普化原理第五章化学平衡

5.15.1Fe2O3+3CO=2Fe+ChapterChemical向进行，这种反应称为可逆反应(reverseN2O4(g)2NO2(g) N2O4(g)2NO2(g)应，才能建立化学平衡—平衡建立的前提；平衡保持不变—建立平衡的标志；1234001.aA(g)1.aA(g)+bB(g)gG(g)+hHKpGpgpa或KC A[A](pressureequilibriumconstantKc：浓度平衡常数(concentrationequilibriumconstant)。上述aA(g)+bB(g)gG(g)+hHpgKpGpaA(或量纲)：(压力单位)(g+h–aaA(aq)+bB(aq)gG(aq)+hHK[G][H[A](或量纲)：(浓度单位)(g+h–aKp与KcpVc=n/V=pgGHKC[G]g[H[RTRTppabGHRTpgABpaARTRT即：KKRT)ghab)KRT p ngas=0Kc=Kp（没有单位或KKRT气相反应：aAgbBggGg)+hHK (p/p)g(p/pGH(pA/p)(pB/p aAaqbBaqgGaq)+hHK G (c/c)(c/c(cA/c)a(cB/c多相反应：aAgbBaqgGg)+hH(p/p)(c/cK (pA/p)a(cB/cNO(g)2NO22KpNON2NO(g)NO22KPNO0.155p12NO(g)4NO22KNO6.462 )2NO22NO2N2O42NO2½½N2O4Kc= =Kc=[NO2]2/[N2O4]=Kc==Kc=[NO2]/[N2O4]½=例5-1：反应N2O4(g)2NO2(g)的Kc=0.024，则反应NO2(g)½N2O4(g)的Kc)1KcKK(RT)K(RT c6.45(0.0831318)c(mol/ R=5.1.3H2SO3H++HSO3Kc1=1.3x10-HSOH++SO33K=6.3x10-HSO+HSO2H+HSO+H2SO32H++SO3Kc=Kc 3 3 [H][HSO][H][SO2 [H2SO3 [HSO3=Kc1Kc28.2N2O(g)+½O22NO(g)Kc=N2(g)+½O2N Kc(2)=2.710+18N2(g)+O22 Kc(3)=4.710-31=[N][O K c[NO][O]½[N][O][N=[N2][O2]½= =1.710- K[CaO][CO]constant[COc[CaCO223固体的浓度项写在其中。即：KcCO2]或Kp=pCO2。平衡是均相的(homogeneous)。例如:CaCO3(s)CaO(s)+CO2(g)Cr2O72-(aq)Cr2O72-(aq)+H2O=2CrO2-(aq)+2H+4K ][H]/[Cr 2c42 例如：C2H5OH+CH3COOHCH3COOC2H5+H2OK[CHCOOCH[CHOH][CHCOOHquotient)Q:aA+bBgG+QpGpgpa或QCcgc AcAQCGcgcApgcacQpapGAQ>Q<当c(Ag+)=1.0010-2mol·dm-3，c(Fe2+)=0.100)Qc(Fe)c(Ag)0.1001.00102c(FeQK,Fe2+(aq)+Ag+(aq)Fe3+(aq)+起始cB/(mol·dm-3)变化cB/(mol·dm-3)- 平衡cB/(mol·dm-3)0.100-x1.0010-2-x1.0010-[Fe3K3.2[Fe2][Ag1.00103(0.100x)(1.001023.2x21.352x+2.210-3=x=1.610-3(moldm-3)[Ag+]=8.410-3mol·dm-3n0Bnn0解：设平衡时有xmolL-1的CO2生成，则有：CO+H2OH2+ 0.020 平衡(moldm-3)0.020-x0.020- (0.020x)(0.020x0.015100%GHTSUpV理想气体：S>SGGRTln例：N2(g)3H2(g)rG2GNHGNGGRTp2 p3)GRTlnr下的反应方向，已知1mol反应的G=-33.3kJ。