   # Consider two reaction vessels, one containing A and the other containing B, with equal concentrations at t = 0. If both substances decompose by first-order kinetics, where k A = 4.50 × 10 − 4 s − 1 k B = 3.70 × 10 − 3 s − 1 how much time must pass to reach a condition such that [A] = 4.00[B]? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 11, Problem 85AE
Textbook Problem
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## Consider two reaction vessels, one containing A and the other containing B, with equal concentrations at t = 0. If both substances decompose by first-order kinetics, where k A =   4.50   ×   10 − 4 s − 1 k B =   3.70   ×   10 − 3 s − 1 how much time must pass to reach a condition such that [A] = 4.00[B]?

Interpretation Introduction

Interpretation: The rate constants of two first order reactions that have equal concentrations at t=0 are given. The time is to be calculated for the given condition.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction.

To determine: The time that must pass to reach a condition of [A]=4.00[B] .

### Explanation of Solution

Given

The rate constant of a reaction that contains A is 4.50×104s1 .

The rate constant of a reaction that contains B is 3.70×103s1 .

The relation between concentration of A compound and B compound is,

[A]=4.00[B] (1)

Formula

The integral rate law equation of first order reaction is,

ln[A]=kt+[A]0 (2)

Where,

• k is rate constant.
• t is time.
• [A]0 is initial concentration.
• [A] is concentration at time t .

Substitute the values of k for compound A in the above equation.

ln[A]=kt+ln[A]0ln[A]=4.50×104s1×t+ln[A]0 .(3)

Substitute the values of k for compound B in equation (1).

ln[B]=kt+ln[B]0ln[B]=3.70×103s1×t+ln[B]0

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