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# The reaction $2\mathrm{NO}+{\mathrm{O}}_{2}\stackrel{k}{⟶}2{\mathrm{NO}}_{2}$ follows the mechanism (in rapid equilibrium)${\mathrm{N}}_{2}{\mathrm{O}}_{2}+{\mathrm{O}}_{2}\stackrel{{k}_{2}}{⇌}2{\mathrm{NO}}_{2}$ (slow)The rate law of the reaction will be

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a
$r=k\left[\mathrm{NO}{\right]}^{2}\left[{\mathrm{O}}_{2}\right]$ with $k={k}_{2}{K}_{\mathrm{eq}}$
b
$r=k{\left[{\mathrm{NO}}_{2}\right]}^{2}\left[{\mathrm{O}}_{2}\right]$ with
c
$r=k{\left[{\mathrm{NO}}_{2}\right]}^{2}\left[{\mathrm{O}}_{2}\right]$ with  $k={K}_{\mathrm{eq}}/{k}_{2}$
d
$r=k\left[\mathrm{NO}{\right]}^{2}\left[{\mathrm{O}}_{2}\right]$ with

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detailed solution

Correct option is A

Since the second step is slow, the rate law is $\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{NO}}_{2}\right]}{\mathrm{d}t}={k}_{2}\left[{\mathrm{N}}_{2}{\mathrm{O}}_{2}\right]\left[{\mathrm{O}}_{2}\right]$ From the fast equilibrium, we have ${K}_{\mathrm{eq}}=\frac{\left[{\mathrm{N}}_{2}{\mathrm{O}}_{2}\right]}{\left[\mathrm{NO}{\right]}^{2}}$ This gives $\left[{\mathrm{N}}_{2}{\mathrm{O}}_{2}\right]={K}_{\mathrm{eq}}\left[\mathrm{NO}{\right]}^{2}$Hence, $\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{NO}}_{2}\right]}{\mathrm{d}t}={k}_{2}{K}_{\mathrm{eq}}\left[{\mathrm{NO}}^{2}\left[{\mathrm{O}}_{2}\right]\right$

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