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Chapter 27: Problem 4

For \(T<500 \mathrm{~K}\), the reaction $$ \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \stackrel{k_{\mathrm{obs}}}{\longrightarrow} \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{NO}(\mathrm{g}) $$ has the rate law $$ \frac{d\left[\mathrm{CO}_{2}\right]}{d t}=k_{\mathrm{obs}}\left[\mathrm{NO}_{2}\right]^{2} $$ Show that the following mechanism is consistent with the observed rate law $$ \begin{aligned} &\mathrm{NO}_{2}(\mathrm{~g})+\mathrm{NO}_{2}(\mathrm{~g}) \stackrel{k_{1}}{\longrightarrow} \mathrm{NO}_{3}(\mathrm{~g})+\mathrm{NO}(\mathrm{g}) \quad(\text { rate determining }) \\ &\mathrm{NO}_{3}(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{NO}_{2}(\mathrm{~g}) \end{aligned} $$ Express \(k_{\text {obs }}\) in terms of \(k_{1}\) and \(k_{2}\).

Short Answer

Expert verified
The mechanism is consistent with \(k_{\text{obs}} = k_1\).

Step by step solution

01

Analyze the Rate-Determining Step

Identify the slowest step in the reaction mechanism, which is considered the rate-determining step. According to the given mechanism, the reaction \( \text{NO}_2(\text{g}) + \text{NO}_2(\text{g}) \to \text{NO}_3(\text{g}) + \text{NO}(\text{g}) \) is the rate-determining step. Its rate is expressed as: \( \text{Rate} = k_1 [\text{NO}_2]^2 \).
02

Relate Rate Law to Mechanism

Compare the rate expression obtained from the rate-determining step, \( k_1 [\text{NO}_2]^2 \), to the given rate law of the overall reaction, \( \frac{d[\text{CO}_2]}{dt} = k_{\text{obs}}[\text{NO}_2]^2 \). Since the first step is the rate-determining step, the rate law from this step must match the observed rate law.
03

Identify Consistency Between Mechanism and Rate Law

Realize that the first step defines the rate of the overall reaction because it is slow compared to the second step. The presence of \([\text{NO}_2]^2\) in both the rate from the mechanism and the observed rate law confirms their consistency. Hence, \(k_{\text{obs}} = k_1\).
04

Determine Overall Rate Constant

The second step of the mechanism quickly consumes \(\text{NO}_3\) and thus its concentration remains low. Since the first step provides \(\text{NO}_3\) for the second reaction, the mechanism ensures that \(k_{\text{obs}} = k_1\). The consistency between the inertia of the first reaction and the overall rate confirms that no additional terms (e.g., involving \(k_2\)) appear in \(k_{\text{obs}}\).

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Key Concepts

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

Rate Law
In the world of chemical kinetics, the rate law is an expression that relates the reaction rate to the concentration of reactants. For any given reaction, the rate law can be expressed in terms of the concentration of reactants raised to a power, which corresponds to the order of the reaction with respect to each reactant.

For the reaction \[\text{NO}_2(\text{g}) + \text{CO}(\text{g}) \rightarrow \text{CO}_2(\text{g}) + \text{NO}(\text{g}),\]the observed rate law is \[\frac{d[\text{CO}_2]}{dt} = k_{\text{obs}}[\text{NO}_2]^2.\]This rate law indicates that the rate of the reaction depends quadratically on the concentration of \([\text{NO}_2]\).

Understanding this relationship is crucial because it provides insights into the reaction's mechanism and which steps are important in controlling the overall rate. When comparing the experimental rate law to a proposed mechanism, chemists ensure that the overall rate law and observed concentrations align with predicted steps.
Reaction Mechanism
A reaction mechanism is a sequence of elementary steps that describe how a reaction occurs at the molecular level. These steps form a pathway from reactants to products.

In our discussed reaction, the proposed mechanism consists of the following two steps:
  • Step 1: \(\text{NO}_2(\text{g}) + \text{NO}_2(\text{g}) \rightarrow \text{NO}_3(\text{g}) + \text{NO}(\text{g})\)
  • Step 2: \(\text{NO}_3(\text{g}) + \text{CO}(\text{g}) \rightarrow \text{CO}_2(\text{g}) + \text{NO}_2(\text{g})\)
These steps suggest that \(\text{NO}_2\) molecules first interact to form \(\text{NO}_3\) and \(\text{NO}\), a crucial intermediate in the process, before \(\text{NO}_3\) reacts with \(\text{CO}\) in the next step.

