C= k一C12 H2 Cl2 2HCI ka+k2CH2 (1)C2 2C1 (2)C1+H2 HCI+H (3)H+C2 HCI+CI ka (4)C1+C 1→Cl2 C1,H为中间物 dCa=kCo-k.CcCm+kCuCo.-k.C-0 dt dCu-kCeCn,-kyCuCc.=0 dt 4 2 2 3 2 2 1 I H I C k k C k C H2 + Cl2 → 2HCl (1)Cl2 → 2Cl (2)Cl + H2 → HCl + H (3)H + Cl2 → HCl + Cl (4)Cl + Cl →Cl2 k1 k2 k3 k4 Cl,H 为中间物 0 2 1 2 3 4 2 2 2 Cl Cl H H Cl Cl Cl k C k C C k C C k C dt dC 0 2 2 2 Cl H 3 H Cl H k C C k C C dt dC