Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 43

As described in Exercise 14.41 , 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 \(4.5 \times 10^{-2} \mathrm{s}^{-1}\) .half-life for this reaction? (b) If you start with 0.050\(M \mathrm{I}_{2}\) at this temperature, how much will remain after 5.12 s assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2}\) ?

Short Answer

Expert verified
The half-life of the decomposition of sulfuryl chloride (SO2Cl2) at 660 K is 15.4 seconds. After 5.12 seconds, the concentration of I2 remaining is 0.0249 M.

Step by step solution

01

Determine the half-life of the first-order reaction

We will use the half-life formula for a first-order reaction: \[t_{1/2} = \frac{0.693}{k}\] Where \(t_{1/2}\) is the half-life, and k is the rate constant. Given the rate constant, k = \(4.5 \times 10^{-2}\, s^{-1}\), we can find the half-life: \[t_{1/2} = \frac{0.693}{4.5 \times 10^{-2} \,s^{-1}}\]
02

Calculate the half-life

Plug in the rate constant value into the equation and solve for the half-life: \[t_{1/2} = \frac{0.693}{4.5 \times 10^{-2}\, s^{-1}} = 15.4\, s\] So, the half-life of the reaction is 15.4 seconds.
03

Use the integrated rate law for a first-order reaction

To determine the amount of I2 remaining after 5.12 s, we will use the integrated rate law for a first-order reaction: \[ ln[\frac{A}{A_0}] = -kt\] Where A is the final concentration of I2, \(A_0\) is the initial concentration of I2, k is the rate constant, and t is the time. We are given the initial concentration of I2: \(A_0 = 0.050\, M\), the rate constant k, and the time: t = 5.12 s. Now we can solve for the final concentration, A.
04

Calculate the final concentration of I2

Rearrange the integrated rate law equation to solve for A: \[A = A_0 e^{-kt}\] Plug in the given values: \[A = 0.050\, M \times e^{-4.5 \times 10^{-2}\, s^{-1} \times 5.12\, s}\]
05

Find the remaining amount of I2

Evaluate the expression to find A: \[A = 0.050\, M \times e^{-(4.5 \times 10^{-2}\, s^{-1})(5.12\, s)} = 0.0249\, M\] So, after 5.12 seconds, there will be 0.0249 M of I2 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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

For the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) the activation energy \(\left(E_{a}\right)\) and overall \(\Delta E\) are 154 \(\mathrm{kJ} / \mathrm{mol}\) and 136 \(\mathrm{kJ} / \mathrm{mol}\) , respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\) . (b) What is the activation energy for the reverse reaction?

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?

Indicate whether each statement is true or false. \(\begin{array}{l}{\text { (a) If you measure the rate constant for a reaction at different}} \\ {\text { temperatures, you can calculate the overall }} \\ {\text { enthalpy change for the reaction. }} \\ {\text { (b) Exothermic reactions are faster than endothermic }} \\ {\text { reactions. }} \\ {\text { (c) If you double the temperature for a reaction, you cut }} \\ {\text { the activation energy in half. }}\end{array}\)

You have studied the gas-phase oxidation of HBr by \(\mathrm{O}_{2}\) : $$4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g)$$ You find the reaction to be first order with respect to HBr and first order with respect to \(\mathrm{O}_{2}\) . You propose the following mechanism: $$ \begin{array}{c}{\operatorname{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{HOOBr}(g)} \\ {\operatorname{HOOBr}(g)+\operatorname{HBr}(g) \longrightarrow 2 \mathrm{HOBr}(g)} \\\ {\operatorname{HOBr}(g)+\operatorname{HBr}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g)}\end{array}$$ (a) Confirm that the elementary reactions add to give the overall reaction. (b) Based on the experimentally determined rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

(a) In which of the following reactions would you expect the orientation factor to be least important in leading to reaction: \(\mathrm{NO}+\mathrm{O} \longrightarrow \mathrm{NO}_{2}\) or \(\mathrm{H}+\mathrm{Cl} \longrightarrow \mathrm{HCl}\) ? (b) Does the orientation factor depend on temperature?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

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.

Sign-up for free