/*! 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 86 The decomposition of \(\mathrm{N... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The decomposition of \(\mathrm{N}_{2} \mathrm{O}\) to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) is a first-order reaction. At \(730^{\circ} \mathrm{C}\) the half-life of the reaction is \(3.58 \times 10^{3}\) min. If the initial pressure of \(\mathrm{N}_{2} \mathrm{O}\) is 2.10 atm at \(730^{\circ} \mathrm{C},\) calculate the total gas pressure after one half-life. Assume that the volume remains constant.

Short Answer

Expert verified
The total gas pressure after one half-life is 2.625 atm.

Step by step solution

01

Understand the Problem

The problem involves the decomposition of nitrous oxide (\( \mathrm{N}_{2} \mathrm{O} \)) into nitrogen (\( \mathrm{N}_{2} \)) and oxygen (\( \mathrm{O}_{2} \)). The total gas pressure after one half-life is to be calculated. This occurs in a constant volume at a given initial pressure and temperature.
02

Reaction and Stoichiometry

The reaction is given by \( 2 \mathrm{N}_{2} \mathrm{O} \rightarrow 2 \mathrm{N}_{2} + \mathrm{O}_{2} \). For every 2 moles of \( \mathrm{N}_{2} \mathrm{O} \) that decompose, 2 moles of \( \mathrm{N}_{2} \) and 1 mole of \( \mathrm{O}_{2} \) are produced.
03

Determine Remaining Reactant

After one half-life, the amount of \( \mathrm{N}_{2} \mathrm{O} \) left is half of its initial amount due to the first-order kinetics. Therefore, the remaining \( \mathrm{N}_{2} \mathrm{O} \) pressure is \( \frac{2.10}{2} = 1.05 \) atm.
04

Calculate Total Pressure Increase

At the half-life, the decomposed \( \mathrm{N}_{2} \mathrm{O} \) is 1.05 atm. This produces 1.05 atm of \( \mathrm{N}_{2} \) and 0.525 atm of \( \mathrm{O}_{2} \) from the stoichiometry: \( 1 \) mole of \( \mathrm{O}_{2} \) for every 2 moles \( \mathrm{N}_{2} \mathrm{O} \).
05

Calculate Total Gas Pressure

The total pressure after one half-life is the sum of the pressures of all gases present: \( 1.05 \) atm (remaining \( \mathrm{N}_{2} \mathrm{O} \)) + \( 1.05 \) atm (produced \( \mathrm{N}_{2} \)) + \( 0.525 \) atm (produced \( \mathrm{O}_{2} \)). Total pressure = \( 2.625 \) atm.

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.

Understanding Chemical Kinetics in First-Order Reactions
In chemical kinetics, the focus is on the rate at which a chemical reaction proceeds. A first-order reaction is one where the rate depends linearly on the concentration of a single reactant. For the decomposition of nitrous oxide \( (\mathrm{N}_{2}\mathrm{O}) \), this means that its rate decreases as \( \mathrm{N}_{2} \mathrm{O} \) decreases. The concept of half-life, commonly used in first-order reactions, is the time it takes for half of the reactant to decompose. For \( \mathrm{N}_{2}\mathrm{O} \) at \(730^{\circ} \mathrm{C} \), this half-life is \(3.58 \times 10^{3}\) minutes. The half-life remains constant throughout the reaction, which helps in predicting when half of any current amount of \( \mathrm{N}_{2} \mathrm{O} \) will decompose. This understanding is crucial in determining how much reactant remains and how much product forms after a given time, like one half-life.
The Role of Gas Pressure in Chemical Reactions
Gas pressure is a key factor in gas-phase reactions. It reflects the force exerted by gas molecules hitting the walls of a container. In the decomposition reaction of \( \mathrm{N}_{2} \mathrm{O} \), pressure changes help monitor the progress of the reaction. At the start, the pressure is determined solely by \( \mathrm{N}_{2} \mathrm{O} \). As the reaction proceeds, \( \mathrm{N}_{2} \) and \( \mathrm{O}_{2} \) are formed, altering the pressure. After one half-life, the pressure due to \( \mathrm{N}_{2} \mathrm{O} \) is reduced to half its initial value because half of it has decomposed. The produced gases \( \mathrm{N}_{2} \) and \( \mathrm{O}_{2} \) increase the total pressure. Thus, the total pressure at this point is the sum of the pressures exerted by \( \mathrm{N}_{2} \mathrm{O} \), \( \mathrm{N}_{2} \), and \( \mathrm{O}_{2} \). This principle shows how pressure data can be applied to deduce the extent of reaction progress without directly measuring concentrations of individual gases.
Interpreting Reaction Stoichiometry
Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. For \( \mathrm{N}_{2} \mathrm{O} \) decomposition, the balanced chemical equation is \(2 \mathrm{N}_{2} \mathrm{O} \rightarrow 2 \mathrm{N}_{2} + \mathrm{O}_{2}\). This indicates that two moles of \( \mathrm{N}_{2} \mathrm{O} \) produce two moles of \( \mathrm{N}_{2} \) and one mole of \( \mathrm{O}_{2} \). Stoichiometry enables us to calculate the amounts of products formed from a known quantity of reactants. After one half-life in our reaction, 1.05 atm of \( \mathrm{N}_{2} \mathrm{O} \) decomposes to produce equivalent 1.05 atm of \( \mathrm{N}_{2} \) and a corresponding \( 0.525 \) atm of \( \mathrm{O}_{2} \). Understanding stoichiometry is essential as it not only facilitates calculations of pressure changes in gas-phase reactions but also plays a critical role in scaling reactions for industrial 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

The reaction \(\mathrm{H}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2}+\mathrm{H}\) has been studied for many years. Sketch a potential-energy versus reaction progress diagram for this reaction.

A flask contains a mixture of compounds \(\mathrm{A}\) and \(\mathrm{B}\). Both compounds decompose by first-order kinetics. The half-lives are 50.0 min for \(\mathrm{A}\) and 18.0 min for \(\mathrm{B}\). If the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are equal initially, how long will it take for the concentration of \(\mathrm{A}\) to be four times that of \(\mathrm{B}\) ?

The following scheme in which \(\mathrm{A}\) is converted to \(\mathrm{B}\), which is then converted to \(\mathrm{C}\), is known as a consecutive reaction: \(\mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C}\) Assuming that both steps are first order, sketch on the same graph the variations of \([\mathrm{A}],[\mathrm{B}],\) and \([\mathrm{C}]\) with time.

In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is \(10.0 \mathrm{~cm}^{3} .\) Calculate the surface area of the catalyst. If the sphere is broken down into eight smaller spheres, each having a volume of \(1.25 \mathrm{~cm}^{3},\) what is the total surface area of the spheres? Which of the two geometric configurations of the catalyst is more effective? (The surface area of a sphere is \(4 \pi r^{2}\), where \(r\) is the radius of the sphere.) Based on your analysis here, explain why it is sometimes dangerous to work in grain elevators.

When methyl phosphate is heated in acid solution, it reacts with water: $$ \mathrm{CH}_{3} \mathrm{OPO}_{3} \mathrm{H}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}_{3} \mathrm{PO}_{4} $$ If the reaction is carried out in water enriched with \({ }^{18} \mathrm{O},\) the oxygen- 18 isotope is found in the phosphoric acid product but not in the methanol. What does this tell us about the mechanism of the reaction?

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.