Chemistry - Chemical Kinetics Question with Solution | TestHub

Step i: ${\mathrm{O}}_3(\mathrm{g})\underset{{\mathrm{k}}_2}{\stackrel{{\mathrm{k}}_1}{\rightleftharpoons }}{\mathrm{O}}_2(\mathrm{g})+\mathrm{O}(\mathrm{g})$ fast

Step ii: ${\mathrm{O}}_3(\mathrm{g})+\mathrm{O}(\mathrm{g})\stackrel{{\mathrm{k}}_3}{\to }2{\mathrm{O}}_2(\mathrm{g}{)}^{\text{slow }}$

Overall: $2{\mathrm{O}}_3(\mathrm{g})⟶3{\mathrm{O}}_2(\mathrm{g})$

$$-\frac{\mathrm{d}\left[{\mathrm{O}}_3\right]}{\mathrm{dt}}={\mathrm{k}}_3\left[{\mathrm{O}}_3\right][\mathrm{O}]$$From the fast step, we can write the equilibrium constant as:

$$\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{\left[{\mathrm{O}}_2\right][\mathrm{O}]}{\left[{\mathrm{O}}_3\right]}\Rightarrow [\mathrm{O}]=\frac{{\mathrm{k}}_1\left[{\mathrm{O}}_3\right]}{{\mathrm{k}}_2\left[{\mathrm{O}}_2\right]}$$

Substituting the expression for $[\mathrm{O}]$ into the rate law for the slow step:

$$-\frac{\mathrm{d}\left[{\mathrm{O}}_3\right]}{\mathrm{dt}}=\frac{{\mathrm{k}}_3{\mathrm{k}}_1{\left[{\mathrm{O}}_3\right]}^2}{{\mathrm{k}}_2\left[{\mathrm{O}}_2\right]}$$