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Well, consider this argument on the undoubtedly actual expanded system, which would commence with the action of strong UV on a Chlorine gas in presence of Methane, and seemingly would proceed based on random kinetics, as follows: \begin{align} \ce{Cl2 + UV &-> Cl^. + Cl^.}\\ \ce{CH4 + Cl^. &-> CH3^. + HCl}\\ \ce{CH3^. + Cl2 &-> CH3Cl + ...


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Your interpretation is interesting. As you state, the free radical $\ce{Cl·}$ is regenerated at the end of the reaction. It behaves like Vanadium(V) which oxidizes $\ce{SO2}$ into $\ce{SO3}$. And once reduced into Vanadium(IV), it gets reoxidized by $\ce{O2}$ into Vanadium(V), according to the two equations $$\ce{V2O5 + 2 SO2 -> 2 VO2 + 2 SO3}$$ $$\ce{2 ...


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