/*! 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 58 Dinitrogen pentoxide, \(\mathrm{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 58

Dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), decomposes when heated in carbon tetrachloride solvent. $$ \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}(g) $$ If the rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(6.2 \times\) \(10^{-4} / \mathrm{min}\), what is the half-life? (The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5} .\) ) How long would it take for the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to decrease to \(25 \%\) of its initial value? to \(12.5 \%\) of its initial value?

Short Answer

Expert verified
Half-life is 1118.55 min. Time to reach 25%: 2237.1 min; time to reach 12.5%: 3355.65 min.

Step by step solution

01

Write the Formula for Half-life of a First-order Reaction

For a first-order reaction, the half-life \( t_{1/2} \) is given by the formula: \[t_{1/2} = \frac{0.693}{k},\]where \( k \) is the rate constant.
02

Calculate the Half-life Using the Given Rate Constant

The rate constant \( k \) is given as \(6.2 \times 10^{-4} \text{ min}^{-1}\). Plug this value into the half-life formula: \[t_{1/2} = \frac{0.693}{6.2 \times 10^{-4}} \approx 1118.55 \text{ min}.\]
03

Determine Time to Decrease to 25% of Initial Concentration

For a first-order reaction, the concentration at any time \( t \) is given by \[ [A]_t = [A]_0 e^{-kt}, \]where \([A]_t\) is the concentration at time \( t \) and \([A]_0\) is the initial concentration. For \([A]_t = 0.25[A]_0\): \[0.25 = e^{-kt}.\]Take the natural logarithm on both sides: \[\ln(0.25) = -kt. \] Thus: \[t = \frac{\ln(0.25)}{-k} \approx \frac{-1.3863}{-6.2 \times 10^{-4}} \approx 2237.1 \text{ min}.\]
04

Determine Time to Decrease to 12.5% of Initial Concentration

Similarly, for \([A]_t = 0.125[A]_0\): \[0.125 = e^{-kt}.\]Take the natural logarithm on both sides: \[\ln(0.125) = -kt.\]Thus: \[t = \frac{\ln(0.125)}{-k} \approx \frac{-2.0794}{-6.2 \times 10^{-4}} \approx 3355.65 \text{ min}.\]

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 91影视!

Key Concepts

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

First-order Reactions
A first-order reaction is a type of chemical reaction where the rate is directly proportional to the concentration of a single reactant. This means that as the concentration of the reactant decreases, the reaction rate also decreases.
  • The mathematical expression for a first-order reaction is given by the rate law: \ \( \text{Rate} = k[A] \), where \( [A] \) is the concentration of the reactant and \( k \) is the rate constant.
  • In these reactions, the rate of decomposition can be explained through the integrated rate law: \ \( [A]_t = [A]_0 e^{-kt} \), where \( [A]_t \) is the concentration at time \( t \) and \( [A]_0 \) is the initial concentration.
The key characteristic of first-order reactions is their exponential decay of concentration over time, which makes them predictable and easier to analyze mathematically in kinetics studies.
Half-life Calculation
The half-life of a reaction is the time required for half of the reactant to decompose. In first-order reactions, the half-life is constant and does not depend on the initial concentration. This is a distinctive feature of first-order reactions.
  • The formula for the half-life \( (t_{1/2}) \) of a first-order reaction is given by: \ \[ t_{1/2} = \frac{0.693}{k} \], where \( k \) is the rate constant.
  • This formula arises from the natural logarithm of 2, which reflects the time at which the reactant concentration drops to half its original value.
Using the given rate constant from the exercise, the calculation of the half-life showed that for the decomposition of dinitrogen pentoxide, the half-life is approximately 1118.55 minutes. This regularity provides a straightforward way to predict how the concentration of reactants in first-order reactions will change over time.
Rate Constant
The rate constant (\( k \)) is a crucial element in understanding the kinetics of a reaction. It is a measure of the speed of a reaction at a given temperature.
  • For first-order reactions, \( k \) carries the units of \( \text{time}^{-1} \), such as \( \text{min}^{-1} \), reflecting how often the reaction proceeds per unit time.
  • In this context, the larger the value of the rate constant, the faster the reaction occurs.
In the documented exercise, a rate constant of \( 6.2 \times 10^{-4} \, \text{min}^{-1} \) was provided, indicating a moderately slow yet steady decomposition rate for \( N_2O_5 \). The rate constant is an essential parameter for plugging into various kinetic equations to predict concentration changes over time or to determine half-life, as demonstrated in the solutions.

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

Nitric oxide, \(\mathrm{NO}\), is believed to react with chlorine according to the following mechanism: $$ \mathrm{NO}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{NOCl}_{2} $$ (elementary reaction) $$ \mathrm{NOCl}_{2}+\mathrm{NO} \longrightarrow 2 \mathrm{NOCl} \quad \text { (elementary reaction) } $$ Identify any reaction intermediate. What is the overall equation?

The reaction \(\mathrm{A}(g) \longrightarrow \mathrm{B}(g)+\mathrm{C}(g)\) is known to be first order in \(\mathrm{A}(g)\). It takes \(25 \mathrm{~s}\) for the concentration of \(\mathrm{A}(g)\) to decrease by one-half of its initial value. How long does it take for the concentration of \(\mathrm{A}(g)\) to decrease to one-fourth of its initial value? to one-eighth of its initial value?

In experiments on the decomposition of azomethane, $$ \mathrm{CH}_{3} \mathrm{NNCH}_{3}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{N}_{2}(g) $$ the following data were obtained: $$ \begin{array}{lll} & \text { Initial } & \\ & \begin{array}{l} \text { Concentration } \\ \text { of Azomethane } \end{array} & \text { Initial Rate } \\ \text { Exp. 1 } & 1.13 \times 10^{-2} M & 2.8 \times 10^{-6} \mathrm{M} / \mathrm{s} \\ \text { Exp. } 2 & 2.26 \times 10^{-2} M & 5.6 \times 10^{-6} \mathrm{M} / \mathrm{s} \end{array} $$ What is the rate law? What is the value of the rate constant?

What is the molecularity of each of the following elementary reactions? a. \(\mathrm{O}+\mathrm{O}_{2}+\mathrm{N}_{2} \longrightarrow \mathrm{O}_{3}+\mathrm{N}_{2}^{*}\) b. \(\mathrm{NO}_{2} \mathrm{Cl}+\mathrm{Cl} \longrightarrow \mathrm{NO}_{2}+\mathrm{Cl}_{2}\) c. \(\mathrm{Cl}+\mathrm{H}_{2} \longrightarrow \mathrm{HCl}+\mathrm{H}\) d. \(\mathrm{CS}_{2} \longrightarrow \mathrm{CS}+\mathrm{S}\)

In the presence of a tungsten catalyst at high temperatures, the decomposition of ammonia to nitrogen and hydrogen is a zero-order process. If the rate constant at a particular temperature is \(3.7 \times 10^{-6} \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})\), how long will it take for the ammonia concentration to drop from an initial concentration of \(5.0 \times 10^{-4} M\) to \(5.0 \times 10^{-5} M ?\) What is the half-life of the reaction under these conditions?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation
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.