Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 11, Problem 14QRT

(a)

Interpretation Introduction

Interpretation:

It has to be shown that the reaction obeys the rate law as given below.

  N2O52NO2+12O2Rate=Δ[N2O5]Δt=k[N2O5]

Concept Introduction:

In a first-order reaction, the graph of concentration vs. rate is linear.

(a)

Expert Solution
Check Mark

Explanation of Solution

Average concentrations can be calculated as given below.

From 0to0.50h:

  Averageconcentration=(0.849mol/L)+(0.733mol/L)2=0.791mol/L

From 0.50to1.00h:

  Averageconcentration=(0.733mol/L)+(0.633mol/L)2=0.683mol/L

From 1.00to2.00h:

  Averageconcentration=(0.633mol/L)+(0.472mol/L)2=0.553mol/L

From 2.00to3.00h:

  Averageconcentration=(0.472mol/L)+(0.352mol/L)2=0.412mol/L

From 3.00to4.00h:

  Averageconcentration=(0.352mol/L)+(0.262mol/L)2=0.307mol/L

From 4.00to5.00h:

  Averageconcentration=(0.262mol/L)+(0.196mol/L)2=0.229mol/L

The average rate and concentration is given in the table below.

Average concentration (mol/L)Average rate (molL1h1)
0.7910.23
0.6830.20
0.5530.161
0.4120.120
0.3070.090
0.2290.066

A graph of concentration vs. rate is given below.

Chemistry: The Molecular Science, Chapter 11, Problem 14QRT

Figure 1

The above graph is linear which satisfies the rate law.

(b)

Interpretation Introduction

Interpretation:

The rate constant k has to be evaluated as an average of the values obtained for the six intervals.

(b)

Expert Solution
Check Mark

Answer to Problem 14QRT

The average value of k is 0.29h-1.

Explanation of Solution

The rate constant k as an average of the values obtained for the six intervals can be calculated as given below.

From 0to0.50h:

Average rate is 0.23molL1h1.

Average concentration is 0.791molL1.

Then, the rate constant can be calculated as given below.

  k=Rate[N2O5]=0.23molL1h10.791molL1=0.29h1.

From 0.50to1.00h:

Average rate is 0.20molL1h1.

Average concentration is 0.683molL1.

Then, the rate constant can be calculated as given below.

  k=Rate[N2O5]=0.20molL1h10.683molL1=0.29h1.

From 1.00to2.00h:

Average rate is 0.161molL1h1.

Average concentration is 0.553molL1.

Then, the rate constant can be calculated as given below.

  k=Rate[N2O5]=0.161molL1h10.553molL1=0.291h1.

From 2.00to3.00h:

Average rate is 0.120molL1h1.

Average concentration is 0.412molL1.

Then, the rate constant can be calculated as given below.

  k=Rate[N2O5]=0.120molL1h10.412molL1=0.291h1.

From 3.00to4.00h:

Average rate is 0.090molL1h1.

Average concentration is 0.307molL1.

Then, the rate constant can be calculated as given below.

  k=Rate[N2O5]=0.090molL1h10.307molL1=0.29h1.

From 4.00to5.00h:

Average rate is 0.066molL1h1.

Average concentration is 0.229molL1.

Then, the rate constant can be calculated as given below.

  k=Rate[N2O5]=0.066molL1h10.229molL1=0.29h1.

The average value of k is 0.29h-1.

(c)

Interpretation Introduction

Interpretation:

The reaction rate exactly 2.5h from the start has to be calculated.

(c)

Expert Solution
Check Mark

Answer to Problem 14QRT

The reaction rate exactly 2.5h from the start is 0.12molL-1h-1.

Explanation of Solution

The initial concentration of N2O5 is 0.849mol/L.  The rate constant is 0.29h1.

The rate law describes that the reaction is a first-order reaction.  The integrated first-order rate law is given below.

  ln[N2O5]t=kt+ln[N2O5]0=(0.29h1)×(2.50h)+ln(0.849)ln[N2O5]t=0.889[N2O5]t=0.411mol/L.

Now, the rate of the reaction at 2.5h can be calculated as follows,

  Rate=k[N2O5]=(0.29h1)×(0.411mol/L)=0.12molL-1h-1.

Therefore, the reaction rate exactly 2.5h from the start is 0.12molL-1h-1.

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Chapter 11 Solutions

Chemistry: The Molecular Science

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Kinetics: Initial Rates and Integrated Rate Laws; Author: Professor Dave Explains;https://www.youtube.com/watch?v=wYqQCojggyM;License: Standard YouTube License, CC-BY