A) increases the rate of reaction for a brief period of time
B) decreases the rate of reaction for a brief period of time
C) does not affect the rate of reaction
D) completely stops the reaction
Correct Answer: B
Solution :
During chlorination of methane, a small amount of oxygen acts as radical inhibitor. \[{{O}_{2}}\] combines with \[^{*}C{{H}_{3}}\] free radical to form methyl peroxy free radicals which are much less reactive than \[C{{H}_{3}}\] free radicals and hence the reaction slows down. After the inhibitor has been consumed, the reaction proceeds normally. \[\underset{\begin{smallmatrix} methyl \\ radical \end{smallmatrix}}{\mathop{^{*}C{{H}_{3}}+O-O}}\,\xrightarrow{{}}\underset{\begin{smallmatrix} methyl \\ peroxy\,radical \end{smallmatrix}}{\mathop{C{{H}_{3}}-O-{{O}^{\bullet }}}}\,\]
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