/*! 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 As described in Exercise \(14.43... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 45

As described in Exercise \(14.43,\) 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 450 torr, what is the pressure of this substance after \(60 \mathrm{~s} ?\) (b) At what time will the pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decline to one-tenth its initial value?

Short Answer

Expert verified
(a) After 60 seconds, the pressure of \(\mathrm{SO_2Cl_2}\) is approximately 69.88 torr. (b) The time taken for the pressure to decline to one-tenth of the initial value is approximately 51.29 seconds.

Step by step solution

01

Write down the given information

We are given the following information: Initial pressure of SO2Cl2: \(P_0 = 450\) torr Rate constant: \(k = 4.5 \times 10^{-2} \mathrm{s^{-1}}\) Temperature: \(660 K\) Time for part (a): \(t = 60 \mathrm{s}\)
02

Use the first-order kinetic equation

Since the decomposition of SO2Cl2 is a first-order process, we can use the first-order kinetic equation: \(P(t) = P_0 \times e^{-kt}\) where \(P(t)\) is the pressure at time \(t\), \(P_0\) is the initial pressure, \(k\) is the rate constant, and \(t\) is the time.
03

Calculate the pressure of SO2Cl2 after 60 s (a)

Using the given values and the first-order kinetic equation, we can find the pressure of SO2Cl2 after 60 s: \(P(60) = 450 \times e^{-(4.5 \times 10^{-2})(60)}\) Solving the above equation, we find the pressure after 60 s is: \(P(60) \approx 69.88\) torr
04

Find the time for the pressure to decline to one-tenth its initial value (b)

Now, we need to find the time taken for the pressure of SO2Cl2 to be reduced to one-tenth of its initial value: \(P(t) = \frac{1}{10} \times P_0\) Plugging \(P_0 = 450\) torr and using the first-order equation, we get: \(45 = 450 \times e^{-kt}\) Divide both sides by 450: \(\frac{1}{10} = e^{-kt}\) Now, take the natural logarithm of both sides to isolate \(t\): \(\ln(0.1) = -kt\) Rearrange the equation to find the time: \(t = \frac{\ln(0.1)}{-k}\) Finally, plug in the given rate constant: \(t = \frac{\ln(0.1)}{-(4.5 \times 10^{-2})}\) Solving the equation, we find the time taken for the pressure to reduce to one-tenth of its initial value: \(t \approx 51.29 \mathrm{s}\) So, after following these steps, we find that (a) the pressure of SO2Cl2 is approximately 69.88 torr after 60 s, and (b) the time taken for the pressure to decline to one-tenth of the initial value is approximately 51.29 s.

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 First-Order Reaction Kinetics
In the realm of chemical kinetics, the rate constant is a crucial value that determines the speed at which a reaction progresses. In the context of a first-order reaction, this constant directly correlates with the reaction's half-life; it remains constant irrespective of the reactant concentrations.

Take for instance the decomposition of sulfuryl chloride, which adheres to first-order kinetics according to our textbook exercise. The rate constant here, denoted as \(k\), not only helps in calculating how fast the substance breaks down but also in predicting concentrations, or in our case, pressures at any given time.

When measuring a first-order reaction, the formula \(P(t) = P_0 \times e^{-kt}\) is essential. In this expression, \(P_0\) signifies the initial pressure, \(P(t)\) is the pressure at time \(t\), and \(e\) stands for the mathematical constant that is the base of natural logarithms. It's imperative to understand that the rate constant, \(k\), is unique for each reaction and is influenced by various factors such as temperature and the presence of catalysts, making it a pivotal factor in our reaction's investigation.
Decomposition of Sulfuryl Chloride
The decay of sulfuryl chloride (\(SO_2Cl_2\)) serves as a good illustration of first-order kinetics. This reaction involves the breakdown of sulfuryl chloride into sulfur dioxide (\(SO_2\)) and chlorine gas (\(Cl_2\)), without the need for its concentration to dictate the rate at which this happens.

Under the set experimental conditions, which include a specific temperature of 660 K, the decomposition rate is determined solely by the rate constant. This type of reaction is characterized by its independence from reactant concentration, making it simpler to analyze compared to other reaction orders.

In practical terms, if you started with a certain amount of sulfuryl chloride, you could predict how much of it would remain after a period of time, which is precisely what the textbook problem requires us to do. This predictability is what makes studying chemical kinetics so valuable for scientists and engineers planning experiments or designing chemical processes.
Pressure Change Over Time
Changes in pressure over time can provide insightful information about the progress of a chemical reaction, especially when dealing with gases. In our exercise, we use the reduction of pressure as an indirect method of measuring the decomposition of sulfuryl chloride.

By applying the first-order kinetic equation, we can determine the pressure of \(SO_2Cl_2\) at any moment throughout the reaction. This approach is particularly beneficial as it allows us to anticipate how pressure decreases as the reaction advances and how this decrease is exponentially related to time due to the nature of first-order kinetics.

It's crucial for students to grasp that in the case of a first-order reaction, the pressure does not drop linearly but exponentially鈥攁 fact often corroborated by a negative exponential graph. This ties back to the constant proportionality which is the rate constant \(k\). The larger its value, the sharper the decline in pressure over time, illustrating the swift progression of the reaction.

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

(a) What is a catalyst? (b) What is the difference between a homogeneous and a heterogeneous catalyst? (c) Do catalysts affect the overall enthalpy change for a reaction, the activation energy, or both?

(a) Define the following symbols that are encountered in rate equations for the generic reaction \(\mathrm{A} \longrightarrow \mathrm{B}:[\mathrm{A}]_{0}, t_{1 / 2}[\mathrm{~A}]_{t}, k\) (b) What quantity, when graphed versus time, will yield a straight line for a first-order reaction? (c) How can you calculate the rate constant for a first- order reaction from the graph you made in part (b)?

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: \(\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-}\). This rapid reaction gives the following rate data: $$ \begin{array}{lll} \hline\left[\mathrm{OCl}^{-}\right](M) & {\left[I^{-}\right](M)} & \text { Initial Rate }(M / s) \\ \hline 1.5 \times 10^{-3} & 1.5 \times 10^{-3} & 1.36 \times 10^{-4} \\ 3.0 \times 10^{-3} & 1.5 \times 10^{-3} & 2.72 \times 10^{-4} \\ 1.5 \times 10^{-3} & 3.0 \times 10^{-3} & 2.72 \times 10^{-4} \\ \hline \end{array} $$ (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{OCl}^{-}\right]=2.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=5.0 \times 10^{-4} \mathrm{M}\)

Consider two reactions. Reaction (1) has a constant half-life, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B},\) what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part (a)? (c) How do the half-lives of first-order and second-order reactions differ?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

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.