“…For the sake of completeness, numerical calculations show that (i) the reactions (5), (6), (10), and (12) tend rapidly towards a steady state of quasiequilibrium; (ii) the amounts of the side products Cl 2 , CHClFCH 3 , CCl 3 H, and C 2 Cl 6 formed in steps (10) to (13), respectively, are at least 10 3 times lower than that of the reaction product CClF " CH 2 ; and (iii) CCl 4 is almost not consumed in the course of the reaction.…”