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Homework Help: Born Haber Cycle (help please)

  1. Dec 1, 2017 #1
    1. The problem statement, all variables and given/known data
    calculate enthalpy of formation for lithium fluoride... LiF

    2. Relevant equations
    Given from the worksheet is:

    Lithium sublimation = 129 kj/mol
    Lithium first ionization = 520 kj/mol
    bond-dissociation energy of fluorine is 159 kj/mol F2
    electron affinity for fluorine is -328 kj/mol
    lattice energy of LiF = -1047 kj/mol


    3. The attempt at a solution
    The chemical equations i wrote/figured out are:

    Li(s) -> Li(g) 159 kj
    Li(g) -> Li+(g) + e- 520 kj
    1/2F2(g) - > F(g) -159 kj
    F(g) + e- -> F-(g) -328 kj
    Li+(g) + F-(g) -> LiF(s) -1047

    When i added it all up to find enthalpy formation , i get -537 kj but the answer is -617. what did i do wrong?
     
    Last edited by a moderator: Dec 1, 2017
  2. jcsd
  3. Dec 1, 2017 #2

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    Hey Lori!

    Shouldn't we have F2(g) - > 2F(g) -159 kJ, so that it only costs half?
    And shouldn't Li+(g) + F-(g) -> LiF(s) have a positive energy that is released when we bring oppositely charged particles together?
    And shouldn't F(g) + e- -> F-(g) have a positive energy as well, since the electron bonds naturally?
    We should be consistent with the minus signs, shouldn't we? :rolleyes:
     
  4. Dec 1, 2017 #3
    I'm confused why there sublimation would cost half though ;(
     
  5. Dec 1, 2017 #4

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    Aren't we talking about the bond-dissociation energy of F2?
     
  6. Dec 1, 2017 #5
    Oops I meant the bond dissociation . Yeah
     
  7. Dec 1, 2017 #6

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    Isn't bond dissociation for F2, leaving us with 2 F atoms?
    We can make 2 LiF with that.
    So the cost for 1 LiF is half of that.
     
  8. Dec 1, 2017 #7
    Oh it makes sense now. It's cause we only needed 1 so I took one half of the F so I should also take one half of the cost
     
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