1现代基础化学第二版课后答案第三版也可用化学工业出版社

现代基础化学答案 化学工业出版社第一章，习题答案，P44-461. E = -B/n2 (B= 2.18 10-18 JDE = -B/52 + B/22 = 4.58 10-19 Jhv = DE, ln = cl = c h DE = 434 nm3不存在；存在4Ni 3s2 3p6 3d8 4s2的四个量子数是：3s2: n= 3, l=0, m=0, ms = +1/2; n= 3, l=0, m=0, ms = -1/23p6: n= 3, l=1, m=-1, ms = +1/2; n= 3, 1=1, m=-1, ms = -1/2; n= 3, 1=1, m=0, ms = +1/2n= 3, 1=1, m=0, ms = -1/2; n= 3, 1=1, m=1, ms = +1/2; n= 3, 1=1, m=1, ms = -1/23d8: n= 3, 1=2, m=-2, ms = +1/2; n= 3, 1=2, m=-2, ms = -1/2; n= 3, 1=2, m=-1, ms = +1/2n= 3, 1=2, m=-1, ms = -1/2; n= 3, 1=2, m=0, ms = +1/2; n= 3, 1=2, m=0, ms = -1/2n= 3, 1=2, m=1, ms = +1/2; n= 3, 1=2, m=2, ms = +1/24s2: n= 4, 1=0, m=0, ms = +1/2; n= 4, 1=0, m=0, ms = -1/25 3d 4f 2s 3d 1s 3p; 结果： = 3；3d l = 1；2p m= -1 /0/+1; 3p ms = +1/2 或-1/2, 4s948Cd 4s24p64d105s2.10有三个19K；24Cr；29Cu均第四周期, K: s区，IA族；Cr: d区VIB族；Cu: ds区IB族11. (1) Ar: 1s22s22p63s23p6 (2)Fe: 1s22s22p63s23p64s23d6 (3) I: 1s22s22p63s23p63d104s24p64d105s25p5 ; (4) Ag: 1s22s22p63s23p63d104s24p64d105s112(1)Ca: s区, 第四周期，IIA族；(2) Cl: p区，第三周期，VIIA；(3) Ti: d区，第四周期，IVB；(4) Hg: ds区, 第六周期, IIB族 Ca: +2，Cl: +7，Ti: +4，Hg: +2 (1)(3)(4)(2)13铁原子3d64s2；26Fe1448；Kr4d105s2, IIB族；48Cd15甲：3s23p5，VIIA，非金属，电负性高；乙：3d24s2，IVB，金属，电负性低。16Cl；Cs；Be；Cl；Mn1922Ti3+ Ar3d1 1个未成对；24Cr2+ Ar3d4 4个未成对；27Co3+ Ar3d6 4个未成对；46Pd2+ Kr4d8 2个未成对；58Ce3+ Xe4f1 1个未成对；57La3+ Xe5d0 无未成对；第二章习题答案，P81-821 BeH2: sp杂化，直线型 SiH4: sp3杂化，正四面体。 BF3: sp2杂化，平面三角形 CO2: sp杂化，直线型。2BF3采用sp2杂化；BF4-采用sp33H2O sp3不等性杂化；BeH2 sp杂化4H2O sp3不等性杂化，104.5度，CH4 sp3杂化，109.5度；CO2 sp杂化，180度。11H2色散O2色散H2O色散诱导取向氢键H2S色散诱导取向H2S-H2O色散诱导取向H2O- O2色散诱导HCl-H2O色散诱导取向CH3Cl色散诱导取向12乙醇高，有氢键 甲、乙、丙醇分子结构相似，分子量依次增大，色散力递增，熔沸点依次升高。丙三醇分子量大，熔沸点高，HF有氢键，熔沸点高。13CH4,CCl4,CI4分子量递增，色散力增大，熔沸点依次升高。H2O有氢键，CH4无氢键。