   # A certain first-order reaction is 45.0% complete in 65 s. What are the values of the rate constant and the half-life for this process? ### 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 49E
Textbook Problem
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## A certain first-order reaction is 45.0% complete in 65 s. What are the values of the rate constant and the half-life for this process?

Interpretation Introduction

Interpretation: The time of 45% completion of first order reaction is given. The rate constant and half life of this process are to be calculated.

Concept introduction: The half-life is defined as the time required for the concentration of reactant to be reduced to one-half of its initial value.

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 rate constant of first order reaction.

### Explanation of Solution

Given

The first order reaction is completed 45% in 65s .

Formula

The integral rate law equation of first order reaction is,

kt=ln([A]0[A]) (1)

Where,

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

The first order reaction is completed 45% in 65s . It means that this amount of reactant is consumed in the reaction. The percentage of reactant concentration left is,

100%45%=55%

The percentage composition of reactant left in a reaction is calculated by using the formula,

The %reactant left=[A][A]0×100%

Substitute the value of percentage of reactant concentration left in above equation.

The %reactant left=[A][A]0×100%55%=[A][A]0×100%[A]0[A]=1.82

Substitute the values of [A]0[A] and t in the equation (1)

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