/*! 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 28 The compound \(\mathrm{Xe}\left(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 28

The compound \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) decomposes in a firstorder reaction to elemental Xe with a half-life of 30\. minutes. If you place \(7.50 \mathrm{mg}\) of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) in a flask, how long must you wait until only \(0.25 \mathrm{mg}\) of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) remains?

Short Answer

Expert verified
It takes 147.2 minutes for the compound to decompose to 0.25 mg.

Step by step solution

01

Understand the First-Order Kinetics

The decomposition of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) follows first-order kinetics. In first-order reactions, the rate of reaction is proportional to the concentration of the reactant. The formula to determine the half-life \(t_{1/2}\) for first-order reactions is given by \(t_{1/2} = \frac{0.693}{k}\), where \(k\) is the rate constant.
02

Calculate the Rate Constant

Using the given half-life of 30 minutes, we can calculate the rate constant \(k\):\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{30} = 0.0231 \, \text{min}^{-1} \]
03

Apply First-Order Integrated Rate Law

The integrated rate law for a first-order reaction is \( \ln \left( \frac{[A]_t}{[A]_0} \right) = -kt \), where \([A]_0\) is the initial concentration and \([A]_t\) is the concentration at time \(t\). Begin by substituting the initial and final concentrations and the rate constant into this formula:\[ \ln \left( \frac{0.25}{7.50} \right) = -0.0231t \]
04

Solve for Time (t)

First, calculate the natural logarithm:\[ \ln \left( \frac{0.25}{7.50} \right) = \ln (0.0333) = -3.4012 \]Now, solve for \(t\):\[ -3.4012 = -0.0231t \]\[ t = \frac{-3.4012}{-0.0231} = 147.2 \, \text{min} \]
05

Verify the Calculation

Check the calculation to ensure accuracy. The given half-life and calculated time should coincide with the proper rate decay expected for first-order reactions. Repeat calculation if necessary to confirm that the time for \(7.50 \, \text{mg}\) to decompose to \(0.25 \, \text{mg}\) is about 147.2 minutes.

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 chemistry, understanding the rate at which a reaction occurs is crucial. For a first-order reaction, like the one involving the decomposition of \(\mathrm{Xe(CF_3)_2}\), the rate constant \(k\) plays a significant role. It dictates how fast the reactant concentration decreases over time. In first-order reactions, the rate of reaction is directly proportional to the concentration of the single reactant involved.
To determine the rate constant \(k\), we use the half-life \(t_{1/2}\), which is the time taken for the concentration of a reactant to reduce to half its initial value. The formula connecting half-life to rate constant in first-order reactions is \(t_{1/2} = \frac{0.693}{k}\). By inserting the half-life of 30 minutes into this equation, the rate constant \(k\) is calculated to be \(0.0231 \, \text{min}^{-1}\). This value helps predict how quickly the substance will decompose over a given period.
Integrated Rate Law
The integrated rate law is a powerful tool in kinetics, allowing us to relate concentrations of reactants to time elapsed in a reaction. For first-order reactions, the integrated rate law is expressed as \( \ln \left( \frac{[A]_t}{[A]_0} \right) = -kt \), where \([A]_0\) is the initial concentration and \([A]_t\) is the concentration at time \(t\).
This formula helps in finding out how long it will take for a certain amount of reactant to convert into products. Substituting initial concentration \(7.50\, \text{mg}\), final concentration \(0.25\, \text{mg}\), and rate constant \(0.0231\, \text{min}^{-1}\) into the integrated rate law, we set up the equation \[ \ln \left( \frac{0.25}{7.50} \right) = -0.0231t \].
After solving the natural logarithm and rearranging, we find the time \(t\) it takes for \(\mathrm{Xe(CF_3)_2}\) to decompose to the desired concentration.
Half-Life Calculation
Half-life is a central concept when dealing with first-order reactions. It represents the time required for the concentration of the reactant to be reduced by half. In our reaction involving \(\mathrm{Xe(CF_3)_2}\), understanding half-life helps us predict how the reactant concentration decreases over multiple intervals.
For first-order reactions, the half-life is constant and can be calculated using the formula \(t_{1/2} = \frac{0.693}{k}\). Given a rate constant of \(0.0231 \, \text{min}^{-1}\), the half-life is 30 minutes.
This steady half-life allows us to easily predict concentrations at different times, as it does not depend on initial concentration. Calculating multiple half-lives indicates the rapidity of decomposition within the given timeframe, making it essential for understanding reaction kinetics.

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

Formic acid decomposes at \(550^{\circ} \mathrm{C}\) according to the equation $$ \mathrm{HCO}_{2} \mathrm{H}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) $$ The reaction follows first-order kinetics. In an experiment, it is determined that \(75 \%\) of a sample of \(\mathrm{HCO}_{2} \mathrm{H}\) has decomposed in 72 seconds. Determine \(t_{1 / 2}\) for this reaction.

A reaction that occurs in our atmosphere is the oxidation of NO to the brown gas \(\mathrm{NO}_{2}\) $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) $$ The mechanism of the reaction is thought to be Step \(1: \quad 2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2}(\mathrm{g})\) rapidly established equilibrium Step \(2: \quad \mathrm{N}_{2} \mathrm{O}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) \quad\) slow Which is the rate determining step? Is there an intermediate in the reaction? If this is the correct mechanism for this reaction, what is the experimentally determined rate law?

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$ \mathrm{sO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ is first-order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), and the reaction has a halflife of 245 minutes at 600 K. If you begin with \(3.6 \times 10^{-3}\) mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a \(1.0-\mathrm{L}\) flask, how long will it take for the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease to \(2.00 \times 10^{-4}\) mol?

You want to study the hydrolysis of the beautiful green, cobalt-based complex called transdichlorobis-(ethylenediamine)cobalt(III) ion, In this hydrolysis reaction, the green complex ion trans- \(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) forms the red complex ion \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}\) as a \(\mathrm{Cl}^{-}\) ion is replaced with a water molecule on the \(\mathrm{Co}^{3+}\) ion (en \(=\) \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) trans-\(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) Reactions such as this have been studied extensively, and experiments suggest that the initial, slow step in the reaction is the breaking of the Co-Cl bond to give a five-coordinate intermediate. The intermediate is then attacked rapidly by water. Slow: \(\left.\quad \text { trans-ICo(en) }_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq}) \rightarrow\) $$ \left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) $$ Fast: \(\quad\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ \left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq}) $$ (a) Based on the reaction mechanism, what is the predicted rate law? (b) As the reaction proceeds, the color changes from green to red with an intermediate stage where the color is gray. The gray color is reached at the same time, no matter what the concentration of the green starting material (at the same temperature). How does this show the reaction is first-order in the green form? Explain. (c) The activation energy for a reaction can be found by plotting In \(k\) versus \(1 / T .\) However, here we do not need to measure \(k\) directly. Instead, because \(k=-(1 / t) \ln \left([\mathrm{R}] /[\mathrm{R}]_{0}\right),\) the time needed to achieve the gray color is a measure of \(k\). Use the data below to find the activation energy. green $$ \left.\underset{\text { red }}{\operatorname{Co}(\mathrm{en})_{2}}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) $$ The reaction progress is followed by observing the color of the solution. The original solution is green, and the final solution is red, but at some intermediate stage when both the reactant and product are present, the solution is gray. The shape in the middle of the beaker is a vortex that arises because the solutions are being stirred using a magnetic stirring bar in the bottom of the beaker.

For a first-order reaction, what fraction of reactant remains after five half- lives have elapsed?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

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.