Chemical Principles: The Quest for Insight
Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
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Chapter 7, Problem 7B.3E

(a)

Interpretation Introduction

Interpretation:

The rate constant for the given first order reaction has to be determined.

Concept Introduction:

According to the integrated rate law for the first order reaction, the concentration of reactant is the exponential function of time.  The two equations that represent the integrated rate law for the first order kinetics is shown below.

    ln[A]t[A]0=krt[A]t=[A]0ekrt

(a)

Expert Solution
Check Mark

Answer to Problem 7B.3E

The rate constant of the given first order reaction is 6.93×10-4s-1_.

Explanation of Solution

The given first order reaction is shown below.

  AB

As per the data given in the question, the time taken for the decomposition of reactant to half of its initial value is 1000s.

The final concentration of A at half life is represented as shown below.

    [A]t=12[A]0

The relation between the changes in the concentration of reactant after time t for the first order reaction is shown below.

    ln[A]t[A]0=krt        (1)

Where,

  • [A]t is the concentration of A at time t.
  • [A]0 is the initial concentration of A.
  • t is the time taken.
  • kr is the rate constant of the reaction.

The value of [A]t is 12[A]0.

The value of t is 1000s.

Substitute the value of t and [A]t in equation (1).

  ln12[A]0[A]0=kr×(1000s)ln12=kr×(1000s)0.693=kr×(1000s)kr=0.6931000s

On further calculation the rate constant of the give reaction is calculated as shown below.

    kr=6.93×104s1

Thus, the rate constant of the given first order reaction is 6.93×10-4s-1_.

(b)

Interpretation Introduction

Interpretation:

The rate constant for the given first order reaction has to be determined.

Concept Introduction:

Same as part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 7B.3E

The rate constant of the given first order reaction is 9.37×10-3s-1_.

Explanation of Solution

The given first order reaction is shown below.

  AB

As per the data given in the question, the initial concentration of A is 0.67molL1 and gets reduced to 0.53molL1 in 25s.

The value of [A]t is 0.53molL1.

The value of [A]0 is 0.67molL1.

The value of t is 1000s.

Substitute the value of [A]0, [A]t and t in equation (1).

  ln0.53molL10.67molL1=kr×(25s)ln0.7910=kr×(25s)0.2344=kr×(25s)kr=0.234425s

On further calculation the rate constant of the give reaction is calculated as shown below.

    kr=9.37×103s1

Thus, the rate constant of the given first order reaction is 9.37×10-3s-1_.

(c)

Interpretation Introduction

Interpretation:

The rate constant for the given first order reaction has to be determined.

Concept Introduction:

Same as part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 7B.3E

The rate constant of the given first order reaction is 5.11×10-3s-1_.

Explanation of Solution

The given first order reaction is shown below.

  2AB+C

According to the given chemical reaction, 2.0molL1 of A gets decomposed to form 1.0molL1 of B.

As per the data given in the question, the initial concentration of A is 0.153molL1 and concentration of rises to 0.034molL1 in 115s.  Therefore the left over concentration of A after 115s is calculated by the expression  shown below.

    [A]t=[A]02[B]        (2)

Where,

  • [A]t is the concentration of A at time t.
  • [A]0 is the initial concentration of A.
  • [B] is the concentration of B after time t.

The value of [A]0 is 0.153molL1.

The value of [B] is 0.034molL1.

Substitute the value of [A]0 and [B] in equation (2).

  [A]t=0.153molL12×0.034molL1=0.085molL1

Therefore the final concentration of A after 115s is 0.085molL1.

The value of [A]t is 0.085molL1.

The value of [A]0 is 0.153molL1.

The value of t is 115s.

Substitute the value of [A]0, [A]t and t in equation (1).

  ln0.085molL10.153molL1=kr×(115s)ln0.5555=kr×(115s)0.5878=kr×(115s)kr=0.5878115s

On further calculation the rate constant of the give reaction is calculated as shown below.

    kr=5.11×103s1

Thus, the rate constant of the given first order reaction is 5.11×10-3s-1_.

