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    How the hell do I calculate the average bond enthalpy of this bond?!

    The standard molar enthalpy change of formation of H20 is -242kJmol^-1. Calculate the average bond enthalpy of the O-H bond.

    O=O = 496
    H-H = 436
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    Enthalpy of formation = total bonds broken - total bonds made



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    So first you need to write out the equation for the formation of water and then draw out the bonds included on each side of the equation so you know what is being broken/formed. Then you can work backwards from there. And if you get a negative value you know to try again as bond enthalpies are endothermic


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    1/2 O2 + H2 -> H2O so the bond enthalpy is 1/2 * 496 + 436 - 242 = 442 I think. However, I have a feeling that the signs should be reversed and bond enthalpies should be negative (as bonds are more energetically stable than radicals and the like) but maybe someone who's done AS more recently can confirm.
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    (Original post by AmateurRocketman)
    1/2 O2 + H2 -> H2O so the bond enthalpy is 1/2 * 496 + 436 - 242 = 442 I think. However, I have a feeling that the signs should be reversed and bond enthalpies should be negative (as bonds are more energetically stable than radicals and the like) but maybe someone who's done AS more recently can confirm.
    I got +463 as the answer


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    (Original post by AmateurRocketman)
    1/2 O2 + H2 -> H2O so the bond enthalpy is 1/2 * 496 + 436 - 242 = 442 I think. However, I have a feeling that the signs should be reversed and bond enthalpies should be negative (as bonds are more energetically stable than radicals and the like) but maybe someone who's done AS more recently can confirm.
    That's what I did and it was wrong according to the book
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    (Original post by VokeA)
    I got +463 as the answer


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    That's right! How did you get that answer?
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    (Original post by AndrewKn0x)
    That's right! How did you get that answer?
    Name:  ImageUploadedByStudent Room1460285975.563277.jpg
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    (Original post by AndrewKn0x)
    How the hell do I calculate the average bond enthalpy of this bond?!

    The standard molar enthalpy change of formation of H20 is -242kJmol^-1. Calculate the average bond enthalpy of the O-H bond.

    O=O = 496
    H-H = 436
    so the equation: 2H2O -----> 2H2 + O2

    H-O-H H-H O=O
    H-O-H H-H

    standard enthalpy change of formation = enthalpy of products - enthalpy of reactants
    -242 = +1368 - x (total enthalpy of reactants)
    x = +1368 + 242 = +1610
    O-H = +1610 / 2 (because 2 moles in equation)
    = +805 kJmol^-1
    Honestly, I don't know if this is correct or not - it may actually be +805 / 2 = +402.5 kJmol^-1 = +403 kJmol^-1. Hope this helps though!
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    Ok thanks! I see it now. Looking at the spec I think we should've been taught this, but we haven't but I get the idea now.
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    (Original post by AndrewKn0x)
    Ok thanks! I see it now. Looking at the spec I think we should've been taught this, but we haven't but I get the idea now.
    You're welcome. And sorry for my handwriting. Its not usually like that


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    (Original post by VokeA)
    You're welcome. And sorry for my handwriting. Its not usually like that


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    Your writing is v neat don't worry
 
 
 
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