第六章 化学平衡常数 习题

PCl3 + Cl2 PCl5 达平衡时物质的量/mol (1.00-x) (1.00-x)  x 物质的总量 = (2.00-x) mol

1.00?xxkPa pPCl5?100?kPa

2.00?x2.00?xx/(2.00?x) K?== 0.54 p

(1.00?x)2/(2.00?x)2 pCl2?pPCl3?100? 解得 x = 0.19 xCl2?xPCl3? pPC5l= 0.10

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PCl5(g) PCl3 + Cl2(g) 初始 n /mol 2.00 1.00

平衡 n /mol 2.00-x 1.00+x  x  n总=3.00+x

按Dalton分压定律: pi = p总

1.00?x= 0.45

2.00?xxPCl5?pPCl52.00?x1.00?xx,xPCl3?,xCl2?

3.00?x3.00?x3.00?x?xPCl5p,pPCl3?xPCl3p,pCl2?xCl2p

p1.00?xpx??pp3.00?xp3.00?xp?PCl3Cl2?

Kp=?2.00?xppPCl53.00?xp?(1.00?x)xp?

??=K?p p= 100 kPa

(3.00?x)(2.00?x)p 解得 x = 1.82

所以pPCl5= 7.5 kPa,pPCl3= 117 kPa,pCl2= 75.5 kPa

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CaCO3 CaO + CO2↑

Kp = pCO2,在800℃时pCO2= 1.16 ? 102 kPa 加入的CaCO3为 20/100 = 0.20 mol

设在800℃达到平衡时CO2的物质的量为n 则由 pV = nRT n =

pV1.16?10.0= 0.130 mol ?RT0.0831?1073此0.130 mol CO2是由0.20 mol CaCO3分解得到的,未反应的CaCO3百分率为 [(0.20-0.13) / 0.20] ? 100%=35%

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在平衡态 p = 220.3 kPa V = 0.4720 dm3 T = 1000 K 则n总=

pV220.3?0.4720= 0.01251 (mol) ?RT8.314?1000 初态nCOCl2= 0.6310 / 98.9 = 0.00638 (mol)

设有x mol的COCl2分解为CO和Cl2

COCl2 CO + Cl2 达平衡时物质的量/mol 0.00638-x   x x

n总= 0.00638 + x = 0.01251 (mol) 所以 x = 0.00613

[CO] = 0.0130 mol·dm-3 [Cl2] = 0.0130 mol·dm-3

[COCl2] = 5.3 ? 10-4 mol·dm-3

[CO][Cl2] 所以 Kc == 0.32

[COCl2] 27 (1)

pV18.41.0?6.0= 0.40 mol NO2 n = = 0.24(mol) ?4.6RT0.083?300 2NO2(g) O4(g)

平衡时物质的量/mol 0.40-2 x 0.40-2x + x = 0.24 x = 0.16 mol(N2O4) 0.40-2x = 0.40 - 2 ? 0.16 = 0.08 (mol)

(0.16/0.24)?1.0 K?== 6 p

[(0.08/0.24)?1.0]2 (2) 温度升高到111℃,平衡常数变小,所以此反应为放热反应。 28

PCl5 PCl3 + Cl2 初始压力/100 kPa 0 p0 p0 平衡压力/100 kPa 1.00 p0 -1.00  p0 -1.00

(p0?1.00)2 所以Kp== 1.85 p0 = 2.36 ? 100 kPa

1.00 n0=

p0V2.36?5.0= 0.27 (mol) ?RT0.083?523 2NH3

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(1) N2+ 3H2

pNH332pH2pN2= Kp= 0.10

pNH3321.0?10= 0.10

1.0?4.92=0.10 (mol)

RT0.083?600 nHCl = 1.000 ? 0.10 mol 所以NH3和HCl完全反应,生成0.10 mol的NH4Cl。  

pNH3= 1.0 ? 100 kPa nNH3=

pNH3V? [H+] =(Kw/Kb)?cNH??(1.0?10?14/1.8?10?5)?0.10= 7.5 ? 10-6 (mol·dm-3)

4 pH = 5.12

(2) 加入甲基橙后,呈黄色。

(3) 若在合成氨时减小压力,平衡向减小氨的方向移动,NH3的物质的量减小,有过量的HCl存在,则溶液的pH值下降。 30

该反应在恒容恒温下进行,各物质的分压力变化与浓度的变化一样也正比于物质的量的变化,故可直接根据方程式来确定分压力的变化。 2NO(g) + O2(g)2NO2(g) 起始分压力/100 kPa 1.00 3.00 0 平衡分压力/100 kPa 1.00-x  3.00-x  x

因为pNO2/p?= 12 / 100 = 0.12 = x 所以 pNO/p?= (100-12)/100 = 0.88 pO2/p?= (300-1? 12) / 100 = 2.94 2 而K?p

?(pNO2/p?)2pO2p?(pNOp?)20.122= 6.3 ? 10-3 ?22.94?0.88 31 Cl2(l)

Cl2(g) K?p= pCl2(g) /100 kPa

??= -?fGm(Cl2, l) = -RT ln?rGmK?p

??

