$=-\frac{1}{\text { stoichiometric coefficient of reactant }} \frac{\text {d[reactant]}}{\mathrm{d} t}$
$=+\frac{1}{\text { stoichiometric coefficient of product }} \frac{\text {d[product]}}{\mathrm{dt}}$
For the reaction
$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$
$-\frac{\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{dt}}=+\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{dt}}$
$=+\frac{2 \mathrm{d}\left[O_{2}\right]}{\mathrm{dt}}$
therefore, $\frac{\mathrm{d}\left[ \mathrm{NO}_{2}\right]}{\mathrm{dt}}=-2 \frac{\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{dt}}$
$=2 \times 6.25 \times 10^{-3}\, \mathrm{mol} \,L^{-1}\, \mathrm{s}^{-1}$
$=12.5 \times 10^{-3} \,\mathrm{mol} \,\mathrm{L}^{-1}\, \mathrm{s}^{-1}$
$=1.25 \times 10^{-2}\, \mathrm{mol} \,\mathrm{L}^{-1}\, \mathrm{s}^{-1}$
$\frac{\mathrm{d}\left[O_{2}\right]}{\mathrm{dt}}=-\frac{\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{d} t} \times \frac{1}{2}$
$\frac{6.25 \times 10^{-3} \,\mathrm{mol} \,\mathrm{L}^{-1} \,\mathrm{s}^{-1}}{2}$
$=3.125 \times 10^{-3} \,\mathrm{mol}\, \mathrm{L}^{-1}\, \mathrm{s}^{-1}$
$\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$
$23 \mathrm\ {sec}$ પછી જો વાયુઆનું કુલ દબાણ $200\ torr$ મળી આવેલ હોય અને ખુબજ લાંબા સમય બાદ $A$ નાં સંપૂર્ણ વિધટન પર $300\ torr$ મળી આવેલ હોય તો આપેલ પ્રક્રિયા નો વેગ અચળાંક ......... $\times 10^{-2} \mathrm{~s}^{-1}$ છે. [આપેલ : $\left.\log _{10}(2)=0.301\right]$