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Thread: Boiling points

  1. #1 Boiling points 
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    Hi guys, i have been asked a question why 1-hexanol has a higher boiling point than 1-pentanol. Is the greater surface area giving stronger dispersion forces the only reason ?


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    Quote Originally Posted by tecnica View Post
    Hi guys, i have been asked a question why 1-hexanol has a higher boiling point than 1-pentanol. Is the greater surface area giving stronger dispersion forces the only reason ?
    Yes, I think so, more or less. I used to think that the mass of the molecule itself had an effect but my current understanding is that this is not so.


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    boiling point of straight chain hydrocarbons is a function of the molecular weight,
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    Quote Originally Posted by fizzlooney View Post
    boiling point of straight chain hydrocarbons is a function of the molecular weight,
    But WHY?, is the question.

    It is apparently not due to greater mass of the molecules per se. It is because larger molecules experience greater dispersion forces between them. Or so I understand.
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by fizzlooney View Post
    boiling point of straight chain hydrocarbons is a function of the molecular weight,
    But WHY?, is the question.

    It is apparently not due to greater mass of the molecules per se. It is because larger molecules experience greater dispersion forces between them. Or so I understand.
    This is true but the mass alone also has an effect (although often not as large as the effect of dispersion forces), if there were no inter-molecular forces you would still expect a higher molecular mass compound to have a higher boiling point due to the higher temperature required to give the molecule enough kinetic energy to travel fast enough to escape the liquid phase.
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    Quote Originally Posted by PhDemon View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by fizzlooney View Post
    boiling point of straight chain hydrocarbons is a function of the molecular weight,
    But WHY?, is the question.

    It is apparently not due to greater mass of the molecules per se. It is because larger molecules experience greater dispersion forces between them. Or so I understand.
    This is true but the mass alone also has an effect (although often not as large as the effect of dispersion forces), if there were no inter-molecular forces you would still expect a higher molecular mass compound to have a higher boiling point due to the higher temperature required to give the molecule enough kinetic energy to travel fast enough to escape the liquid phase.
    That's what I used to think but now I'm not sure it is so. Do more massive molecules have less kinetic energy for a given temperature? Or are we confusing that with lower speed. I'd have thought it would be the k.e. that is relevant to whether or not a certain proportion of the molecules can escape the dispersion force "bonding" of their neighbours in the liquid. But to be honest I've forgotten the relationship between temperature and r.m.s. kinetic energy - or is it speed? - in stat TD. Can you remind me?

    P.S. I may have the answer to my own question here: http://en.wikipedia.org/wiki/Kinetic_theory

    From this I understand that the average kinetic energy is proportional to temperature and the mass of the molecule does not enter into it. So I think I'm tempted to conclude that BP should be independent of the mass of the molecule.

    But I'd welcome a check of this logic.
    Last edited by exchemist; October 8th, 2014 at 12:07 PM. Reason: note added later
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    You could be right, the logic certainly makes sense at first glance, stat mech was a long time ago..., I didn't really think it through before my last post... Although a quick wiki search gave
    The distribution is a probability distribution for the speed of a particle within the gas - the magnitude of its velocity. This probability distribution indicates which speeds are more likely: a particle will have a speed selected randomly from the distribution, and is more likely to be within one range of speeds than another. The distribution depends on the temperature of the system and the mass of the particle
    for Maxwell-Boltzmann distribution.
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  9. #8  
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    Quote Originally Posted by PhDemon View Post
    You could be right, the logic certainly makes sense at first glance, stat mech was a long time ago..., I didn't really think it through before my last post... Although a quick wiki search gave
    The distribution is a probability distribution for the speed of a particle within the gas - the magnitude of its velocity. This probability distribution indicates which speeds are more likely: a particle will have a speed selected randomly from the distribution, and is more likely to be within one range of speeds than another. The distribution depends on the temperature of the system and the mass of the particle
    for Maxwell-Boltzmann distribution.
    Acc. my Wiki link, the average k.e. of a molecule is 3/2 kT, i.e. independent of mass. Sounds familiar, come to think of it.

    If this is right, our questioner can earn extra points, if he or she is in education, by deploying Richard Wayne's, "It's very interesting you say that, because it's ballocks.", when someone says it's to do with mass. Personally, I would not be able to resist it.
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    I'll get into trouble if I use it! (although I was tempted with some of the answers 9C gave yesterday )
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    Informative thread. My intuition was to go with mass as the controlling factor. Amazing how something so basic could be invisible to us.
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    Quote Originally Posted by John Galt View Post
    Informative thread. My intuition was to go with mass as the controlling factor. Amazing how something so basic could be invisible to us.
    Yes, in practice it is true that BP does go up with mass, but it is due to the greater dispersion forces in larger molecules, rather than anything to do with mass per se. Or at least that is what kinetic theory seems to me to be saying.

