Question

1. use the steady state approximation to derive a rate law for the following two step reaction: o₃ ⇌ o₂ + o; o + o₃ → 2o₂. the observed rate law is rate = k o₃²/o₂. can you obtain the observed rate law from the one that you derived? what does that mean about the relative rates of the elementary steps?

Explanation:

Step1: Define rate - equations for elementary steps

Let the forward and reverse rate constants for the first step $O_3
ightleftharpoons O_2 + O$ be $k_1$ and $k_{ - 1}$ respectively, and the rate constant for the second step $O+O_3
ightarrow 2O_2$ be $k_2$.
The rate of formation of $O$ in the first step is $r_{1f}=k_1[O_3]$, and the rate of consumption of $O$ in the first - step reverse reaction is $r_{1r}=k_{ - 1}[O_2][O]$, and the rate of consumption of $O$ in the second step is $r_{2}=k_2[O][O_3]$.

Step2: Apply steady - state approximation

In the steady - state approximation, $\frac{d[O]}{dt}=0$. So, $k_1[O_3]-k_{ - 1}[O_2][O]-k_2[O][O_3]=0$.
Solve for $[O]$:
\[

$$\begin{align*} k_1[O_3]&=(k_{ - 1}[O_2]+k_2[O_3])[O]\\ [O]&=\frac{k_1[O_3]}{k_{ - 1}[O_2]+k_2[O_3]} \end{align*}$$

\]

Step3: Determine the rate of the overall reaction

The rate of the overall reaction is determined by the rate of the second step (since $O$ is an intermediate), $r = k_2[O][O_3]$.
Substitute $[O]$ into the rate - equation:
\[
r=\frac{k_1k_2[O_3]^2}{k_{ - 1}[O_2]+k_2[O_3]}
\]

Step4: Analyze to get the observed rate law

If $k_{ - 1}[O_2]\gg k_2[O_3]$, then $r=\frac{k_1k_2}{k_{ - 1}}\frac{[O_3]^2}{[O_2]}$. Let $k = \frac{k_1k_2}{k_{ - 1}}$, we get the observed rate law $r = k\frac{[O_3]^2}{[O_2]}$. This implies that the reverse reaction of the first step is much faster than the second step.

Answer:

The derived rate law is $r=\frac{k_1k_2[O_3]^2}{k_{ - 1}[O_2]+k_2[O_3]}$. When $k_{ - 1}[O_2]\gg k_2[O_3]$, the observed rate law $r = k\frac{[O_3]^2}{[O_2]}$ can be obtained, which means the reverse reaction of the first step is much faster than the second step.