第三章习题答案，P1191NaF AgBr NH3 O2离子晶体 过渡晶体 含氢键20.225 r+/r- (ZnS) 0.414, 0.414 r+/r (NaCl) 0.732, 0.732 r+/r (CsCl) Sn2+ Fe2+ Sr2+ S2-O2-F- 第四章习题答案，P1511配合物中心离子配体配位数配离子电荷名称Cu(NH3)4(OH)2Cu2+NH3 42+氢氧化四氨合铜（II）CrCl(NH3)5Cl2Cr3+Cl-,NH3 62+二氯化一氯。五氨合铬（III）CoCl(NH3)(en)2Cl2Co3+Cl-, NH3,en 62+氯化二氯一氨二（乙二氨）合钴(III)Pt Cl2(OH)2(NH3)2Pt4+Cl-,OH-, NH3 60二氯二羟基。四氨合铂（IV）Ni(CO)4NiCO 40五羰基合镍K3Fe(CN)5(CO)Fe2+CN-,CO 63-五氰根。一羰基合铁（II）酸钾2配合物中心离子配体配位数配离子电荷名称(NH4)3 SbCl6Sb3+Cl- 63-六氯合锑（III）酸铵CoCl(NH3)5Cl2Co3+Cl-, NH3 62+氯化一氯五氨合钴(III)Pt Cl(OH)(NH3)2Pt2+Cl-,OH-, NH3 40一氯一羟基。二氨合铂(II)Cr(NH3)2(H2O)43+Cr3+NH3H2O 63+二氨四水合铬（III）PtCl3(OH)(NH3)2Pt4+Cl-,OH-, NH3 60三氯。一羟基。二氨合铂(IV)3名称化学式内界外界单基配体多基配体1CoCl2(NH3)3(H2O)ClCoCl2(NH3)3(H2O)+Cl-Cl-,NH3,H2O2Na2Ca(EDTA)Ca(EDTA)2-Na+EDTA3(NH4) Cr(SCN)4 (NH3)2 Cr(SCN)4 (NH3)2- NH4+NH3，SCN-4Cr(OH)3(H2O)(en)Cr(OH)3(H2O)(en)无OH-，H2Oen5K2PtCl6PtCl62-K+Cl-4. Co (NH3)6Cl3； CoCl(NH3)5Cl2； CoCl2(NH3)4Cl； CoCl3(NH3)38. Ni(CN)42-为dsp2杂化， Ni(NH3)42+为sp3杂化， Fe(CN)64-为d2sp3杂化，无单电子；FeF63-为sp3 d2杂化，有五个单电子。9. 名称磁矩电子分布空间构型内外构型FeF63-5.90sp3 d2 正八面体外轨型Fe(CN)64-0d2sp3 E E E - -正八面体内轨型Fe(H2O)62+5.30sp3 d2 E 正八面体外轨型Co(NH3)63+0d2sp3 E E E - -正八面体内轨型Co(NH3)62+4.26sp3 d2 E E 正八面体外轨型Mn(CN)64-1.80d2sp3 E E - -正八面体内轨型第五章习题答案1. W = -P(V2 - V1)=-P2(P1V1/P2 - V1)=-1105(5 2 10-3 2 10-3)=-800J2. U=Q+W=Q+p(V2-V1)=6.48+-93.3(150-50)=4.68+(-9.33)=-2.85kJ5. H = Qp = 20 858 = 17.16 103 J = 17.16 kJW = -PoutV = -101325 20 607 10-6 = -1230J= -1.230 kJU = Qp + W = 17.16 1.23 = 15.93 kJ7. 2(2) (1) - 5/2 (3)得，fHm (C2H2) = 90.2 2 -(-1255.6) - 483.7 2.5 = 226.75 kJ/mol8. 1/2+- rHm= 0.5-234.7-184.6-30.59+285.85= -127.02kJ/mol9. cHm= 2fHm(CO2) + 3fHm(H2O) -fHm(C2H5OH) = 2(-393.51) + 3(-285.85) (-276.98) = -1367.59 kJ/mol；92 克乙醇燃烧放出的热量为因为92/46 = 2mol, rHm=2(-1366.75)= -2733.5 kJ12. fHm(C4H10) = 4fHm(CO2) + 5fHm(H2O) cHm(C4H10) = 4(-393.51) + 5(-285.85) - (-2878.51) = -124.78 kJ/mol七 