The mechanism helps chemists understand why certain reactant concentrations appear in the rate law while others do not. It bridges the gap between the observed rate law and the apparent complexity of multi-step reactions. Understanding the mechanism provides insights into which species are transient and have low concentrations during the reaction.
Rate-Determining Step
The rate-determining step (RDS) is the slowest step in a reaction mechanism, which limits or 'determines' the speed at which the overall reaction proceeds.

In our mechanism for the reaction involving \(\text{NO}_2\), the first step:\[\text{NO}_2(\text{g}) + \text{NO}_2(\text{g}) \rightarrow \text{NO}_3(\text{g}) + \text{NO}(\text{g})\]is the rate-determining step. Rate law derived from this step gives the expression\[\text{Rate} = k_1 [\text{NO}_2]^2,\]which matches the observed rate law, indicating that this step controls the overall reaction rate.

Because the RDS is slower than subsequent steps, it essentially forms a bottleneck, dictating the reaction pace. Understanding the RDS is vital for optimizing reactions and designing catalysts to expedite the slower steps, thus enhancing the efficiency of a reaction's pathway.

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Most popular questions from this chapter

Antibiotic-resistant bacteria have an enzyme, penicillinase, that catalyzes the decomposition of the antibiotic. The molecular mass of penicillinase is \(30000 \mathrm{g} \cdot \mathrm{mol}^{-1}\). The turnover number of the enzyme at \(28^{\circ} \mathrm{C}\) is \(2000 \mathrm{s}^{-1}\). If \(6.4 \mu \mathrm{g}\) of penicillinase catalyzes the destruction of \(3.11 \mathrm{mg}\) of amoxicillin, an antibiotic with a molecular mass of \(364 \mathrm{g} \cdot \mathrm{mol}^{-1}\) in 20 seconds at \(28^{\circ} \mathrm{C},\) how many active sites does the enzyme have?

Show that the Michaelis-Menton mechanism for enzyme catalysis gives \(v=(1 / 2) v_{\max }\) when \([\mathrm{S}]_{0}=K_{m}\).

Nitramide \(\left(\mathrm{O}_{2} \mathrm{NNH}_{2}\right)\) decomposes in water according to the chemical equation $$\mathrm{O}_{2} \mathrm{N} \mathrm{NH}_{2}(\mathrm{aq}) \stackrel{k_{\mathrm{obs}}}{\rightarrow} \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$ The experimentally determined rate law for this reaction is $$\frac{d\left[\mathrm{N}_{2} \mathrm{O}\right]}{d t}=k_{\mathrm{obs}} \frac{\left[\mathrm{O}_{2} \mathrm{NNH}_{2}\right]}{\left[\mathrm{H}^{+}\right]}$$ A proposed mechanism for this reaction is, Is this mechanism consistent with the observed rate law? If so, what is the relationship between \(k_{\text {obs }}\) and the rate constants for the individual steps of the mechanism?

Consider the reaction mechanism $$A+B \stackrel{k_{1}}{\rightleftharpoons} C$$ $$\mathrm{C} \stackrel{k_{2}}{\Rightarrow} \mathrm{P}$$ Write the expression for \(d[\mathrm{P}] / d t,\) the rate of product formation. Assume equilibrium is established in the first reaction before any appreciable amount of product is formed, and thereby show that $$\frac{d[\mathrm{P}]}{d t}=k_{2} K_{c}[\mathrm{A}][\mathrm{B}]$$ where \(K_{c}\) is the equilibrium constant for step ( 1 ) of the reaction mechanism. This assumption is called the fast- equilibrium approximation.

Consider the following mechanism for the recombination of bromine atoms to form molecular bromine $$\begin{array}{c} 2 \operatorname{Br}(\mathrm{g}) \stackrel{k_{1}}{\rightleftharpoons} \operatorname{Br}_{2}^{*}(\mathrm{g}) \\ \mathrm{Br}_{2}^{*}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{Br}_{2}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \end{array}$$ The first step results in formation of an energized bromine molecule. This excess energy is then removed by a collision with a molecule \(\mathrm{M}\) in the sample. Show that if the steady-state approximation is applied to \(\mathrm{Br}_{2}^{*}(\mathrm{g}),\) then $$\frac{d[\mathrm{Br}]}{d t}=-\frac{2 k_{1} k_{2}[\mathrm{Br}]^{2}[\mathrm{M}]}{k_{-1}+k_{2}[\mathrm{M}]}$$ Determine the limiting expression for \(d[\mathrm{Br}] / d t\) when \(v_{2} \gg v_{-1} .\) Determine the limiting expression for \(d[\operatorname{Br}] / d t\) when \(v_{2} \ll v_{-1}\).
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