The mechanism of the reaction 2NO_2 + F_2 to | vedclass

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(C) The rate-determining step is the slow step,which is $(i)$ $NO_2 	o NO + O$.The rate law is determined by this slow step: $Rate = k[NO_2]^1$.Since

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The reaction is zero order with respect to $F_2$.

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(C) The rate-determining step is the slow step,which is $(i)$ $NO_2 	o NO + O$.The rate law is determined by this slow step: $Rate = k[NO_2]^1$.Since the rate depends only on the concentration of $NO_2$,the order of the reaction is $1$.For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k}$,which is independent of the initial concentration of the reactant.Therefore,the reaction is zero order with respect to $F_2$ because the rate expression does not contain $[F_2]$.

The mechanism of the reaction $2NO_2 + F_2 \to 2NO_2F$ is given by:
$(i)$ $NO_2 \xrightarrow{slow} NO + O$
$(ii)$ $F_2 + O + NO \xrightarrow{fast} NO_2F + F$
$(iii)$ $F + NO_2 \xrightarrow{fast} NO_2F$
Select the correct statement.

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The mechanism of the reaction $2NO_2 + F_2 \to 2NO_2F$ is given by:$(i)$ $NO_2 \xrightarrow{slow} NO + O$$(ii)$ $F_2 + O + NO \xrightarrow{fast} NO_2F + F$$(iii)$ $F + NO_2 \xrightarrow{fast} NO_2F$Select the correct statement.