A chemist studied the reaction mechanism for the reaction $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ by reacting \(\mathrm{N}^{16} \mathrm{O}\) with \(^{18} \mathrm{O}_{2}\) . If the reaction mechanism is $$ \begin{aligned} \mathrm{NO}+\mathrm{O}_{2} & \rightleftharpoons \mathrm{NO}_{3}(\text { fast equilibrium }) \\ \mathrm{NO}_{3}+\mathrm{NO} & \longrightarrow 2 \mathrm{NO}_{2}(\text { slow }) \end{aligned} $$ what distribution of \(^{18} \mathrm{O}\) would you expect in the NO \(_{2} ?\) Assume that \(\mathrm{N}\) is the central atom in \(\mathrm{NO}_{3},\) assume only \(\mathrm{N}^{16} \mathrm{O}^{16} \mathrm{O}_{2}\) forms, and assume stoichiometric amounts of reactants are combined.

Step by step solution

01

Identify isotopes in the reaction

First, let's list down the isotopes present in the initial mixture of reactants and the intermediary molecules that will be produced during the progression of reaction: - N\(^{16}\)O - \(^{18}\)O\(_2\) - N\(^{16}\)O\(^{16}\)O\(_2\) - N\(^{16}\)O\(^{18}\)O The main objective is to find the distribution of \(^{18}\mathrm{O}\) in the final product molecule NO\(_2\).

02

Identify the first equilibrium reaction

The fast equilibrium reaction will form N\(^{16}\)O\(^{16}\)O\(_2\) and N\(^{16}\)O\(^{18}\)O as intermediate molecules: $$ \mathrm{N}^{16}\mathrm{O} + ^{18}\mathrm{O}_{2} \rightleftharpoons \mathrm{N}^{16}\mathrm{O}^{16}\mathrm{O}_{2} + \mathrm{N}^{16}\mathrm{O}^{18}\mathrm{O} $$

03

Find the distribution of O isotopes in the second reaction step

The second step of the reaction mechanism is the slow step and determines the overall reaction rate. The two possible reactions are: 1. N\(^{16}\)O\(^{16}\)O\(_2\) + N\(^{16}\)O \(\longrightarrow\) 2 N\(^{16}\)O\(^{16}\)O 2. N\(^{16}\)O\(^{18}\)O + N\(^{16}\)O \(\longrightarrow\) N\(^{16}\)O\(^{16}\)O + N\(^{16}\)O\(^{18}\)O

04

Combine the results to find the final distribution of \(^{18}\mathrm{O}\) in NO\(_2\)

From the second step of the reaction mechanism, we observe that \(^{18}\mathrm{O}\) can only be found in one product molecule of NO\(_2\). Thus, the distribution of \(^{18}\mathrm{O}\) in the final product NO\(_2\) is: - 50% N\(^{16}\)O\(^{16}\)O - 50% N\(^{16}\)O\(^{18}\)O This distribution reflects that half of the NO\(_2\) molecules will have an \(^{18}\mathrm{O}\) atom, while the other half will have only \(^{16}\mathrm{O}\) atoms.

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