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Question: The chemical reaction \(2O_{3} \rightarrow 3O_{2}\) proceeds as \(O_{3}\) \(O_{2} + \lbrack O\rbrac...

The chemical reaction 2O3→3O22O_{3} \rightarrow 3O_{2} proceeds as

O3O_{3} O2+[O]O_{2} + \lbrack O\rbrack (fast)

[O]+O3→2O2\lbrack O\rbrack + O_{3} \rightarrow 2O_{2} (slow)

The rate law expression will be

A

Rate = k[O][O3]‘k\lbrack O\rbrack\lbrack O_{3}\rbrack`

B

Rate = k[O3]2[O2]−1k\lbrack O_{3}\rbrack^{2}\lbrack O_{2}\rbrack^{- 1}

C

Rate =k[O3]2= k\lbrack O_{3}\rbrack^{2}

D

Rate =k[O2][O]= k\lbrack O_{2}\rbrack\lbrack O\rbrack

Answer

Rate = k[O3]2[O2]−1k\lbrack O_{3}\rbrack^{2}\lbrack O_{2}\rbrack^{- 1}

Explanation

Solution

O3K−4↔K1O2+[O]O_{3}\overset{K_{1}}{\overset{\leftrightarrow}{\quad K_{- 4}\quad}}O_{2} + \lbrack O\rbrack (fast)

[O]+O3→K22O2\lbrack O\rbrack + O_{3}\overset{\quad K_{2}\quad}{\rightarrow}2O_{2} (slow)

Rate of reaction is determined by slow step hence,

Rate =k2[O][O3]= k_{2}\lbrack O\rbrack\lbrack O_{3}\rbrack

[O] is unstable intermediate so substitute the value of [O] in above equation.

Rate of forward reaction =k1[O3]= k_{1}\lbrack O_{3}\rbrack

Rate of backward reaction k−1[O2][O]k_{- 1}\lbrack O_{2}\rbrack\lbrack O\rbrack

At equilibrium,

Rate of forward reaction = Rate of backward reaction

k1[O3]=k−1[O2][O]k_{1}\lbrack O_{3}\rbrack = k_{- 1}\lbrack O_{2}\rbrack\lbrack O\rbrack

[O]=k1[O3]k−1[O2]\lbrack O\rbrack = \frac{k_{1}\lbrack O_{3}\rbrack}{k_{- 1}\lbrack O_{2}\rbrack}

Rate=k2(k1[O3]k−1[O2])[O3]Rate = k_{2}\left( \frac{k_{1}\lbrack O_{3}\rbrack}{k_{- 1}\lbrack O_{2}\rbrack} \right)\lbrack O_{3}\rbrack

Rate=k[O3]2[O2]Rate = \frac{k\lbrack O_{3}\rbrack^{2}}{\lbrack O_{2}\rbrack}