Question

Addition of small amount of $\,{\left( {{{\mathbf{C}}_2}{{\mathbf{H}}_5}} \right)_4}{\mathbf{Pb}}\,$ to a mixture of methane and chlorine, starts the reaction at $\,{140^o}\,$instead of the usual minimum $\,{250^o}C\,$.
If true enter $\,1\,$ ,if false enter $\,0\,$ .

Answer 1

Here, actually the chlorination of the methyl group is taking place. This is actually a radical chain mechanism. So, we have to analyse the temperatures at which this reaction properly take place.

Complete step by step answer:
Let us first understand the chlorination of methane and following reactions;
Chloromethane is the product as chlorine and methane are combined. A chlorine atom has substituted one of the hydrogen atoms in methane, so this is a replacement reaction. The reaction does not end there, though, and all of the hydrogens in methane can be substituted by chlorine atoms in response. Substitution reactions arise in which chlorine atoms are substituted one at a time by hydrogen atoms in the methane.
The reaction continues through the process of the radical chain. Three stages describe the radical chain mechanism: initiation, propagation and termination. Initiation requires energy input, but the response is self-sustaining after that. One of the products from initiation is used in the first propagation step, and another is created in the second propagation step, so the loop can proceed forever.
Now, coming to the question;
Methane is chlorinated at $\,{250^o}C\,$ . The method of reaction was selected by free radicals. At lower temperatures, Tetraethyllead $\,\,{\left( {{{\mathbf{C}}_2}{{\mathbf{H}}_5}} \right)_4}{\mathbf{Pb}}\,\,$ dissociates and forms free radicals. That, in fact, induces free radical chlorine formation. For the formation of $\,C{l^*}\,$ free radical, the reaction does not require bond dissociation energy, and the halogenation reaction occurs at $\,{140^o}C\,$ instead of $\,{250^o}C\,$.
So, this statement is true hence, we can enter $\,1\,$ here.

Note:
After one chlorination, the chlorination of methane does not necessarily cease. In fact, having monosubstituted chloromethane can be very difficult. Di-,tri-and even tetra-chloromethanes are produced instead. The use of a much higher methane concentration in contrast to chloride is one way to prevent this issue. This lowers the risk that a chlorine radical can run into a chloromethane and start the mechanism to form a dichloromethane again.