物理化学 傅献彩 第六章 化学平衡 下载本文

$$$K$)K$p(5)?Kp(1p(2)Kp(3)Kp(4)

(3)?rGm$(5)??rGm$(1)??rGm$(2)??rGm$(3)??rGm$(4)???rGm?34K?exp???2.58?10?RT?$p

22、800K,100kPa时,C6H5C2H5(g)==C6H5C2H3(g)+H2(g)的Kp=0.05,试计算: (1) 平衡时乙苯的解离度α;

(2) 若在原料中添加水蒸气,使乙苯和水蒸气的摩尔比为1:9,总压仍为100kPa,求此时乙苯的解离度α。 解:

(1) C6H5C2H5(g)==C6H5C2H3(g)+H2(g)

起始时p/kPa: p 0 0 平衡时p/kPa: p(1-α) pα pα p总=p+ pα=100,p=100/(1+α)

Kp=(pα/100)(pα/100)/{p(1-α)/100}=0.05 pα2/(1-α)=5,α2/(1-α2)=0.05

α2=0.05-0.05α2;5.05 α2=0.05;α2=9.425×10-4;α=0.0307 (2) C6H5C2H5(g)==C6H5C2H3(g)+H2(g) H2O(g)

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起始时p/kPa: p 0 0 9p 平衡时p/kPa: p(1-α) pα pα 9p p总=p+ pα+9p=100,p=100/(10+α) Kp= (pα/100)(pα/100)/{p(1-α)/100}=0.05

pα2/(1-α)=5,α2/{(1-α)(10+α)=0.05,α2=0.05(10-9α-α2) 1.05α2+0.45α=0.5,α2+0.428571α=0.47619

(α+0.214286)2=0.47619+0.2142862=0.522108 α+0.214286=0.722571;α=0.508285

24、一个可能大规模制备氢气的方法是:将CH4(g) +H2O(g)的混合气通过灼热的催化床,若原料气体组成的摩尔比为nH2O:nCH4=5:1,温度为873K,压力为100kPa,并假设只发生如下两个反应:

(1) CH4(g) +H2O(g)==CO(g)+H2(g) △r1Gm=4.435 kJ·mol-1 (2) CO(g) +H2O(g)==CO2(g)+H2(g) △r1Gm=-6.633kJ·mol-1

试计算达到平衡并除去H2O(g)后,平衡干气的组成,用摩尔分数表示。 解:

△r1Gm= -RTlnKp1,lnKp1=-△r1Gm /(RT)= -4.435×103/(8.314×873)=-0.61104 Kp1=0.542786

△r2Gm= -RTlnKp2,lnKp2=-△r2Gm /(RT)= 6.633×103/(8.314×873)=0.913873 Kp2=2.493962 (1)

CH4(g) + H2O(g) == CO(g) +3H2(g)

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起始时kPa: 1 5 0 0 平衡时kPa: 1-n1 5-n1-n2 n1-n2 3n1+n2 (2)

CO(g) + H2O(g)==CO2(g)+H2(g)

起始时kPa: 0 5 0 0 平衡时p/kPa:n1-n2 5-n1-n2 n2 3n1+n2

n总=nCO+nCO2+nH2O+nH2+nCH4 =n1-n2+ n2 +5-n1-n2+3n1+ n2+x-n1= 6+2n1

Kp1=(3n1+n2)3(n1-n2)/{( 1-n1)( 5-n1-n2)}/(p)2=0.542786 Kp2=(3n1+n2)n2 /{(n1-n2)( 5-n1-n2)=2.493962 解得n1=0.91 n2=0.65

水的含量为5-n1-n2=3.44mol 除去水后n干总 =6+2n1-n水=4.38mol xCO (干)=(n1-n2)/ n干总=5.94% xCO2(干)=n2/n干总=14.8% xH2(干)=(n1+n2)/ n干总=77.17% xCH4(干)=( x-n1)/ n干总=2.05%

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