Question

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?

Short answer

In summary: 1) The overall reaction obtained from the proposed mechanism is 2NO(g) + H2(g) → N2O(g) + H2O(g), which is balanced. 2) The rate laws for each elementary reaction are: Rate 1 = k₁[NO]² and Rate 2 = k₂[N2O2][H2]. 3) The intermediate species in this mechanism is N2O2(g). 4) Comparing the observed rate law (rate = k[NO]²[H2]) with the elementary reactions' rate laws, we can conclude that the second elementary reaction (involving N2O2 and H2) is much faster than the first elementary reaction involving two NO molecules. The first reaction, involving NO molecules, is the rate-determining (slow) step in this reaction mechanism.

Step-by-step solution

Overall Reaction

To find the overall reaction, let's add the two elementary reactions: 1) NO(g) + NO(g) → N2O2(g) 2) N2O2(g) + H2(g) → N2O(g) + H2O(g) Adding reactions 1 and 2: NO(g) + NO(g) + N2O2(g) + H2(g) → N2O2(g) + N2O(g) + H2O(g) Now, we can cancel the intermediate N2O2 from both sides of the equation, resulting in: NO(g) + NO(g) + H2(g) → N2O(g) + H2O(g) We can simplify it further into: 2NO(g) + H2(g) → N2O(g) + H2O(g) The overall reaction is a balanced equation, and thus, the mechanism is consistent.

Writing Rate Laws

For each elementary step, we need to write its corresponding rate law: For reaction 1: NO(g) + NO(g) → N2O2(g) Rate 1 = k₁[NO]² For reaction 2: N2O2(g) + H2(g) → N2O(g) + H2O(g) Rate 2 = k₂[N2O2][H2]

Identify Intermediates

Intermediates are species that appear in the mechanism but not in the overall reaction. For this exercise, we have only one intermediate species: Intermediate: N2O2(g)

Compare Rate Laws & Determine Reaction Speeds

Based on the given observed rate law: \(rate = k[NO]^2[H2]\) Comparing the observed rate law with the rate laws obtained from the elementary steps, we can observe that Rate 1=k₁[NO]² contributed to the part [NO]² in the observed rate law. Then, we also have Rate 2=k₂[N2O2][H2], which contributed the [H2] part in the observed rate law. One key point is that the reaction intermediate N2O2 does not show up in the observed rate law. What this suggests is that the second elementary reaction (involving N2O2 and H2) is much faster than the first elementary reaction involving two NO molecules. It means that the concentration of N2O2 does not significantly build up in the reaction mechanism, which in turn implies that the first reaction (involving NO molecules) is the rate-determining (slow) step in this reaction mechanism.