Chemistry - Chemical Kinetics Question with Solution | TestHub

Key Idea Rate of disappearance of reactant $=$ rate of appearance of product $$\begin{array}{rl}& \text{ or }-\frac{1}{\text{ stoichiometric coefficient }}\frac{\mathrm{d}\text{ [reactant }]}{\mathrm{dt}}\\ & \text{ of reactant }\\ & =+\frac{1}{\text{ stoichiometric }}\frac{\mathrm{d}\text{ [product }]}{\mathrm{dt}}\\ & \text{ coefficient of product }\\ & \end{array}$$ For the reaction, $\begin{array}{rl}& {\mathrm{N}}_2{\mathrm{O}}_5(\mathrm{g})⟶2{\mathrm{NO}}_2(\mathrm{g})+\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})\\ & \frac{-\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{dt}}=+\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{dt}}\\ & =+\frac{2\mathrm{d}\left[{\mathrm{O}}_2\right]}{\mathrm{dt}}\\ & \therefore \frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{dt}}=-2\frac{\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{dt}}\\ & =2\times 6.25\times 10^{-3}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}\\ & =12.5\times 10^{-3}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}\\ & =1.25\times 10^{-2}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}\\ & \end{array}$ $$\begin{array}{rl}\frac{\mathrm{d}\left[{\mathrm{O}}_2\right]}{\mathrm{dt}}& =-\frac{\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{dt}}\times \frac{1}{2}\\ & =\frac{6.25\times 10^{-3}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}}{2}\\ & =3.125\times 10^{-3}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}\end{array}$$