Chemistry - Chemical Kinetics Question with Solution | TestHub

${\mathrm{N}}_2{\mathrm{O}}_5\to 2\mathrm{N}{\mathrm{O}}_2+\frac{1}{2}{\mathrm{O}}_2$
As$\frac{-\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{d}\mathrm{t}}=6.25\times 10^{-3}\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$\frac{-\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{d}\mathrm{t}}=\frac{1}{2}\frac{\mathrm{d}\left[\mathrm{N}{\mathrm{O}}_2\right]}{\mathrm{d}\mathrm{t}}=\frac{2\mathrm{d}\left[{\mathrm{O}}_2\right]}{\mathrm{d}\mathrm{t}}$
$\frac{\mathrm{d}\left[\mathrm{N}{\mathrm{O}}_2\right]}{\mathrm{d}\mathrm{t}}=\frac{2\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{d}\mathrm{t}}=2\times 6.25\times 10^{-3}=1.25\times 10^{-2}\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
As$2\frac{\mathrm{d}\left[{\mathrm{O}}_2\right]}{\mathrm{d}\mathrm{t}}=\frac{\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{d}\mathrm{t}}$
$\frac{\mathrm{d}{\mathrm{O}}_2}{\mathrm{d}\mathrm{t}}=\frac{6.25\times 10^{-3}}{2}=3.125\times 10^{-3}$$\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$=3.125\times 32\times 10^{-3}$$\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$