物理化学(复旦大学药学院)第三章习题答案.pdf

H H 第三章习题解答1.先求 H2O(g)H2O(l) 的相变热：J85593166lno1pRTGG2=0G3=Vl(po-3166)J75131669971018o3.)p(G=G1+G2+G3= 8557 JO(l)HO(g)H(g)(g)O21(g)H22omr22GGrGom(l)= rGom(g)+G= 228.57 8.557 = 237.13 kJ2. 反应C(s)+2H2(g)=CH4(g) rGom=19290 Jmol1摩尔分数0.8 0.1(1) T=1000K时，0980100019290expexpomro.RRTGKp156080102o.QpQopKop反应不会正向进行2) 设压力为p，当098080101o2.pp.Qp时，即p1.59po时，反应才能进行3. SO2Cl2(g) + SO2(g) Cl2(g)反应前压力 (kPa) 44.786 47.836平衡时压力 (kPa) x44.786-x47.836-xp总=x+(44.786-x)+( 47.836-x)=86.096 kPa x=6.526 kPa422(100)6.5266.526)8366.526)(47.(44.786)(1oo.pKKpp4. H2(g) + I2(g) 2HI(g)开始 (mol) 7.94 5.3平衡 (mol) 2947x.235x.x=0H2O(g) poH2O(l) poGG3 G2 G1 H2O(g) 3166 Pa H2O(l) 3166 Pa T=298K H H 15023529472o.x.x.xKKxpx=9.478 mol (另一根x=19.30舍去 )5. A(g) + B(g) AB(g)开始 (mol) 1 1平衡 (mol) 1-0.4 1-0.4 0.4 n总=1.6 molRTGppKKxpomrooexpRpp././.3008368exp616061401o2p=0.06206po=6206 Pa6. A(g) B(g)平衡压力10popo10101o.KprGom= RTlnKop=5708 JrGm(1)= rGomRTlnQop(1)0J398921ln2985708反应不会自发进行。

rGm(2)= rGomRTlnQop(2)0J17141005ln2985708反应自发进行7. N2(g) + 3H2(g) 2NH3(g)开始 (mol) 1 3平衡 (mol) 21x213xxn总=4-x xx42o32oo42334214ppx/xx/xxxppKKxp(1) 当总压p=10 po, =3.85% 时，x=0.1483mol , 代入上式得Kop=1.6410 4(2) 当=5% 时，x=0.1905mol，Kx=0.029112ooppKKxp2o402911010641pp.p=13.323 po =1332 kPa8. 0dl1pVGG2=0 G总4490 JCH3OH(l) poCH3OH(g) poGG3 G2 G1 CH3OH(l) 16343Pa CH3OH(g) 16343Pa T=298K H H 449034316100ln3.nRTGH H G总=fG om(g) -fGom(l)即 4.49= 161.96 -fGom(l)fGom(l)= 166.45 kJmol19. 先计算丁二酸（m=1 molkg ）的fGom：1)(OHC0.715)(饱饱和OHC0(纯纯态OHC46424641464mGmGfGom(m=1)= fGom(s)+G1+G2= 748099+0+71501ln.RT= 747268 Jmol1一级电离C4H6O4(m=1)C4H5O4(m=1) + H+(m=1)fGom(kJmol1) 747.268 723.037 0 rGom=24.23 kJmol15omr106965229824230expexp.R.RTGKm10. NH4HS(s) NH3(g) + H2S(g)平衡 (kPa) 66.662166.66211111033332oo.p.Kp(1) 原有 H2S，平衡p39.99+p11110)()9939(2oo.pp.pKp解得p=18.87 kPa p总=2p+39.99=77.73 kPa(2) 形成固体的条件：QpKop设 H2S(g)的压力为p11110)(66662.p.pop166.6 kPa11. A(g) + B(g) C(g) + D(g)(1) 平衡时 (mol) 1/3 1/3 2/3 2/3(2) 设 C 为x mol 1x2xxx(3) 设生成 C 为x mol 1x1x0.5+xx(4) 设 C 减少x mol xx1x2x由(1) =0 4313222o/KKxpH H 由(2) 4)2)(1(2oxxxKp 3x212x+8=0 x=0.845 mol由(3) 4)1 (502oxxx.Kp3x28.5x+4=0 x=0.596mol C总 量 =1.096 mol由(4) 4212oxxxKp3x2+3x2=0 x=0.457 mol C的量=1x=0.543 mol12. 