Chemistry
Chemistry
13th Edition
ISBN: 9781259911156
Author: Raymond Chang Dr., Jason Overby Professor
Publisher: McGraw-Hill Education
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Question
Chapter 13, Problem 13.84QP
Interpretation Introduction

Interpretation:

From the given information, how long it take for the concentration of A to be four times that of B has to be determined.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

  • Half-life of a reaction is represented by the symbol as t12
  • Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
  • The half-life period is then shortened as half-life in early 1950s.

Order of a reaction:  The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

Expert Solution & Answer
Check Mark

Answer to Problem 13.84QP

The time taken for the concentration of A to be four times that of B is t=56.4min

Explanation of Solution

From the given information, A and B both were decomposed by first-order kinetics.

Therefore, we can represent a first-order rate equation for A and also one for B as follows.

ln[A]t[A]0=kAtln[B]t[B]0=kBt

[A]t[A]0=ekAt[B]t[B]0=ekBt

[A]t=[A]0ekAt[B]t=[B]0ekBt

Now, we should calculate each of the rate constant as kAandkB from their corresponding half-lives as follows.

kA=0.69350.0min=0.0139min1kB=0.69318.0min=0.0385min1

Then, the initial concentrations for A and B are equal.  So, [A]0 = [B]0

Therefore, from the first order rate expressions we can write as

[A]t[B]t = 4[A]0ekAt[B]0ekBtekAtekBt=e(kBkA)t=e(0.03850.0139)t

4=e0.0246t

ln4=0.0246t

t=56.4min

Conclusion

The time taken for the concentration of A to be four times that of B was t=56.4min

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Chapter 13 Solutions

Chemistry

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