Question

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.

Short answer

In the final product NO\(_2\), the distribution of \(^{18}\mathrm{O}\) is: - 50% N\(^{16}\)O\(^{16}\)O - 50% N\(^{16}\)O\(^{18}\)O This means 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.

Step-by-step solution

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\(_2\).

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: $$ \mathrm{N}^{16}\mathrm{O} + ^{18}\mathrm{O}_{2} \rightleftharpoons \mathrm{N}^{16}\mathrm{O}^{16}\mathrm{O}_{2} + \mathrm{N}^{16}\mathrm{O}^{18}\mathrm{O} $$

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

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.

Key concepts

Equilibrium Reactions

In chemistry, an equilibrium reaction is a dynamic state where the rate of the forward reaction equals the rate of the reverse reaction. These reactions do not proceed to completion but instead establish a balance between reactants and products. In our given mechanism, the reaction \( \mathrm{NO} + \mathrm{O}_2 \rightleftharpoons \mathrm{NO}_3 \) is identified as a fast equilibrium. This means that the reactants \( \mathrm{NO} \) and \( \mathrm{O}_2 \) rapidly convert to \( \mathrm{NO}_3 \) and vice versa.

Equilibrium reactions are important because they maintain a constant concentration of intermediates, facilitating a balance that allows the reaction to proceed efficiently. In this case, the equilibrium allows for the steady formation of \( \mathrm{NO}_3 \), which then participates in the subsequent slow step of the mechanism, affecting the overall reaction rate.

Isotopic Labeling

Isotopic labeling is a technique used to track the passage of an isotope through a reaction mechanism. In the exercise, the chemist uses \(^{18}\mathrm{O} \) to label the oxygen in \( \mathrm{O}_2 \). This method helps determine how oxygen atoms are distributed in the final product, \( \mathrm{NO}_2 \).

By using isotopes like \(^{18}\mathrm{O} \), scientists can observe which pathways reactants follow and identify intermediates formed during the reaction. It is particularly helpful when the reaction involves complex mechanisms or multiple steps. Here, it highlights that in the equilibrium step, different isotopic forms of NO molecules form, which then allow for an analysis of which oxygen atoms end up in \( \mathrm{NO}_2 \).

Reaction Intermediates

Throughout chemical reactions, short-lived species known as intermediates are formed. They are crucial for understanding the mechanism of a reaction. In this scenario, \( \mathrm{NO}_3 \) serves as an intermediate, formed during the fast equilibrium step. It plays a pivotal role as it transitions through the mechanism before the reaction concludes with the formation of \( \mathrm{NO}_2 \).

Intermediates can be elusive because their concentrations are typically very low during the course of the reaction. However, they are necessary for the conversion of reactants to products and often affect the reaction path significantly. The presence of \( \mathrm{NO}_3 \) as an intermediate indicates that it is part of the complex interplay that delivers the final product.

Rate-Determining Step

The rate-determining step in a reaction mechanism is the slowest step, which controls the overall rate of the reaction. In the given example, the step \( \mathrm{NO}_3 + \mathrm{NO} \rightarrow 2 \mathrm{NO}_2 \) is labeled as the slow step, making it the rate-determining step.

This means that despite the fast formation of intermediates in previous steps, the conversion to the final products is bottlenecked by the reaction between \( \mathrm{NO}_3 \) and \( \mathrm{NO} \). Understanding this concept is crucial because the rate-determining step forms the basis for calculating the reaction rate and provides insights into which steps in a mechanism require more energy or are more complex. This is why chemists often seek ways to alter or speed up this step through catalysts or changing reaction conditions.