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Experimental analysis shows that the homogeneous decomposition of ozone proceeds with a rate

Experimental analysis shows that the homogeneous decomposition of ozone proceeds with a rate

$$\mathrm {-r_{O_3}=k[ O_3]^2[O_2 ]^{-1}}$$

$$\mathrm {-r_{O_3}=kC^2_{O_3}/C_{O_2}}$$

1. Suggest a two-step mechanism to explain this rate.

2. What is the overall order of reaction?

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Solution :

The homogeneous decomposition of ozone proceeds as

$$2 \mathrm{O}_{3} \rightarrow 2 \mathrm{O}_{2}$$

and follows the rate law

$$-\mathrm{r}_{\mathrm O_{3}}=\mathrm k\left[\mathrm{O}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{-1}$$

The two-step mechanism, consistent with the rate, suggested is

Step 1: (fast, at equilibrium)

$$\mathrm{O}_{3} \underset{\mathrm{k}_{2}}{\stackrel{\mathrm{k}_{1}}{\rightleftharpoons}} \mathrm{O}_{2}+\mathrm{O} \quad$$

Step 2: (slow)

$$\quad \mathrm{O}+\mathrm{O}_{3} \stackrel{\mathrm{k}_{3}}{\longrightarrow} 2 \mathrm{O}_{2} \quad$$

The step 2 is the slower, rate determining step and accordingly, the reaction rate is given by,

$$-\mathrm{r}_{\mathrm{O}_{3}}=\frac{-\mathrm{d}\left[\mathrm{O}_{3}\right]}{\mathrm{dt}}=\mathrm{k}_{3}\left[\mathrm{O}_{3}\right][\mathrm O] ...(i)$$

The step 1 is fast and reversible. For this equilibrium step, we have

\begin{aligned} \mathrm{K} &=\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}=\frac{\left[\mathrm{O}_{2}\right][\mathrm{O}]}{\left[\mathrm{O}_{3}\right]} \\ {[\mathrm{O}] } &=\mathrm{K}\left[\mathrm{O}_{3}\right] /\left[\mathrm{O}_{2}\right] ...(ii) \end{aligned}

Putting value of $[\mathrm O]$ from equation (ii) into equation (i), we get,

Let

\begin{aligned} -\mathrm{r}_{\mathrm{O}_{3}} &=\frac{-\mathrm{d}\left[\mathrm{O}_{3}\right]}{\mathrm{dt}}=\mathrm{k}_{3} \cdot \mathrm{K}\left[\mathrm{O}_{3}\right]^{2} /\left[\mathrm{O}_{2}\right] \\ \mathrm{k} &=\mathrm{k}_{3} \mathrm{~K} \\ -\mathrm{r}_{\mathrm{O}_{3}} &=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{-1} \\ \end{aligned}

Overall order of reaction is 2-1=1