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Calculating lattice enthalpy of FeF2

  1. Oct 21, 2011 #1
    Calculate the lattice enthalpy of FeF2 using the following data:
    Hlatt (FeCl2) = 2569 kJ/mol
    r(Cl-) = 181 pm
    r(F-) = 133 pm

    I started with Hlatt is proportional to [absolute value of the product of charge on cation and anion] divided by the distance between their nuclei. Since the charge on F and Cl are the same, you could equate 2 proportionality equations to get Hlatt FeCl2/Hlatt FeF2 = 133 pm / 181 pm. I realized that this wouldn't work because the proportionality is for the distance between nuclei, not the radius of the anion.

    So I tried to use the Kapustinskii equation, thinking that I could find the radius of iron. This didn't turn out well because I don't know how to isolate the distance between nucleus.

    My prof said that this is an easy question and you're not allowed to use any other data (ex. enthalpy of sublimation, ionization, etc).
     
  2. jcsd
  3. Oct 21, 2011 #2

    Borek

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    Staff: Mentor

    That's a correct approach - and to be honest, I have no idea where the problem is. Looks to me like just a quadratic equation (after rearranging). Complete the square or start with discriminant - that's a basic HS algebra.
     
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