復(fù)旦 物理化學(xué) 第三章 習(xí)題答案.doc

第三章習(xí)題解答H2O(g)pH2O(l)pDGDG3DG2DG1H2O(g)3166PaH2O(l)3166PaT=298K1. 先求H2O(g)H2O(l)的相變熱：DG2=0DG3=Vl(p-3166)DG=DG1+DG2+DG3=8590.9 JDrGm(l)= DrGm(g)+DG=228.578.59 =237.16 kJ2.反應(yīng)C(s)+2H2(g)=CH4(g)DrGm=19397 Jmol1摩爾分?jǐn)?shù)0.80.1(1) T=1000K時(shí)，Qp>Kp反應(yīng)不會(huì)正向進(jìn)行。(2) 設(shè)壓力為p，當(dāng)時(shí)，即p>1.61p時(shí)，反應(yīng)才能進(jìn)行。3. SO2Cl2(g) + SO2(g) Cl2(g)反應(yīng)前壓力(kPa)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) xDn=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=0.06206p=6288 Pa6.A(g) B(g)平衡壓力10ppDrGm=RTlnKp=5708 JDrGm(1)= DrGmRTlnQp(1)反應(yīng)不會(huì)自發(fā)進(jìn)行。DrGm(2)= DrGmRTlnQp(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=10p, a=3.85%時(shí)，x=0.1483mol , 代入上式得 Kp=1.64104(2) 當(dāng) a=5%時(shí)，x=0.1905mol，Kx=0.02911p=13.323p=1350 kPaCH3OH(l)pCH3OH(g)pDGDG3DG2DG1CH3OH(l)16343PaCH3OH(g)16343PaT=298K8.DG2=0DG總4523 JDG總=DfGm(g) -DfGm(l)即4523=161.92 -DfGm(l)DfGm(l)=166.44 kJmol19.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：Qp>Kp 設(shè)H2S(g)的壓力為pp>166.6 kPa10.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 mol1-x2-xxx(3) 設(shè)生成C為x mol1-x1-x0.5+xx(4) 設(shè)C減少x molxx1-x2-x由(1)Dn=0由(2)3x2-12x+8=0x=0.845 mol由(3)3x2-8.5x+4=0x=0.596molC總量=1.096 mol由(4)3x2+3x-2=0x=0.457 molC的量=1-x=0.543 mol11.設(shè)開(kāi)始為1 mol，解離度為 aPCl5(g) PCl3(g) + Cl2(g)平衡時(shí)(mol)1-a a an總=1+a當(dāng)a=0.5，n總=1.5 mol，p=p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 a 的改變：可得壓力 ，代入：得 a =0.618(2) 通入N2，使體積增加1倍，計(jì)算 a 的改變：可得氣體摩爾 n總=n2=3 mol得 a =0.618(3) 通入N2，使壓力p增加1倍，計(jì)算 a 的改變：可得氣體摩爾 n總=n2=3 mol得 a =0.50(4) 通入Cl2，使壓力p增加1倍，計(jì)算 a 的改變：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-a a a+xn總=1+a+x=3nCl2=a+x=2 mol得 a =0.2012.(1) DrGm=RTlnKp=Jmol1 kJmol1 Jk1mol1(2) T=573K時(shí)，I2 + 環(huán)戊烯 2HI + 環(huán)戊二烯開(kāi)始0.5p0.5p平衡pp2(0.5p-p)(0.5p-p)解此三次方程，p=34.95 kPa(3) 同理，起始?jí)毫?0p時(shí)， 解得p=423.73 kPa13先計(jì)算反應(yīng)前：，(1) 610K時(shí)，xH2O=0.02nH2O=0.50.02=0.01 molCO2(g) + H2S(g) COS(g) + H2O(g)平衡時(shí)(mol)0.09 0.39 0.01 0.01Dn=0(2) DrGm(610K)=RTlnKp=29.72 kJmol1(3) 620K時(shí)，xH2O=0.023nH2O=0.50.03=0.015 molCO2(g) + H2S(g) COS(g) + H2O(g)平衡時(shí)(mol)0.0985 0.385 0.015 0.015Dn=0解得DrHm=277.0 kJmol114.(1) j FeO(s)+H2(g)=Fe(s)+H2O(g)k H2O(g)=H2(g)+0.5O2(g)l=j+kFeO(s)=Fe(s)+ 0.5O2(g)Kp(3)= Kp(1) Kp(2)=4.96710-7解得分解壓 pO2=2.5010-11 kPa(2)j FeO(s)+CO(g)=Fe(s)+CO2(g)Kp(1)=？k CO2(g)=CO(g)+0.5O2(g)l=j+kFeO(s)=Fe(s)+ 0.5O2(g)Kp(3)= Kp(1) Kp(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)Sm=130.68，pCH3OH (g)pDSDS3DS2DS1CH3OH (l)16.59 kPaCH3OH (g)16.59 kPaT=298K15.先計(jì)算CH3OH(g)的Sm(g)：DS10DS=DS1+DS2+DS3=112.39 JK1mol1Sm(g)= Sm(l)+DS=239.39 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)DfHm(Jmol1)-110.520-200.7DrHm=90.18 kJmol1Sm(JK1mol1)197.67130.68239.39DrSm=219.64 JK1mol1DrGm=DrHmTDrSm=90.18298.15(219.64)103=24.68 kJmol116.CH3COOH(l) + C2H5OH(l) = CH3COOC2H4(l) + H2O(l)DfGm/kJmol1395.8175.1341.1236.6DrGm = S(DfGm)產(chǎn)物S(DfGm)反應(yīng)物 = (-341.1236.6)(395.8175.1) = 6.8 kJmol117Kp,2=4.305106 Pa1