CaO(s) = Ca2+ (g)＋O2﹣(g)
所以：U (CaO) = 1735 + 172 + 737 + 249 –(–635.5) = + 3538.5 kJ.mol﹣1
2-63.已知：锂的升华热(S)为159 kJ·mol-1，锂的第一电离势(I1)为520.2 kJ·mol-1， F2的离解能(D)为155.5kJ·mol﹣1，F的电子亲合势(E)为 – 349.3kJ·mol﹣1，氟
化锂的生成热(?H?fm)为 – 612kJ·mol﹣1，.求LiF(s)的晶格能(U)是多少?
解： 依题意有：
反应
?H?rm
Li (g ) = Li (s ) –159 kJ·mol–1 Li + (g ) +e – = Li (g ) – 520.2 kJ·mol–1 F– (g ) = F (g ) + e – +349.3 kJ·mol–1 F (g ) = 12F2 (g ) –155.5×1 kJ·mol–12
Li (s ) +
1–1
2F2 (g ) = LiF (s ) – 612 kJ·mol将上面各式相加，得： Li + (g ) + F– (g ) = LiF (s )
??rHm= –159–520.2 +349.3 –155.5×12– 612 = –1019.65 kJ·mol –1所以：U(LiF) =+1019.65 kJ·mol –1
2-64.根据下列数据，求N—H键能。
??fHm(NH3,g)＝–46kJ·mol
﹣1
，
D(H-H )＝436kJ·mol﹣1， D(N≡N )＝946kJ·mol﹣1?
解： 依题意，得：
反应
?H?rm
31
12N2 (g ) +
32H2 (g ) = NH3 (g ) – 46 kJ·mol﹣1
N2 (g ) = 2N (g ) +946 kJ·mol﹣1 H2 (g ) = 2H (g ) +436 kJ·mol﹣1
整理，得 ：
NH3 (g ) =
1212N2 (g ) +
32H2 (g ) 46 kJ·mol﹣1
12N2 (g ) = N (g ) 946×
2kJ·mol﹣1
32H2 (g ) = 3H (g ) 436×3kJ·mol﹣1
将上面各式相加，得
NH3 (g ) = N (g ) + 3H (g ) 46 + 946×
12+ 436×3 = 1173 kJ·mol﹣1
2N—H的键能D (N—H )为： 1173 ÷3 = 391 kJ·mol﹣1
2-65. 已知下列数据，求D(N—N)为多少？
??fHm(H2N—NH2)＝+95 kJ·mol??fHm(N,g )＝473 kJ·mol
﹣1
? D(N—H )＝391 kJ·mol﹣1?
??fHm ? ?rHm﹣1
?
(H,g)＝218 kJ·mol﹣1
解： 依题意有：
2H2 (g ) + N2 (g ) = H2N—NH2 (g ) 95 kJ·mol﹣1
1212H2 (g ) = H (g ) 218 kJ·mol﹣1 N2 (g ) = N (g ) 473 kJ·mol﹣1
整理，得
4H (g ) = 2H2 (g ) – 218×4 kJ·mol﹣1 2N (g ) = N2 (g ) – 473×2 kJ·mol﹣1 2H2 (g ) + N2 (g ) = H2N—NH2 (g ) 95 kJ·mol﹣1
将上面各式相加，得
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4H (g ) + 2N (g ) = H2N—NH2 (g ) 95 – 218×4 – 473×2 = –1723 kJ·mol
﹣1
另外，已知N—H键能为391 kJ·mol﹣1，生成1mol H2N—NH2分子(即1mol N—N键和4mol N—H键)时放出的热量为 1723 kJ ， 所以生成1mol N—N键放出的热量为 1723 – 391×4 = 159 kJ·mol﹣1。 即 N—N单键的键能为 159 kJ·mol﹣1。
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