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OBJECTIVE. Predict the experimental rate law from the mechanism and differentiate among possible reaction mechanisms by examining experimental rate data. Evaluate each of the following proposed mechanisms to determine whether it is consistent with the experimentallydetermined stochiometry and rate law, and identify intermediates, if any. $$ 2 \mathrm{NO}_{2}+\mathrm{O}_{3} \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}+\mathrm{O}_{2} $$ rate \(=k\left[\mathrm{NO}_{2}\right]\left[\mathrm{O}_{3}\right]\) $$ \text { (a) } 2 \mathrm{NO}_{2} \underset{k_{1}}{\stackrel{k_{1}}{\rightleftarrows}} \mathrm{N}_{2} \mathrm{O}_{4} $$ Fast, reversible $$ \mathrm{N}_{2} \mathrm{O}_{4}+\mathrm{O}_{3} \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}+\mathrm{O}_{2} $$ (b) \(\mathrm{NO}_{2}+\mathrm{O}_{3} \rightarrow \mathrm{NO}_{3}+\mathrm{O}_{2}\) Slow $$ \mathrm{NO}_{3}+\mathrm{NO}_{2} \rightarrow \mathrm{N}_{2} \mathrm{O}_{5} $$ Fast

Short Answer

Expert verified
Mechanism (b) is consistent with the experimental rate law.

Step by step solution

01

Identify the Overall Reaction

The overall reaction is given as \(2 \text{NO}_2 + \text{O}_3 \rightarrow \text{N}_2\text{O}_5 + \text{O}_2\). This serves as the basis to evaluate the proposed mechanisms.
02

Examine Proposed Mechanism (a)

For mechanism (a), the steps are:1. \(2 \text{NO}_2 \rightleftarrows \text{N}_2\text{O}_4\) (fast, reversible)2. \(\text{N}_2\text{O}_4 + \text{O}_3 \rightarrow \text{N}_2\text{O}_5 + \text{O}_2\) (slow)Intermediates: \(\text{N}_2\text{O}_4\) is formed in the first step and consumed in the second step.
03

Determine Rate Law for Mechanism (a)

Since the second step is the slow step, it determines the rate law. According to the rate-determining step:\[ \text{Rate} = k_2[\text{N}_2\text{O}_4][\text{O}_3] \]Using the equilibrium of the fast, reversible first step, we know:\[ K_{eq} = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2} \]This implies that \([\text{N}_2\text{O}_4] = K_{eq}[\text{NO}_2]^2\). Substituting back into the rate law:\[ \text{Rate} = k_2 K_{eq} [\text{NO}_2]^2 [\text{O}_3] \]
04

Compare with Experimental Rate Law

The experimental rate law is \(\text{Rate} = k[\text{NO}_2][\text{O}_3]\). Mechanism (a) predicts a rate law of \(\text{Rate} = k_2 K_{eq} [\text{NO}_2]^2 [\text{O}_3]\), which does not match the experimental rate law. Thus, mechanism (a) is inconsistent with the experimental data.
05

Examine Proposed Mechanism (b)

For mechanism (b), the steps are:1. \(\text{NO}_2 + \text{O}_3 \rightarrow \text{NO}_3 + \text{O}_2 \) (slow)2. \(\text{NO}_3 + \text{NO}_2 \rightarrow \text{N}_2\text{O}_5 \) (fast)Intermediates: \(\text{NO}_3\) is formed in the first step and consumed in the second step.
06

Determine Rate Law for Mechanism (b)

Since the first step is the slow step, it determines the rate law. The rate law for the slow step is:\[ \text{Rate} = k_1 [\text{NO}_2][\text{O}_3] \]
07

Compare with Experimental Rate Law

The determined rate law from mechanism (b) matches the experimental rate law: \(\text{Rate} = k[\text{NO}_2][\text{O}_3]\). This indicates that mechanism (b) is consistent with the experimental findings.

