Asked by Garcia

Consider an ionic compound, MX2, composed of generic metal M and generic halogen X.

The enthalphy of formation of MX2= -891.
The enthalphy of sublimation of M= 141.
The first and second ionization energies of M are 605 and 1392.
The electron affinity of X= -339.
The bond energy of X2= 177.

I used the Born-Haber cycle and came up with:
-Lattice energy of MX2=-891-(141+88.5+605-678) to come up with the answer that the -lattice energy of MX2 is -1047.5 but i'm getting marked wrong. Where did I go wrong?????
Answered by DrBob222
I just took a brief look; where is the 1392 for second ionization potential? I don't see that in your calculation.
Answered by Garcia
Where would I add that in to my calculation? Like would it look like this:
-Lattice energy of MX2=-891-(141+88.5+605+1392-678)
Answered by DrBob222
Yes, that should do it.
Also, I wonder about taking 1/2 x 177. If you do Cl2 bond dissociation for say NaCl, then you take 1/2 Bond energy because you're using only 1/2 of it to make Cl^-. But in this case you're using both parts of the X2 bond energy to make 2X^-. You multiplied 2 x electron affinity (and that is proper).
Answered by Jordynn
where is the -678 coming from?
Answered by Matt
-678 is electron affinity * 2
Answered by Thien
Shouldn't it be +678 because the formula is E= deltaHf-(H sub + IE + HBE/2)- HEA?
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