    I must say I do not think this ever came up at university in stark terms and it was only recently that I started to question what I had always lazily assumed. Let's see if anyone comes out of the woodwork to say I'm wrong.
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    i just want to say i am relatively new to chemistry and am really starting to enjoy it. discussions like this makes understanding concepts a lot easier so thanks
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    i am having trouble explaining how dispersion forces affect boiling points, could some one help?
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    Quote Originally Posted by tecnica View Post
    i am having trouble explaining how dispersion forces affect boiling points, could some one help?
    OK. I'd have thought that should not be too hard, in principle at least. I don't know how much you already know, so please excuse me if I get the level of explanation a bit wrong for you - we can always correct for that subsequently.

    The boiling point of a liquid is the temperature at which vapour can form throughout the liquid, instead of just via evaporation from the surface. This is possible once the vapour pressure of the liquid equals that of the ambient pressure above it. (This is why boiling points vary with atmospheric pressure - you cannot brew a decent cup of tea on top of Mt. Everest, for example. Boiling points are normally quoted at a standard, sea level atmospheric pressure.) So boiling point is a measure of how easily the substance can release enough molecules to create a vapour pressure of one atmosphere.

    The vapour pressure reflects the proportion of the molecules that have enough energy to break completely free from the forces of attraction (dispersion or other) that bond molecules together in the liquid state. If you have strong intermolecular attraction, the molecules need more thermal energy to overcome it and break away than if the forces are weak. Consider a few examples:

    Helium boils at -269C, or 4K, .e. only 4 degrees above absolute zero. "Molecules" of helium are single, very tiny atoms, with almost no intermolecular (interatomic) attraction between them.

    Nitrogen, which forms diatomic molecules from pairs of fairly small atoms but larger than those of helium, has a boiling point of -196C.

    Carbon dioxide, which forms triatomic molecules from 3 atoms a similar size to those of nitrogen, has a boiling point of -59C. But in CO2, the oxygen atoms carry a permanent slight partial -ve charge and the carbon atom in the middle has a slight +ve charge: CO2 is a bit "polar". So these permanent partial charges, or "dipoles" are an extra source of attraction, on top of regular dispersion forces. So this raises the boiling point compared to a molecule of similar size with no polarity.

    Water, which is also a triatomic molecule, smaller than CO2, has a boiling point of 100C, because in addition to dispersion forces, there are, famously, hydrogen bonds between the molecules, which are an entire order of magnitude stronger than dispersion forces.

    If we focus on dispersion forces, these arise from the polarisability of the electron cloud in the molecule. You get temporary dipoles arising by chance and these can induce opposite dipoles in neighbouring molecules, creating an attractive force. The bigger the molecule, the more opportunities there are for all or part of it to become polarised. So the dispersion forces are greater and the boiling points go up.

    There is an easy to follow explanation of intermolecular forces here: intermolecular bonding - van der Waals forces

    Do feel free to come back on this if you want to.
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    Quote Originally Posted by PhDemon View Post
    the mass alone also has an effect
    The isotope effect, that higher mass isotopes form stronger bonds than their lower mass counterparts, is a mass effect.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    Quote Originally Posted by KJW View Post
    Quote Originally Posted by PhDemon View Post
    the mass alone also has an effect
    The isotope effect, that higher mass isotopes form stronger bonds than their lower mass counterparts, is a mass effect.
    True, but would this be significant for dispersion forces, do you think? Most examples I can recall involve substituting deuterium for hydrogen in the covalent bonds of organic compounds, leading to a slightly lower vibrational ground state (lower zero point energy of vibration of the bond).
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by KJW View Post
    Quote Originally Posted by PhDemon View Post
    the mass alone also has an effect
    The isotope effect, that higher mass isotopes form stronger bonds than their lower mass counterparts, is a mass effect.
    True, but would this be significant for dispersion forces, do you think? Most examples I can recall involve substituting deuterium for hydrogen in the covalent bonds of organic compounds, leading to a slightly lower vibrational ground state (lower zero point energy of vibration of the bond).
    I don't know how significant this would be (I guess a comparison of the boiling points of methane and tetradeuteromethane would be in order here). However, I believe that the isotope effect would be pertinent for any type of bonding regardless of how weak or non-covalent. This is because we are still dealing with a potential well in which we would have a quantum harmonic oscillator along with the associated zero-point energy.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    Quote Originally Posted by KJW View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by KJW View Post
    Quote Originally Posted by PhDemon View Post
    the mass alone also has an effect
    The isotope effect, that higher mass isotopes form stronger bonds than their lower mass counterparts, is a mass effect.
    True, but would this be significant for dispersion forces, do you think? Most examples I can recall involve substituting deuterium for hydrogen in the covalent bonds of organic compounds, leading to a slightly lower vibrational ground state (lower zero point energy of vibration of the bond).
    I don't know how significant this would be (I guess a comparison of the boiling points of methane and tetradeuteromethane would be in order here). However, I believe that the isotope effect would be pertinent for any type of bonding regardless of how weak or non-covalent. This is because we are still dealing with a potential well in which we would have a quantum harmonic oscillator along with the associated zero-point energy.
    Indeed. Though my guess would be that it would be a fairly insignificant effect for dispersion forces, especially when atoms of atomic mass >>1 are concerned. Given Tecnica's recent arrival in the wonderful world of chemistry, I'd be tempted to tell him or her not to worry about this particular refinement.

    But thanks for raising: a nice excuse to take a short trip down memory lane…...
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