Posted by Garcia on Thursday, November 29, 2012 at 1:05pm.
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?????
- Chemistry - DrBob222, Thursday, November 29, 2012 at 4:12pm
I just took a brief look; where is the 1392 for second ionization potential? I don't see that in your calculation.
- Chemistry - Garcia, Thursday, November 29, 2012 at 7:00pm
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)
- Chemistry - DrBob222, Thursday, November 29, 2012 at 10:41pm
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).
- Chemistry - Jordynn, Sunday, February 16, 2014 at 7:26pm
where is the -678 coming from?
- Chemistry - Matt, Wednesday, October 29, 2014 at 1:14pm
-678 is electron affinity * 2
- Chemistry - Thien, Saturday, November 15, 2014 at 7:19pm
Shouldn't it be +678 because the formula is E= deltaHf-(H sub + IE + HBE/2)- HEA?
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