In: Chemistry
The standard heat of formation of CaBr2 is -675 kJ/mol. The first ionization energy of Ca is 590 kJ/mol and its second ionization energy is 1145 kJ/mol. The heat of sublimation of Ca[Ca(s)→Ca(g)] is 178 kJ/mol. The bond energy of Br2 is 193 kJ/mol, the heat of vaporization of Br2(l) is 31 kJ/mol, and the electron affinity of Br is -325 kJ/mol.
Calculate the lattice energy of CaBr2.
heat of romation
Ca(s) + Br2(l) ----------------->
CaBr2(s) H =
-675Kj/mole
heat of sublimation
Ca(s) ---------------->
Ca(g)
H1 =
178Kj/mole
The first ionization energy of Ca
Ca(g) ---------------> Ca^+(g) +
e^-
H2 =
590Kj/mole
The second ionization energy of Ca
Ca^+(g) ---------------> Ca^2+(g) +
e^-
H3 =
1145Kj/mole
heat of vaporization of Br2(l)
Br2(l) ----------------->
Br2(g)
H4 =
31Kj/mole
The bond energy of Br2
Br2(g) ------------->
2Br(g)
H5 =
193Kj/mole
the electron affinity of Br
2Br + 2e^- ------------->
2Br^-
H6 =
2*-325 = 650KJ/mole
lattice energy
Ca^2+ (g) + 2Br^-(g) ---------------->
CaBr2(s) H7 =
?
from Hess law
H =
H1 +
H2 +
H3 +
H4 +
H5 +
H6
-675 = 178+590+ 1145+31+193-650 +
H7
H7
= -2162KJ/mole
lattice energy of CaBr2 = -2162KJ/mole