Consider the following reaction, $2{\mathrm{NO}}_2⟶2\mathrm{NO}+{\mathrm{O}}_2$
If rate of disappearance of NO₂ is $6.0\times {10}^{-12}\mathrm{mol} {\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$then the rate of appearance of O₂ is 

 

For the reaction, $2{\mathrm{NO}}_2⟶2\mathrm{NO}+{\mathrm{O}}_2$
Rate = $-\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{NO}\right]}{2\mathrm{dt}}=\frac{\mathrm{d}\left[{\mathrm{O}}_2\right]}{\mathrm{dt}}$
$\because -\frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{dt}}=6\times {10}^{-12}{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}$
$\therefore \frac{d\left[{\mathrm{O}}_2\right]}{dt}=-\frac{1}{2}\frac{d\left[{\mathrm{NO}}_2\right]}{dt}=\frac{1}{2}\times 6\times {10}^{-12}$
                    $=3\times {10}^{-12}\mathrm{mol} {\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$