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NO catalyzes the decomposition of \(\mathrm{N}_{2} \mathrm{O},\) possibly by the following mechanism: $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) & \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) & \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Why is NO considered a catalyst and not an intermediate? (c) If experiments show that during the decomposition of \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism? If you think not, suggest what might be going on.

Short Answer

Expert verified
The overall reaction equation for the decomposition of N2O is N2O(g) → N2(g) + O2(g). NO is considered a catalyst, as it is involved in the reaction but not consumed by the end. The low accumulation of NO2 does not rule out the proposed mechanism but supports it, as NO2 could be rapidly reacting, keeping its concentration low.

Step by step solution

01

Determine the overall reaction equation

To find the overall chemical equation, we need to add both steps of the given mechanism and simplify. Given steps: 1. NO(g) + N2O(g) → N2(g) + NO2(g) 2. 2NO2(g) → 2NO(g) + O2(g) Add both equations together: NO(g) + N2O(g) + 2NO2(g) → N2(g) + NO2(g) + 2NO(g) + O2(g) Now cancel out the species appearing on both sides: N2O(g) → N2(g) + O2(g)
02

Determine if NO is a catalyst or an intermediate

A catalyst is a substance that is involved in a reaction and helps speed up the reaction, but is not consumed in the process. An intermediate is a species that is formed and consumed during the overall reaction. Since NO appears in both steps, but not in the overall reaction equation: N2O(g) → N2(g) + O2(g) NO is considered a catalyst, not an intermediate, because it is involved in the reaction but is not consumed by the end of the process.
03

Discuss whether the experimental evidence rules out the proposed mechanism or not

The experimental evidence shows that NO2 does not accumulate in measurable quantities during the decomposition of N2O, which is consistent with this proposed mechanism since NO2 is an intermediate. This observation does not rule out the proposed mechanism but supports it. It is possible that the reaction occurs rapidly, so the concentration of NO2 remains low with little to no accumulation.

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Key Concepts

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

Catalysts in Chemistry
In chemistry, a catalyst is a remarkable substance. It helps speed up a reaction without being consumed in the process. Think of it as a helper that pushes a sled down a hill faster but doesn't end up at the bottom of the hill itself.

In the given reaction, nitrogen monoxide (NO) acts as a catalyst. It initially reacts with dinitrogen monoxide (N2O) to produce nitrogen (N2) and nitrogen dioxide (NO2). Yet, as seen from the steps, NO is regenerated.
  • First, NO and N2O react.
  • Then, some of the NO is produced again from NO2 in the second step.
At the end of both steps, the same amount of NO exists as initially.

Because NO is not used up by the end of the reaction, it is not part of the overall chemical equation: \[N_{2}O(g) \rightarrow N_{2}(g) + O_{2}(g) \]Thus, NO fulfills the role of a catalyst.
Chemical Kinetics
Chemical kinetics is like the pulse of a reaction. It studies how fast reactions proceed and the factors affecting these speeds. It's akin to knowing how fast different cakes bake under varying oven conditions.

In reaction mechanisms, kinetics can explain why some steps are slow or fast. Importantly, the proposed decomposition mechanism involves two steps. The first step is the union of NO and N2O, and the second step breaks NO2 back into NO and releases O2.
  • The speed of the first step can dictate the visible progress of the reaction.
  • Each step may occur at different rates, impacting the overall reaction.
The detailed kinetics provide clues on the reaction progress and mechanism, making it easier to confirm models like the one presented. Understanding these kinetics helps in identifying the role of intermediates and catalysts within the process.
Reaction Intermediates
Reaction intermediates are like transient guests in a chemical reaction party. They appear in some steps but are not present in the final exit.

In the mechanism involving the decomposition of N2O, nitrogen dioxide (NO2) is a reaction intermediate. It is formed in the first step but consumed in the second.
  • NO2 forms, then quickly converts back to NO and O2.
  • Its presence is brief, thus not showing up in the overall chemical equation.
The absence of NO2 accumulation in experimental observations aligns with it being an intermediate.

Its rapid consumption suggests the steps proceed swiftly, such that NO2 levels remain negligible. This supports the proposed mechanism's credibility, showing how intermediates, despite their elusive nature, help piece together the reaction puzzle thoroughly.

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Most popular questions from this chapter

Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law. (b) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\), what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathrm{c}) \mathrm{What}\) is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of \(\mathrm{NO}\) is increased to \(0.10 \mathrm{M},\) while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M}\) ?

Consider the following reaction: $$ \mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q) $$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-}\). When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M},\) the reaction rate at \(298 \mathrm{~K}\) is \(0.0432 \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) What would happen to the rate if the concentration of \(\mathrm{OH}^{-}\) were tripled? (d) What would happen to the rate if the concentration of both reactants were tripled?

Sketch a graph for the generic first-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) that has concentration of \(\mathrm{A}\) on the vertical axis and time on the horizontal axis. (a) Is this graph linear? Explain. (b) Indicate on your graph the half-life for the reaction.

The following is a quote from an article in the August 18,1998 , issue of The New York Times about the breakdown of cellulose and starch: "A drop of 18 degrees Fahrenheit [from \(77^{\circ} \mathrm{F}\) to \(\left.59^{\circ} \mathrm{F}\right]\) lowers the reaction rate six times; a 36 -degree drop [from \(77^{\circ} \mathrm{F}\) to \(\left.41^{\circ} \mathrm{F}\right]\) produces a fortyfold decrease in the rate." (a) Calculate activation energies for the breakdown process based on the two estimates of the effect of temperature on rate. Are the values consistent? (b) Assuming the value of \(E_{a}\) calculated from the 36 -degree drop and that the rate of breakdown is first order with a half-life at \(25^{\circ} \mathrm{C}\) of 2.7 years, calculate the half-life for breakdown at a temperature of \(-15^{\circ} \mathrm{C}\).

The temperature dependence of the rate constant for a reaction is tabulated as follows: $$ \begin{array}{lc} \hline \text { Temperature (K) } & k\left(M^{-1} \mathrm{~s}^{-1}\right) \\ \hline 600 & 0.028 \\ 650 & 0.22 \\ 700 & 1.3 \\ 750 & 6.0 \\ 800 & 23 \\ \hline \end{array} $$ Calculate \(E_{a}\) and \(A\).

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