/*! 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 26 The table below gives the concen... [FREE SOLUTION] | 91影视

91影视

The table below gives the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g)\) as a function of time for the reaction described by the equation \begin{tabular}{lc} \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\) \\ \hline\(\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right] / \mathrm{M}\) & \(t / \mathrm{min}\) \\ \hline \(0.0345\) & 0 \\ \(0.0245\) & \(3.8\) \\ \(0.0212\) & \(5.6\) \\ \(0.0154\) & \(9.8\) \\ \(0.0103\) & \(14.0\) \\ \hline \end{tabular} Verify that this reaction is a first-order reaction by plotting \(\ln \left(\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]=\mathrm{M}\right)\) versus time and determine the value of the rate constant.

Short Answer

Expert verified
The reaction is first-order with rate constant approximately 0.086 min鈦宦.

Step by step solution

01

Prepare Data for Plotting

First, we need to calculate the natural logarithm (ln) of the concentration \[\left(\left[\text{SO}_2 \text{Cl}_2\right]\right)\] at each given time. Here are the conversions: \( \ln(0.0345) = -3.367 \), \( \ln(0.0245) = -3.704 \), \( \ln(0.0212) = -3.854 \), \( \ln(0.0154) = -4.174 \), and \( \ln(0.0103) = -4.575 \).
02

Create the Plot

Plot each calculated \( \ln \left(\left[\text{SO}_2\text{Cl}_2\right]\right) \) value against the corresponding time in minutes. The x-axis represents time \( t \) and the y-axis represents \( \ln \left(\left[\text{SO}_2\text{Cl}_2\right]\right) \).
03

Verify Linearity

Check that the resulting plot is a straight line. A straight line suggests that the reaction is first-order with respect to \( \text{SO}_2 \text{Cl}_2 \).
04

Calculate the Rate Constant

The slope of the line in a plot of \( \ln \left(\left[\text{SO}_2\text{Cl}_2\right]\right) \) versus time gives the rate constant \( k \). Use the formula for the slope \( m \) of a line \( m = \frac{\Delta y}{\Delta x} \), where \( y \) is \( \ln \left(\left[\text{SO}_2\text{Cl}_2\right]\right) \) and \( x \) is time. Calculate the slope between each pair of points, ensuring that it remains constant: \( m = \frac{-4.575 - (-3.367)}{14.0 - 0} = \frac{-1.208}{14.0} \approx -0.086 \). Thus, the rate constant \( k \approx 0.086 \text{ min}^{-1} \).

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.

Reaction Kinetics
Understanding reaction kinetics is crucial for mastering chemistry concepts, particularly because it helps us know how fast a reaction occurs. When studying reactions, we focus on rates 鈥 how quickly reactants are consumed and products are formed. For a reaction, such as the decomposition of \( \text{SO}_2\text{Cl}_2 \), we observe how the concentration of reactants decreases over time. To analyze this, chemists use the rate law, which mathematically describes the relationship between reactant concentration and reaction rate.
This particular exercise involves the rate law for a first-order reaction, where the rate depends linearly on the concentration of a single reactant. A common tool for investigating reaction kinetics is plotting concentration data versus time, which helps identify the reaction order. First-order kinetics shows a linear relationship when plotting the natural log of concentration against time. This simplicity in pattern derives from the specific characteristics of first-order reactions
and is a distinctive way to differentiate it from zero-order or second-order reactions.
Rate Constant Calculation
Calculating the rate constant \( k \) in a first-order reaction can be straightforward once you've visualized your data. The rate constant provides insights into the speed of the reaction. It's an integral part of the rate equation for first-order reactions, given by: \[Rate = k[ ext{A}]\]When the natural logarithm of concentration \( \ln([\text{SO}_2\text{Cl}_2]) \) is plotted against time, the slope of the resulting line represents the negative value of the rate constant \( k \). Hence, you calculate \( k \) as:
\[k = -\text{slope}\]
In this example, after plotting \( \ln([\text{SO}_2\text{Cl}_2]) \) against time and ensuring a linear relationship, you find the slope between two points. This involves applying the formula \( m = \frac{\Delta y}{\Delta x} \), where \( y \) is \( \ln([\text{SO}_2\text{Cl}_2]) \) change and \( x \) is time change between two consecutive data points.
It's crucial that the slope remains constant across all point intervals, affirming that the reaction exhibits first-order kinetics and the calculated \( k \) is reliable.
Chemical Reaction Rate
The rate of a chemical reaction refers to how quickly the reactants are transformed into products. It is determined by factors such as temperature, concentration, and the presence of catalysts. With first-order reactions, the rate depends solely on the concentration of one reactant.
This makes the study of chemical reaction rates considerably easier for first-order reactions since it commonly involves exponential decay. Recognizing first-order reactions is crucial because they occur in many real-world scenarios, such as radioactive decay and certain chemical decompositions.
In this exercise, the given reaction follows first-order kinetics, as shown by the linear relationship of \( \ln([\text{SO}_2\text{Cl}_2]) \) versus time. By understanding the rate, chemists can predict how long it takes for a reaction to reach a certain point. It also helps to modify conditions to optimize production in industrial processes. Therefore, knowing how to determine and interpret the rate of a chemical reaction is a fundamental skill in chemistry. Understanding these foundational concepts allows one to gain deeper insight into the complex processes behind chemical reactions.

