第五章 化学热力学基础 习题(8)

无机化学课外习题

18

(C) 19

(C) 20

(B) 21

Cl(g)+F(g) ① ClF(g) ② ③

21

= 253 kJ·mol-1 rHm

=2DCl─Cl = 119.5kJ·mol-1 rHm

1

=DF─F rHm

Cl(g) Cl2(g)

F(g) F2(g)

ClF(g) ② + ③ - ① Cl2(g) +F(g) 22

所以 fHm(ClF, g)= -50.6 kJ·mol-1

-50.6 = 119.5 +DF—F - 253 2所以 DF—F = 166 kJ·mol-1 22

H= 丙烷燃烧热 =

② H(g) +O(g) = HO(l) H(HO, l) = -286.0 kJ·mol ③ CH(g) + 5O(g) = 3CO(g) + 4HO(l) Q= -2213.0 kJ·mol ④ CH(g) + H(g) = CH(g) H= -123.9 kJ·mol

① C3H6(g) + 2O2(g) 3CO2(g) + 3H2O(l)

2

r

m1

2

22

r

m

2

-1

33

86

222v

-1

238

r

m

-1

④ + ③ - ② = ①

所以 rHm(1) = rHm(4) + rHm(3) – rHm(2)

-3-1 因为 rHm(3) =ΔU +Δ(pV) = -2213.0 + (-2.5) (8.314 298 10) = -2219.2 (kJ·mol) -1 所以 rHm(1) = -123.9 - 2219.2 - (-286.0) = -2057.1 (kJ·mol) 即： C3H6(g) 的燃烧焓为 -2057.1 kJ·mol-1

而 rHm(1) = 3 fHm(H2O, l) + 3 fHm(CO2, g) – fHm(C3H6, g) 所以 -2057.1 = 3(-286.0) + 3(-393.5) – fHm(C3H6, g) fHm(C3H6, g) = -18.6 kJ·mol-1

23

-1

(1) H2(g) +1O2(g) H2O(g) rHm= -242 kJ·mol

=

=

(3) O(g) =O(g) (1) + (2) + (3) 2H + O = HO

122

(2) 2H(g) H2(g) rHm= -436 kJ·mol-1

= -250 kJ·mol-1 rHm

= -928 kJ·mol-1 rHm

所以O—H键能为

1

928 = 464 (kJ·mol-1) 2

24

CO2(g) + 2H2O(l) rHm 由 (1) CH4(g) + 2O2(g) = -890 kJ·mol-1