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## Q. 13.9

Reaction Mechanisms
Ozone naturally decomposes to oxygen by the reaction:
2$O_{3}(g) → 3 O_{2}$(g)
The experimentally observed rate law for this reaction is:
Rate = k$[O_{3}]^{2}[O_{2}]^{-1}$
Show that the following proposed mechanism is consistent with the experimentally observed rate law.

$O_{3}(g) \xrightleftharpoons[k_{-1}]{k_{1}} O_{2}(g)+O(g)$     Fast

$O_{3}(g)+O(g) \xrightleftharpoons[k_{2}]{} 2O_{2}(g)$          Slow

## Verified Solution

 To determine whether the mechanism is valid, you must first determine whether the steps sum to the overall reaction. Since the steps do indeed sum to the overall reaction, the first condition is met. $O_{3}(g) \xrightleftharpoons[k_{-1}]{k_{1}} O_{2}(g)+\cancel{O}(g)$ $\underline{O_{3}(g)+\cancel{O}(g)\xrightarrow[k_{2}]{}2O_{2}(g)}$ 2$O_{3}(g)→3O_{2}(g)$ The second condition is that the rate law predicted by the mechanism must be consistent with the experimentally observed rate law. Since the second step is rate limiting, write the rate law based on the second step. Rate = k$_{2}[O_{3}$][O] Because the rate law contains an intermediate (O), you must express the concentration of the intermediate in terms of the concentrations of the reactants of the overall reaction. To do this, set the rates of the forward reaction and the reverse reaction of the first step equal to each other. Solve the expression from the previous step for [O], the concentration of the intermediate. Rate (forward) = Rate (backward) k$_{1}[O_{3}] = k_{-1}[O_{2}$][O] [O] =$\frac{ k_{1}[O_{3}]}{k_{-1}[O_{2}]}$ Finally, substitute [O] into the rate law predicted by the slow step. Rate = k$_{2}[O_{3}$][O] = $k_{2}[O_{3}]\frac{k_{1}[O_{3}]}{k_{-1}[O_{2}]}$ = $k_{2}\frac{k_{1}[O_{3}]}{k_{-1}[O_{2}]}$ = k$[O_{3}]^{2}[O_{2}]^{-1}$ CHECK Since the two steps in the proposed mechanism sum to the overall reaction, and the rate law obtained from the pro posed mechanism is consistent with the experimentally observed rate law, the proposed mechanism is valid. The -1 reac-tion order with respect to [O$_{2}$] indicates that the rate slows down as the concentration of oxygen increases—oxygen inhibits, or slows down, the reaction.