Calculate the lattice energy of CaBr2.?

The standard heat of formation of CaBr2 is -675 . The first ionization energy of Ca is 590 and its second ionization energy is 1145 . The heat of sublimation of Ca is 178 . The bond energy of Br2 is 193 , the heat of vaporization of Br2 is 31 , and the electron affinity of Br is -325 .

I suppose these are all kJ/mole. First, gasefy solid Ca:

Ca(s) ===> Ca(g) +178

Next take the first electron off Ca:

Ca(g) ===> Ca+(g) + e- +590

Then the second electron:

Ca+(g) ===> Ca2+(g) +1145

178 + 590 + 1145 = +1913 to get to Ca2+(g) ions

Now vaporize the Br2:

Br2(l) ===> Br2(g) +31

Next atomize Br2

Br2(g) ===> 2Br(g) + 193

Next form bromide ions:

2Br(g) + 2e- ===> 2Br-(g) 2 x -325 = -650

31 + 193 - 650 = -426 to make Br

So far, the process has cost 1913 - 650 = +1263

The heat of formation is -675. So the lattice energy is -1938

this answer was wrong so I went in and realized I didn't use -426 so I tried again and got -1512 and the program said it was wrong...what am I doing wrong?

Can anyone help me please

Thanks in advance