dot{Cl} + CH_4 rightarrow A + BA and B in the | vedclass

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(C) The reaction between a chlorine radical $(\dot{Cl})$ and methane $(CH_4)$ is a propagation step in the free radical substitution mechanism.The chl

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$\dot{CH}_3$ and $HCl$

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(C) The reaction between a chlorine radical $(\dot{Cl})$ and methane $(CH_4)$ is a propagation step in the free radical substitution mechanism.The chlorine radical abstracts a hydrogen atom from methane to form a methyl radical $(\dot{CH}_3)$ and hydrogen chloride $(HCl)$.The reaction is: $\dot{Cl} + CH_4 \rightarrow \dot{CH}_3 + HCl$.Therefore,$A = \dot{CH}_3$ and $B = HCl$.

$\dot{Cl} + CH_4 \rightarrow A + B$
$A$ and $B$ in the above atmospheric reaction step are.

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$\dot{Cl} + CH_4 \rightarrow A + B$$A$ and $B$ in the above atmospheric reaction step are.