For N₂O₅→ 2NO₂+1/2O₂, it is found that $\frac{-\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]={\mathrm{k}}_1\left[{\mathrm{N}}_2{\mathrm{O}}_5\right],\frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{NO}}_2\right]={\mathrm{k}}_2\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$ $\frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{O}}_2\right]={\mathrm{k}}_3\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$then

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​.$$\begin{gathered} {\mathrm{N}}_2{\mathrm{O}}_5\leftrightarrow 2{\mathrm{NO}}_2+\frac{1}{2}{\mathrm{O}}_2\mathrm{\_\_\_\_\_}1 \\ -\frac{\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{dt}}={\mathrm{K}}_1\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]\mathrm{\_\_\_\_\_\_\_}2 \\ \frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{dt}} ={\mathrm{K}}_2\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]\mathrm{\_\_\_\_\_\_\_\_}3 \\ \frac{\mathrm{d}\left[{\mathrm{O}}_2\right]}{\mathrm{dt}} ={\mathrm{K}}_3\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]\mathrm{\_\_\_\_\_\_\_\_\_\_}4 \\ \mathrm{from} \mathrm{equation} \left(1\right) \\ -\frac{\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{dt}}=\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{dt}}=2\frac{\mathrm{d}\left[{\mathrm{O}}_2\right]}{\mathrm{dt}} \\ \mathrm{from} 1,3 \mathrm{and} 4 \\ {\mathrm{K}}_1\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]=\frac{1}{2}{\mathrm{K}}_2\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]=2{\mathrm{K}}_3\left[{\mathrm{N}}_2{\mathrm{O}}_5\right] \\ \mathrm{multiply} \mathrm{by} 2 \\ 2{\mathrm{K}}_1\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]={\mathrm{K}}_2\left[{\mathrm{N}}_2{\mathrm{O}}_5\right] =4{\mathrm{K}}_3\left[{\mathrm{N}}_2{\mathrm{O}}_5\right] \\ \mathrm{divided} \mathrm{by} \left[{\mathrm{N}}_2{\mathrm{O}}_5\right] \\ 2{\mathrm{K}}_1={\mathrm{K}}_2=4{\mathrm{K}}_3 \end{gathered}$$