/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 107 The following mechanism has been... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
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 by step solution

01

(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.
02

(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.
03

(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: \(N_{2}O_{2}\)
04

(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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 Hâ‚‚ to form Nâ‚‚O and Hâ‚‚O involves two steps:
  • First, two NO molecules react to form Nâ‚‚Oâ‚‚.
  • Next, Nâ‚‚Oâ‚‚ reacts with Hâ‚‚ to produce Nâ‚‚O and Hâ‚‚O.
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, Nâ‚‚Oâ‚‚ is the intermediate. It is produced in the first reaction but used up in the second:
  • Formed by 2NO combining to produce Nâ‚‚Oâ‚‚.
  • Consumed when Nâ‚‚Oâ‚‚ reacts with Hâ‚‚ to form Nâ‚‚O and Hâ‚‚O.
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 Nâ‚‚Oâ‚‚ and Hâ‚‚, reflecting the observed rate law.
  • This means the first step, forming Nâ‚‚Oâ‚‚, 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.

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Most popular questions from this chapter

Consider two reactions. Reaction (1) has a half-life that gets longer as the reaction proceeds. Reaction (2) has a half-life that gets shorter as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

Consider the following reaction between mercury(II) chloride and oxalate ion: $$ 2 \mathrm{HgCl}_{2}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+2 \mathrm{CO}_{2}(g)+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) $$ The initial rate of this reaction was determined for several concentrations of \(\mathrm{HgCl}_{2}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), and the following rate data were obtained for the rate of disappearance of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\) : $$ \begin{array}{llll} \hline \text { Experiment } & {\left[\mathrm{HgCl}_{2}\right](M)} & {\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right](M)} & \text { Rate }(M / \mathrm{s}) \\ \hline 1 & 0.164 & 0.15 & 3.2 \times 10^{-5} \\ 2 & 0.164 & 0.45 & 2.9 \times 10^{-4} \\ 3 & 0.082 & 0.45 & 1.4 \times 10^{-4} \\ 4 & 0.246 & 0.15 & 4.8 \times 10^{-5} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(\mathrm{HgCl}_{2}\) is \(0.100 \mathrm{M}\) and that of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is \(0.25 \mathrm{M}\) if the temperature is the same as that used to obtain the data shown?

Consider a hypothetical reaction between \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) that is zero order in A, second order in B, and first order in C. (a) Write the rate law for the reaction. (b) How does the rate change when [A] is tripled and the other reactant concentrations are held constant? (c) How does the rate change when [B] is doubled and the other reactant concentrations are held constant? (d) How does the rate change when [C] is tripled and the other reactant concentrations are held constant? (e) By what factor does the rate change when the concentrations of all three reactants are doubled? (f) By what factor does the rate change when the concentrations of all three reactants are cut in half?

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q)(\text { slow }) \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) \quad(\text { fast }) \end{aligned} $$ (a) Write the chemical equation for the overall process. (b) Identify the intermediate, if any, in the mechanism. (c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g), \quad\) at \(\quad 70^{\circ} \mathrm{C}\) is \(6.82 \times 10^{-3} \mathrm{~s}^{-1}\). Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) in a volume of \(2.0 \mathrm{~L} .(\mathbf{a})\) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min} ?\) (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010 \mathrm{~mol}\) ? (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70{ }^{\circ} \mathrm{C}\) ?

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