/*! 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 42 Explain why a zeroth-order react... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 42

Explain why a zeroth-order reaction cannot be zerothorder for three complete half-lives.

Short Answer

Expert verified
A zeroth-order reaction completes after two half-lives as the reactant is fully consumed.

Step by step solution

01

Understanding Zeroth-Order Reactions

A zeroth-order reaction is one where the rate of reaction is constant and independent of the concentration of the reactant. The rate law for a zeroth-order reaction is given as \( r = k \), where \( k \) is the rate constant. This implies that the concentration of the reactant decreases linearly over time.
02

Derive the Zeroth-Order Half-Life Expression

For a zeroth-order reaction, the concentration of the reactant \( [A] \) at time \( t \) is given by \( [A] = [A]_0 - kt \), where \( [A]_0 \) is the initial concentration. The half-life \( t_{1/2} \) is the time required for the concentration to reach half its initial value: \( [A]_{1/2} = \frac{[A]_0}{2} \). Setting this in the formula gives \( t_{1/2} = \frac{[A]_0}{2k} \).
03

Calculate Time for Complete Consumption

Since the reaction is zeroth-order, we can continue to track the concentration decrease linearly from \([A]_0\) to zero. The time \( t \) required for the reactant to be fully consumed (i.e., \([A] = 0\)) is given by \( t = \frac{[A]_0}{k} \).
04

Compare with Half-Life for Multiple Phases

For the first half-life \( t_{1/2} \), \([A] = \frac{[A]_0}{2}\). After the second half-life \( 2t_{1/2} \), \([A] = 0\), completely consuming the reactant. This shows that a zeroth-order reaction cannot undergo a third complete half-life because there is no reactant left to undergo further half-lives.

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.

Rate Constant
In chemical kinetics, the rate constant plays a crucial role, especially in zeroth-order reactions. The rate constant, denoted by the symbol \( k \), reflects the speed at which reactions occur.
  • For a zeroth-order reaction, the reaction rate is independent of the reactant concentration. This means that \( k \) has units of concentration/time, typically mol/L/s.
  • The rate law for a zeroth-order reaction is written as \( r = k \). Here, \( r \) represents the rate of reaction and is constant throughout the process.
  • The constancy of \( k \) implies that the reaction proceeds at a uniform rate, unaffected by changes in the amount of reactant present.
In sum, the rate constant helps determine how quickly a substance is transformed during a reaction. It provides essential information for calculating the time needed for various degrees of reactant conversion.
Half-Life
The half-life of a reaction is an important aspect of reaction kinetics, indicating the time required for half of the reactant to be consumed.
  • In a zeroth-order reaction, the formula for half-life \( t_{1/2} \) is \( t_{1/2} = \frac{[A]_0}{2k} \), where \([A]_0\) is the initial concentration and \( k \) is the rate constant.
  • This expression shows that the half-life is directly proportional to the initial concentration. More reactant initially means a longer half-life.
  • Unlike first-order reactions where the half-life is constant, here the half-life decreases as the reaction proceeds, because it's dependent on \([A]_0\).
Understanding half-life in zeroth-order reactions highlights why they cannot sustain multiple half-lives, as the reaction completes faster after lesser fractions are used.
Reaction Kinetics
Reaction kinetics is the study of the speed and path of chemical reactions. It offers insights into how reactants transform into products over time.
  • Kinetics helps us understand different types of reactions, such as zeroth-order, and how they progress based on their rate laws.
  • In zeroth-order kinetics, the reaction velocity is constant, meaning it doesn't depend on the reactant concentration.
  • This constant rate leads to a straightforward consumption of the reactant over time, described by the equation: \([A] = [A]_0 - kt\).
Zeroth-order kinetics are simpler compared to other orders of reactions, yet they also illustrate the unique characteristics that make them intriguing to study, such as constant rate and linear concentration decrease.
Reactant Concentration
In the context of reaction kinetics, reactant concentration is a key factor that influences how a reaction proceeds.
  • For zeroth-order reactions, an interesting feature is that the rate does not change with reactant concentration. This is unlike higher-order reactions, where concentration significantly affects the rate.
  • The formula \([A] = [A]_0 - kt\) shows how reactant concentration decreases in a linear fashion over time for zeroth-order reactions.
  • This linear decrease simplifies predictions about how long a reaction will last or when the reactant will be totally used up. For example, the reaction stops entirely when \( [A] = 0 \), as no reactant remains for further reaction phases.
Understanding the relationship between reaction rate and reactant concentration is crucial in studying kinetics, offering insights into controlling and predicting chemical reactions effectively.

