Assume that Ag2S forms a rutile structure with A = 2.406
Fig. Born Haber cycle for Ag2S
-∆Hf0 + 2∆Hsub(Ag) + ∆Hsub(S) + 2I.EAg+ + E.A1 + E.A2 +∆HL = 0
∆HL = ∆Hf0 - 2∆Hsub(Ag) - ∆Hsub(S) - 2I.EAg+ - E.A1 - E.A2
∆HL = -31.8 -2(255) -9.8 – 2(731)- 200+ 456
∆HL = -31.8-510-9.8- 1462- 200-456
∆HL = -2669.6 KJ/mol
∆G = -ANz2e24πϵor(1- -1n)
∆G = - 2.4 x (6.023 x10+23) x 1x 2 x(1.6x10-19)24 x 3.14 x 8.854x 10-12 x 2.406x 10-10X(1-110)
∆G = - 2489.5 KJ/mol
∆G = -NMz2e24πϵr(1- -34.5 x 10-12r)
∆G = -6.023 x10+23x 2.4 x 1x 2 x(1.6x10-19)24 x 3.14 x 8.854 x 10-12 x 2.406x 10-10X(1-34.5 x 10-122.406x 10-10)
∆G = - 2370 KJ/mol
UL = 1.202X10-4 nZ+Z-r 1-34.5 x 10-12r
UL = - 2568 KJ/mol
Method |
Lattice energy (KJ/mol) |
Born Haber Cycle |
-2669.6 |
Born Lande equation |
-2489.5 |
Born Mayer equation |
- 2370 |
Kaputinskii Equation |
-2568 |
Born Haber cycle is most accurate among these methods because it considers all the possible physical and chemical changes during the formation of the lattice.
Kaputinskii method is the easiest to carry out.
Sources
Wikipedia
Periodictable.com
Chemlibretexts.org
Assume that Ag2S forms a rutile structure with A = 2.406 1. Calculate AHL of Ag2S...