4. rGm=rHm-TrSm= -91.92-298.15(-198.3)10-3=-32.79kJ/mol rGm=rHm-TrSm=181.42-298.15.216=117.0kJ/mol rGm=rHm-TrSm= -890.36-298.15(-0.243)= -817.9kJ/mol6. rGm=(-569.6)+2(-228.59)-(805.0)-4(-95.27)= 159.5kJ/mol rGm= -394.38-(-137.3)-(228.59)= -28.49kJ/mol rGm= 3(-394.38)-(-741.0)-3(-137.30)= -30.24kJ/mol7. K25=exp-(rHm-TrSm)/RT =exp-(-211.5103-298.15116.9)/(8.314298.2)=exp99.384=1.451043 K100=exp-(-211.5103-373.15116.9)/(8.314373.15)=exp82.23=5.17510358. rGm= -RTlnK= -8.3143500ln8.28= -61510.7J/mol= -61.51kJ/mol10. rGm=37.22-298.150.09496=8.908kJ/mol0； T1rHm/rSm=37220/94.96= 319.95KrGm=299.8-298.150.3633= 191.48kJ/mol0； T2rHm/rSm=299.8/0.3633= 825.2K 第一种好。17. rGm=-402-298.150.1891= -345.6kJ/mol 0 正方向进行rGm=rHm-TrSm因rSm0，T，rGm也，降低温度有利于反应进行。反应正向进行的最低温度：T=rHm/rSm=-402103/-189.1= 2125.9K八. 思6 条件PAP总TCatalyst速率k不变不变6. v=kCaCc2 三级； k=0.05/(0.010.012)=5104(mol/L)-2.s-1 v=51040.53=62507. C0/C=v0/v=110-5/3.2610-6=3.067; k=(1/t)ln(C0/C)=(1/3600)ln3.067=3.1110-4/s t1/2=ln2 /k=0.693/k=0.693/3.1110-4=2226.26s C=v/k=10-5/3.1110-4=0.03212mol/L10. k=Ae-Ea/RT Ea=RT1T2ln(k2/k1)/(T2-T1)=(8.314300310ln2)/10=53594J=53.59kJ15. k=v/(CA2CB)=2.510-3/0.013=2500(mol/L)-2.s-1 v=25000.01520.030=0.01688mol/L.s16. Q=176-12=164kJ/mol； Ea+=176kJ/mol；Ea-=12kJ/mol 在曲线最高处有活化络合物存在。 是放热反应九. 1ChemicalskindConjugate baseConjugate acidChemicalskindConjugate baseConjugate acidHCOOHCHCl2COOHH2C2O4 H2OH2PO4- AAADD HCOO-CHCl2COO-HC2O4- HO-HPO42-H3O+H3PO4HPO42- HS-CO32- CN-DDBBPO43- S2-H2PO4- H2SHCO3- HCN2. ChemicalsConjugated acid equilibrium constantsConjugated base equilibrium constants S2-Ka2(H2S) = 1.26 10-13HCO3-Ka1(H2CO3) = 4.47 10-7Kw / Ka2 = 2.14 10-4H2PO4-Ka1(H3PO4) = 6.92 10-3Kw / Ka2 = 1.62 10-7HPO42-Ka2(H3PO4-) = 6.17 10-8Kw / Ka3 = 2.09 10-2Ac-1.74 10-53.x = 4.03 x 10-2H+ = SO42- = 