Calculating lattice energy on ionic compound

  • Thread starter flemonster
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Homework Statement



Given the following thermodynamic data, calculate the lattice energy of CaBr2(s) caculate the lattice enegy:

(A) Δ°Hf CaBr2(s) = -675 kJ/mol
(B) Δ°Hf Ca(g) = 179 kJ/mol
(C) Δ°Hf Br(g) = 112 kJ/mol
(D) 1st ionization energy of Ca = 590 kJ/mol
(E) 2nd ionization energy of Ca = 1145 kJ/mol
(F) Electron affinity of Br = -325 kJ/mol


Homework Equations



Heats of Formation + ionization energies + electron affinity - lattice energy = heat of formation



The Attempt at a Solution



(B) + (C) + (D) + (E) + (F) + Lattice Energy = -675

solving for lattice energy:

[179 + 112 + 590 + 1145 + -325] + Lattice Energy = -675

Lattice energy = -675 - 1701

Lattice energy = -2376 kJ/mol

I think I am going wrong with something in the calculations of the energy bromine contributes to the equation. I know bromine is diatomic but I'm not sure how to incorporate that into the equation. I don't want you to give me the answer because I really want to be able to do this on my own but anything you can point out that can get me in the right direction are appreciated.
 

Answers and Replies

  • #2
Borek
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You are not following stoichiometry - you need to deal with two Br atoms all the time. 112 kJ/mol for Δ°Hf Br(g) is a dissociation enthalpy - but it produces just one Br atom, and you need two.

It should became obvious if you will try to write a "reaction equation" - after all, Hess law is just a way of dealing with the energy conservation. Mass conservation and charge conservation are used to balance reaction equation, same thinking can be applied to the energy conservation.
 
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