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