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Lattice Energy and the Born Haber Cycle

  1. Apr 29, 2007 #1
    Ok most questions (that I have come across) when dealing with the Born Haber Cycle give the Sublimation energy, the Dissociation energy, the Electron Affinity, the Ionization Energy, and the Formation Energy. I know that to get the Formation Energy it is:

    Hf= Hs+Hi.e.+1/2Hd+He.a.+U (Lattice Energy)

    My problem is that if everything but the lattice energy is given what is the calculations to get the lattice energy. I know the lattice energy for some examples like NaCl but that's just from memory, I just don't know how to get the lattice energy (assuming that I didn't memorize it) by using a calculation. Can anyone shed some light on this for me? Would be much appreciated.
  2. jcsd
  3. May 1, 2007 #2


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    Answered on another thread
  4. May 2, 2007 #3
    using born haber cycle....right

    to find lattice energy of NaCl
    you should know....
    1. Atomisation energy of Na....dH1
    2. First Ionisation energy of Na.......dH2
    3. Atomistaion Energy of Cl2......dH3
    4. First Electron Affinity of Cl.......dH4
    5. enthalpy of formation NaCl......dH5

    lattice energy will be dH6

    one route to form NaCl is via the enthalpy of formation of NaCl.... which is -424kJmol-1

    a second route is by:

    1. Atomising Na(s) Na(s) ------> Na(g) dH1: +105kJmol-1
    2. First Ionising Na(g) Na(g) -------> Na+(g) + e- dH2: +494kJmol-1
    3. Atomising Cl2(g) 1/2Cl2(g) -------> Cl(g) dH3: +122kJmol-1
    4. First e- affinity Cl(g) Cl(g) + e- -------> Cl-(g) dH4: -364kJmol-1
    5. Lattice Energy Na+(g) + Cl-(g) -------> NaCl dH6

    dH5 = dH1 + dH2 + dH3 + dH4 + dH6
    using the above values.... dH6 = -781kJmol-1

    hope i helped you....

    note: from the eqn above(which i didn't completely understand ;-) ) by making the lattice energy subject of formula you can easily find it.... (if i'm not wrong!!)

    U (Lattice Energy) = Hf - Hs - Hi.e - 1/2Hd - He.a
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