D02 所以 [a]=a,J=V0.54 =1.36P9 由 pV=nRT p=(n/V)RT c=PRTc为浓度 得平衡浓度[Pc,]= Pecls 100kPa RT 8.314kPa.L.K-.mol- =0.023mol.L 1.36×100kPa [Pc]=[Pa,]= RT 8.314kPa.L.K-.mol-x523K =0.031mol.L1 起始浓度 Cpa,=cal=0.031molL+0.023molL=0.054mol.L-   P P P P 1.36 0.54 [ ] [ ] 2 PCl Cl 3 2 所以 = = = 得平衡浓度 1 1 1 1 PCl PCl 0.023mol L 8.314kPa L K mol 100kPa [ ] 5 5 − − − − =     = = RT P P 1 1 1 1 PCl PCl Cl 0.031mol L 8.314kPa L K mol 523K 1.36 100kPa [ ] [ ] 3 3 2 − − − − =       = = = RT P P P 起始浓度 1 1 1 PCl Cl 0.031mol L 0.023mol L 0.054mol L 3 2 − − − c = c =  +  =  由 pV = nRT p = ( n / V ) RT c = P/RT c 为浓度