Problem 107 - Chemistry: The Central Science in SI Units, Global Edition

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 anyintermediatesin 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?

Step-by-Step Solution

Answer

In summary: (a) The elementary reactions add up to the balanced equation: \(NO(g) + NO(g) + H_{2}(g) \rightarrow N_{2}O(g) + H_{2}O(g)\). (b) Rate laws for elementary reactions: \(rate_1 = k_1[NO]^2\) and \(rate_2 = k_2[N_{2}O_{2}][H_{2}]\). (c) The intermediate in the mechanism is \(N_{2}O_{2}\). (d) Based on the observed rate law \(rate = k[NO]^2[H_{2}]\) and the derived rate laws, we conclude that the first reaction is fast and the second reaction is slow.

Step 1: (a) Verify the elementary reactions

First, we need to add up the elementary reactions in the given mechanism to check if they yield a balanced equation. Given reactions: 1st reaction: \(NO(g) + NO(g) \rightarrow N_{2}O_{2}(g)\) 2nd reaction: \(N_{2}O_{2}(g) + H_{2}(g) \rightarrow N_{2}O(g) + H_{2}O(g)\) Let's add them: \(NO(g) + NO(g) + N_{2}O_{2}(g) + H_{2}(g) \rightarrow N_{2}O_{2}(g) + N_{2}O(g) + H_{2}O(g)\) Now, we can remove the species found on both sides of the equation: \(NO(g) + NO(g) + H_{2}(g) \rightarrow N_{2}O(g) + H_{2}O(g)\) This is the balanced equation for the reaction.

Step 2: (b) Rate Law for Each Elementary Reaction

For the given elementary reactions, we need to write the rate laws. 1st reaction: \(NO(g) + NO(g) \rightarrow N_{2}O_{2}(g)\) Rate law: \(rate_1 = k_1[NO]^2\), where \(k_1\) is the rate constant for the first reaction. 2nd reaction: \(N_{2}O_{2}(g) + H_{2}(g) \rightarrow N_{2}O(g) + H_{2}O(g)\) Rate law: \(rate_2 = k_2[N_{2}O_{2}][H_{2}]\), where \(k_2\) is the rate constant for the second reaction.

Step 3: (c) Identify Intermediates

In a reaction mechanism, intermediates are those substances that are produced and then consumed during the sequence of elementary reactions. In our case, N2O2 is an intermediate, as it is formed in the first reaction and consumed in the second reaction. Intermediate: \(N_{2}O_{2}\)

Step 4: (d) Conclusions About Relative Speeds of Reactions

The observed rate law is given as \(rate = k[NO]^2[H_{2}]\). Comparing this with the rate laws derived for the elementary reactions, we can infer that the second reaction is the rate-determining step since it involves the concentrations of \(NO\) and \(H_{2}\) while the intermediate (\(N_{2}O_{2}\)) is not present. In order for the rate-determining step to match the overall rate law, we can deduce that the first reaction must be faster than the second reaction. As such, we can conclude that the first reaction is fast and the second reaction is slow, in accordance with the proposed mechanism.

Key Concepts

Reaction Mechanism

Understanding a reaction mechanism is like looking at a detailed map of a journey. It breaks down a complex reaction into simpler steps called elementary reactions. Each elementary reaction represents a single stage in the transformation of reactants into products.

In our example, the mechanism for the reaction between NO and H₂ to form N₂O and H₂O involves two steps:

• First, two NO molecules react to form N₂O₂.
• Next, N₂O₂ reacts with H₂ to produce N₂O and H₂O.

By combining these steps and eliminating compounds that appear on both sides, we verify they lead to the overall balanced equation. Reaction mechanisms help chemists understand the pathway and sequence of events in a chemical reaction. They are essential for predicting the behavior of chemical processes.

Rate Law

The rate law of a reaction describes how the reaction rate depends on the concentration of reactants. For elementary reactions, the rate law can be written directly from the stoichiometry.

For the given mechanism:

• The first step: \(NO(g) + NO(g) \rightarrow N_{2}O_{2}(g)\) has the rate law \(rate_1 = k_1[NO]^2\).
• The second step: \(N_{2}O_{2}(g) + H_{2}(g) \rightarrow N_{2}O(g) + H_{2}O(g)\) is expressed as \(rate_2 = k_2[N_{2}O_{2}][H_{2}]\).

The overall reaction rate is determined by these rate laws. Analyzing the rate laws helps chemists identify which reactants influence the reaction's speed and by how much.

Reaction Intermediates

Reaction intermediates are species that form in one step of a mechanism and are consumed in another. They do not appear in the overall balanced equation because they are not present at the start or end of the reaction.

In the proposed mechanism, N₂O₂ is the intermediate. It is produced in the first reaction but used up in the second:

• Formed by 2NO combining to produce N₂O₂.
• Consumed when N₂O₂ reacts with H₂ to form N₂O and H₂O.

Identifying intermediates helps chemists understand the transition states and energy changes involved in chemical reactions. Tracking intermediates is crucial in experimental chemistry for verifying reaction mechanisms.

Rate-Determining Step

The rate-determining step is the slowest step in a reaction mechanism. It acts like a bottleneck, controlling the overall reaction rate. In general, this step's rate law matches the observed rate law for the reaction.

In the problem, the experimentally observed rate law is \(rate = k[NO]^2[H_{2}]\). This indicates that the second reaction is the rate-determining step:

• The slow step involves N₂O₂ and H₂, reflecting the observed rate law.
• This means the first step, forming N₂O₂, must be fast.

Understanding the rate-determining step allows chemists to focus on factors that slow down the reaction, providing targets for making reactions faster or more efficient.