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 rate constant for the first-order reaction described by the equation CC1CC1 \((g)\) C=CC \((g)\) cyclopropane \(\quad\) propene at \(500^{\circ} \mathrm{C}\) is \(5.5 \times 10^{-4} \mathrm{~s}^{-1} .\) Calculate the half-life of \(\mathrm{cy}\) clopropane at \(500^{\circ} \mathrm{C}\). Given an initial cyclopropane concentration of \(1.00 \times 10^{-\mathrm{s}} \mathrm{M}\) at \(500^{\circ} \mathrm{C}\), calculate the concentration of cyclopropane that remains after \(2.0\) hours.

Calculate the time required for the concentration to decrease by \(10.0 \%\) of its initial value for a firstorder reaction with \(k=10.0 \mathrm{~s}^{-1}\).

Uranyl nitrate decomposes according to $$ \mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q) \rightarrow \mathrm{UO}_{5}(s)+2 \mathrm{NO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) $$ The rate law is first order in the concentration of uranyl nitrate. The following data were recorded for the reaction at \(25.0^{\circ} \mathrm{C}\). \begin{tabular}{cc} \hline \(\mathrm{t} / \mathrm{min}\) & {\(\left[\mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2}\right] / \mathrm{M}\)} \\\ \hline 0 & \(0.01418\) \\ \(20.0\) & \(0.01096\) \\ \(60.0\) & \(0.00758\) \\ \(180.0\) & \(0.00302\) \\ \(360.0\) & \(0.00055\) \\ \hline \end{tabular} Calculate the value of the rate constant for this reaction at \(25.0^{\circ} \mathrm{C}\).

The decomposition of \(\mathrm{NO}_{2}(g)\) as described by the equation $$ 2 \mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ is second order in \(\mathrm{NO}_{2}(g)\). The rate constant for the reaction at \(300^{\circ} \mathrm{C}\) is \(0.54 \mathrm{M}^{-1 . \mathrm{s}^{-1}} .\) If \(\left[\mathrm{NO}_{2}\right]_{0}=1.25 \mathrm{M}\) what is the value of \(\left[\mathrm{NO}_{2}\right]\) after the reaction has run for \(2.0\) minutes?

Sulfuryl chloride decomposes according to the equation \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\) Using the following initial-rate data, determine the order of the reaction with respect to \(\mathrm{SO}_{2} \mathrm{Cl}_{2}:\) \begin{tabular}{cc} \hline & \multicolumn{2}{l} { Initial rate of reaction of } \\ {\(\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]_{0} / \mathrm{mol} \cdot \mathrm{L}^{-1}\)} & \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) / \mathrm{mol} \cdot \mathrm{L}^{-1} \cdot \mathrm{s}^{-1}\) \\ \hline \(0.10\) & \(2.2 \times 10^{-6}\) \\ \(0.20\) & \(4.4 \times 10^{-6}\) \\ \(0.30\) & \(6.6 \times 10^{-6}\) \\ \(0.40\) & \(8.8 \times 10^{-6}\) \\ \hline \end{tabular} Calculate the value of the rate constant.

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.