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Nitramide, \(\mathrm{NO}_{2} \mathrm{NH}_{2},\) decomposes slowly in aqueous solution according to the following reaction: $$\mathrm{NO}_{2} \mathrm{NH}_{2}(\mathrm{aq}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ The reaction follows the experimental rate law $$\text { Rate }=\frac{k\left[\mathrm{NO}_{2} \mathrm{NH}_{2}\right]}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}$$ (a) What is the apparent order of the reaction in a buffered solution? (b) Which of the following mechanisms is the most appropriate for the interpretation of this rate law? Explain. Mechanism 1 \(\mathrm{NO}_{2} \mathrm{NH}_{2} \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) Mechanism 2 \(\begin{array}{l}\mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{3} \mathrm{O}^{+} \frac{k_{2}}{k_{2}^{\prime}} \mathrm{NO}_{2} \mathrm{NH}_{3}^{+}+\mathrm{H}_{2} \mathrm{O} \\\\\mathrm{NO}_{2} \mathrm{NH}_{3}+\stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2}\mathrm{O}+\mathrm{H}_{3} \mathrm{O}^{+}(\text {rapid equilibrium })\end{array}\) Mechanism 3 \(\mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \frac{k_{4}}{k_{4}^{\prime}} \mathrm{NO}_{2} \mathrm{NH}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}\) \(\mathrm{NO}_{2} \mathrm{NH}^{-} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{OH}^{-} \quad\) (rapid equilibrium) \(\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-} \stackrel{k_{6}}{\longrightarrow} 2 \mathrm{H}_{2} \mathrm{O} \quad\) (very fast reaction) (c) Show the relationship between the experimentally observed rate constant, \(k\), and the rate constants in the selected mechanism. (Note that when writing the expression for \(K\), the equilibrium constant, \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) is not involved. \(\rightarrow\) Chapter \(16 .\) ) (d) Show that hydroxide ions catalyze the decomposition of nitramide.

Short Answer

Expert verified
(a) First-order in buffered solution. (b) Mechanism 2 matches rate law. (c) \(k = \frac{k_2 k_3}{k_2'}\). (d) \(\mathrm{OH}^-\) enhances decomposition.

Step by step solution

01

Determine the Apparent Order in Buffered Solution

In a buffered solution, the concentration of \([\mathrm{H}_3\mathrm{O}^+]\) is constant. Therefore, the rate law becomes \(\text{Rate} = k' [\mathrm{NO}_2 \mathrm{NH}_2]\), where \(k' = \frac{k}{[\mathrm{H}_3\mathrm{O}^+]}\). This indicates that the reaction is first-order with respect to \(\mathrm{NO}_2 \mathrm{NH}_2\) in a buffered solution.
02

Analyze Mechanism 1

Mechanism 1 suggests a simple decomposition without involvement of \([\mathrm{H}_3\mathrm{O}^+]\), contradicting the dependence of the rate law on \([\mathrm{H}_3\mathrm{O}^+]\). Thus, Mechanism 1 is inconsistent with the observed rate law.
03

Analyze Mechanism 2

In Mechanism 2, the first step involves \(\mathrm{H}_3\mathrm{O}^+\) to form \(\mathrm{NO}_2\mathrm{NH}_3^+\), and the second step regenerates \(\mathrm{H}_3\mathrm{O}^+\). Applying the steady-state approximation for \(\mathrm{NO}_2\mathrm{NH}_3^+\), we can derive that the rate law matches the observed form. Mechanism 2 is consistent with the observed rate law.
04

Analyze Mechanism 3

Mechanism 3 introduces \(\mathrm{OH}^-\) in the equilibrium and involves \(\mathrm{H}_3\mathrm{O}^+\) in the rapid step; however, it doesn't directly explain the inverse dependence on \([\mathrm{H}_3\mathrm{O}^+]\) in the rate law. Mechanism 3 is less likely to match the observed rate law as well as Mechanism 2.
05

Show Relationship of Observed k with Mechanism 2 Constants

For Mechanism 2, the steady-state approximation gives us \(k = \frac{k_2 k_3}{k_2'}\). This results from assuming the intermediate \(\mathrm{NO}_2\mathrm{NH}_3^+\) is in rapid equilibrium compared to the slow step that controls the rate.
06

Demonstrate Hydroxide Ion Catalysis

Hydroxide ions can increase the rate of decomposition by neutralizing \(\mathrm{H}_3\mathrm{O}^+\), thus shifting the equilibrium in Mechanism 3 and promoting the breakdown of \(\mathrm{NO}_2\mathrm{NH}_2\) via increasing the proportion of deprotonated intermediates. This enhances the decomposition of nitramide.

