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For \(T<500 \mathrm{K},\) the reaction $$\mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \stackrel{k_{\text {obs }}}{\longrightarrow} \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g})$$ 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 $$\mathrm{NO}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \stackrel{k_{1}}{\longrightarrow} \mathrm{NO}_{3}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) \quad(\text { rate determining })$$ $$\mathrm{NO}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g})$$ Express \(k_{\text {obs }}\) in terms of \(k_{1}\) and \(k_{2}\).

Short Answer

Expert verified
The rate law is consistent; \(k_{\text{obs}} = k_1\).

Step by step solution

01

Identify Rate-Determining Step

The given mechanism suggests that the first step involving \(\mathrm{NO}_2(\mathrm{g}) + \mathrm{NO}_2(\mathrm{g}) \stackrel{k_{1}}{\longrightarrow} \mathrm{NO}_3(\mathrm{g}) + \mathrm{NO}(\mathrm{g})\) is the rate-determining step. This means that the rate of this slow step dictates the overall rate of the reaction.
02

Write the Rate Law for Rate-Determining Step

For the rate-determining step, the rate law is decided solely by the reactants involved in this step. So the rate of formation of \(\mathrm{NO}_3\) and \(\mathrm{NO}\) is \(\text{Rate} = k_1 [\mathrm{NO}_2]^2\).
03

Consider the Fast Step Equilibrium

The second step, \(\mathrm{NO}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g})\), is fast, and can be assumed to proceed to completion almost as soon as \(\mathrm{NO}_3\) is formed.
04

Relate the Mechanism to Observed Rate Law

According to the mechanism, \(\mathrm{NO}_{3}\) is a reactive intermediate and its concentration can be approximated as being in a rapid dynamic equilibrium due to its quick reactivity in the presence of \(\mathrm{CO}\). Hence the observed rate \( \frac{d[\mathrm{CO}_2]}{dt} = k_1[\mathrm{NO}_2]^2\) aligns with the overall rate determined by the slow step.
05

Express \(k_{\text{obs}}\) in Terms of \(k_1\) and \(k_2\)

Given that the step \(\mathrm{NO}_3 + \mathrm{CO} \rightarrow \mathrm{CO}_2 + \mathrm{NO}_2\) proceeds immediately after \(\mathrm{NO}_3\) forms, and the slow step is rate-determining, we equate \(k_{\text{obs}} = k_1\), since the overall rate is governed by the rate of formation of \(\mathrm{NO}_3\) and \(\mathrm{NO}\).

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Key Concepts

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

Rate Determining Step
In any multi-step reaction mechanism, not all steps occur at the same speed. The slowest step is called the rate-determining step. It acts as a bottleneck that limits the speed of the entire reaction. Understanding which step is the rate-determining step helps in predicting the overall rate of the reaction.

For the reaction involving nitrogen dioxide \(\left( \text{NO}_2 \right)\) and carbon monoxide \(\left( \text{CO} \right)\), the first step of the mechanism, which is the reaction of two \(\text{NO}_2\) molecules to form \(\text{NO}_3\) and \(\text{NO}\), is identified as the rate-determining step. This implies that the rate of this slow step, being the slowest, dictates the speed of the entire reaction.

In kinetics, the slower the step, the more it restricts progressing to subsequent steps. Hence, understanding the rate-determining step allows chemists to control the reaction speed more efficiently.
Rate Law
The rate law of a chemical reaction provides an equation that relates the rate of the reaction to the concentration of its reactants. This serves as a mathematical expression showing how the concentration of reactants affects the speed of the reaction.

For the given exercise, the rate law derived from the rate-determining step is \( \text{Rate} = k_1 [\text{NO}_2]^2 \). This expression indicates that the reaction rate is directly proportional to the square of the concentration of \(\text{NO}_2\).

The square exponent signifies that the rate of reaction is particularly sensitive to changes in \(\text{NO}_2\) concentration. If you double the concentration of \(\text{NO}_2\), the rate of the reaction increases fourfold. This shows the importance of determining the correct rate law for predicting how the reaction will proceed under various conditions.
Reactive Intermediate
In multi-step reactions, intermediates are species that are formed in one step and consumed in a subsequent step. They do not appear in the overall balanced equation of the reaction, but they play a critical role in the mechanism.

In the given problem, \(\text{NO}_3\) is a reactive intermediate. It is produced in the rate-determining step (step 1) and immediately consumed in the fast, subsequent step. The fleeting existence of \(\text{NO}_3\) highlights its unstable nature, causing it to rapidly react with \(\text{CO}\) to form \(\text{CO}_2\) and regenerate \(\text{NO}_2\).