K?p= 6.95, Kp=pCl2/p,pCl2?4.794?103 ) = 0.842

2.30?8.31?298= 6.95 ? 100 kPa

lgK?p= - (

32

??? 由 ?rGm= 2?fGm (SO3)- 2?fGm(SO2) = 2 ? (-370.4)-2 ? (-300.4) = -140.0 (kJ·mol-1)

? 及 ?rGm= -2.303RTlgKp(298K)

得 lgK?p(298K) =

?140.0?1000= 24.52

?2.303?8.314?298.224.52

所以 K? p(298K) = 10

??? 又 ?rHm= 2?fHm(SO3) - 2?fHm(SO2) = 2 ? (-395.2)-2 ? (-296.9) = -196.6 (kJ·mol-1)

所以 lg

K?p(298K)K?p(773K)??196.6?1000298?773()= 21.18

2.303?8.31298?773K?p(298K)K?p(773K)3.34

= 1021.18 所以K? = 2.2 ? 103 p(773K) = 10

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???? ?rGm= 2?fGm(CO, g)-?fGm(C, 石) - ?fGm(CO2, g) = 2 (-137.3) – 0 - (-394.4) = 119.8 (kJ·mol-1)

? 因为 ?rGm> 0,所以反应不能自发进行。 ??rGm119.8?103?? 因为 = -21.0

2.30RT2.30?8.31?298-21-22

所以 K?p= 1 ? 10 = 9 ? 10

lgK?p=? 34

(1) 2NO2(g)N2O4(g)

设平衡时NO2的分压力为(1.000 – x) ? 100 kPa,N2O4的分压力为x ? 100 kPa

x = 6.06 x = 0.668

(1.000?x)2 所以 pN2O= 0.668 ? 100 kPa pNO2= 0.332 ? 100 kPa

4 转化率? =

2?0.668= 80.1%

0.332?0.668?2 (2) lgK?p2K?p1??rHm?100011?(?) 2.303?8.314300310 lgK?p2=

?57.5?100011(?)+ lg6.06

2.303?8.314300310 K?p2= 2.90

-3-1? ?rGm= -RT lnK?p2= -8.31 ? 10 ? 310 ? 2.30 ? lg2.90 = -2.7 (kJ·mol)

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(1) 设原有PCl5 n mol,分解达平衡时,体系中共有n总 mol气体 PCl5(g) PCl3(g) + Cl2(g) 平衡时/mol n(1-? ) n? n? n总= n (1+? ) 1.00 ? 1.00 =

2.695(1+? ) ? 0.0831 ? 523, ? = 78.0% 208.5p??2 (2) = 1.55 21??-1? ?rGm = -RTlnK?p= -1.90 (kJ·mol)

K?p=

36

(1) N2O4 2NO2

平衡时物质的量/mol no(1-? ) 2no? n总= no(1+? ) MN2O4= 92 g·mol-1 MNO2= 46 g·mol-1 M?92?1??2? = 61.2 ?46?1??1??(p 所以 ? = 50%

2?2)pNO24?2p4?0.502?1.001???

??? Kp== 1.3 221??pN2O41??1?0.50p1??K?4?'21.3p (2) 当p = 0.100 ? 100 kPa时,= 13 ??p0.1001??'2 所以 ? = 87% 37

(1) C5H10(l) C5H10(g)

??? ?rHm = ?fHm(C5H10, g) -?fHm(C5H10, l) = -77.2 -(-105.9) = 28.7 (kJ·mol-1)

2

?因为?rSm

??rHm-1-1=? 88 (J·mol·K)

Tb 所以 Tb =

??rHm??rSm?28.7= 3.3 ? 102(K) 0.088??? (2) ?rGm = ?rHm - T ?rSm= 28.7 - (298 ? 0.088) = 2.5 (kJ·mol-1) ??rGm?2.5?103? = -0.44 K?p= 0.36

2.30RT2.30?8.31?298 2.30RT 2.30 ? 8.31 ? 298 p = 0.36 ? 100 kPa = 36 kPa

lgK?p=? 38

(1) 2NO(g) + F2(g)2NOF(g)

? 2?fHm(NOF) = -312.96 + 90.37 ? 2 = -132.22 (kJ·mol-1)

? 所以 ?fHm(NOF) = -66.11 kJ·mol-1 (2) 2NO(g) + F2(g)2NOF(g)

? ?rSm= 2 ? 248 -203.3 -2 ? 210.6 = -128.5 (J·mol-1·K-1)

??? ?rGm= ?rHm - T ?rSm = (-312.96) -298 ? (-128.5) ? 10-3 = -274.7 (kJ·mol-1) 48?? ?rGm= -RTlnK?p Kp= 2 ? 10

39 lg lgK?p2K?p1??rHmT?T1?(2) 2.30RT2T1?312.96500?298() 482.30?0.00831298?5001.37?1025

所以 K?p2 = 9 ? 10

? 温度升高,平衡向逆方向移动。 40

△ NH4Cl(s)NH3(g) + HCl(g) 平衡压力/kPa p1 p2 平衡总压力 p = p1+ p2,且 p1 = p2 =1p 2 所以 Kp= p1 p2 = 1/4p2 所以 427℃时:K?p= 9.24

? ?rGm= -12.9 kJ·mol-1

K?p2 459℃时: K?p= 31.0

? ?rGm= -20.9kJ·mol-1

因为lgK?p2K?p1??rHmT?T1? =1.6 ? 102 (kJ·mol-1) ?(2) ?rHm2.30RT2?T1??? 427℃时:?rGm=?rHm- T ?rSm

?=?rSm???(?rGm??rSm)= 0.25 (kJ·mol-1·K-1)

T 41

C2H5OH(l)C2H5OH(g)

? ?rHm= -235.4 - (-277.6) = 42.2 (kJ·mol-1)

? ?rSm= 282 - 161 = 121 (J·mol-1·K-1)

??? ?rGm=?rHm - T ?rSm = 42.2 – 298 (0.121) = 6.1 (kJ·mol-1) ???rGm?6.1?103? = -1.07

2.30RT2.30?8.31?298 K?p= 0.085

lgK?p=

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