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 \(\mathrm{NO}_{2} ? \)Assume that \(\mathrm{N}\) is the central atom in \(\mathrm{NO}_{3},\) assume only \(\mathrm{N}^{16} \mathrm{O}^{18} \mathrm{O}_{2}\) forms, and assume stoichiometric amounts of reactants are combined.

Short answer

Given the reaction mechanism and assumptions, we can expect the distribution of \(^{18}\mathrm{O}\) in the final product, \(\mathrm{NO}_{2}\), to be roughly: - 50% (half) of the product composed of \(\mathrm{N}^{16}\mathrm{O}^{18}\mathrm{O}\) molecules - 50% (half) of the product composed of \(\mathrm{N}^{16}\mathrm{O}^{16}\mathrm{O}\) molecules This results in an even distribution of the \(^{18}\mathrm{O}\) isotope in the final product.

Step-by-step solution

Determine the species involved in the reaction

Write the reaction mechanism using the isotopes information given: 1. \(\mathrm{N}^{16}\mathrm{O} + ^{18}\mathrm{O}_{2} \rightleftharpoons \mathrm{N}^{16}\mathrm{O}^{18}\mathrm{O}_{2}\) (fast equilibrium) 2. \(\mathrm{N}^{16}\mathrm{O}^{18}\mathrm{O}_{2} + \mathrm{N}^{16}\mathrm{O} \longrightarrow 2 \mathrm{NO}_{2}\) (slow) Since we only need to consider \(\mathrm{N}^{16}\mathrm{O}^{18}\mathrm{O}_{2}\) formation, we can now examine the possibilities for the distribution of \(^{18}\mathrm{O}\) in the final products, \(\mathrm{NO}_{2}\).

Analyze the reaction mechanism and consider the possible outcomes

In the first step, fast equilibrium, we have the formation of \(\mathrm{N}^{16}\mathrm{O}^{18}\mathrm{O}_{2}\). The second step, slow, we have the reaction: \[\mathrm{N}^{16}\mathrm{O}^{18}\mathrm{O}_{2} + \mathrm{N}^{16}\mathrm{O} \longrightarrow 2 \mathrm{NO}_{2}\] When the intermediate \(\mathrm{N}^{16}\mathrm{O}^{18}\mathrm{O}_{2}\) reacts with \(\mathrm{N}^{16}\mathrm{O}\), the \(^{18}\mathrm{O}\) can be distributed in two different ways: 1. The \(^{18}\mathrm{O}\) combines with one of the \(\mathrm{N}^{16}\mathrm{O}\), forming two molecules of \(\mathrm{N}^{16}\mathrm{O}^{18}\mathrm{O}\). 2. The \(^{18}\mathrm{O}\) stays with one of the oxygen atoms making a \(\mathrm{N}^{16}\mathrm{O}^{18}\mathrm{O}\) molecule and the other \(\mathrm{N}^{16}\mathrm{O}\) will form a \(\mathrm{N}^{16}\mathrm{O}^{16}\mathrm{O}\).

Compare the expected distribution of the products

Given the assumptions and the reaction mechanism, we can expect that the distribution of \(^{18}\mathrm{O}\) in the final product, \(\mathrm{NO}_{2}\), will be a mixture of both isotope combinations: - Roughly 50% (half) of the product will be composed of molecules with the formula: \(\mathrm{N}^{16}\mathrm{O}^{18}\mathrm{O}\) - Roughly 50% (half) of the product will be composed of molecules with the formula: \(\mathrm{N}^{16}\mathrm{O}^{16}\mathrm{O}\) Thus, we expect a roughly even distribution of the \(^{18}\mathrm{O}\) isotope in the final product.

Key concepts

Isotopic Labeling

Isotopic labeling is a powerful tool used by chemists to track atoms through reaction mechanisms and understand the paths molecules take during chemical transformations. In our exercise, isotopic labeling involves using an oxygen isotope, ^{18}O, which is heavier than the most common oxygen isotope, ^{16}O. By substituting ^{18}O for ^{16}O in one of the reactants, researchers can trace the movement of oxygen atoms during the chemical reaction.

Isotopic labeling allows chemists to confirm or refute proposed reaction mechanisms by observing the distribution of isotopes in the product molecules. If a proposed mechanism predicts certain isotopic patterns in the products, and these predictions match the actual distribution after the reaction, it lends support to the mechanism. Conversely, if the distributions differ, the mechanism may need to be revised. This technique is essential in elucidating complex reaction pathways and identifying the roles of various molecular species in a reaction.

For example, in the provided exercise, we expect an equal mix of isotopes, ^{18}O and ^{16}O, in the NO_2 product due to the reaction mechanism and isotopic labeling of the reactants.

Chemical Kinetics

Chemical kinetics deals with the rates at which chemical reactions occur and the steps involved in these reactions. Fundamental to this field is the concept of reaction rates and the various factors that influence them, such as temperature, concentration, and the presence of catalysts. The exercise above introduces the notion of a reaction being broken down into multiple steps, with different rates: a fast initial equilibrium step followed by a slower step.

In kinetic studies, the rate of a chemical reaction is often determined by the slowest step, known as the rate-determining step. This is because this step limits the overall rate at which the reaction can proceed. The reason for a particular step to be slow can be related to a higher energy barrier that the reactants need to overcome to form the products. Understanding the kinetics of a reaction is crucial for controlling the reaction conditions to optimize yield and selectivity.

The exercise demonstrates how knowledge of kinetics can also be used to deduce the likely reaction mechanism, since the formation of intermediates and their consumption will influence how fast the observed products can form.

Intermediate Species

Intermediate species are molecules or ions that are formed during the middle steps of a reaction sequence but do not appear in the overall final equation. They are usually very reactive and have short lifetimes, making them difficult to detect directly. However, they are crucial in determining the pathway and rate of a reaction.

In the reaction outlined in our exercise, NO_3 is an intermediate species that forms in a fast equilibrium with NO and O_2 before reacting with another NO to produce NO_2 in a slower step. Assumptions about how it forms and reacts are important for predicting the distribution of isotopes in the products. Intermediates often hold the key to understanding the detailed steps of complex reactions.

The ability to identify and propose intermediates allows chemists to design experiments for probing reaction mechanisms, such as by using isotopic labeling to trace the fate of atoms as intermediates form and react. The eventual distribution of isotopes in the product gives valuable information about the intermediates' structures and the sequence of bond-making and breaking events that have occurred.