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.4E

(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.4E

The rate constant of the given first order reaction is 1.10×10-2min-1_.

Explanation of Solution

The given first order reaction is shown below.

  AB+C

As per the data given in the question, the time taken for the decomposition of reactant to one-fourth of its initial value is 125min.

The final concentration of A after 125min is represented as shown below.

    [A]t=14[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 14[A]0.

The value of t is 125min.

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

  ln14[A]0[A]0=kr×(125min)ln14=kr×(125min)1.3862=kr×(125min)kr=1.3862125min

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

    kr=1.10×102min1

Thus, the rate constant of the given first order reaction is 1.10×10-2min-1_.

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 7B.4E

The rate constant of the given first order reaction is 1.028×10-3s-1_.

Explanation of Solution

The given first order reaction is shown below.

  2AD+E

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

As per the data given in the question, the initial concentration of A is 0.0421molL1 and concentration of rises to 0.00132molL1 in 63s.  Therefore the left over concentration of A after 63s is calculated by the expression  shown below.

    [A]t=[A]02[D]        (2)

Where,

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

The value of [A]0 is 0.0421molL1.

The value of [D] is 0.00132molL1.

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

  [A]t=0.0421molL12×0.00132molL1=0.03946molL1

Therefore the final concentration of A after 63s is 0.03946molL1.

The value of [A]t is 0.03946molL1.

The value of [A]0 is 0.0421molL1.

The value of t is 63s.

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

  ln0.03946molL10.0421molL1=kr×(63s)ln0.9372=kr×(63s)0.0648=kr×(63s)kr=0.064863s

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

    kr=1.028×103s1

Thus, the rate constant of the given first order reaction is 1.028×10-3s-1_.

(c)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 7B.4E

The rate constant of the given first order reaction is 7.25×10-2min-1_.

Explanation of Solution

The given first order reaction is shown below.

  3AF+G

According to the given chemical reaction, 3.0molL1 of A gets decomposed to form 1.0molL1 of F.

As per the data given in the question, the initial concentration of A is 0.080molL1 and concentration of F rises to 0.015molL1 in 11.4min.  Therefore the left over concentration of A after 11.4min is calculated by the expression shown below.

    [A]t=[A]03[F]        (3)

Where,

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

The value of [A]0 is 0.080molL1.

The value of [F] is 0.015molL1.

Substitute the value of [A]0 and [F] in equation (3).

  [A]t=0.080molL13×0.015molL1=0.035molL1

Therefore the final concentration of A after 11.4min is 0.035molL1.

The value of [A]t is 0.035molL1.

The value of [A]0 is 0.080molL1.

The value of t is 11.4min.

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

  ln0.035molL10.080molL1=kr×(11.4min)ln0.4375=kr×(11.4min)0.8266=kr×(11.4min)kr=0.826611.4min

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

    kr=7.25×102min1

Thus, the rate constant of the given first order reaction is 7.25×10-2min-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