最新-天津大学无机化学第五版习题答案 精品

第1章 化学反应中的质量关系和能量关系 习题参考答案

1.解:1.00吨氨气可制取2.47吨硝酸。 2.解:氯气质量为2.9×10g。 3.解:一瓶氧气可用天数

n1(p?p1)V1(13.2?103-1.01?103)kPa?32L??9.6d ?n2p2V2101.325kPa?400L ? d-13

4.解:T?pVMpV ?nRmR = 318 K ?44.9℃ 5.解:根据道尔顿分压定律

pi?p(N2) = 7.6?104 Pa

p(O2) = 2.0?104 Pa p(Ar) =1?103 Pa

nip n6.解:(1)n(CO2)? 0.114mol; p(CO2)? 2.87 ? 104 Pa

(2)p(N2)?p?p(O2)?p(CO2)?3.79?104Pa (3)

n(O2)p(CO2)2.67?104Pa???0.286 np9.33?104Pa7.解:(1)p(H2) =95.43 kPa (2)m(H2) =

pVM = 0.194 g RT8.解:(1)? = 5.0 mol

(2)? = 2.5 mol

结论: 反应进度(?)的值与选用反应式中的哪个物质的量的变化来进行计算无关,但与反应式的写法有关。

9.解:?U = Qp ? p?V = 0.771 kJ 10.解: (1)V1 = 38.3?10 m= 38.3L

-3

3

(2) T2 =

pV2= 320 K nR(3)?W = ? (?p?V) = ?502 J (4) ?U = Q + W = -758 J (5) ?H = Qp = -1260 J

11.解:NH3(g) +

5O(g) ???3??1298.15K??NO(g) + H2O(g) ?rHm= ? 226.2 kJ·mol 2标准态4212.解:?rHm= Qp = ?89.5 kJ ?rUm= ?rHm? ?nRT

= ?96.9 kJ

13.解:(1)C (s) + O2 (g) → CO2 (g)

?? ?rHm = ?fHm(CO2, g) = ?393.518 kJ·mol?

1

1CO(g) + 1C(s) → CO(g)

2

221? = 86.229 kJ·mol? ?rHm CO(g) +

1FeO(s) → 2Fe(s) + CO(g)

232

331

1

? ?rHm = ?8.3 kJ·mol?

??各反应?rHm之和?rHm= ?315.6 kJ·mol?。

(2)总反应方程式为

3C(s) + O(g) + 1FeO(s) → 3CO(g) + 2Fe(s)

2232

23231? = ?315.5 kJ·mol? ?rHm由上看出:(1)与(2)计算结果基本相等。所以可得出如下结论:反应的热效应只与反应的始、终态有关,而与反应的途径无关。

???14.解:?rHm(3)=?rHm(2)×3-?rHm(1)×2=?1266.47 kJ·mol?

1

???15.解:(1)Qp =?rHm== 4?fHm(Al2O3, s) -3?fHm(Fe3O4, s) =?3347.6 kJ·mol?

1

(2)Q = ?4141 kJ·mol?

1

???16.解:(1)?rHm =151.1 kJ·mol?(2)?rHm = ?918.47 kJ·mol?(3)?rHm =?71.7

1

1

kJ·mol?

1

?????17.解:?rHm=2?fHm(AgCl, s)+?fHm(H2O, l)??fHm(Ag2O, s)?2?fHm(HCl, g) ? ?fHm(AgCl, s) = ?127.3 kJ·mol?

1

18.解:CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

???? ?rHm = ?fHm(CO2, g) + 2?fHm(H2O, l) ??fHm(CH4, g)

= ?890.36 kJ·mo ?

1

Qp = ?3.69?10kJ

第2章 化学反应的方向、速率和限度 习题参考答案

??1.解:?rHm = ?3347.6 kJ·mol?;?rSm = ?216.64 J·mol?·K?;?rGm = ?3283.0

1

4

?11

kJ·mol?< 0

1

该反应在298.15K及标准态下可自发向右进行。

?2.解: ?rGm = 113.4 kJ·mol? > 0

1

该反应在常温(298.15 K)、标准态下不能自发进行。

?? (2)?rHm = 146.0 kJ·mol?;?rSm = 110.45 J·mol?·K?;?rGm = 68.7 kJ·mol?

1

?111

> 0

该反应在700 K、标准态下不能自发进行。

???3.解:?rHm = ?70.81 kJ·mol? ;?rSm = ?43.2 J·mol?·K?;?rGm = ?43.9 kJ·mol?

1

1

1

1

(2)由以上计算可知:

?? ?rHm(298.15 K) = ?70.81 kJ·mol?;?rSm(298.15 K) = ?43.2 J·mol?·K?

1

1

1

??? ?rGm = ?rHm ? T ·?rSm ≤ 0

T ≥

??rHm(298.15 K)??rSm(298.15 K) = 1639 K

33p (CO) ? p (H2) ?c (CO) ? c (H2) ?4.解:(1)Kc = Kp = p (CH4) p (H2O)c (CH4) c (H2O) K?

?p (CO) / p?? p (H) / p ? =

?p (CH)/p?? p (HO) / p???32??42 (2)Kc =

?c (N2)?? c (H2) ?c (NH3) 12 32 Kp =

2)/??p (N2)?? p (H2) ?p (NH3) 12 32

K? =

?p (N2)/1?2p ?? p (Hp ??32p (NH3) /p

(3)Kc =c (CO2) Kp =p (CO2) K? =p (CO2)/p? (4)Kc =

?

? c (H2O) ? ? c (H2) ?33 Kp =

? p (H2O) ? ? p (H2) ?33

K

?=

? p (H p (H2O)/p? 2)/p?? ?33

??5.解:设?rHm、?rSm基本上不随温度变化。

??? = ?rHm ? T ·?rSm ?rGm?(298.15 K) = ?233.60 kJ·mol? ?rGm1

?(298.15 K) = ?243.03 kJ·mol? ?rGm1

lgK?(298.15 K) = 40.92, 故 K?(298.15 K) = 8.3?10

40

34

lgK?(373.15 K) = 34.02,故 K?(373.15 K) = 1.0?10

??6.解:(1) ?rGm=2?fGm(NH3, g) = ?32.90 kJ·mol? <0

1

该反应在298.15 K、标准态下能自发进行。

(2) lgK?(298.15 K) = 5.76, K?(298.15 K) = 5.8?118

??7. 解:(1) ?rGm(l) = 2?fGm(NO, g) = 173.1 kJ·mol?

1

?= lgK1???fGm(1)31? = ?30.32, 故 K1= 4.8?10?

2.303 RT1

??(2)?rGm(2) = 2?fGm(N2O, g) =218.4 kJ·mol?

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