a.pNH1.00bar,pN1.47bar,pHb.pNH1.00bar,pN1.00bar,pHQ NH 6.80pGGRTln33.3kJ/mol8.314298ln(6.80105)2b.QGGRTlnQG33.3kJ/对于任一气相反应：aAbBmMrG(mGMnGN)(aGAbGB(mGRTlnpmnGRTlnpnM N(aGRTlnpabGRTlnpbA B pm bG)RT paArGrGRTlnGGRTlnK GRTlnKGRTlnKRTlnQRTlnKQ<KG<0,正方向Q=KG=0,平衡态Q>KG>0,逆方向GfCHOH(lfCO(g2fH(g166kJ(137kJ)029kJ/molGGRTln1GRTpCO291048.314298 5.038104J/mol38kJ/mol)+ S(JK-1mol-GRTlnKGHT00[2(826)00]1000298[29042731.490106J8.314298lnKlnKKNi(aq)Ni(aq)6NH(aq)Ni(NH)33Ni2(aq)3en(aq)Ni(en)23enH2N-CH2-CH2-解：GRTlnK GHTHSOH++HSO 3KHSOH++SO 1K2HSO+HSO2H+HSO+H2SO32H++SO3K3GRTlnKGRTlnKGRTlnKGRTlnK1 3GG RTlnKRTlnKRTlnK123KKK 5.3N2(g)+3H2(g)Canthisbe5.5.3.1N2(g)+3H2(g)在上述的平衡体系中加入H2，浓度商Qc将H2，生成产物，直到达到新的因此，体系中[H2[N2]例5-8：CO变换反应COH例5-8：CO变换反应COH2OH2CO2，在773K时，Kc=9，如初始时CO的浓度是0.020molL-1，H2OCO+H2OH2+起始(moldm-3)0.0200.0800平衡(moldm-3)0.020-x0.080-xx(0.020x)(0.080x0.0194100%aA(g)+bB(g)gG(g)+hHpgKpGpaA>(xp)g(xpQpxpA)(xp x(gKxB5.3.25.3.2分压，使Qp增大，导致Qp>Kp，平衡向5.3.35.3.3GRTlnKHTlnKHSH1 RTRy=ax+b:HaR2.2104K8.314J/(Kmol)184kJ/SbR3.48.314J/(Kmol)28.3J/(Kmol缩，n0，平衡向气体分子数减小的方向移动；n=0，平衡不移动。例5-9：298K时反应N2O4(g)2(g)K解:解: N2O4(g)2NO2 1- 为离解度，平衡时总物质的量为1+pNO2=[2/(1+)]p总pN2O4=[(1-)/(1+)]p总K=(pNO2/p)2/(pN2O4/p)=42p总/[p(1当p总=p时，K=42/(1-2)==同理，当p总=0.5p时,K22120.155=pNO2=[2/(2ppN2O4=[(1)/(2p总K=4a2p总/[p(1-)(2+)]当p总p时K421)(2=0.255 lnK 1 lnK2 1 H11 RT2T1 8.314JK 550K900K 111lnK2K28.210HO(g)ClO(g)2HOCl(22KS=266.1J/(Kmol)；H2O(g)的fH=-242kJ/mol,S=189J/(Kmol)估算HOCl(g)的fH和S，(e)计算该反应在500K时的K值；(f)在25C，当pH2O=18torr,pCl2O=2.0torr,pHOCl=01.0torr。(b)OHOOClO2HO(c)GHTSHG06.010320.(J/K (d)H02 [1(242)1f 81(kJ/molf20.218(J/KmolG(500K)0500(20.)1.0104(JGRTlnK KGGRTln2GRTPHOPCl6.01038.314298214(kJ/molVan’t如果反应是放热的，则rH为负值，温度升高时(T2>T1)，K2<K1，即平衡常数随着如果反应是吸热的，则rH为正值，