The rate law for the production of \(O_2(g)\) is \(r_{\text{O}_2} = k_2[N_2O_5][NO_2]\).
To derive the rate law for the production of \(O_2(g)\), we can use the steady-state approximation for the intermediate species \([NO_3]\). Steady-state conditions assume that the rate of formation of an intermediate is equal to its rate of consumption.
Let's denote the rate of the overall reaction as \(r\), and the rate of the formation of \(O_2(g)\) as \(r_{\text{O}_2}\).
The suggested mechanism is given by:
\[
\begin{align*}
\text{(1)} & \quad N_2O_5(g) \underset{k_2}{\overset{\text{(2)}}{\rightleftharpoons}} NO_2(g) + NO_3(g) \\
\text{(3)} & \quad NO(g) + N_2O_5(g) \rightarrow 3NO_2(g) \\
\text{(2)} & \quad NO_2(g) + NO_3(g) \xrightarrow{k_3} NO_2(g) + O_2(g) + NO(g)
\end{align*}
\]
Now, we can write the rate expressions for each elementary step:
\[
\begin{align*}
\text{(1)} & \quad r_1 = k_1[N_2O_5] \\
\text{(3)} & \quad r_3 = k_3[NO_2][NO_3] \\
\text{(2)} & \quad r_2 = k_2[N_2O_5] - k_3[NO_2][NO_3]
\end{align*}
\]
Since the intermediate \([NO_3]\) is governed by steady-state conditions, we can set \(r_3\) to zero:
\[
0 = k_3[NO_2][NO_3]
\]
Now, solve for \([NO_3]\):
\[
[NO_3] = \frac{k_2}{k_3}[N_2O_5]
\]
Now, substitute this expression for \([NO_3]\) into the rate expression for the production of \(O_2(g)\):
\[
r_{\text{O}_2} = k_3[NO_2][NO_3] = k_3[NO_2] \left(\frac{k_2}{k_3}[N_2O_5]\right) = k_2[N_2O_5][NO_2]
\]
So, the rate law for the production of \(O_2(g)\) is \(r_{\text{O}_2} = k_2[N_2O_5][NO_2]\).