Question

The following mechanism has been proposed to account for the rate law of the decomposition of ozone to ${\mathit{O}}_{\mathbf{2}}$(g):

$$\begin{gathered} {\mathit{O}}_{\mathbf{3}}\mathbf{+}\mathit{M}\underset{{\mathbf{K}}_{\mathbf{-}\mathbf{1}}}{\overset{{\mathbf{K}}_{\mathbf{1}}}{\mathbf{⟷}}}{\mathit{O}}_{\mathbf{2}}\mathbf{+}\mathit{O}\mathbf{+}\mathit{M} \\ \mathit{O}\mathbf{+}{\mathbf{O}}_{\mathbf{3}}\stackrel{{\mathbf{K}}_{\mathbf{2}}}{\mathbf{\to }}\mathbf{2}{\mathit{O}}_{\mathbf{2}} \end{gathered}$$

Apply the steady-state hypothesis to the concentration of atomic oxygen, and derive the rate law for the decomposition of ozone. ($\mathit{M}$ stands for an atom or molecule that can exchange kinetic energy with the particles undergoing the chemical reaction) .

Short answer

Rate= $\frac{K_1{K}_2{\left[O_3\right]}^2}{K^{-1}\left[O_2\right]+K_2\left[O_3\right]}$ is the rate law expression for the decomposition of ozone.

Step-by-step solution

Reaction mechanism for the rate law of the decomposition of ozone to O2 has been explained in this step.

$$\begin{gathered} O_3+M\rightleftarrows O_2+O+M \\ O+O_3\rightleftarrows 2O_2 \end{gathered}$$

Where$M$stands for an atom or molecule that can exchange kinetic energy with the particles undergoing chemical reaction.

The intermediate in the decomposition of ozone to $O_2$is $O.$ Applying the steady rate approximation to the reaction mechanism, the concentration of $O$will be constant as shown below:

$\frac{d\left[O\right]}{dt}=O$

The intermediate $O$ is only produced in forward reaction of the first elementary step.

$$\begin{gathered} O_3+M\rightleftarrows O_2+O+M \\ {\text{Rate law:k}}_{\text{1}}\left[{\text{O}}_{\text{3}}\right] \end{gathered}$$

The intermediate $O$ is consumed in the reverse reaction of the first elementary step.

$O_3+M\rightleftarrows O_2+O+M$

Also collision between $B$molecules are occurring much faster than decomposition of $B^{*}$. This is the physical significance of the result from part $B$.

Writing the overall reaction for decomposition of ozone →O2 .

$Rate={\text{K}}_{\text{- 1}}\left[O_2\right]\left[O\right]$

And in second elementary step,

$O+O_3\to 2O_2$

And the rate law is Rate= $K_2\left[O\right]\left[O_3\right]$

$\therefore K_1\left[O_3\right]=K_{-1}\left[O_2\right]\left[O\right]+K_2\left[O\right]\left[O_3\right]$

The overall reaction for the decomposition of ozone to $O_2$is $2O_3\to 3O_2$

Rate of decomposition of ozone=$\frac{-d\left[O_3\right]}{dt}=K_2\left[O_3\right]\left[O\right]$

Substitution of  [O] to the rate law equation.

The rate of production and consumption of $O$ can be arranged to calculate $\left[O\right]$

as shown below.

$$\begin{gathered} K_1\left[O_3\right]=K_{-1}\left[O_2\right]\left[O\right] \\ \left[O\right]=\frac{K_1{O}_3}{K_{-1}\left[O_2\right]+K\left[O_3\right]} \end{gathered}$$

The $\left[O\right]$will be substituted to the rate law as shown below:

Rate of reaction= $\frac{-d\left[O_3\right]}{dt}=K_{\partial }\left[O_3\right]\left[O\right]$

Rate of reaction= $\frac{-d\left[O_2\right]}{dt}=K_2\left[O_3\right]\left(\frac{K_1\left[O_3\right]}{K_{-1}\left[O_2\right]+K_2\left[O_3\right]}\right)$

Rate of reaction= $\frac{-d\left[O_3\right]}{dt}=\frac{K_1{K}_2{\left[O_3\right]}^2}{K_{-1}\left[O_2\right]+K_2\left[O_3\right]}$