Chemistry - Chemical Kinetics Question with Solution | TestHub

$\begin{array}{rl}& \frac{-\Delta \left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\Delta \mathrm{t}}=x\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}\\ & 2{\mathrm{N}}_2{\mathrm{O}}_{5(\mathrm{g})}\to 4{\mathrm{NO}}_{2(\mathrm{g})}+{\mathrm{O}}_{2(\mathrm{g})}\end{array}$ Average rate of reaction $\begin{array}{c}=-\frac{1}{2}\frac{\Delta \left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\Delta \mathrm{t}}=\frac{1}{4}\frac{\Delta \left[{\mathrm{NO}}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta \left[{\mathrm{O}}_2\right]}{\Delta \mathrm{t}}\\ \therefore \frac{\Delta \left[{\mathrm{NO}}_2\right]}{\Delta \mathrm{t}}=-\frac{4}{2}\frac{\Delta \left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\Delta \mathrm{t}}=\frac{4}{2}\times x\\ =2x\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}\end{array}$ $\therefore$ Average rate of formation of ${\mathrm{NO}}_{2(\mathrm{g})}$ $=2x\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$ [Note: In the question, $\frac{\Delta \left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\Delta \mathrm{t}}$ is changed to $\frac{-\Delta \left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\Delta \mathrm{t}}$ to apply appropriate textual concepts.]