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第三章习题解答
1.先求H2OgH2Ol的相变热G2=0G3=Vlpo-3166G=G1+G2+G3=–8557JrGoml=rGomg+G=–
228.57–
8.557=–
237.13kJ
2.反应Cs+2H2g=CH4grGom=19290Jmol1摩尔分数
0.
80.11T=1000K时,QopKop反应不会正向进行2设压力为p,当时,即p
1.59po时,反应才能进行
3.SO2Cl2g+SO2gCl2g反应前压力kPa
44.
78647.836平衡时压力kPax
44.786-x
47.836-xp总=x+
44.786-x+
47.836-x=
86.096kPax=
6.526kPa
4.H2g+I2g2HIg开始mol
7.
945.3平衡molx=0x=
9.478mol另一根x=
19.30舍去
5.Ag+BgABg开始mol11平衡mol1-
0.41-
0.
40.4n总=
1.6molp=
0.06206po=6206Pa
6.AgBg平衡压力10poporGom=–RTlnKop=5708JrGm1=rGom–RTlnQop1反应不会自发进行rGm2=rGom–RTlnQop2反应自发进行
7.N2g+3H2g2NH3g开始mol13平衡molxn总=4-x1当总压p=10po=
3.85%时,x=
0.1483mol代入上式得Kop=
1.6410–42当=5%时,x=
0.1905mol,Kx=
0.02911p=
13.323po=1332kPa
8.G2=0G总=4490JG总=fGomg-fGoml即
4.49=–
161.96-fGomlfGoml=–
166.45kJmol
19.先计算丁二酸(m=1molkg)的fGomfGomm=1=fGoms+G1+G2=–748099+0+=–747268Jmol1一级电离C4H6O4m=1C4H5O4m=1+H+m=1fGomkJmol1–
747.268–
723.0370rGom=
24.23kJmol
110.NH4HSsNH3g+H2Sg平衡kPa1原有H2S,平衡p
39.99+p解得p=
18.87kPap总=2p+
39.99=
77.73kPa2形成固体的条件QpKop设H2Sg的压力为pp
166.6kPa
11.Ag+BgCg+Dg1平衡时mol1/31/32/32/32设C为xmol1x2xxx3设生成C为xmol1x1x
0.5+xx4设C减少xmolxx1x2x由1=0由23x212x+8=0x=
0.845mol由33x2
8.5x+4=0x=
0.596molC总量=
1.096mol由43x2+3x2=0x=
0.457molC的量=1x=
0.543mol
12.设开始为1mol,解离度为PCl5gPCl3g+Cl2g平衡时mol1n总=1+当=
0.5,n总=
1.5mol,p=po时,代入上式,可得平衡常数1降低总压p,使体积增加1倍,计算的改变可得压力,代入得=
0.618解离度增加2通入N2,使体积增加1倍,p=po,计算的改变可得气体摩尔n总=n2=3mol得=
0.618解离度增加3通入N2,使压力p增加1倍,V1=V2,计算的改变可得气体摩尔n总=n2=3mol得=
0.50解离度不变4通入Cl2,使压力p增加1倍,计算的改变如3计算n总=3mol,设Cl2的加入量xmol,计算Cl2总量PCl5gPCl3g+Cl2g1-+xn总=1++x=3nCl2=+x=2mol得=
0.20解离度减少
13.1rGom=–RTlnKop=Jmol–1kJmol–1Jk–1mol–12T=573K时,I2+环戊烯2HI+环戊二烯开始
0.5po
0.5po平衡pp
20.5pop
0.5pop解此三次方程,p=
34.50kPa3同理,起始压力为10po时,解得p=
418.20kPa14.先计算反应前,,1610K时,xH2O=
0.02nH2O=
0.4929
0.02=
0.009859molCO2g+H2SgCOSg+H2Og平衡时mol
0.
090140.
38300.
0098590.009859=02rGom610K=–RTlnKop=
29.78kJmol–13620K时,xH2O=
0.03nH2O=
0.4929
0.03=
0.01479molCO2g+H2SgCOSg+H2Og平衡时mol
0.
085210.
37810.
014790.01479=0解得rHom=
276.9kJmol–
115.1FeOs+H2g=Fes+H2OgH2Og=H2g+
0.5O2g=+FeOs=Fes+
0.5O2gKop3=Kop1Kop2=
4.96710-7解得分解压pO2=
2.4810-11kPa2FeOs+COg=Fes+CO2gKop1=?CO2g=COg+
0.5O2g=+FeOs=Fes+
0.5O2gKop3=Kop1Kop2=
4.96710-7得Kp1=
0.42计算CO的用量FeOs+COg=Fes+CO2g起始molx0平衡molx-11x=
3.38mol
16.先计算CH3OHg的SomgS10S=S1+S2+S3=
112.51JK–1mol–1Somg=Soml+S=
239.3JK–1mol–1计算反应的热力学函数和平衡常数COg+2H2gCH3OHgfHomJmol1-
110.5250-
200.67rHom=–
90.145kJmol–1SomJK1mol
1197.
674130.
684239.3rSom=-
219.74JK–1mol–1rGom=rHom–TrSom=–
90.145–
298.15–
219.74103=–
24.63kJmol–
117.CH3COOHl+C2H5OHl=CH3COOC2H4l+H2OlfGom/kJmol1–
389.9–
168.49–
332.55–
237.129rGom=nfGom产物–nfGom反应物=-
332.55–
237.129––
389.9–
168.49=–
11.289kJmol118.Kp2=
4.305106Pa1H2OgpoH2OlpoGG3G2G1H2Og3166PaH2Ol3166PaT=298KCH3OHlpoCH3OHgpoGG3G2G1CH3OHl16343PaCH3OHg16343PaT=298KCH3OHlSom=
126.8,poCH3OHgpoSS3S2S1CH3OHl
16.59kPaCH3OHg
16.59kPaT=298K。