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

The distribution of \(^{18}O\) in the \(NO_2\) molecules will be approximately 50%, as half of the produced \(NO_2\) molecules will contain \(^{18}O\) and the other half will hold \(^{16}O\).

Step-by-step solution

Understand the reaction mechanism

Given the mechanism in the problem, it happens in two steps. In the first step, the reaction between \(NO\) and \(O_2\) forms \(NO_3\). Then, the \(NO_3\) reacts with another \(NO\) molecule to produce 2 \(NO_2\) molecules. The given mechanism can be summarized as follows: \begin{aligned} N^{16}O+^{18}O_{2} &\rightleftharpoons N^{16}O^{18}O_2 (\text{fast equilibrium}) \\ N^{16}O^{18}O_2 + N^{16}O &\longrightarrow 2 NO_2(\text{slow}) \end{aligned}

Form the possible \(NO_2\) products

Knowing that \(N\) is the central atom in \(NO_3\), we can expect that the oxygen atoms can exchange with the other oxygen atoms in the reaction. Let's write down the possible \(NO_2\) products: 1. \(N^{16}O^{18}O\): When \(NO_3\) loses one \(^{18}O\) and forms \(NO_2\), 2. \(N^{16}O^{16}O\): When \(NO_3\) loses one \(^{16}O\) and forms \(NO_2\).

Analyze the stoichiometry of the reaction

The given balanced reaction has stoichiometric coefficients: $$ 2\text{ } NO(g) + O_{2}(g) \longrightarrow 2 \text{ } NO_{2}(g) $$ Since we are assuming the reactants are combined in stoichiometric amounts, we can expect that after the reaction reaches equilibrium, we will have roughly equal amounts of \(N^{16}O^{18}O\) and \(N^{16}O^{16}O\) in the final products.

Determine the distribution of \(^{18}O\) in the \(NO_2\)

Since we are using stoichiometric amounts of reactants, we will have equal concentrations of \(N^{16}O^{18}O\) and \(N^{16}O^{16}O\) in the final products. Because the \(^{18}O\) is distributed between two \(NO_2\) molecules, half of the produced \(NO_2\) molecules will contain \(^{18}O\), and the other half will hold \(^{16}O\). Therefore, the distribution of \(^{18}O\) in \(NO_2\) will be approximately 50%.

Key concepts

Understanding Chemical Kinetics

Chemical kinetics is the study of reaction rates, how quickly reactions occur, and the factors that affect these rates. When studying chemical reactions, understanding the mechanism—a step-by-step sequence of elementary reactions by which the overall chemical change occurs—is crucial. A mechanism describes the exact process and order in which bonds are broken and formed, and how reactive intermediates, such as free radicals or complexes, are involved.

In the study of the reaction between nitrogen monoxide (o2) and oxygen (o2), the chemist is specifically looking at how these gases interact to produce nitrogen dioxide (o2). Here, kinetics plays a key role in identifying the 'fast' initial reaction establishing a quick equilibrium between o and o2, forming o3, and the 'slow', rate-determining step where o3 reacts with another o molecule to form the final product, o2. The rate at which these steps occur helps in understanding the reaction mechanism. It is the slowest step in this sequence that determines the overall reaction rate, thus underlining the importance of chemical kinetics in the study of reaction mechanisms.

The Role of Isotopic Labeling

Isotopic labeling is a technique used to track the movement of atoms through a reaction mechanism. By replacing a common isotope of an element with a detectable one, chemists can ascertain the fate of atoms in the products. This is particularly useful in complex reactions where multiple pathways and intermediates are involved.

In the given reaction, isotopic labeling involves the use of o and o2, where the oxygen-18 (o) isotope is used as a tracer. Since oxygen generally forms two bonds, the labeled oxygen from o2 can incorporate into the intermediates and final products. By examining the distribution of o in the final nitrogen dioxide molecules, one can confirm the pathway predicted by the proposed mechanism. This hands-on confirmation using isotopic labeling displays the powerful interplay between experimental techniques and theoretical predictions in understanding chemical processes.

Stoichiometry in Reaction Mechanisms

Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. In a chemical equation, the stoichiometric coefficients represent the number of moles of each substance that react to form the products. It is essential for predicting the quantities of products formed and the amount of reactants needed.

In the context of the given problem where nitrogen monoxide reacts with oxygen to form nitrogen dioxide, stoichiometry dictates that two moles of o will react with one mole of o2 to produce two moles of o2, based on the balanced chemical equation. When the reaction proceeds under stoichiometric conditions—that is, the reactants are combined in the exact ratios as depicted by the balanced equation—we can predict the amounts of each possible product, which is vital for interpreting results such as the distribution of isotopes in the products. Stoichiometry, in this case, assures us that if the reaction goes to completion under the given conditions, we should expect equal amounts of the two different isotopologues of o2, thereby giving a 50% distribution of the o isotope among the o2 molecules.