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Bonding energy?

  1. Feb 7, 2005 #1
    Could some kind chemistry pro here tell me the energy required for molecular separation of water into H and O2?
    Thxs. :smile:
     
  2. jcsd
  3. Feb 7, 2005 #2

    dextercioby

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    Shouldn't it be minus the enthaply of creation in the reaction

    [tex] 2H_{2}_{g}+O_{2}_{g}\rightarrow 2H_{2}O_{vap} [/tex] ??


    Daniel.
     
  4. Feb 7, 2005 #3

    chem_tr

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    What do you mean by molecular separation? Some kind of electrolysis-related phenomenon?

    [tex]H_2O \rightarrow H_2 + O_2[/tex]

    The energy required to undergo redox can be found by the following half reactions:

    [tex]4H^+ + 4e^- \rightarrow 2H_2[/tex]
    [tex]2O^{2-} \rightarrow O_2 + 4e^-[/tex]

    Find the energy amounts required (or given out) for these half reactions, and add them; I love this property of thermodynamics :wink:
     
  5. Feb 7, 2005 #4

    Gokul43201

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    If you are looking for the bonding energy of water (since your thread is titled so), you must use Hess' Law, with the following reactions :

    [tex]2H_2 + O_2 \longrightarrow 2H_2O[/tex]
    [tex]O_2 \longrightarrow 2O[/tex]
    [tex]H_2 \longrightarrow 2H [/tex]

    Dexter's equation tells you the enthalpy of formation, but this does not give you the bond energy. To get the bond energies, you must also consider the dissociation enthalpies of oxygen and hydrogen (and remember that there are 2 bonds per water molecule).
     
    Last edited: Feb 7, 2005
  6. Feb 7, 2005 #5
    t
    Yes, exactly; I was asked by a friend, and not being a chemist, (my expertise being far greater in physics), I was hoping someone here could save me from appearing too ignorant. :wink:
    Electrolysis is what was in question. However, I made a mistake on the 1st post. :blushing: I need the energy required for 'disassociation' of water into H2 and O per molecule; can you give it to me in ev?

    Thanks again, Chem_tr., that's the chemistry logic I was missing. So (rewritten for one oxygen) this energy would account for the amount of energy necessary for breaking the 2 'hydrogen' bonds, correct?
    So my question is still, 1st, what is the total E required in terms of electron volts (per molecule), and secondly, how can this account for the total when there are still 2 covalent bonds energetically unaccounted for in the tetrahedral arrangement in water? :confused:
     
    Last edited: Feb 7, 2005
  7. Feb 7, 2005 #6
    Ok, thnks for the reply Gokul; I crossed over you in replying to Chem_tr.

    Yes, you are correct in what I was looking for. So the same question I wrote to him apply.
    So, again, 1.what are the values of those bonding energies in ev;
    and 2). why don't I need to account for the other two (intermolecular) covalent bonds?
     
    Last edited: Feb 7, 2005
  8. Feb 7, 2005 #7

    Gokul43201

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    Unless you better specify EXACTLY what you are looking for, it would be hard for us to help.

    You can not get bond energies from electrolysis. The relevant energy there is the ionization energy, and the products are charged ions rather than uncharged atoms.
     
  9. Feb 7, 2005 #8

    GCT

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    I'm guessing what she/he wants is the energy (bond energy being partly relevant to this of course) relevant to this process as a whole, in this case it would probably be better to solve it in the realm of EMF and relate this value to the free energy if required (there should be simple equations your text showing you how to get from Voltage to free energy).
     
  10. Feb 7, 2005 #9

    GCT

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    In fact, in relevance to the answer your friend is searching for, I would state the free energy value, you don't need to think about any complications, just solve for this value and it will be a complete answer to the question.
     
  11. Feb 7, 2005 #10

    Gokul43201

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    We don't even know what the question really is, do we ? GeneralChem, what is your understanding of the question ?

    Also, from the EMF, you can get to the enthalpy change, only the free energy change. But the energy required to perform any reaction is the enthalpy. And without a knowledge of the entropy change, you can not determine the enthalpy from the EMF.

    But if it's the free energy change that is required, then I agree that the best way to find that would be from

    [tex]\Delta G^o = -nF\Delta E^o [/tex]

    [tex]\Delta E^o = -1.23 V [/tex]

    for the reaction (I think...)

    [tex]H_2O \longrightarrow H_2 + \frac{1}{2} O_2 [/tex]
     
    Last edited: Feb 7, 2005
  12. Feb 8, 2005 #11

    GCT

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    Well technically, if this was in reference to a chemistry class, perhaps this is a specific reference to enthalpy. However, enthalpy questions are quite specific and will frequently mention a specific situation;when such questions are worded broadly as "molecular" separation I am assuming that if it came out of a text for instance, that it is referring to free energy...if it was a general question from a layman then I would suggest it in terms of free energy also. This is just from my small experience with these classes.

    If you wish to find the enthalpy of the process, it is very simple. From what I remember, there are tables in the text which lists the enthalpy of formation values and I'm quite positive that diatomic hydrogen and oxygen are in there as well as water, in fact the enthalpy of formation for the two might be zero.
     
  13. Feb 8, 2005 #12

    GCT

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    all of this referring to standard conditions
     
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