydrocarbon chains, the same concepts apply. The more surface area for dispersion forces, the higher the boiling point. Similarly, the heavier the molecule, the higher the boiling point. These are related concepts regardless of whether it's applied to a hydrocarbon or N2/O2 molecules.

 

[anegron519]

.

 

[QrtrLifeCrisis]

I think that's why I already said he was correct.

Just saying that you should do a little fact check before implying that my explanation was "overcomplicating" things.

 

[anegron519]

.

 

[QrtrLifeCrisis]

I think it's more like you shouldn't take things so personally? He pointed out that I was wrong and I didn't get upset, I just realized I was wrong. It's a good trait to have and obviously you lack it.

Not at all taking personally. I apologize if it seems that way. That comment probably came off a little snarkier than I intended. Yours, on the otherhand...

 

[gettheleadout]

Okay chill guys haha we all understand each other.

 

[TalinAagam]

Hey guys sorry about this, but I still don't get it...

The reason the boiling point of O2 is higher is not because of increased van der Waals interactions, but simple physics. The mass of a molecule of O2 is greater than that of a molecule of N2, so the molecule of O2 traveling at a speed sufficient to break out of the liquid phase has a greater kinetic energy than an analogous N2 molecule.

The net effect is that more energy must be distributed throughout a sample of O2 to achieve a given vapor pressure (in this case equal to atmospheric pressure) than for a sample of N2. More energy means greater temperature.

*I'm making some simplifying assumptions here about van der Waals interactions and heat capacities, but this isn't terribly important for this problem given the gases in question.

Im still a little confused about the bolded and underlined parts. so if O2 is larger in mass, then why does that imply it is carrying more kinetic energy to get out of the liquid phase?

I can see how that works if O2 and N2 have exactly the same intermolecualr forces holding them togather but I don't see how more mass = more energy needed to remove them from their intermolecualr forces can happen, because they will have differnet inter molecular forces.

Can you explain a little more?

OR
The other explanation that I see in this thread is that O2 has more surface area than N2, giving it more intermolecular forces but How do we know that?
are we also saying that the larger in mass the molecule = the larger surface area = the larger van der waals forces?, which we learned works for organic molecules but also works for these.

edit: gettheleadout, I just saw your assumption about intermolecular forces, so are you assuming that N2 and O2 molecules will have same intermolecular forces? I can see the answer if I make that assumption but I didn't know I should make that assumption.. now that I think about it, if we were dealing with Ideal Gases, we can make that assumption. Perhaps thats the key!!!

 

[QrtrLifeCrisis]

I think the consensus was that because oxygen molecules are larger/heavier than nitrogen molecules, they require more energy to go into gaseous form and thus, have higher boiling points. Differences in dispersion forces are negligible because these two molecules don't differ very much in terms of surface area so you can ignore that. You can also note that that the difference in boiling point is not big so that slight difference in mass accounts for only a slight difference in boiling point. That's all the MCAT expects you to reason for this question.

 

[Illuminateur]

O2 has a smaller surface area than N2, not larger.

The reason is that despite the greater polarizability of each electron due to the weaker hold of the less electronegative nitrogen, because O2 has 2 more total electrons, the same electric field will induce greater polarization and thus intermolecular forces. It is the intermolecular forces that determine boiling point, not molecular weight, because you’re trying to separate molecules from each other, not lift them up against gravity. You can have a gas that lies on the ground, with insufficient energy to lift the molecular weight, but enough energy to break the intermolecular attractions enough to put it in the gas phase.