ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
7th Edition
ISBN: 9781319403959
Author: ATKINS
Publisher: MAC HIGHER
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Chapter 7, Problem 7B.7E

(a)

Interpretation Introduction

Interpretation:

The time taken by A to get reduced into one-eighth of its initial concentration 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 equation that represents the integrated rate law for the first order kinetics is shown below.

    [A]t=[A]0ekrt

The half-life of the particular chemical reaction is the time in which exactly half of the reactant gets consumed.  The mathematical expression for the half-life for a reaction follows first order kinetics is shown below.

  t1/2=ln2kr

(a)

Expert Solution
Check Mark

Answer to Problem 7B.7E

The time taken by A to get reduced into one-eighth of its initial concentration is 1.44×103min_.

Explanation of Solution

The decomposition of A follows the first order kinetics and having the half-life equals to 355s.

The relation between the half-life and the rate constant for the first order is shown below.

    t1/2=ln2kr        (1)

Where,

  • t1/2 is the half-life.
  • kr  is the order rate constant.

The value of t1/2 is 355s.

Substitute the value of kr in equation (1).

    kr=ln2355s=1.95×103s1

Therefore, the rate constant for the decomposition of A is 1.95×103s1.

The concentration of A after time t is one-eighth of the initial concentration.  Mathematically the final concentration, [A]t in terms of initial concentration, [A]0 is shown below.

    [A]t=18[A]0

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

  [A]t=[A]0ekrt        (2)

Where,

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

The value kr is 1.95×103s1.

The value of [A]t is 18[A]0.

Substitute the value of kr and [A]t in equation (2).

    18[A]0=[A]0e(1.95×103s1)t18=e(1.95×103s1)t(1.95×103s1)t=ln18(1.95×103s1)t=2.0794

On further calculation the required time taken is calculated as shown below.

    t=2.07941.95×103s1=1.06×103s

Thus, the time taken by A to get reduced into one-eighth of its initial concentration is 1.06×103s_.

(b)

Interpretation Introduction

Interpretation:

The time taken by A to get reduced into one-fourth of its initial concentration has to be determined.

Concept Introduction:

Same as part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 7B.7E

The time taken by A to get reduced into one-fourth of its initial concentration is 710.8s_.

Explanation of Solution

As per the given data the decomposition of A follows the first order kinetics and having the rate constant equals to 1.95×103s1.

The concentration of A after time t is one-fourth of the initial concentration.  Mathematically the final concentration, [A]t in terms of initial concentration, [A]0 is shown below.

    [A]t=14[A]0

The value kr is 1.95×103s1.

The value of [A]t is 14[A]0.

Substitute the value of kr and [A]t in equation (2).

    14[A]0=[A]0e(1.95×103s1)t14=e(1.95×103s1)t(1.95×103s1)t=ln14(1.95×103s1)t=1.3862

On further calculation the required time taken is calculated as shown below.

    t=1.38621.95×103s1=710.8s

Thus, the time taken by A to get reduced into one-fourth of its initial concentration is 710.8s_.

(c)

Interpretation Introduction

Interpretation:

The time taken by A to get reduced into 15% of its initial concentration has to be determined.

Concept Introduction:

Same as part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 7B.7E

The time taken by A to get reduced into 15% of its initial concentration is 972.8s_.

Explanation of Solution

As per the given data the decomposition of A follows the first order kinetics and having the rate constant equals to 1.95×103s1.

The final concentration of A is 15% of the initial value.  Mathematically the final value, [A]t in terms of initial value, [A]0 is shown below.

    [A]t=15%[A]0=15100[A]0=0.15[A]0

The value kr is 1.95×103s1.

The value of [A]t is 0.15[A]0.

Substitute the value of kr and [A]t in equation (2).

    0.15[A]0=[A]0e(1.95×103s1)×t0.15=e(1.95×103s1)×t(1.95×103s1)t=ln0.15(1.95×103s1)t=1.8971

On further calculation the required time taken is calculated as shown below.

    t=1.89711.95×103s1=972.8s

Thus, the time taken by A to get reduced into 15% of its initial concentration is 972.8s_.

(d)

Interpretation Introduction

Interpretation:

The time taken by A to get reduced into one-ninth of its initial concentration has to be determined.

Concept Introduction:

Same as part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 7B.7E

The time taken by A to get reduced into one-ninth of its initial concentration is 1.12×103s_.

Explanation of Solution

As per the given data the decomposition of A follows the first order kinetics and having the rate constant equals to 1.95×103s1.

The concentration of A after time t is one-ninth of the initial concentration.  Mathematically the final concentration, [A]t in terms of initial concentration, [A]0 is shown below.

    [A]t=19[A]0

The value kr is 1.95×103s1.

The value of [A]t is 19[A]0.

Substitute the value of kr and [A]t in equation (2).

    19[A]0=[A]0e(1.95×103s1)t19=e(1.95×103s1)t(1.95×103s1)t=ln19(1.95×103s1)t=2.1973

On further calculation the required time taken is calculated as shown below.

    t=2.19731.95×103s1=1.12×103s

Thus, the time taken by A to get reduced into one-ninth of its initial concentration is 1.12×103s_.

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Chapter 7 Solutions

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

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