Posted by Crystal on Wednesday, February 2, 2011 at 10:12am.
I keep finding all sorts of different enthalpy of vaporizations for bromine. I used: 254.66
So factoring 254.66/2 into my equation below i ended up with a lattice energy of -863.33.
Can you confirm this?
____________________________________________________________
What is the enthalpy change (in kJ/mol) for the following process?
Na+(g) + Br-(g) --> NaBr (s)
Hint: use the Born-Haber cycle
Note: the reference state for Br2 is Br2(l) not Br2(g)
Data (in kJ/mol):
enthalpy of sublimation of Na(s): 107
standard enthalpy of formation of Br2(g): 32
bond dissociation energy of Br2(g): 194
ionization energies (1, 2, 3) of Na(g): 496
ionization energies Br(g): 140, 103, 473
first electron affinity of Br(g): -326
standard enthalpy of formation of NaBr(s): -351
Sooo..
deltaH1=+107
deltaH2=194/2=+97
deltaH3=+32
deltaH4=+496
deltaH5=-326
deltaH6=lattice energy=?
deltaHf= -351
Sooo..
-351=107+97+32+496-326+lattice energy
lattice energy= -757
This is not the correct answer.
Can someone please tell me where I've gone wrong please!
Chemistry - DrBob222, Wednesday, February 2, 2011 at 12:24am
The problem gave you a hint that the standard state for Br2 was liquid and not gas. Therefore, you must include enthalpy for Br2(l) to Br2(gas) (and 1/2 that)
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