/*! 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 71 The decomposition of gaseous dim... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The decomposition of gaseous dimethyl ether at ordinary pressures is first order. Its half-life is 25.0 min at \(500^{\circ} \mathrm{C}\) $$\mathrm{CH}_{3} \mathrm{OCH}_{3}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ (a) Starting with \(8.00 \mathrm{g}\) of dimethyl ether, what mass remains (in grams) after 125 min and after 145 min? (b) Calculate the time in minutes required to decrease \(7.60 \mathrm{ng}\) (nanograms) to 2.25 ng. (c) What fraction of the original dimethyl ether remains after 150 min?

Short Answer

Expert verified
(a) 1.00 g after 125 min, 0.47 g after 145 min. (b) 42.49 min for reduction. (c) 4.89% remains after 150 min.

Step by step solution

01

Understand First Order Reactions

A first-order reaction depends linearly on the concentration of a single reactant. In such reactions, the rate is directly proportional to the concentration of one reactant. The half-life for first-order reactions is constant, meaning it does not depend on the initial concentration.
02

Use the Half-Life Formula

The half-life of a first-order reaction is given by the formula: \[ t_{1/2} = \frac{0.693}{k} \] where \( k \) is the rate constant. Given that the half-life is 25.0 min, substitute to find \( k \): \[ k = \frac{0.693}{25.0} \approx 0.02772 \text{ min}^{-1} \].
03

Calculate Remaining Mass Using First Order Kinetics

The amount remaining at any time for a first-order reaction is described by: \[ [A]_t = [A]_0 e^{-kt} \], where \([A]_0\) is the initial concentration, \([A]_t\) is the concentration at time \( t \), and \( k \) is the rate constant. First, calculate for 125 min:\([A]_{125} = 8.00 \times e^{-0.02772 \times 125} \approx 1.00 \text{ g}\).Now for 145 min:\([A]_{145} = 8.00 \times e^{-0.02772 \times 145} \approx 0.47 \text{ g}\).
04

Calculate Time for Concentration Reduction

Using the formula \( ln(\frac{[A]_t}{[A]_0}) = -kt \), calculate the time required to reduce from 7.60 ng to 2.25 ng:\( ln(\frac{2.25}{7.60}) = -0.02772 \times t \), solve for \( t \):\( t = \frac{ln(0.2961)}{-0.02772} \approx 42.49 \text{ min}\).
05

Calculate Fraction Remaining After 150 min

To find the fraction of dimethyl ether remaining after 150 min, use the formula:\( Fraction = e^{-kt} = e^{-0.02772 \times 150} \approx 0.0489 \).This fraction represents about 4.89% of the original amount remaining.

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.

First-Order Reactions
In chemical kinetics, a first-order reaction is a process where the rate of reaction is directly proportional to the concentration of one reactant. This means that as the concentration of the reactant decreases, the rate of the reaction also decreases. This kind of reaction is crucial to understanding how chemical substances change over time, especially in processes such as the decomposition of compounds. A key feature of first-order reactions is that the half-life is independent of the initial concentration. This property makes it relatively simple to study and predict the behavior of such reactions once we know the half-life. Knowing that dimethyl ether decomposes in a first-order manner allows us to apply mathematical models to calculate remaining quantities of reactant, as well as predict how long it takes for a substance to decompose by a given percentage.
Half-Life Calculations
Half-life ( denoted as \( t_{1/2} \)) is the period required for the concentration of a reactant to decrease to half its initial value. For first-order reactions, this value is computed through a specific formula: \( t_{1/2} = \frac{0.693}{k} \), where \( k \) is the rate constant of the reaction. This constant nature of half-life simplifies calculations and predictions regarding the reaction's progress.Given that the half-life of the decomposition of dimethyl ether is 25.0 minutes at \(500^{\circ}C\), we can determine the rate constant to be approximately 0.02772 min^{-1} based on this given half-life. Knowing this rate constant is crucial as it serves as a foundational step in further calculations, such as determining the remaining mass of the reactant after a certain time period or the time required to achieve a specific amount of decomposition. Being conversant with half-life calculations aids in understanding how quickly a chemical process reaches its endpoint.
Rate Constant Determination
The rate constant \( k \) in a first-order reaction is a crucial value as it links the reaction's rate to the concentration of the reactant. It provides a quantitative measure of how fast a reaction occurs. As illustrated in the example involving dimethyl ether decomposition, the rate constant was determined using the half-life formula \( k = \frac{0.693}{t_{1/2}} \). This relationship allows us to find \( k \) with the given half-life, thereby allowing us to calculate other relevant time-dependent changes in the reaction. Understanding how to determine the rate constant allows a student to find out more about the reactivity and stability of a substance under set conditions.A practical understanding of \( k \) enables chemists to predict how long a reaction will take to complete, how much reactant remains after a specific time, or even to estimate unknown values in similar reactions with comparable conditions. The clarity provided by the rate constant greatly enhances our ability to model chemical reactions in a predictive manner.

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 decomposition of nitrogen dioxide at a high temperature $$\mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ is second order in this reactant. The rate constant for this reaction is \(3.40 \mathrm{L} / \mathrm{mol} \cdot\) min. Determine the time needed for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(2.00 \mathrm{mol} / \mathrm{L}\) to \(1.50 \mathrm{mol} / \mathrm{L}.\)

Common sugar, sucrose, breaks down in dilute acid solution to form glucose and fructose. Both products have the same formula, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}.\) $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})$$ The rate of this reaction has been studied in acid solution, and the data in the table were obtained. $$\begin{array}{cc}\hline \begin{array}{c}\text { Time } \\\\(\text { min })\end{array} & \begin{array}{c}{\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathbf{0}_{11}\right]} \\\\(\mathrm{mol} / \mathrm{L})\end{array} \\\\\hline 0 & 0.316 \\\39 & 0.274 \\\80 & 0.238 \\\140 & 0.190 \\\210 & 0.146 \\\\\hline\end{array}$$ (a) Plot ln [sucrose] versus time and \(1 /\) [sucrose] versus time. What is the order of the reaction? (b) Write the rate equation for the reaction, and calculate the rate constant, \(k.\) (c) Estimate the concentration of sucrose after 175 min.

Give the relative rates of disappearance of reactants and formation of products for each of the following reactions. (a) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NOBr}(\mathrm{g})\) (b) \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\)

The mechanism for the reaction of \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{HBr}\) is believed to involve two steps. The overall reaction is exothermic. Step 1 \(\quad\) Fast, endothermic $$\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}_{2}^{+}$$ Step 2 Slow $$\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O}$$ (a) Write an equation for the overall reaction. (b) Draw a reaction coordinate diagram for this reaction. (c) Show that the rate law for this reaction is \(-\Delta\left[\mathrm{CH}_{3} \mathrm{OH}\right] / \Delta t=k\left[\mathrm{CH}_{3} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{Br}^{-}\right]\)

After five half-life periods for a first-order reaction, what fraction of reactant remains?

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.