## Q. 12.16

Supporting a Mechanism Given the Rate Law: Reactions with an Initial Fast Step

The experimental rate law for the decomposition of ozone is second order in ozone and inverse first order in molecular oxygen:

2 $O_3$(g) → 3 $O_2$(g)     Rate =$– \frac{Δ[O_3]}{Δt} = k \frac{[O_3]²}{[O_2]}$

Show that the following mechanism is consistent with the experimental rate law, and relate the observed rate constant k to the rate constants for the elementary reactions:

$\mathrm{O}_3(g) \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} \mathrm{O}_2(g)+\mathrm{O}(g)$    Faster, reversible

$\mathrm{O}(g)+\mathrm{O}_3(g) \stackrel{k_2}{\longrightarrow} 2 \mathrm{O}_2(g)$    Slower, rate-determining

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$2 \mathrm{O}_3(g) \longrightarrow 3 \mathrm{O}_2(g)$    Overall reaction

STRATEGY

To show that the mechanism is consistent with experiment, we must derive the rate law predicted by the mechanism and compare it with the experimental rate law. If we assume that the faster, reversible step is at equilibrium, we can eliminate the concentration of the intermediate O atoms from the predicted rate law.

## Verified Solution

The rate law for the rate-determining step is rate = $k_2[O][O_3]$, but the stoichiometry of the reaction indicates that the overall rate of consumption of ozone is twice the rate of the rate-determining step:

$\text { Rate }=-\frac{\Delta\left[\mathrm{O}_3\right]}{\Delta t}=2 k_2[\mathrm{O}]\left[\mathrm{O}_3\right]$

(When one O atom and one $O_3$ molecule react in the rate-determining step, two $O_3$ molecules are consumed in the overall reaction.)

The rates of the forward and reverse reactions in the faster, reversible step are given by

$\text { Rate }_{\text {forward }}=k_1\left[\mathrm{O}_3\right] \quad \text { Rate }_{\text {reverse }}=k_1\left[\mathrm{O}_2\right][\mathrm{O}]$

Assuming that the first step is at equilibrium, we can equate the rates of the forward and reverse reactions and then solve for the concentration of the intermediate O atoms:

$k_1\left[\mathrm{O}_3\right]=k_{-1}\left[\mathrm{O}_2\right][\mathrm{O}] \quad \text { so, }[\mathrm{O}]=\frac{k_1\left[\mathrm{O}_3\right]}{k_{-1}\left[\mathrm{O}_2\right]}$

Substituting this expression for [O] into the predicted rate law for overall consumption of ozone gives the predicted rate law in terms of only reactants and products:

$\text { Rate }=-\frac{\Delta\left[\mathrm{O}_3\right]}{\Delta t}=2 k_2[\mathrm{O}]\left[\mathrm{O}_3\right]=2 k_2 \frac{k_1}{k_{-1}} \frac{\left[\mathrm{O}_3\right]^2}{\left[\mathrm{O}_2\right]}$

Because the predicted and experimental rate laws have the same reaction orders in $O_3$ and $O_2$, the proposed mechanism is consistent with the experimental rate law and is a plausible mechanism for the reaction. Comparison of the predicted and experimental rate laws indicates that the observed rate constant k equals 2$k_2k_1/k_{-1}$.