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Key Concepts

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

Rate Law
In chemistry, the rate law is essential for understanding how reaction rates relate to the concentration of reactants. It expresses this relationship through a mathematical equation. The rate law for a reaction is determined from experimental data and identifies which reactants influence the rate of reaction. The expression typically looks like:
\[\text{Rate} = k [A]^x[B]^y\]where:
  • \(k\) is the rate constant.
  • \([A]\) and \([B]\) are the concentrations of the reactants, and \(x\) and \(y\) are the orders with respect to each reactant.
In the context of this exercise, the experimental rate law is given as:\[\text{Rate} = k [\text{NO}_2][\text{O}_3]\]This indicates that the reaction rate is dependent on the concentration of nitrogen dioxide \(\text{NO}_2\) and ozone \(\text{O}_3\), both to the first power, meaning the reaction is first-order with respect to each reactant. Recognizing the rate-determining step within a mechanism is crucial as it allows us to connect the mechanism to the observed rate law.
Intermediates
In multi-step reactions, intermediates are species that form in one elementary step and are consumed in another. They are not present in the overall balanced equation, as they do not appear as final products or reactants. Instead, they play a vital role by facilitating the transition between reactants and products. When evaluating potential reaction mechanisms, identifying intermediates helps to differentiate steps within a process, shedding light on each elementary reaction's progression.
For example, in mechanism (a), \(\text{N}_2\text{O}_4\) is identified as an intermediate. It forms during the reversible conversion of \(\text{NO}_2\) and is then used up in the reaction with \(\text{O}_3\). Similarly, in mechanism (b), the intermediate \(\text{NO}_3\) is generated in the first slow reaction step and utilized quickly in a subsequent reaction with \(\text{NO}_2\). Recognizing intermediates helps us deduce that mechanism (b) aligns with the experimental rate law, as it accurately captures the involvement of these transient species in the rate-determining step.
Stoichiometry
Stoichiometry is the calculation of reactants and products in chemical reactions. It involves using the balanced chemical equation to determine the proportions of each substance involved in a reaction. This concept ensures that the law of conservation of mass is followed, meaning no atoms are lost or gained during a reaction; they're merely rearranged.
To evaluate the consistency of a proposed mechanism with the experimental data, one must compare the stoichiometry of the mechanism's steps with the overall reaction. In this exercise, the overall reaction is:\[2 \text{NO}_2 + \text{O}_3 \rightarrow \text{N}_2\text{O}_5 + \text{O}_2\]In mechanism (b), this stoichiometry is upheld through the steps \(\text{NO}_2 + \text{O}_3 \rightarrow \text{NO}_3 + \text{O}_2\) followed by \(\text{NO}_3 + \text{NO}_2 \rightarrow \text{N}_2\text{O}_5\). Each step respects the overall stoichiometry, where the usage of reactants and formation of products ultimately account for the consumption and generation seen in the overall balanced reaction. Proper stoichiometric analysis can thus confirm that a mechanism is valid if both stoichiometry and the rate law align.

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Most popular questions from this chapter

OBJECTIVE. Write the rate law from an elementary step and determine its molecularity. Assuming that each reaction is elementary, predict the rate law and molecularity. (a) \(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})\) (b) \(\mathrm{O}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{~g}) \rightarrow 2 \mathrm{O}_{2}(\mathrm{~g})\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(\mathrm{aq}) \rightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\) (d) \(2 \mathrm{HI}(\mathrm{g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\)

OBJECTIVE. Draw energy-level diagrams for catalyzed and uncatalyzed reactions. A catalyst decreases the activation energy of a particular endothermic reaction by \(50 \mathrm{~kJ} / \mathrm{mol}\), from 140 to \(90 \mathrm{~kJ} / \mathrm{mol}\). Assuming that the reaction is endothermic, that the mechanism has only one step, and that the products differ from the reactants by \(20 \mathrm{~kJ}\), sketch approximate energy-level diagrams for the catalyzed and uncatalyzed reactions.

OBJECTIVE. Calculate the concentration-time behavior for a first-order reaction from the rate law and the rate constant. When formic acid is heated, it decomposes to hydrogen and carbon dioxide in a first-order decay. $$ \mathrm{HCOOH}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) $$ At \(550^{\circ} \mathrm{C}\), the half-life of formic acid is 24.5 minutes. (a) What is the rate constant, and what are its units? (b) How many seconds are needed for formic acid, initially \(0.15 M\), to decrease to 0.015 M?

OBJECTIVE. Draw energy-level diagrams for catalyzed and uncatalyzed reactions. A catalyst decreases the activation energy of a particular exothermic reaction by \(15 \mathrm{~kJ} / \mathrm{mol}\), from 40 to \(25 \mathrm{~kJ} / \mathrm{mol}\). Assuming that the reaction is exothermic, that the mechanism has only one step, and that the products differ from the reactants by \(40 \mathrm{~kJ}\), sketch approximate energy-level diagrams for the catalyzed and uncatalyzed reactions.

OBJECTIVE. Use stoichiometry to relate the rate of reaction to changes in the concentrations of reactants and products. For the reaction, $$ 2 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{3}(\mathrm{~g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) $$ the dinitrogen pentoxide appears at a rate of \(0.0055 \mathrm{M} / \mathrm{s}\). Calculate the rate at which the \(\mathrm{NO}_{2}\) disappears and the rate of the reaction.

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