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Key Concepts

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

Rate Laws
Rate laws describe how the concentration of reactants affects the rate of a chemical reaction. In this context, the rate law for nitramide decomposition is given by:
\[\text{Rate} = \frac{k[\mathrm{NO}_2\mathrm{NH}_2]}{[\mathrm{H}_3\mathrm{O}^+]}\]
This formula indicates that the rate of reaction is directly proportional to the concentration of nitramide \([\mathrm{NO}_2\mathrm{NH}_2]\) and inversely proportional to the concentration of hydronium ions \([\mathrm{H}_3\mathrm{O}^+]\). Thus, even though hydronium ions are not reactants in the overall reaction, they significantly influence the reaction rate.
  • Order of Reaction: The reaction is first-order concerning nitramide, meaning if the concentration of nitramide doubles, the reaction rate also doubles.
  • Equilibrium Impact: Keep in mind that in a buffered solution, the \([\mathrm{H}_3\mathrm{O}^+]\) remains constant, reinforcing the observation that the reaction appears first-order relative to nitramide.
Understanding these interactions is crucial for predicting how changes in concentration affect reaction speeds.
Equilibrium in Reactions
In many chemical reactions, temporary states of balance, known as equilibria, are established. These include rapid formations and breakdowns of intermediates before reaching the final products. This concept is expertly illustrated in the proposed mechanism 2 for nitramide decomposition.
This mechanism describes rapid equilibrium:
  • Initially, nitramide and hydronium ions establish an equilibrium to form an intermediate \(\mathrm{NO}_2\mathrm{NH}_3^+\).
  • This intermediate then quickly decomposes back to the hydronium ion and the final product, nitrous oxide \(\mathrm{N}_2\mathrm{O}\).
The equilibrium concept helps to understand why hydronium ions are inversely affecting the rate of reaction in the rate law. By maintaining constant hydronium ion concentration, any absorbed hydronium ions are rapidly released back into the solution following this equilibrium. Recognizing these rapid equilibria is central to understanding more complex kinetics.
Chemical Kinetics
Chemical kinetics is a domain of chemistry focused on understanding the speeds of reactions and the conditions affecting those speeds. For nitramide, the decomposition reaction delineates several key principles:
- **Rate-Determining Steps**: The slowest step in a reaction mechanism, often determines the overall reaction rate. In mechanism 2, the formation and decomposition of the intermediate \(\mathrm{NO}_2\mathrm{NH}_3^+\) highlight this principle.
- **Steady-State Approximation**: This useful assumption states that the concentration of an unstable intermediate remains relatively constant throughout the reaction. This assumption helps break down complex mechanisms into simpler rate laws.
- **Catalysis**: Introduction of a catalyst, such as hydroxide ions in mechanism 3, can modify the reaction pathway, effectively increasing the rate by altering rate-determining steps or shifting equilibria.In chemical kinetics, understanding the interplay of these factors allows chemists to predict and control reaction rates effectively, an essential for optimizing reactions in both industrial and laboratory settings.

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Most popular questions from this chapter

For a first-order reaction, what fraction of reactant remains after five half- lives have elapsed?

The decomposition of gaseous dimethyl ether at ordinary pressures is first order. Its half-life is 25.0 minutes at \(500^{\circ} \mathrm{C}\) $$\mathrm{CH}_{3} \mathrm{OCH}_{3}(\mathrm{g}) \rightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ (a) Starting with \(8.00 \mathrm{g}\) of dimethyl ether, what mass remains (in grams) after 125 minutes and after 145 minutes? (b) Calculate the time in minutes required to decrease \(7.60 \mathrm{ng}\) (nanograms) to 2.25 ng. (c) What fraction of the original dimethyl ether remains after 150 minutes?

