The following mechanism has been proposed for the reaction of \(\mathrm{NO}\) with \(\mathrm{H}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}\) and \(\mathrm{H}_{2} \mathrm{O}\) : $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{NO}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g) \\ \mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ (a) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reaction. (b) Write a rate law for each elementary reaction in the mechanism. (c) Identify any intermediates in the mechanism. (d) The observed rate law is rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]\). If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?

To verify if the proposed mechanism adds up to a balanced chemical equation, we need to add the elementary reactions and check if the sum of the reactants equals the sum of the products: \[\mathrm{NO}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}\mathrm{O}_{2}(g)\] \[\mathrm{N}_{2}\mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}\mathrm{O}(g)+\mathrm{H}_{2}\mathrm{O}(g)\] Adding the reactions gives: \[2\,\mathrm{NO}(g) + \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}\mathrm{O}(g) + \mathrm{H}_{2}\mathrm{O}(g)\] The reaction is balanced: \(2\,\mathrm{NO}\) molecules and one \(\mathrm{H}_{2}\) molecule react to form one molecule of \(\mathrm{N}_{2}\mathrm{O}\) and one molecule of \(\mathrm{H}_{2}\mathrm{O}\). Step 2: Write the rate law for each elementary reaction

For each elementary reaction, the rate law can be written as follows: Reaction 1: \[rate_1 = k_1 [\mathrm{NO}]^2\] Reaction 2: \[rate_2 = k_2 [\mathrm{N}_{2}\mathrm{O}_{2}] [\mathrm{H}_{2}]\] Step 3: Identify the Intermediate in the Mechanism

An intermediate is a species that is formed and consumed in the mechanism but does not appear in the overall balanced equation. In this case, the intermediate is \(\mathrm{N}_{2}\mathrm{O}_{2}\). It is formed in the first elementary reaction and consumed in the second elementary reaction, but it does not appear in the overall balanced equation. Step 4: Compare the Observed Rate Law to the Proposed Mechanism

The observed rate law is given by: \[rate = k[\mathrm{NO}]^2[\mathrm{H}_{2}]\] To relate this observed rate law to the proposed mechanism, we can consider that the first reaction reaches equilibrium and that the second reaction is slow compared to the first one. In such a case, we can assume that \[\frac{rate_1}{rate_2} \approx [\mathrm{N}_{2}\mathrm{O}_{2}]\] Since \(rate_1 = k_1 [\mathrm{NO}]^2\), we can calculate the concentration of \(\mathrm{N}_{2}\mathrm{O}_{2}\) as: \[[\mathrm{N}_{2}\mathrm{O}_{2}] = \frac{rate_1}{rate_2} \approx \frac{k_1[\mathrm{NO}]^2}{k_2 [\mathrm{H}_{2}]}\] Now we can substitute the concentration of \(\mathrm{N}_{2}\mathrm{O}_{2}\) in the rate law for Reaction 2: \[rate_2 = k_2 \times \frac{k_1[\mathrm{NO}]^2}{k_2 [\mathrm{H}_{2}]} \times [\mathrm{H}_{2}]\] \[rate_2 = k_1k_2\frac{[\mathrm{NO}]^2[\mathrm{H}_{2}]}{k_2}\] The observed rate law and the proposed mechanism rate law are equal, which means that the proposed mechanism with Reaction 1 being fast and in equilibrium and Reaction 2 being slow is consistent with the observed rate law.