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\[^{\bullet }C{{H}_{3}}\]free radical to form methyl peroxy free radicals which are much less reactive than\[^{\bullet }C{{H}_{3}}\]free radicals and hence the reaction slows down. After the inhibitor has been consumed, the reaction proceeds normally. \[\underset{\begin{smallmatrix} methyl \\ radical \end{smallmatrix}}{\mathop{^{\bullet }C{{H}_{3}}}}\,+O-O\xrightarrow{{}}\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\,methyl \\ peroxy\text{ }radical \end{smallmatrix}}{\mathop{C{{H}_{3}}-O-{{O}^{\bullet }}}}\,\]
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