/*! 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 45 In a first-order decomposition r... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 45

In a first-order decomposition reaction, \(50.0 \%\) of a compound decomposes in \(10.5 \mathrm{~min}\). (a) What is the rate constant of the reaction? (b) How long does it take for \(75.0 \%\) of the compound to decompose?

Short Answer

Expert verified
The rate constant is approximately 0.066 min^{-1}. It takes about 21.0 minutes for 75.0% decomposition.

Step by step solution

01

Understanding First-Order Reactions

First-order reactions follow the rate law: \[ \text{rate} = k[A] \] where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.
02

Use the First-Order Integrated Rate Law

The integrated rate law for a first-order reaction is: \[ \text{ln}([A]_t/[A]_0) = -kt \] where \( [A]_0 \) is the initial concentration, \( [A]_t \) is the concentration at time \( t \), and \( k \) is the rate constant.
03

Calculate the Rate Constant

Given that 50.0% of the compound decomposes in 10.5 minutes, \( \frac{[A]_t}{[A]_0} = 0.50 \). Substitute into the integrated rate law: \[ \text{ln}(0.50) = -k(10.5) \] Solve for \( k \): \[ k = -\frac{\text{ln}(0.50)}{10.5} \] Calculate \( k \): \[ k \approx 0.066 \thinspace \text{min}^{-1} \]
04

Determine the Time for 75% Decomposition

For 75.0% decomposition, \( \frac{[A]_t}{[A]_0} = 0.25 \). Using the integrated rate law: \[ \text{ln}(0.25) = -0.066t \] Solve for \( t \): \[ t = -\frac{\text{ln}(0.25)}{0.066} \] Calculate \( t \): \[ t \approx 21.0 \thinspace \text{min} \]

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.

rate constant
The rate constant, denoted as 'k', is a crucial factor in chemical kinetics. It indicates how quickly a reaction progresses. For first-order reactions, the rate constant can be determined using the integrated rate law.

In the context of a first-order decomposition reaction, the rate constant is calculated based on the time it takes for a given percentage of the compound to decompose. In our exercise, 50% decomposition occurs in 10.5 minutes.

Using the integrated rate law for a first-order reaction: \ \[\text{ln}( [A]_t / [A]_0 ) = -kt\] \ we start by substituting the known values into the equation:

\[\text{ln}(0.50) = -k(10.5)\]

Solving for 'k', we get:

\[k = - \frac{\text{ln}(0.50)}{10.5} \approx 0.066 \text{ min}^{-1} \]

This tells us the reaction rate is 0.066 per minute.
integrated rate law
The integrated rate law connects the concentration of reactants to time, making it easier to predict how long a reaction will take given specific conditions.

For first-order reactions, the integrated rate law is expressed as:

\[\text{ln}( [A]_t / [A]_0 ) = -kt\]

where:\ * \([A]_0\) is the initial concentrationewline* \([A]_t\) is the concentration at time 't'ewline* 'k' is the rate constant

This law allows us to calculate the reaction time or the remaining concentration at any moment during the reaction. For instance, to find out how long it takes for 75% of a compound to decompose (leaving 25% of the original concentration), we use:

\[\text{ln}(0.25) = -kt\]

Rearranging to solve for time, 't', we get:

\[t = - \frac{\text{ln}(0.25)}{0.066} \approx 21.0 \text{ min} \]

This tells us that it will take about 21 minutes for 75% of the compound to decompose.
reaction kinetics
Reaction kinetics is the study of how fast chemical reactions proceed and the factors that influence these rates. Several core concepts in reaction kinetics include reaction order, rate laws, and reaction mechanisms.

First-order reactions, like the one in our exercise, have a reaction rate that depends linearly on the concentration of a single reactant. The rate law for a first-order reaction is given by:

\[\text{rate} = k[A]\]

To understand how reactants convert to products over time, we use the integrated rate law. This helps us relate the concentration of reactants at any time to the initial concentration, allowing predictions about the progress of the reaction.

Knowing the rate constant 'k' and applying the integrated rate law, we can determine both the speed and extent of the reaction under different conditions. This information is invaluable for predicting how a reaction will behave in practical scenarios, such as in industrial processes or biochemical pathways.

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

In the lower troposphere, ozone is one of the components of photochemical smog. It is generated in air when nitrogen dioxide, formed by the oxidation of nitrogen monoxide from car exhaust, reacts by the following mechanism: Assuming the rate of formation of atomic oxygen in step 1 equals the rate of its consumption in step \(2,\) use the data below to calculate (a) the concentration of atomic oxygen [O] and (b) the rate of ozone formation. $$ \begin{array}{lr} k_{1}=6.0 \times 10^{-3} \mathrm{~s}^{-1} & {\left[\mathrm{NO}_{2}\right]=4.0 \times 10^{-9} \mathrm{M}} \\ k_{2}=1.0 \times 10^{6} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s} & {\left[\mathrm{O}_{2}\right]=1.0 \times 10^{-2} \mathrm{M}} \end{array} $$

How does an increase in temperature affect the rate of a reaction? Explain the two factors involved.

16.103 Even when a mechanism is consistent with the rate law, later work may show it to be incorrect. For example, the reaction between hydrogen and iodine has this rate law: rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]\). The long-accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) In the 1960 s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism: (1) \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) [fast] (2) \(\mathrm{H}_{2}(g)+\mathrm{I}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{I}(g)\) [fast] (3) \(\mathrm{H}_{2} \mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) \quad[\) slow \(]\) Show that this mechanism is consistent with the rate law.

The proposed mechanism for a reaction is (1) \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{X}(g) \quad\) [fast \(]\) (2) \(\mathrm{X}(g)+\mathrm{C}(g) \longrightarrow \mathrm{Y}(g) \quad\) [slow] (3) \(\mathrm{Y}(g) \longrightarrow \mathrm{D}(g) \quad\) [fast (a) What is the overall equation? (b) Identify the intermediate(s), if any. (c) What are the molecularity and the rate law for each step? (d) Is the mechanism consistent with the actual rate law: Rate \(=k[\mathrm{~A}][\mathrm{B}][\mathrm{C}] ?\) (e) Is the following one-step mechanism equally valid? \(\mathrm{A}(g)+\mathrm{B}(g)+\mathrm{C}(g) \longrightarrow \mathrm{D}(g) ?\)

You are studying the reaction \(\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \longrightarrow 2 \mathrm{AB}(g)\) to determine its rate law. Assuming that you have a valid experimental procedure for obtaining \(\left[\mathrm{A}_{2}\right]\) and \(\left[\mathrm{B}_{2}\right]\) at various times, explain how you determine (a) the initial rate, (b) the reaction orders, and (c) the rate constant.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Organic 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.