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

Many gas-phase reactions require some inert body. usually represented as \(\mathrm{M}\), to absorb or supply energy in a collision in order to proceed. In the spontaneous decomposition of ozone, \(\mathrm{O}_{2}\), we can suggest the mechanism $$ \begin{aligned} \mathrm{O}_{3}+\mathrm{M} & \longrightarrow \mathrm{O}_{3} *+\mathrm{M} \\ \mathrm{O}_{3} * & \longrightarrow \mathrm{was}_{2}+\cdot \mathrm{O}^{-} \\ -\mathrm{O}-+\mathrm{O}_{3} & \longrightarrow 2 \mathrm{O}_{2} \end{aligned} $$ for the overall reaction $$ 2 \mathrm{O}_{3} \longrightarrow 3 \mathrm{O}_{2} $$ In the mechanism, \(\mathrm{O}_{2}\) * refers to an ozone molecule in some energetically excited state that can react spontaneously to form \(\mathrm{O}_{2}\) and \(\mathrm{O}\) atoms. Determine the rate law of the proposed mechanism in terms of \(\mathrm{O}_{3}\) and \(\mathrm{M}\), where the second step is the rate-determining step. Will adding an inert gas like Ar to a sample of ozone increase or decrease the rate of the reaction?

Chemical processes that are triggered by photons are also understood using the kinetics concepts mentioned in this chapter. The chemistry of vision is one example. The currently accepted mechanism for vision involves a compound called modopsin, which is composed of a protein molecule (opsin) attached to a colored polyene molecule called c/s-retinal. (Cls-retinal is chemically related to a class of molecules called carotenes, which are highly colored compounds responsible for the colors of carrots and tomatoes. Eating carrots does help vision, specifically in low light.) The vision process begins when the cls-retinal absorbs a photon and is isomerized about one of its double bonds to make trans-retinal: $$ h v+c / s-r e t i n a l \longrightarrow \text { trans-retinal } $$ The reaction is energetically faworable; the photon's energy is apparently necessary only to overcome an activation energy barrier. (a) The least energetic photon that is considered visible light has a wavelength of \(750 \mathrm{~nm}\). What is the activation energy for the isomerization reaction? (b) If the rate constant at \(37^{-} \mathrm{C}\) is \(3 \times 10^{11} \mathrm{~s}^{-1}\), what is the pre-exponential factor \(A\) for this reaction? (c) The Antarctic icefish lives in Southern Hemisphere waters that can reach \(-2^{\circ} \mathrm{C}\). Assuming that this chemical process is the same in an icefish eye, what is \(k\) at this temperature?

When ionic compounds crystallize from a supersaturated solution, the crystallization front (that is, the barrier between the crystalline solid and the supersaturated solution) typically travels at a constant speed through the solution until it reaches the boundaries of the system, then stops. What order of rate law is described by this behavior?

A researcher determined the rate law $$ \text { rate }=k \cdot[\mathrm{A}]^{2} $$ for a simple chemical reaction. If the rate was \(2.44 \times 10^{-4} \mathrm{M} / \mathrm{s}\) when \([\mathrm{A}]\) was \(0.167 \mathrm{M}\), what would \([\mathrm{A}]\) be when the rate of the reaction is \(1.55 \times 10^{-6} \mathrm{M} / \mathrm{s}\) ?

Most halogenation reactions of hydrocarbons proceed via a free-radical chain reaction mechanism. Of the halogens \(\mathrm{Cl}_{2}, \mathrm{Br}_{2}\), and \(\mathrm{I}_{2}\) which initiation reaction should proceed most easily? Explain your answer.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

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.