4.03 x 10-2 mol / LHSO4- = 0.2 x = 0.16 mol/L4. H+=HCO3-=(0.0254.3610-7)1/2=1.0610-4 mol/L CO32-=4.6810-11; H2CO3=0.0256. pKa1 = 2.16, pKa2 = 7.21, pKa3 = 12.32 pH = 1, pH H2PO4- HPO42- PO43- pH = 5, pKa1 pH H2PO4- H3PO4 PO43- pH = 10, pKa2 pH PO43- H2PO4- PO43- pH = 14, pH pKa3, PO43- HPO42- H2PO4- H3PO4 7. HCl=0.025, pH=1.61 mol/L NH4Cl=0.05; H+=(0.0510-14/1.7410-5)1/2=5.3510-6; pH=5.27HAc=0.05; NaAc=0.05; H+=KCa/Cb=1.7510-5; pH=4.76HAc=0.05; NaAc=0.05; H+=KCa/Cb=1.7510-5; pH=4.7610. H+=KCa/Cb=10-4.5=3.1610-5; CNaAc=0.45mol/L; 0.450.5136 = 30.6g;HAc: 0.82500/6 = 68.3ml11. H+=1.7510-50.02/0.8=4.3810-7; pH=6.36HAc=(0.110/55)+0.15/55=0.027; NaAc=0.8-0.15/55=0.718; H+=1.7510-50.027/0.718=6.5810-7; pH=6.18 pH=0.18HAc=(0.110/55)-0.15/55=0.009; NaAc=(0.140/55)+0.15/55=0.736; H+=1.7510-50.009/0.736=2.14810-7; pH=6.67 pH=0.3112. pH = pKa + lg(Ac- / HAc) = 2.68(1) pH = pKa + lg(Ac- / HAc) = 2.38(2) pH = pKa + lg(Ac- / HAc) = 2.8613. K = Ka2/Kw = 4.68 103; K = Ka2/Kw = 6.17 106;K = Ka1/Kw = 1.41 1012;K = Ka2/Ka3 = 1.29 10515. 1.3710-4 / 811.5 = 1.68810-7; Ksp = (3 1.688 10-7)3 (2 1.688 10-7)2 = 1.48 10-3216. PbSO4, Pb2+ = = = 1.59 10-4 mol / L PbS, Pb2+ = = = 2.83 10-14 mol / L17. Pb2+=0.0050; I-=0.005; Pb2+I-2 = (0.0050)3 = 1.3 10-7 Ksp(9.8 10-9)；有沉淀。Ba2+ = 0.020; CO32- = 0.30; Ba2+CO32- = 0.0200.30 = 6.0 10-3 Ksp(2.5810-9)；有沉淀。Ag+ = 0.010; Cl- = 0.10; Ag+Cl- = 0.0100.100 = 1.010-3 Ksp(1.7710-10)；有沉淀。OH-=5.010-5; Mg2+OH-2=10-7 (5.010-5)2=2.510-16 Ksp, 产生Mg(OH)2沉淀不产生Mg(OH)2沉淀的条件是Mg2+OH-2Ksp, OH- Ksp,PbS(810-28); 有沉淀。十. 1 N = 0.0406 mol; t = nZF / I = (0.0406X2X96485/5) = 1567 s = 0.44 h4. 