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How does heat affect bonds and their motion?

  1. Mar 28, 2013 #1
    If I had a some covalent molecules with a total intermolecular bond strength of 100J and total covalent bond strength of 100J and when I supply 100J of heat into those molecules, how are the bonds going to break? Will both bonds break simultaneously or will one bond break in preference to the other? I don't think its possible to break both with just 100J of heat because in those molecules I'd have -200J of potential energy and by putting 100J into that system, it would still have 100J worth of bond energy. So I think they'd be a preference of which bond gets broken or perhaps the 100J would spread themselves equally between the 2 bonds?

    if I had a bunch of covalent molecules with a total intermolecular bond strength of 100J and a total covalent bond strength of only 50J and I were to supply 50J of heat, so assuming that heat prefers to break covalent bonds, all that heat would be used to destroy that bond. And as a result of that the larger covalent molecule would be broken into smaller particles with different intermolecular bond energies. But still the total bond energies must add up to 100J?

    Lastly, I thought of these factors when I came across the rotational kinetic energy and translational energy of monatomic and polyatomic elements/molecules. For polyatomic molecules by heating it, it equates to increasing both rotational and translational kinetic energy. So this shows us that heat chooses to increase various parties' kinetic energy so I was wondering about how this would correlate to how heat affects the various bonds.

    Thanks so much for the help :smile:
  2. jcsd
  3. Mar 28, 2013 #2


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    Staff: Mentor

    I'm not sure I understand what you mean. So lets try to break it down (pun intended)!

    What do you call a covalent molecule? Do you mean a covalent bond inside a molecule?

    What do you mean by intermolecular bond? Do you mean molecule-molecule interaction?

    And bond strength are not expressed in units of energy. You should talk about dissociation energy. And 100 J is a gigantic amount of energy for a molecule.

    It depends by which mecanism they break.

    If you do not supply enough energy to break both bonds, obviously only one will get broken.

    Again, it depends on how that energy got there. With lasers, it is possible to selectively break bonds, even break bonds that are stronger while leaving weaker bonds alone. In most cases, as you point out, even if you can supply the energy to a particular bond, what will happen is that the energy will spread to all bonds more quicly than you can excite a single one.

    There is something call the equipartition theorem that tells you (for an ideal case) how energy will be distributed between the different degrees of freedom of the molecule.

    First, heat doesn't "choose" anything. Second, don't think in terms of kinetic energy, but rather total energy (potential energy plays an equal role). Generally speaking, increasing the temperature increases the vibration of molecular bonds. But be aware that at room temperature, most molecules are not vibrationally excited (there is a threshold for excitation, due to quantum mechanics, which is usually not reached at room temperature).
  4. Mar 29, 2013 #3
    Sorry for being vague in my post. I meant having a bunch of molecules that have covalent bonds. Like water. So using water as an example, at room temperature and I heat it up to 100 degrees and let it boil the amount of energy increases by mcΔT. But where does the energy go to? Does some go to breaking the intermolecular bonds as well as some of the covalent bonds in it?

    So actually what does temperature tell us? We learn that heating increases the average kinetic energy what what is the kinetic energy of a particle?

    Thanks for the help :)
  5. Mar 29, 2013 #4


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    Staff: Mentor

    When water is heating up, the energy goes mostly into the relative motion of the molecules (where there is again kinetic and potential energy). At the boiling point, first note that since temperature is constant, you cannot use the heat capacity anymore, but have to consider the latent heat. During the phase transformation, part of the energy goes to combatting the intermolecular attraction, part to translational kinetic energy and part to rotational kinetic energy.

    Have a look at the equipartition theorem.

    You're welcome!
  6. Mar 29, 2013 #5
    Hi :)

    The heat goes into the relative motion of the molecules - does this mean that the heat goes into breaking the intermolecular bond only? Or does it do any work on the intramolecular covalent bond in it?

    Thanks for the link :) I read through the basics and it seems that the temperature is dependent on half translational and half rotational kinetic energy. So now that I put in energy to the molecules as a whole they have more kinetic energy (as well as potential as they constantly switch?) so now they can move about further from each other. But I can't seem to link this to the intramolecular bonds..

    I would think that some heat is also used to break those bonds? So by putting in more energy how should I visualize the energy in terms of kinetic energy (and PE) and the both the inter and intra-molecular bonds?

    Thanks again for the great help :)
  7. Mar 29, 2013 #6


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    Staff: Mentor

    If you take two molecules in a liquid, say A and B, there is an attractive interaction between the two, although you don't usally call it a bond, more intermolecular forces (there is the exception of what is called a hydrogen bond). But A and B are in motion, and after a while, they are not in contact anymore, rather A is interacting with molecules C and D, and B connects with E, F and G, and so on.

    Generally speaking, no, covalent bonds are not broken, unless you go to high temperatures. Some molecules are more stable than others.

    Yes, for an ideal gas of polyatomic molecules at room temperature. A diatomic gas (which is mostly what air is) actually has a bit more translational energy than rotational energy at a given temperature. And as the temperature increases, part of the energy goes also to vibrations.

    To pull apart molecules that otherwise have an attractive interaction, you need energy. The more energy you have, the more the molecules will get separated.

    I'm not sure how you can "visualize" it. Energy is more about accounting...
  8. Mar 30, 2013 #7
    Oh so most of the heat goes into the rotational and translational kinetic energy.

    It's just that I can't visualize how energy goes into the molecule. I sorta understand the intermolecular interactions as they gain more KE so are able to move further away from each other. But what about the covalent bonds? Even if they are not 'broken' shouldn't some energy be used to break them? But then I can't visualize the kinetic energy in this sense.

    Previously it was only discreet molecules with more kinetic energy to be converted to potential energy (moves further from each other). But with the actual molecule itself I'm not too sure how are the particles within the molecular during heating.

    Lastly, how do we tell the proportion of energy to each type of bond? Like is there a ratio or formula to tell that?

    Thanks for the help :) I really appreciate it :)
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