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For the following problem:

Example 3: Write the rate law for $\ce{2 O3 -> 3 O2}$ given that the f‌irst step is fast and reversible, and the second step is slow.

$$ \begin{align} &\text{Step 1 (fast, reversible)} &\qquad \ce{O3 &<=>[$k_1$][$k_{-1}$] O2 + O} \\ &\text{Step 2 (slow)} &\qquad \ce{O + O3 &->[$k_2$] O2 + O2} \end{align} $$

The rate is determined by the slowest step.

The rate of formation of $\ce{O2}$ is equal to $2$ times the rate of the slow step $(k_2[\ce{O}][\ce{O3}]),$ since two molecules of $\ce{O2}$ are formed.

Thus, rate of formation of $\ce{O2} = 2k_2[\ce{O}][\ce{O3}],$ but $“\ce{O}”$ is an intermediate, solve for $“\ce{O}”$ in terms of products and reactants and rate constants.

Since the first step is fast and reversible and the second step is slow, the f‌irst step is in equilibrium and we can write

$$\frac{[\ce{O2}][\ce{O}]}{[\ce{O3}]} = \frac{k_1}{k_{-1}} = K_1 \quad\text{or}\quad [\ce{O}] = \frac{k_1[\ce{O3}]}{k_{-1}[\ce{O2}]}$$

Substituting:

$$ \begin{align} \text{rate} &= \frac{2k_2k_1[\ce{O3}]^2}{k_{-1}[\ce{O2}]} \\ \text{rate} &= k_\mathrm{obs}\frac{[\ce{O3}]^2}{[\ce{O2}]} \end{align} $$

The solution indicates that the rate of formation $\ce{O2}$ is $2k_2[\ce{O}][\ce{O3}]$. Why isn't it $2k_2[\ce{O}][\ce{O3}] + k_1[\ce{O3}]$, because the first reaction also produces $\ce{O2}$?

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You should write down the rate equations and set [O] as an intermediate so that $\ce{d[O]/dt} = 0$. This is the steady state approach and leads to $\displaystyle \frac{d\mathrm{[O_2]}} {dt}=2\frac{k_1k_2\mathrm{[O_3]}^2}{k_{-1}\mathrm{[O_2]}+k_2\mathrm{[O_3]}}$ which differs from your answer because when you assume a fast equilibrium you also assumed that $k_{-1} \gg k_1$. You should make this assumption after you have the full equation, if necessary, but not before you do the calculation.

To answer your last point you still have the intermediate [O] which is an unknown and so has to be eliminated as shown by using the steady state approach used above.

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$$ \begin{align} &\text{Step 1 (fast, reversible)} &\qquad \ce{O3 &<=>[$k_1$][$k_{-1}$] O2 + O} \\ &\text{Step 2 (slow)} &\qquad \ce{O + O3 &->[$k_2$] O2 + O2} \end{align} $$

The rate is determined by the slowest step.

$$r=k_2[\ce{O}][\ce{O3}]$$ If the concentration of $\ce{ O}$ which produced from the first step increased, the reaction rate increased in the opposite direction, leading to the arrival of the concentration intermediate material $\ce{ O}$ to a fixed value. This is called Steady State, where the concentration of $\ce{O}$ is constant, The rate of formation of the intermediate material $\ce{O}$ equal the rate of its consumpution, this material produced from the first step in the forward direction and consumed from the first step in the opposite direction, and in the second step also. $$\text{ Production rate of}~ \ce{O} = \text{ Consumption rate of}~ \ce{O}$$ $$k_\mathrm {1}\ce{[O3]}= k_\mathrm{-1}[\ce{O2}][\ce{O}] + k_\mathrm {2} \ce{[O] [O3]}$$ But $k_\mathrm{2} << k_\mathrm{-1}~ \text{and}~ k_\mathrm{1} $ because the second step is the slow step, so the term $k_2\ce{ [O] [O3] }$ can be neglected at the far right of the last equation. $$k_\mathrm {1}\ce{[O3]}= k_\mathrm{-1}[\ce{O2}][\ce{O}]$$ $$[\ce{O}] =\frac{ k_1 [\ce{O3}] }{ k_\mathrm{-1}[\ce{O2}]}$$ The value of$ [\ce{O}]$ from the last equation can now be compensated in the reaction rate equation which represents the slow step.

$$\text{r} = {k_\mathrm{2}[\ce{O}][\ce{O2}] = \frac{k_\mathrm{1}k_\mathrm{2}}{k_\mathrm{-1}}\times\frac{[\ce{O3}][\ce{O3}]}{[\ce{O2}]}} =\frac{k[\ce{O3}]^2}{[\ce{O2}]}=k[\ce{O3}]^2[\ce{O2}]^{-1}$$ In it we see that $k$ expresses the value of the fractional $\frac{k_\mathrm{1}k_\mathrm{2}}{k_\mathrm{-1}}$ and we note that the concentration of $\ce{O2}$ has appeared in the denominator of the reaction rate law , although it is a substance resulting from the reaction. This means that the order of reaction for oxygen gas $\ce{O2}$ is $(-1)$ that the doubling of the concentration of $\ce{O2}$ leads to a reduction in the rate of the reaction by half.

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