2KMnO4+5K2SO3+3H2SO4=6K2SO4+2MnSO4+3H2O 5NaBiO3(s) + 2MnSO4 + 16HNO3 = 2HMnO4 + 5Bi(NO3)3 + 2Na2SO4 + NaNO3 + 7H2O 4Zn + NO3- + 10H+ = 4Zn2+ + NH4+ + 3H2O 3Ag+NO3-+4H+=3Ag+NO+2H2O Cl2+2OH-=Cl-+ClO-+H2O 8Al+3NO3-+5OH-+18H2O=8Al(OH)4-+3NH35 (a) 氧化剂：Ag+, 还原剂：Cu -)CuCu2+(c1) |Ag+ (c2) Ag (+(b) 氧化剂：Pb2+, 还原剂：Cu -) Cu,CuSS2-(c1) |Pb2+ (c2) Pb (+(c) 氧化剂: H+,还原剂: Pb 原电池: -) Pb , PbCl2 | Cl- (c1) | H+ (c2)| H2 , Pt (+6 最强氧化剂最强还原剂1.MnO4-(H+) 1.5VFe2+ 0.771V2.Cr2O72- 1.23Cr(OH)3 -0.133.Fe3+ 0.771Fe -0.4477 能共存，不能共存8 ECu2+/Cu=0.342+0.0296lg0.5=0.333V E=0.342+0.0296lgKsp1/2=-0.179VECuS/Cu=0.342+0.0296lgKsp=-0.700V9. E = -0.0592V E = -0.414 V E = -0.17 V10. 正极：Cu, 负极：Zn正极：Cu2+ + 2e Cu, 负极：Zn - 2e Zn2+电池反应不：Zn + Cu2+ = Zn2+ + CuE=0.342-0.760+0.0296lg0.001 = 0.342-(-0.849) = 1.19 V正极：Pb, 负极：Cu正极：Pb2+ + 2e Pb, 负极：Cu + S2- - 2e CuS电池反应：Cu + Pb2+ + S2- = Pb + CuSE=-0.126+0.0296lg0.1-0.342+0.0296lg(6.3X10-36/0.1)=-0.156-(-0.670)=0.514V正极：Hg, 负极：H2, Pt正极：Hg2Cl2 + 2e 2Hg + 2Cl-, 负极：H2 - 2e 2H+电池反应：H2 + Hg2Cl2- = 2Hg + 2 HClE = 0.268 + 0.327 0 = 0.327 V正极：H2, 负极：Zn正极：2H+ 2e H2, 负极：Zn - 2e Zn2+电池反应：Zn + 2H+ = Zn2+ + H2E=0.05916lg(cKa)1/2-0.760+0.296lg0.1 = -0.17-(-0.790) = 0.62 V11. E=0.799-0.0592 - (-0.76) - 0.0296lgZn2+ = 1.51 Zn2+ = 0.545 V12. E = -0.1375 + 0.1262 - lgKsp = 0.22; Ksp=1.53 10-813. 和正向进行，反向进行14. K=1.41X1041K=0.149 K=2.97X10-5 K=2.14X106216. EBrO-/Be- =0.76; EBrO3-/Br-=0.5217. (3X1.679+2x)/(3+2)=1.507V x=1.249V18. G = -ZFE = -6X96500X(1.23-1.36)X10-3=72.9kJ/molECr2O72-/Cr3+=1.23+(0.0592/6)lg1214; ECl2/Cl-=1.36+0.0296lg12-2=1.29VE=1.38-1.29 = 0.09V19. G = -ZFE=-2X96500X(0.222-0.151)X10-3 = -13.7 kJ/mol20. (1) E=0.236；G=-22.7 kJ/mol;K=9.32X103(2) E=0.236, G=-45.34 kJ/mol;K=8.96X107 十一.1. b6 = 1.0 X 1020, K不稳=1.0X10-20 2. Ag+ = 9.92X10-8 mol/L; NH3=0.3 mol/L; Ag(NH3)2+ = 0.1 mol/L3. 使AgCl溶解，NH3=2.25+0.2=2.45mol/L; . 使AgI溶解，NH3= + 0.2 = 3.23 X 103mol/L, 氨水不能使AgI溶解；使AgI溶解，CN- = 3.0 X 10-4 + 0.2 = 0.2 mol/L;4. (1) Ag+ = 2.85 X 10-9 mol/L; NH3 =1.12 mol/L; Ag(NH3)2+ = 0.04 mol/L(2) Ag+Cl- = 2.85 X 10-10 Ksp, 有AgCl沉淀产生。