物理化學(xué)(復(fù)旦大學(xué)藥學(xué)院) 第三章習(xí)題答案

1、第三章習(xí)題解答H2O(g)poH2O(l)poGG3G2G1H2O(g)3166 PaH2O(l)3166 PaT=298K先求H2O(g)H2O(l)的相變熱：G2=0G3=Vl(po-3166)G=G1+G2+G3=8557 JrGom(l)= rGom(g)+G=228.578.557 =237.13 kJ2.反應(yīng)C(s)+2H2(g)=CH4(g)rGom=19290 Jmol1摩爾分?jǐn)?shù)0.80.1(1) T=1000K時(shí)，QopKop反應(yīng)不會(huì)正向進(jìn)行。(2) 設(shè)壓力為p，當(dāng)時(shí)，即p1.59po時(shí)，反應(yīng)才能進(jìn)行。3. SO2Cl2(g) + SO2(g) Cl2(g)反應(yīng)前壓力(kPa

2、)44.78647.836平衡時(shí)壓力(kPa)x44.786-x47.836-xp總=x+(44.786-x)+( 47.836-x)=86.096 kPax=6.526 kPa4.H2(g) + I2(g) 2HI(g)開(kāi)始(mol)7.94 5.3平衡(mol) x=0 x=9.478 mol(另一根x=19.30舍去)5.A(g) + B(g) AB(g)開(kāi)始(mol)1 1平衡(mol)1-0.41-0.40.4n總=1.6 molp=0.06206po=6206 Pa6.A(g) B(g)平衡壓力10poporGom=RTlnKop=5708 JrGm(1)= rGomRTlnQop

3、(1)反應(yīng)不會(huì)自發(fā)進(jìn)行。rGm(2)= rGomRTlnQop(2)反應(yīng)自發(fā)進(jìn)行。7.N2(g) + 3H2(g) 2NH3(g)開(kāi)始(mol)1 3平衡(mol)xn總=4-x(1) 當(dāng)總壓p=10 po, =3.85%時(shí)，x=0.1483mol , 代入上式得 Kop=1.64104(2) 當(dāng) =5%時(shí)，x=0.1905mol，Kx=0.02911p=13.323 po =1332 kPaCH3OH(l)poCH3OH(g)poGG3G2G1CH3OH(l)16343PaCH3OH(g)16343PaT=298K8.G2=0G總4490 JG總=fG om(g) -fGom(l)即4.49

4、=161.96 -fGom(l)fGom(l)=166.45 kJmol19. 先計(jì)算丁二酸（m=1 molkg）的fGom：fGom(m=1)= fGom(s)+G1+G2=748099+0+=747268 Jmol1一級(jí)電離C4H6O4(m=1)C4H5O4(m=1) + H+(m=1)fGom(kJmol1)747.268723.0370rGom=24.23 kJmol110.NH4HS(s) NH3(g) + H2S(g)平衡(kPa)(1) 原有H2S，平衡p39.99+p解得 p=18.87 kPap總=2p+39.99=77.73 kPa(2) 形成固體的條件：QpKop 設(shè)H2

5、S(g)的壓力為pp166.6 kPa11.A(g) + B(g) C(g) + D(g)(1) 平衡時(shí)(mol)1/31/32/32/3(2) 設(shè)C為x mol1x2xxx(3) 設(shè)生成C為x mol1x1x0.5+xx(4) 設(shè)C減少x molxx1x2x由(1)=0由(2)3x212x+8=0 x=0.845 mol由(3)3x28.5x+4=0 x=0.596molC總量=1.096 mol由(4)3x2+3x2=0 x=0.457 molC的量=1x=0.543 mol12.設(shè)開(kāi)始為1 mol，解離度為 PCl5(g) PCl3(g) + Cl2(g)平衡時(shí)(mol)1 n總=1+當(dāng)

6、=0.5，n總=1.5 mol，p=po時(shí)，代入上式，可得平衡常數(shù) (1) 降低總壓p，使體積增加1倍，計(jì)算 的改變：可得壓力 ，代入：得 =0.618解離度增加(2) 通入N2，使體積增加1倍，p=po，計(jì)算 的改變：可得氣體摩爾 n總=n2=3 mol得 =0.618 解離度增加(3) 通入N2，使壓力p增加1倍，V1=V2，計(jì)算 的改變：可得氣體摩爾 n總=n2=3 mol得 =0.50 解離度不變(4) 通入Cl2，使壓力p增加1倍，計(jì)算 的改變：如(3)計(jì)算n總=3 mol，設(shè)Cl2的加入量x mol，計(jì)算Cl2總量：PCl5(g) PCl3(g) + Cl2(g)1- +xn總=1

7、+x=3nCl2=+x=2 mol得 =0.20 解離度減少13.(1) rGom=RTlnKop=Jmol1 kJmol1 Jk1mol1(2) T=573K時(shí)，I2 + 環(huán)戊烯 2HI + 環(huán)戊二烯開(kāi)始0.5po0.5po平衡pp2(0.5pop)(0.5pop)解此三次方程，p=34.50 kPa(3) 同理，起始?jí)毫?0po時(shí)， 解得p=418.20 kPa14先計(jì)算反應(yīng)前：，(1) 610K時(shí)，xH2O=0.02nH2O=0.49290.02=0.009859 molCO2(g) + H2S(g) COS(g) + H2O(g)平衡時(shí)(mol)0.09014 0.3830 0.00

8、9859 0.009859=0(2) rGom(610K)=RTlnKop=29.78 kJmol1(3) 620K時(shí)，xH2O=0.03nH2O=0.49290.03=0.01479 molCO2(g)+ H2S(g) COS(g) + H2O(g)平衡時(shí)(mol)0.08521 0.3781 0.014790.01479=0解得rHom=276.9 kJmol115.(1) FeO(s)+H2(g)=Fe(s)+H2O(g) H2O(g)=H2(g)+0.5O2(g)=+FeO(s)=Fe(s)+ 0.5O2(g)Kop(3)= Kop(1) Kop(2)=4.96710-7解得分解壓 p

9、O2=2.4810-11 kPa(2) FeO(s)+CO(g)=Fe(s)+CO2(g)Kop(1)=？ CO2(g)=CO(g)+0.5O2(g)=+FeO(s)=Fe(s)+ 0.5O2(g)Kop(3)= Kop(1) Kop(2)=4.96710-7得Kp(1)=0.42計(jì)算CO的用量：FeO(s)+CO(g)=Fe(s)+CO2(g)起始(mol)x0平衡(mol)x-11x=3.38 molCH3OH(l)Som=126.8，poCH3OH (g)poSS3S2S1CH3OH (l)16.59 kPaCH3OH (g)16.59 kPaT=298K16.先計(jì)算CH3OH(g)的Som(g)：S10S=S1+S2+S3=112.51 JK1mol1Som(g)= Som(l)+S=239.3 JK1mol1計(jì)算反應(yīng)的熱力學(xué)函數(shù)和平衡常數(shù)：CO(g) +2H2(g) CH3OH(g)fHom(Jmol1)-110.5250-200.67rHom=90.145 kJmol1Som(JK1mol1)197.674130.684239.3rSom= -219.74 JK1mol1rGom=rHomTrSom=90.145298.15(219.74)103=24.63 kJmol117.CH3COOH(l) + C2H5OH