设开始为 1 mol ，解离度为PCl5(g) PCl3(g) + Cl2(g)平衡时 (mol) 1n总=1+o2o2oo)1(1ppnppn/n/ppKKxp总总总）（当=0.5 ，n总=1.5 mol，p=po时，代入上式，可得平衡常数31opK(1) 降低总压p，使体积增加1 倍，计算的改变：222111nVpnVp1251121oVp.Vp可得压力o231ppp，代入：3131)1)(1()1(oo2o2op/pppnKp总得 =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：222111nVpnVp21o1o251nVp.Vp可得气体摩尔n总=n2=3 mol31)1( 3)1(oo2o2oppppnKp总得 =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：222111nVpnVp21o1o251nVp.Vp可得气体摩尔n总=n2=3 mol312)1( 3)1(oo2o2oppppnKp总得 =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=3 mol，设 Cl2的加入量x mol ，计算 Cl2总量：H H PCl5(g) PCl3(g) + Cl2(g)1-+xn总=1+x=3 nCl2=+x=2 mol3123)1(323ooopp/Kp得 =0.20 解离度减少13. (1) rGom= RTlnKop=989815754510349317T.RTJmol 17492575451034omr.R.HkJmol 16144omromromr.TGHSJk 1mol 1(2) T=573K时，125205739898expexpomro.RRTGKpI2+ 环戊烯 2HI + 环戊二烯开始0.5po0.5po平衡pp2(0.5pop) (0.5pop)12520)()50(4)(1o23ooo.pppp.pKKpp5212)50(423.pp解此三次方程，p=34.50 kPa(3) 同理，起始压力为10po时，12520)()5(41o23oo.ppppKp5212)500(423.pp解得p=418.20 kPa14 先计算反应前：mol1044442CO.n，mol49290610521000.R.RTpVn总，mol392901049290SH2.n(1) 610K时，xH2O=0.02 nH2O=0.49290.02=0.009859 molCO2(g) + H2S(g) COS(g) + H2O(g)平衡时 (mol) 0.09014 0.3830 0.009859 0.009859 =032o108152383000901400098590.KKxp(2) rGom(610K)= RTlnKop=29.78 kJmol 1(3) 620K时，xH2O=0.03 nH2O=0.49290.03=0.01479 molCO2(g) + H2S(g) COS(g) + H2O(g)平衡时 (mol) 0.08521 0.3781 0.01479 0.01479 =0H H 32o10790637810085210014790)620(.KKKxpH H 61016201610620lnorooRH)K(K)K(Kmpp解得rHom=276.9 kJmol 115. (1) FeO(s)+H2(g)=Fe(s)+H2O(g) 85190540460)1(o.KKxpH2O(g)=H2(g)+0.5O2(g) 713o1083151043)2(.Kp=+FeO(s)=Fe(s)+ 0.5O2(g) Kop(3)= Kop(1)Kop(2)=4.96710-7750oo109674)3(2.ppK.Op解得分解压pO2=2.4810-11 kPa(2) FeO(s)+CO(g)=Fe(s)+CO2(g) Kop(1)= ？CO2(g)=CO(g)+0.5O2(g) 612o1018311041)2(.Kp=+FeO(s)=Fe(s)+ 0.5O2(g) Kop(3)= Kop(1)Kop(2)=4.96710-7得Kp(1)=0.42计算 CO 的用量：FeO(s)+CO(g)=Fe(s)+CO2(g) 起始 (mol) x0平衡 (mol) x-1 1 42011(1)o.xKKxpx=3.38 mol16. 先计算 CH3OH(g) 的Som(g) ：S10451271529838000omV2.THS935145916100lnln123.RppRSS=S1+S2+S3=112.51 JK 1mol 1Som(g)= Som(l)+S=239.3 JK 1mol 1计算反应的热力学函数和平衡常数：CO(g) + 2H2(g) CH3OH(g)fHom(Jmol1) -110.525 0 -200.67 rHom=90.145 kJmol 1CH3OH(l) Som=126.8，poCH3OH (g) poSS3 S2 S1 CH3OH (l) 16.59 kPa CH3OH (g) 16.59 kPa T=298K H H Som(JK1mol1) 197.674 130.684 239.3 rSom= -219.74 JKH H 1mol 1rGom=rHomTrSom= 90.145 298.15( 219.74)103= 24.63 kJmol 14omro1006721529824630expexp.R.RTGKp17. CH3COOH(l) + C2H5OH(l) = CH3COOC2H4(l) + H2O(l)fGom/kJmol1 389.9 168.49 332.55 237.129rGom = (nfGom)产物(nfGom)反应物= (-332.55 237.129) ( 389.9 168.49) = 11.289 kJmol196942298314811189expexpom.RTGKx18 12r1211lnTTRHKK,p,p1000111001101891043ln352R.K,pKp,2=4.305106 Pa1欢迎下载，资料仅供参考! 。