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    So here is the question:
    p.s. I dont get why the enthalpy change of formation and the equations dont add up? The figures written next to the equations are they the enthalpy change?

    Given the following data
    Enthalpy change of Formation NF3= -114
    Enthalpy Change of Formation NH3 = -46.2
    Bond Enthalpy N=-N +944
    N(g) + 3F(g) --> NF3 -816
    N + 3H --> NH3 -1173.6

    Calculate:
    a) N-F Bond Energy in NF3
    -114/3?
    b) N-H Bond Energy in NH3
    46.2/3?
    c) F-F Bond Energy
    Can't see where F-F is bonded?
    d) H-H Bond Energy
    Can't see where H-H is bonded??
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    hint: Enthalpy change of formation of NF3 = 0.5N2 + 1.5F2 ---> NF3

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    (Original post by EierVonSatan)
    hint: Enthalpy change of formation of NF3 = 0.5N2 + 1.5F2 ---> NF3

    Thanks, silly mistake - Should remember definitions!
    So is c -1402?

    And a -38
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    (Original post by aimz08)
    Thanks, silly mistake - Should remember definitions!
    So is c -1402?

    And a -38
    you can try compare with literature values here,
    http://www.science.uwaterloo.ca/~cch...20/bondel.html
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    (Original post by shengoc)
    you can try compare with literature values here,
    http://www.science.uwaterloo.ca/~cch...20/bondel.html
    Mine are completely different
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    (Original post by aimz08)
    Mine are completely different
    try this hess cycle,

    1/2N2 + 3/2F2 ----> NF3

    N + 3 F

    with arrows coming down from the LHS and RHS of top line to the bottom line. Label all the states. Then sub in values of enthalpies according to values given. Your unknown in this case is F-F bond dissociation energy, let it be x. If you are breaking 3/2 F-F bonds, enthalpy change = 3/2 x

    Try to complete the hess cycle and do the calculation.

    The same procedure could be used to find H-H bond energy.
 
 
 
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