Since intermediates are often very reactive, they partake in reactions that are fast and tend to establish equilibrium quickly. Understanding reactive intermediates helps to clarify complex reaction mechanisms and allows chemists to modify reactions by stabilizing or destabilizing these unstable species.
Equilibrium
Equilibrium in chemical kinetics refers to a state where the concentrations of reactants and products remain constant over time, typically due to a balance between forward and reverse reactions. In the context of multi-step reactions, although the reaction doesn’t reach a stop, some fast steps may approach dynamic equilibrium very quickly.

For this kinetic problem, the second step in the mechanism is considered to be fast and closely associated with equilibrium. When \(\text{NO}_3\) is formed during the rate-determining step, it quickly reacts with \(\text{CO}\) in the second step. The fast nature of this step suggests that it could be approximated by assuming a dynamic equilibrium between \(\text{NO}_3\) and its reactants and products.

Understanding equilibrium in relation to the fast steps in a reaction mechanism is crucial, as it can help approximate the makeup of transient species in solutions and predict how these components contribute to the overall reaction rate. It further aids in easily deriving expressions connecting different rate constants in mechanism analysis.

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Most popular questions from this chapter

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?

Determine the rate law for the reaction $$ \mathrm{I}(\mathrm{g})+\mathrm{I}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \stackrel{k}{\Longrightarrow} \mathrm{I}_{2}(\mathrm{~g})+\mathrm{M}(\mathrm{g}) $$ where \(M\) is any molecule present in the reaction container. Give the units of \(k\). Determine the molecularity of this reaction. Is this reaction identical to $$ \mathrm{I}(\mathrm{g})+\mathrm{I}(\mathrm{g}) \stackrel{k}{\Longrightarrow} \mathrm{I}_{2}(\mathrm{~g}) $$ Explain.

The protein catalase catalyzes the reaction $$2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{g})$$ and has a Michaelis constant of \(K_{m}=25 \times 10^{-3} \mathrm{mol} \cdot \mathrm{dm}^{-3}\) and a turnover number of \(4.0 \times 10^{7} \mathrm{s}^{-1} .\) Calculate the initial rate of this reaction if the total enzyme concentration is \(0.016 \times 10^{-6} \mathrm{mol} \cdot \mathrm{dm}^{-3}\) and the initial substrate concentration is \(4.32 \times 10^{-6} \mathrm{mol} \cdot \mathrm{dm}^{-3}\) Calculate \(v_{\max }\) for this enzyme. Catalase has a single active site.

The turnover number for acetylcholinesterase, an enzyme with a single active site that metabolizes acetylcholine, is \(1.4 \times 10^{4} \mathrm{s}^{-1} .\) How many grams of acetylcholine can \(2.16 \times 10^{-6} \mathrm{g}\) of acetylcholinesterase metabolize in one hour? (Take the molecular mass of the enzyme to be \(4.2 \times 10^{4} \mathrm{g} \cdot \mathrm{mol}^{-1} ;\) acetylcholine has the molecular formula \(\mathrm{C}_{7} \mathrm{NO}_{2} \mathrm{H}_{16}^{+} .\) )

The \(\mathrm{HF}(\mathrm{g})\) chemical laser is based on the reaction $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HF}(\mathrm{g}) $$ The mechanism for this reaction involves the elementary steps \(-\Delta_{\mathrm{r}} I I^{\circ} / \mathrm{kJ} \cdot \mathrm{mol}^{-1}\) at \(298 \mathrm{~K}\) (1) \(\mathrm{F}_{2}(\mathrm{~g})+\mathrm{M}(\mathrm{g}) \stackrel{\stackrel{k_{1}}{\rightleftarrows}}{\stackrel{k_{-1}}} 2 \mathrm{~F}(\mathrm{~g})+\mathrm{M}(\mathrm{g}) \quad+159\) (2) \(\mathrm{F}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g}) \stackrel{\stackrel{k_{2}}{\rightleftarrows}}{\stackrel{k_{-2}}} \mathrm{HF}(\mathrm{g})+\mathrm{H}(\mathrm{g}) \quad-134\) (3) \(\mathrm{H}(\mathrm{g})+\mathrm{F}_{2}(\mathrm{~g}) \stackrel{k_{3}}{\Longrightarrow} \mathrm{HF}(\mathrm{g})+\mathrm{F}(\mathrm{g}) \quad-411\) Comment on why the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{M}(\mathrm{g}) \rightarrow 2 \mathrm{H}(\mathrm{g})+\mathrm{M}(\mathrm{g})\) is not included in the mechanism of the HF(g) laser even though it produces a reactant that could participate in step (3) of the reaction mechanism. Derive the rate law for \(d[\mathrm{HF}] / d t\) for the above mechanism assuming that the steady-state approximation can be applied to both intermediate species, \(\mathrm{F}(\mathrm{g})\) and \(\mathrm{H}(\mathrm{g})\)

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