Ammonia decomposes when heated according to the equation $$\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{NH}_{2}(\mathrm{g})+\mathrm{H}(\mathrm{g})$$ The data in the table for this reaction were collected at a high temperature. $$\begin{array}{cc}\text { Time (h) } & \text { [NH }\left._{3}\right] \text { (mol/L) } \\\\\hline 0 & 8.00 \times 10^{-7} \\\25 & 6.75 \times 10^{-7} \\\50 & 5.84 \times 10^{-7} \\\75 & 5.15 \times 10^{-7} \\\\\hline\end{array}$$ Plot ln \(\left[\mathrm{NH}_{3}\right]\) versus time and \(1 /\left[\mathrm{NH}_{3}\right]\) versus time. What is the order of this reaction with respect to NH \(_{3} ?\) Find the rate constant for the reaction from the slope.

At \(573 \mathrm{K},\) gaseous \(\mathrm{NO}_{2}(\mathrm{g})\) decomposes, forming \(\mathrm{NO}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) If a vessel containing \(\mathrm{NO}_{2}(\mathrm{g})\) has an initial concentration of \(2.8 \times 10^{-2} \mathrm{mol} / \mathrm{L},\) how long will it take for \(75 \%\) of the \(\mathrm{NO}_{2}(\mathrm{g})\) to decompose? The decomposition of \(\mathrm{NO}_{2}(\mathrm{g})\) is second order in the reactant and the rate constant for this reaction, at \(573 \mathrm{K},\) is \(1.1 \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}.\)

We want to study the hydrolysis of the beautiful green, cobalt-based complex called trans-dichlorobis(ethylenediamine) cobalt(III) ion, (Check your book to see figure) In this hydrolysis reaction, the green complex ion trans\(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) forms the red complex ion \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}\) as a \(\mathrm{Cl}^{-}\) ion is replaced with a water molecule on the \(\mathrm{Co}^{3+}\) ion (en \(=\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) ). The reaction progress is followed by observing the color of the solution. The original solution is green, and the final solution is red, but at some intermediate stage when both the reactant and product are present, the solution is gray. Changes in color with time as \(\mathrm{Cl}^{-}\) ion is replaced by \(\mathrm{H}_{2} \mathrm{O}\) in a cobalt(III) complex. The shape in the middle of the beaker is a vortex that arises because the solutions are being stirred using a magnetic stirring bar in the bottom of the beaker. Reactions such as this have been studied extensively, and experiments suggest that the initial, slow step in the reaction is the breaking of the Co-Cl bond to give a five-coordinate intermediate. The intermediate is then attacked rapidly by water. Slow: $$\begin{aligned}\operatorname{trans}-\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq}) & \rightarrow \\\&\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\end{aligned}$$ Fast: $$\begin{aligned}\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{aq}) & \rightarrow \\\&\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq})\end{aligned}$$ (a) Based on the reaction mechanism, what is the predicted rate law? (b) As the reaction proceeds, the color changes from green to red with an intermediate stage where the color is gray. The gray color is reached at the same time, no matter what the concentration of the green starting material (at the same temperature). How does this show the reaction is first order in the green form? Explain. (c) The activation energy for a reaction can be found by plotting In \(k\) versus \(1 / T .\) However, here we do not need to measure \(k\) directly. Instead, because \(k=-(1 / t) \ln \left([\mathrm{R}] /[\mathrm{R}]_{0}\right),\) the time needed to achieve the gray color is a measure of \(k .\) Use the data below to find the activation energy. $$\begin{array}{cc}\text { Temperature }^{\circ} \mathrm{C} & \text { (for the Same Initial Concentration) } \\\\\hline 56 & 156 \mathrm{s} \\\60 & 114 \mathrm{s} \\\65 & 88 \mathrm{s} \\\75 & 47 \mathrm{s} \\\\\hline\end{array}$$

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