Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

NO catalyzes the decomposition of \(\mathrm{N}_{2} \mathrm{O}\), possibly by the following mechanism: $$ \begin{array}{r} \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{array} $$ (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 chemical equation for the decomposition of N2O is: N2O(g) -> N2(g) + O2(g). NO is considered a catalyst instead of an intermediate because it helps to initiate the chemical reaction without being consumed in the overall reaction. The lack of NO2 accumulation doesn't necessarily rule out the proposed mechanism; it might suggest that the second step is happening quickly or there is another decomposition pathway that doesn't produce NO2.

Step by step solution

01

(a) Finding the overall chemical equation.

We have the following reactions: 1: NO(g) + N2O(g) -> N2(g) + NO2(g) 2: 2NO2(g) -> 2NO(g) + O2(g) To find the overall reaction, we must cancel out intermediate species and add step 1 and step 2. Intermediates are species that appear on both the reactant and product sides of the overall reaction (not necessary within individual steps).
02

(b) Explaining the role of NO as a catalyst.

NO is considered a catalyst rather than an intermediate because it helps initiate the chemical reaction but is not consumed by the overall reaction. Catalysts are substances that speed up a chemical reaction without themselves undergoing any permanent chemical change. We will confirm NO as a catalyst in the overall reaction when we add the above reactions.
03

(c) Discussing whether the lack of NO2 accumulation rules out the proposed mechanism.

If experiments show that NO2 does not accumulate in measurable quantities, we need to analyze the proposed mechanism to determine if it is still viable, and suggest possible reasons for the lack of NO2 accumulation. Now let's find the overall chemical equation and answer the remaining questions.
04

Adding the reaction steps to find the overall equation.

Add step 1 and step 2: 1: NO(g) + N2O(g) -> N2(g) + NO2(g) 2: 2NO2(g) -> 2NO(g) + O2(g) ----------------------------- Overall: N2O(g) -> N2(g) + O2(g) The intermediate, NO2, is canceled out as it appears on both sides of the overall reaction. Furthermore, NO is not consumed in the overall reaction, as it appears on both sides of reaction steps but not in the overall reaction, making it a catalyst. Since we found the overall reaction to be N2O(g) -> N2(g) + O2(g), it is clear that NO is a catalyst and not an intermediate.
05

Analyzing the NO2's non-accumulation.

Now let's address part (c) regarding the lack of NO2 accumulation. The lack of NO2 accumulation does not necessarily rule out the proposed mechanism. It is possible that the second step of the reaction is happening so quickly that the concentration of NO2 remains very low. This would suggest that both reaction steps might be taking place almost simultaneously, which is commonly seen in catalytic cycles. Another possibility is that there is another pathway for the decomposition of N2O that does not produce NO2, and the decomposition occurs through the non-NO2 pathway as well. In this case, the overall reaction would still be supported, but the proposed mechanism would not be the only one responsible for the decomposition. In conclusion, the lack of measurable NO2 does not necessarily rule out the proposed mechanism, but it might indicate other factors that are influencing the decomposition process.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Chemical Kinetics
Chemical kinetics is the study of how fast reactions occur. It involves understanding how different factors affect the speed of a chemical reaction. A catalyst can greatly influence these rates. Catalysts provide an alternative reaction pathway that lowers the activation energy, meaning that less energy is required for the reaction to occur. This can make reactions happen faster.In the context of the decomposition of nitrogen monoxide (\[\text{NO}\]) and dinitrogen oxide (\[\text{N}_2\text{O}\]), introducing \[\text{NO}\] speeds up the process as it catalyzes the reaction. The rate at which the overall reaction occurs can be measured and compared in different scenarios, such as with and without the catalyst.To sum up:
  • Catalysts speed up reactions by reducing activation energy.
  • They are not consumed in the reaction, meaning they can be reused.
  • Studying reaction rates helps understand catalyst effectiveness.
Reaction Mechanism
A reaction mechanism outlines the step-by-step sequence of elementary reactions by which an overall chemical change occurs. Each step in the sequence is known as an elementary step, and together they form the complete reaction pathway.In the given catalytic process involving \[\text{N}_2\text{O}\] dissociation:
  • The first step involves \[\text{NO}\] and \[\text{N}_2\text{O}\] forming \[\text{N}_2\] and \[\text{NO}_2\].
  • The second step is the rapid transformation of \[\text{NO}_2\] back into \[\text{NO}\] and the release of oxygen, \[\text{O}_2\].
Adding both steps provides the overall balanced equation, eliminating intermediates like \[\text{NO}_2\] and showing the direct conversion of \[\text{N}_2\text{O}\] to \[\text{N}_2\] and \[\text{O}_2\].Understanding these steps is crucial because:
  • It helps in identifying catalysts and intermediates.
  • It provides insight into the speed and efficiency of the reaction.
  • It supports verifying the proposed mechanism through experiments.
Intermediates
Intermediates are species that are produced in one step of a reaction mechanism and consumed in another. They do not appear in the overall reaction equation because they are neither the initial reactants nor the final products.In the decomposition of \[\text{N}_2\text{O}\] catalyzed by \[\text{NO}\], \[\text{NO}_2\] is an intermediate. It forms in the first step but is consumed in the second step, hence not showing up in the final balanced equation.Some key points about intermediates:
  • They are essential for completing the overall reaction process.
  • Their presence can be short-lived, making them sometimes undetectable in experiments.
  • They should not be confused with catalysts, which facilitate but are not consumed in reactions.
Understanding intermediates is crucial for chemists to accurately trace reaction pathways and optimize reaction conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You study the effect of temperature on the rate of two reactions and graph the natural logarithm of the rate constant for each reaction as a function of \(1 / T .\) How do the two graphs compare (a) if the activation energy of the second reaction is higher than the activation energy of the first reaction but the two reactions have the same frequency factor, and (b) if the frequency factor of the second reaction is higher than the frequency factor of the first reaction but the two reactions have the same activation energy? [Section 14.5]

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product or disappearance of each reactant: (a) \(2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (c) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

A certain first-order reaction has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if (a) \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ;\) (b) \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol} ?\)

The gas-phase decomposition of \(\mathrm{NO}_{2}, 2 \mathrm{NO}_{2}(g) \longrightarrow\) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\), is studied at \(383^{\circ} \mathrm{C}\), giving the following data: $$ \begin{array}{ll} \hline \text { Time (s) } & {\left[\mathrm{NO}_{2}\right](M)} \\ \hline 0.0 & 0.100 \\ 5.0 & 0.017 \\ 10.0 & 0.0090 \\ 15.0 & 0.0062 \\ 20.0 & 0.0047 \\ \hline \end{array} $$ (a) Is the reaction first order or second order with respect to the concentration of \(\mathrm{NO}_{2} ?\) (b) What is the value of the rate constant?

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride proceeds as follows: \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} .\) The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(64{ }^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M} ?(\mathrm{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M} ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free