The chemical reaction, $2 \mathrm{O}_{3} ightarrow 3 \mathrm{O}_{2}$ proceeds as follows\ $\mathr

Question

The chemical reaction, $2 \mathrm{O}_{3} ightarrow 3 \mathrm{O}_{2}$ proceeds as follows\ $\mathrm{O}_{3} ightleftarrows \mathrm{O}_{2}+\mathrm{O}$ (fast)\ $\mathrm{O}+\mathrm{O}_{3} ightarrow 2 \mathrm{O}_{2}$ (Slow)\ The rate law expression should be\ i) This is treated as a complex reaction in kinetics\ ii) Rate $=\mathrm{k}[\mathrm{O}]\left[\mathrm{O}_{3}ight]$\ iii) the fast step is completely neglected in deriving the final rate law\ iv) Rate $=\mathrm{K}\left[\mathrm{O}_{3}ight]^{2}\left[\mathrm{O}_{2}ight]^{-1}$\ the correct statement/s is/are\

Professor PC Thomas Classes · Answer

Rate $=\mathrm{k}[\mathrm{O}]\left[\mathrm{O}_{3}ight]$ ( slow step ) From $$ \mathrm{O}_{3} ightleftarrows \mathrm{O}_{2}+\mathrm{O} 	ext { (fast) } $$ We get the equilibrium constant\ $\mathrm{K}^{\prime}=\frac{[\mathrm{O}]\left[\mathrm{O}_{2}ight]}{\left[\mathrm{O}_{3}ight]} \quad 	herefore[\mathrm{O}]=\frac{\mathrm{K}^{\prime}\left[\mathrm{O}_{3}ight]}{\left[\mathrm{O}_{2}ight]}$\ Substituting this expression for $[\mathrm{O}]$ in rate equation, $$ 	ext { Rate }=\frac{\mathrm{k} 	imes \mathrm{K}^{\prime} 	imes\left[\mathrm{O}_{3}ight]^{2}}{\left[\mathrm{O}_{2}ight]}=\frac{\mathrm{K}\left[\mathrm{O}_{3}ight]^{2}}{\left[\mathrm{O}_{2}ight]}=\mathrm{K}\left[\mathrm{O}_{3}ight]^{2}\left[\mathrm{O}_{2}ight]^{-1} $$