/*! 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 40 (a) Develop an equation for the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 40

(a) Develop an equation for the half-life of a zero-order reaction. (b) Does the half-life of a zero-order reaction increase, decrease, or remain the same as the reaction proceeds?

Short Answer

Expert verified
(a) The equation for the half-life of a zero-order reaction is \(t_{1/2} = \frac{[A]_0}{2k}\). (b) The half-life of a zero-order reaction decreases as the reaction proceeds.

Step by step solution

01

Write the rate law for a zero-order reaction

A zero-order reaction is one in which the rate of the reaction is independent of the concentration of the reactant. The rate law for a zero-order reaction can be written as: Rate = k where Rate is the rate of reaction, and k is the rate constant. Since Rate is also equal to the change in concentration (鈭哰A]) over change in time (鈭唗): k = -鈭哰A]/鈭唗 Now, let's integrate this equation to find the relationship between the concentration of the reactant and time.
02

Integrate the rate law equation

Integrating the rate law equation, we get: \(\int_{[A]_0}^{[A]} -d[A] = \int_0^t kdt\) Solving this integral: \(-([A]-[A]_0) = kt\) Rearranging the equation for [A]: [A] = [A]_0 - kt Now we can use this equation to find the half-life of the reaction.
03

Calculate the half-life equation

The half-life (t1/2) is the time it takes for the concentration of the reactant to reach half of its initial value. In our equation for [A], let [A] = [A]_0/2, and the time as t = t1/2. Substituting these values and solving for t1/2: [A]_0/2 = [A]_0 - kt1/2 Now, let's isolate t1/2 to obtain the equation for half-life: kt1/2 = [A]_0/2 t1/2 = [A]_0/(2k) This is the equation for the half-life of a zero-order reaction.
04

End Task (a)

We have found the equation for the half-life of a zero-order reaction: t1/2 = [A]_0/(2k) Now, let's move on to part (b) of the exercise.
05

Determine the relationship between half-life and reaction progress

We want to find out if the half-life of a zero-order reaction increases, decreases, or remains the same as the reaction proceeds. For this, let's consider the half-life equation we derived in step 3: t1/2 = [A]_0/(2k) Notice that t1/2 depends on the initial concentration [A]_0. As the reaction proceeds, the concentration of the reactant decreases. Therefore, as [A]_0 decreases, t1/2 will also decrease. #End Task (b)# The half-life of a zero-order reaction decreases as the reaction proceeds.

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 Law for Zero-Order Reactions
In a zero-order reaction, something interesting occurs: the rate of the reaction does not depend on the concentration of the reactant. This means that whether you have a lot or a little of the reactant, the rate stays the same. This is described by the rate law equation. For zero-order reactions, the rate law is expressed as:
\[\text{Rate} = k\]
where \( k \) is the rate constant, a fixed value at a given temperature. Since the rate does not change with concentration, it simplifies the study of kinetics considerably. Another way to express the rate is by noting the change in concentration over time:
\[k = -\frac{\Delta [A]}{\Delta t}\]
This allows us to investigate how much the concentration of a reactant changes over a period of time. As we move forward, this understanding will help us see how the reaction progresses.
Understanding Half-Life in Zero-Order Reactions
The half-life, \( t_{1/2} \), of a reaction is the time it takes for the concentration of the reactant to reduce by half. For a zero-order reaction, the half-life equation is:
\[t_{1/2} = \frac{[A]_0}{2k}\]
This equation shows that the initial concentration \( [A]_0 \) directly affects half-life. Thus, as the concentration decreases during the reaction, the half-life also decreases. Unlike other reactions where half-life may remain constant, in zero-order reactions, it becomes shorter as the reaction proceeds. This happens because there is less reactant available over time, affecting how quickly half the amount is used up.
Reaction Progress in Zero-Order Reactions
The progress of a zero-order reaction can be mapped out using the change in concentration over time. The integrated rate law for this is:
\[[A] = [A]_0 - kt\]
Here, \( [A]_0 \) is the initial concentration, and \( [A] \) is the concentration at time \( t \). This linear equation signifies a steady decrease in concentration as the reaction progresses. Since the rate is constant, every unit of time sees the same amount of reactant being consumed. This straight-line behavior is characteristic of zero-order reactions, making them unique and simpler to predict across a given time frame. Understanding this pattern helps in modeling how substances react and transform efficiently.

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

Based on their activation energies and energy changes and assuming that all collision factors are the same, rank the following reactions from slowest to fastest. (a) \(E_{a}=75 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-20 \mathrm{~kJ} / \mathrm{mol}\) (b) \(E_{a}=100 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=+30 \mathrm{~kJ} / \mathrm{mol}\) (c) \(E_{a}=85 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-50 \mathrm{~kJ} / \mathrm{mol}\)

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2} .\) When \([\mathrm{NO}]=0.040 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M},\) the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s} .(\mathbf{a})\) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of \(1.8 ?\)

As described in Exercise 14.41 , the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at \(660 \mathrm{~K}\) is \(4.5 \times 10^{-2} \mathrm{~s}^{-1}\). (a) If we begin with an initial \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) pressure of \(60 \mathrm{kPa}\), what is the partial pressure of this substance after 60 s? (b) At what time will the partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decline to one-tenth its initial value?

(a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst?

In solution, chemical species as simple as \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) can serve as catalysts for reactions. Imagine you could measure the \(\left[\mathrm{H}^{+}\right]\) of a solution containing an acidcatalyzed reaction as it occurs. Assume the reactants and products themselves are neither acids nor bases. Sketch the \(\left[\mathrm{H}^{+}\right]\) concentration profile you would measure as a function of time for the reaction, assuming \(t=0\) is when you add a drop of acid to the reaction.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

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.