/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 The decomposition of dinitrogen ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The decomposition of dinitrogen pentoxide is described by the chemical equation \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) If the rate of appearance of \(\mathrm{NO}_{2}\) is equal to \(0.560\) mol/min at a particular moment, what is the rate of appearance of \(\mathrm{O}_{2}\) at that moment? a. \(2.70 \mathrm{~mol} / \mathrm{min}\) b. \(3.40 \mathrm{~mol} / \mathrm{min}\) c. \(0.140 \mathrm{~mol} / \mathrm{min}\) d. \(1.14 \mathrm{~mol} / \mathrm{min}\)

Short Answer

Expert verified
The rate of appearance of \(\mathrm{O}_{2}\) is \(0.140 \mathrm{~mol} / \mathrm{min}\).

Step by step solution

01

Understand the Relationship in the Reaction

The given reaction is \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{~g}) + \mathrm{O}_{2}(\mathrm{~g})\). From the chemical equation, the stoichiometry shows that 4 moles of \(\mathrm{NO}_{2}\) are produced for every 1 mole of \(\mathrm{O}_{2}\) formed. Therefore, the rate at which \(\mathrm{NO}_{2}\) appears can be used to determine the rate of appearance of \(\mathrm{O}_{2}\).
02

Determine Proportional Rates

If \(0.560\) mol/min of \(\mathrm{NO}_{2}\) is produced, we use the stoichiometric coefficients to find the rate of \(\mathrm{O}_{2}\). Since \(4\) moles of \(\mathrm{NO}_{2}\) are produced for every \(1\) mole of \(\mathrm{O}_{2}\), the rate of appearance of \(\mathrm{O}_{2}\) will be one-fourth of that of \(\mathrm{NO}_{2}\). Thus, the rate of \(\mathrm{O}_{2}\) appearance is \(0.560 / 4 = 0.140\) mol/min.

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 91Ó°ÊÓ!

Key Concepts

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

Stoichiometry
Stoichiometry is like a recipe book for chemical reactions. It tells us how much of each ingredient, or chemical, is needed to produce certain products. In a chemical equation, stoichiometry is dictated by the coefficients in front of substances. These coefficients show the ratios of molecules or moles that react with each other.

In our exercise, the reaction of dinitrogen pentoxide (\(\mathrm{N}_{2}\mathrm{O}_{5}\)) decomposes to form nitrogen dioxide (\(\mathrm{NO}_{2}\)) and oxygen (\(\mathrm{O}_{2}\)). The balanced equation is:
\[2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{~g}) + \mathrm{O}_{2}(\mathrm{~g})\]

This equation shows:
  • 2 moles of dinitrogen pentoxide decompose to produce 4 moles of nitrogen dioxide and 1 mole of oxygen.
  • The stoichiometric coefficients indicate the proportionate relationship between reactants and products.
  • Understanding these relationships allows us to predict how changes in rate or quantity of one substance will affect the others.
Dinitrogen Pentoxide Decomposition
Dinitrogen pentoxide, a chemical compound with the formula \(\mathrm{N}_{2}\mathrm{O}_{5}\), undergoes decomposition in this reaction. Decomposition reactions involve a single compound breaking down into two or more simpler substances. In this case, \(\mathrm{N}_{2}\mathrm{O}_{5}\) splits into nitrogen dioxide and oxygen.

The decomposition of dinitrogen pentoxide is an interesting reaction because of its applications in understanding chemical kinetics, the study of rate of chemical processes.
  • The reaction is endothermic, requiring energy input to break the bonds in the \(\mathrm{N}_{2}\mathrm{O}_{5}\) molecule.
  • Understanding the decomposition helps chemists to design and control chemical processes where precise amounts of \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\) are essential.
  • This decomposition is crucial in pollution dynamics, particularly involving nitrogen compounds in the atmosphere.
It’s a perfect example of how complex molecules break down, and how rate dynamics are essential in predicting product formation over time.
Mole Ratios
Mole ratios are a vital tool in chemistry that allow us to convert between quantities of reactants and products. In any balanced chemical equation, mole ratios are derived directly from the coefficients. They express the relationship between the amount in moles of any two compounds involved in a chemical reaction.

In the dinitrogen pentoxide decomposition reaction, the mole ratio between \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\) is 4:1, which means 4 moles of \(\mathrm{NO}_{2}\) are produced for every 1 mole of \(\mathrm{O}_{2}\) formed.

Using this mole ratio helps determine the rate of oxygen production when given the rate of nitrogen dioxide production. If \(0.560 \mathrm{mol/min}\) of \(\mathrm{NO}_{2}\) is produced, the rate of \(\mathrm{O}_{2}\) production, according to the 4:1 ratio, will be one-fourth the rate of \(\mathrm{NO}_{2}\) production. Thus, it is \(0.140 \mathrm{mol/min}\).

Mole ratios not only confirm the stoichiometric relationships but also offer a method to transition from theoretical data to practical data in chemical reactions. This capability is essential for planning experiments and industrial chemical applications.

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

In Arrhenius equation: \(\mathrm{K}=\mathrm{Ae}^{-\mathrm{Ea} \mathrm{KT}}\) a. The pre-exponential factor has the units of rate constant of the reaction b. The exponential factor is a dimensionless quantity c. The exponential factor has the units of reciprocal of temperatures d. The pre-exponential factor has the units of rate of the reaction

In Arrhenius equation, \(\mathrm{k}=\mathrm{A} \exp (-\mathrm{Ea} / \mathrm{RT})\). A may be regarded as the rate constant at a. Very high temperature b. Very low temperature c. High activation energy d. Zero activation energy

For this reaction \(\mathrm{X}^{-}+\mathrm{OH}^{-} \rightarrow \mathrm{X}^{-}+\mathrm{XO}^{-}\)in an aque- ons medium, the rate of the reaction is given as \(\frac{\left(\mathrm{d}\left(\mathrm{XO}^{-}\right)\right.}{\mathrm{dt}}=\mathrm{K} \frac{\left[\mathrm{X}^{-}\right]\left[\mathrm{XO}^{-}\right]}{\left[\mathrm{OH}^{-}\right]}\) The overall order for this reaction is a. Zero b. 1 c. \(-1\) d. \(1 / 2\)

Which of the following expressions is/are not correct? a. \(\log \mathrm{k}=\log \mathrm{A}-\frac{\mathrm{Ea}}{2.303 \mathrm{RT}}\). b. \(\operatorname{In} \mathrm{A}=\operatorname{In} \mathrm{k}+\frac{\mathrm{Ea}}{\mathrm{RT}}\). c. \(\mathrm{k}\) Ae \(^{-R T / E a}\) d. In \(\mathrm{k}=\operatorname{In} \mathrm{A}+\mathrm{Ea} / \mathrm{RT}\)

Consider the chemical reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) The rate of this reaction can be expressed in terms of time derivatives of concentration of \(\mathrm{N}_{2}(\mathrm{~g}), \mathrm{H}_{2}\) (g) or \(\mathrm{NH}_{3}(\mathrm{~g})\). Identify the correct relationship amongst the rate expressions. a. rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=1 / 2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) b. rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) c. rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) d. rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\)

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.