物理化学总复习题答案.doc

答案：一、填空题1. 0.5kB= 0.5kC 2. 0.0693 min 3. (AgI)mnI-(n-x)K+ 4. K3 Fe(CN)6 5. E(Cl-/AgCl/Ag)(RT/F)ln 0.01 6. E(Pb2+/Pb)(RT/2F)lna7. 10 mol 8. 0.690 9. 降压, 通入惰性气体10. 降温, 增压 11. 2, 112. 2, 1 二、选择题1. C 2. A 3. C 4. C 5. A 6. B 7. A 8. C 9. B 10. D 11. C 三、计算题1解:(1) t1/2 = A 0/ (2k) k = A 0/ (2t1/2)A = A 0- kt = A 0- A 0t/(2t1/2)( A/A 0)100 = 1-t/(2t1/2)100 = 0已反应完,反应掉 100.(2) k = 0.6932/1000 = 6.93210-4 sln( A/A0) = - kt = - 6.932102000 = - 1.386A/A 0= 0.25(A0-A)/A 0100= (1-0.35)100 = 75(3) x/(1-x) = A 0kt = A 0 1/(A 0t1/2)t = t/t1/2 = 2x = 2/3 = 66.7 2. 解: (1) k = (1/t)ln1/(1-x) = (1/10)ln1/(1-0.75) = 0.139 min当 t = 15 min时, ln1/(1-x) = kt= 0.13915 = 2.085x= 0.88 = 88 (2) k = (1/tAo)x/(1-x) = (1/10Ao)0.75/(1-0.75) = 0.3/Aok = (1/tAo)x/(1-x) = 0.3/Ao1/(15Ao)x/(1-x) = 0.3/Aox= 0.82 = 82 (3) k = xAo/t = 0.75Ao/10 = 0.075Aox= kt/Ao = 0.07515 = 1.13, A 已反应完.3. 解:R = 0.510-3 m = (1-0.8)o = 0.2o = 0.272.75 = 14.55 mN/m泡内压力p p0p p0 4/R 101.325414.5510-3/(0.510-2)10-3 101.337 kPa4. 解: 2M 272.810-31810-3r = = = 54.3 nm RTln(pr/p0) 11038.314293ln1.02p = p0p = p0(2/r) = 101.3272.810-3/(54.310-9)10-3 = 2.78103 kPa5. 解:(1) (-) Fe(s) Fe2+(a=1) + 2e (+) 2Fe3+(a=1) + 2e = 2Fe2+(a=1) 电池 Fe(s) + 2Fe3+(a=1) = 3Fe2+(a=1)(2) Fe3+(a=1) + 3e Fe(s) (a) Fe2+(a=1) + 2e Fe(s) (b) Fe3+(a=1) + e Fe2+(a=1) (c) (a) - (b) = (c), G(c) =G(a) -G(b)E(Fe3+/Fe2+) = 3E(Fe3+/Fe) - 2E(Fe2+/Fe) = 3(-0.04) - 2(-0.409) = 0.698 VE= E(Fe3+/Fe2+) - E(Fe2+/Fe) = 0.698 - (-0.409) = 1.107 V G= zEF = - 2964851.107 = - 213.6 kJ/molS= zF(E/T)p= 2964851.1410-3 = 220 J/(K.mol)H= G+ TS = - 213.6+29822010-3 = - 148 kJ/mol6. 解:(1)(-) Zn(s) Zn2+a(Zn2+) + 2e (+) PbSO4(s) + 2e Pb(s) + SO42-a(SO42-)电池 Zn(s) + PbSO4(s) = ZnSO4(m=0.01mol/kg,= 0.38) + Pb(s)(2) E = E- (RT/zF)lna(Zn2+)a(SO42-) = E- (RT/2F)ln(m)0.5477 = E(SO42-/PbSO4/Pb) - (-0.763) - (8.314298/296485)ln(0.010.38)E(SO42-/PbSO4/Pb) = - 0.359 V(3) (-) Pb(s) = Pb2+( a1) + 2e (+) PbSO4(s) + 2e = Pb(s) + SO42-( a2) 电池 PbSO4(s) = Pb2+( a1) + SO42-( a2)故可设计电池:Pb(s)Pb2+( a1)SO42-( a2)PbSO4(s)Pb(s)7. (1) V1= nRT/p = 18.314273/101.3 = 22.4 dm3W = nRT ln ( V1/V2) = 8.314273 ln (22.4/224) = - 5226 J 恒温变化, U = H = 0Q = - W = 5226 JS = Q/T = 5226/273 = 19.14 J/KS(环境) = - Q/T = - 19.14 J/KA = U - TS = - 27319.14 = - 5226 J G = A = - 5226 J(2) 终态压力为 10.13 kPa时,V2= nRT/p = 8.314273/10.13 = 224.0 dm此过程的始末状态与前一过程的始末状态相同,故U.H.G与S都与前一过程相应量相同.因 p(外) = 0, - W = 0Q = 0S(环) = - Q/T = 08. 解:(1)设始态气体 O2的体积为 V1,始态氮气的体积为 V2,由PV=nRT 可知， 0.2RT = 20.265V 0.8RT = 81.06V V1= V2 V(总) = 2V2= 2(0.2RT)/20265 p(总) = n(总)RT/V(总) = RT/2(0.2RT/20265) = 50663 Pa = 50.7 kPa(2) Q = 0, U = 0, W = 0, H = 0S = 0.2R ln ( 2V1/V1) + 0.8R ln ( 2V2/V2) = 5.76 J.KG = H - TS = - TS = - 2985.76 = - 1716 J9. 解:(1)由题可知: Gm(298K) =Gm(298K) = - 16.8 kJ.mol-1 故Gm =Gm + RT ln (p3/p)2/ (p1/p) (p2/p) = - 16.8 + 8.31410-3298 ln 10.02/ (10.010.0) = - 16.8 kJ.mol-12) ln Kp(298K) = -Gm(298K)/(RT) = 16.8103/ (8.314298) = 6.78Kp(298K) = 881(3) Gm(298K) = - (4) Kp(298K) = (p3/p)2/ (p1/p)(p2/p) = p32/ (p1p2) = 881 p3 = 18.9 Pa,且 p1 = p2 p1 = p2=0.637 Pa10. 解: p(H2O)/p a(Na2S2O7) p(H2O)Kp = = a(NaHSO4)2 p RT1T2 Kp(T2) RT1T2 p( H2O,T2)Hm = ln = ln T2-T1 Kp(T1) T2-T1 p( H2O,T1) 8.314413453 2.38 = ln = 8.23104J/mol = 82.3 kJ/mol 453-413 0.287lnKp(150) = ln Kp(140)Hm(T2-T1) / (RT1T2) = ln 0.2878.231010 /(8.314413423)= 0.681p(150) = 0.506 kPaKp(150) = p(150) / p= 0.506 / 101.3 = 5.010-3Gm = RT ln Kp(150) = -8.314423 ln (5.010-3) = 1.86104 J.mol= 18.6 kJ.molSm = ( H m -Gm) /T =(82.3-18.6)103/423 = 151 J.K-1.mol-111. 解:(1) p = pA+ pB= pA*x+ pB*( 1-xA) = 40.530.5 + 121.590.