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Chapter 7 Solutions

Chemical Principles: The Quest for Insight

Ch. 7 - Prob. 7A.3ECh. 7 - Prob. 7A.4ECh. 7 - Prob. 7A.7ECh. 7 - Prob. 7A.8ECh. 7 - Prob. 7A.9ECh. 7 - Prob. 7A.10ECh. 7 - Prob. 7A.11ECh. 7 - Prob. 7A.12ECh. 7 - Prob. 7A.13ECh. 7 - Prob. 7A.14ECh. 7 - Prob. 7A.15ECh. 7 - Prob. 7A.16ECh. 7 - Prob. 7A.17ECh. 7 - Prob. 7A.18ECh. 7 - Prob. 7B.1ASTCh. 7 - Prob. 7B.1BSTCh. 7 - Prob. 7B.2ASTCh. 7 - Prob. 7B.2BSTCh. 7 - Prob. 7B.3ASTCh. 7 - Prob. 7B.3BSTCh. 7 - Prob. 7B.4ASTCh. 7 - Prob. 7B.4BSTCh. 7 - Prob. 7B.5ASTCh. 7 - Prob. 7B.5BSTCh. 7 - Prob. 7B.1ECh. 7 - Prob. 7B.2ECh. 7 - Prob. 7B.3ECh. 7 - Prob. 7B.4ECh. 7 - Prob. 7B.5ECh. 7 - Prob. 7B.6ECh. 7 - Prob. 7B.7ECh. 7 - Prob. 7B.8ECh. 7 - Prob. 7B.9ECh. 7 - Prob. 7B.10ECh. 7 - Prob. 7B.13ECh. 7 - Prob. 7B.14ECh. 7 - Prob. 7B.15ECh. 7 - Prob. 7B.16ECh. 7 - Prob. 7B.17ECh. 7 - Prob. 7B.18ECh. 7 - Prob. 7B.19ECh. 7 - Prob. 7B.20ECh. 7 - Prob. 7B.21ECh. 7 - Prob. 7B.22ECh. 7 - Prob. 7C.1ASTCh. 7 - Prob. 7C.1BSTCh. 7 - Prob. 7C.2ASTCh. 7 - Prob. 7C.2BSTCh. 7 - Prob. 7C.1ECh. 7 - Prob. 7C.2ECh. 7 - Prob. 7C.3ECh. 7 - Prob. 7C.4ECh. 7 - Prob. 7C.5ECh. 7 - Prob. 7C.6ECh. 7 - Prob. 7C.7ECh. 7 - Prob. 7C.8ECh. 7 - Prob. 7C.9ECh. 7 - Prob. 7C.11ECh. 7 - Prob. 7C.12ECh. 7 - Prob. 7D.1ASTCh. 7 - Prob. 7D.1BSTCh. 7 - Prob. 7D.2ASTCh. 7 - Prob. 7D.2BSTCh. 7 - Prob. 7D.1ECh. 7 - Prob. 7D.2ECh. 7 - Prob. 7D.3ECh. 7 - Prob. 7D.5ECh. 7 - Prob. 7D.6ECh. 7 - Prob. 7D.7ECh. 7 - Prob. 7D.8ECh. 7 - Prob. 7E.1ASTCh. 7 - Prob. 7E.1BSTCh. 7 - Prob. 7E.1ECh. 7 - Prob. 7E.2ECh. 7 - Prob. 7E.3ECh. 7 - Prob. 7E.4ECh. 7 - Prob. 7E.5ECh. 7 - Prob. 7E.6ECh. 7 - Prob. 7E.7ECh. 7 - Prob. 7E.8ECh. 7 - Prob. 7E.9ECh. 7 - Prob. 1OCECh. 7 - Prob. 7.1ECh. 7 - Prob. 7.2ECh. 7 - Prob. 7.3ECh. 7 - Prob. 7.4ECh. 7 - Prob. 7.5ECh. 7 - Prob. 7.6ECh. 7 - Prob. 7.7ECh. 7 - Prob. 7.9ECh. 7 - Prob. 7.11ECh. 7 - Prob. 7.14ECh. 7 - Prob. 7.15ECh. 7 - Prob. 7.17ECh. 7 - Prob. 7.19ECh. 7 - Prob. 7.20ECh. 7 - Prob. 7.23ECh. 7 - Prob. 7.25ECh. 7 - Prob. 7.26ECh. 7 - Prob. 7.29ECh. 7 - Prob. 7.30ECh. 7 - Prob. 7.31E
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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