（3）Ag+ Ksp/ Cl- , Ag+ 1.42 mol/L, NH3始 1.42 + 0.08 = 1.50 mol/L, VNH3 = 1.50 X 100 12 = 12.5 mL.5. Cu2+ = 0.05 / (2.84 X Ks) = 3.89 X 10-17 mol/L;Cu(NH3)42+ = 0.05 mol/L; NH3 = 2.8 mol/LOH- = 0.2 1 21 = 9.52 10-3 mol/LCu2+ OH-2 = 3.89 X 10-17 X (9.52 X 10-3)2 = 3.53 X 10-21 HCl HBr HI, F, Cl, Br, I得电子能力依次减弱。（2）解离能：Cl2 Br2 I2, 原因：它们的化学键能依次减小。（3）沸点：He Ne Ar Ke Xe Rn, 原因：分子晶体，随原子量的增大，分子间作用力依次增加。（4）还原性: HF HCl HBr PH3 AsH3 9. (1) C + H2O = CO + H2 (2) Si + 2NaOH + H2O = Na2SiO3 + 2H2(3) 2B + 3H2SO4 = 2H3BO3 + 3SO2(4) 2HBr + H2SO4 = Br2 + SO2 + 2H2O(5) 3FeS2 + 12C + 8O2 = 3Fe3O4 + 12 CO + 6S(6) 2NaCl + 2H2O = Cl2 + 2NaOH + H2 (电解)十三.7. (3)9. 解：Cr3+ + 3OH- = Cr(OH)3 10-5 x Cr3+ x3 = Ksp = 6.3 10-31 x = OH- = 3.98 10-9 mol/L pH = 14 pOH = 14 8.40 = 5.60 11.（1）2K2CrO4 + H2SO4 = K2Cr2O7 + K2SO4 + H2O(2) K2Cr2O7 + 4AgNO3 + H2O = 2Ag2CrO4 + 2 KNO3 + 2HNO3(3) K2Cr2O7 + 6FeSO4 + 7H2SO4 = 3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 7H2O(5) 2CrCl3 + 10NaOH + 3H2O2 = 2Na2CrO4 + 8H2O + 6NaCl(6) 错误！链接无效。3H2O2+ 4H2SO4 = Cr2(SO4)3 + K2SO4 + 7H2O + 3O215. (1) 2Mn(NO3)2 + 5PbO2 + 6HNO3 = 2HMnO4 + 5Pb(NO3)2 + 2H2O(2) MnO2 + 4HCl （浓）= MnCl2 + Cl2 + 2H2O(3) 3MnO2 + 6KOH + KClO3 = 3K2MnO4 + KCl + 3H2O(4) 2K2MnO4 + Cl2 = 2KMnO4 + 2KCl(5) 3K2MnO4 + 4HAc = 2KMnO4 + MnO2 + 4KAc + 2H2O(6) 2KMnO4 + 16HCl = 2MnCl2 + 5Cl2 + 2KCl + 8H2O(7) 2KMnO4 + 3Na2SO3 + H2O = 2MnO2 + 3Na2SO4 + 2KOH(8) 2KMnO4 + 3KNO2 + H2O = 2MnO2 + 3KNO3 + 2KOH十五。2加成聚合反应和缩合聚合反应，其中加成聚合又分为均聚反应和共聚反应。加聚反应：没有小分子物质生成。缩聚反应：有小分子物质生成。5Tg玻璃化温度，Tf黏流化温度，Tg Tf 范围越宽，高弹态适用的温度范围越宽。11.（1）聚氯乙烯（2）尼龙66 （3）丁苯橡胶（4）丁腈橡胶13. 适合于做塑料的有：（1）聚氯乙烯、（2）尼龙-66、（4）聚苯乙烯和（6）聚酰胺因为它们在室温下是玻璃态。适合于做弹性材料的有：（3）天然橡胶、（5）聚异丁烯、因为它们在-70oC 122oC或更宽的温度范围处于高弹态。14