   Chapter 12, Problem 46E

Chapter
Section
Textbook Problem

# Consider the data plotted in Exercise 45 when answering the following questions.a. What is the concentration of A after 9 s?b. What are the first three half-lives for this experiment?

(a)

Interpretation Introduction

Interpretation: The graph of exercise 45 is to be considered to answer the question. The concentration of A is to be calculated for the given time. The first three half-lives is to be calculated for this experiment.

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. The differential rate law provides the rate of a reaction at specific reaction concentrations.

To determine: The concentration of A after 9s .

Explanation

The given time is 9s .

The initial concentration is 0.1mol/L .

The slope of second order is equal to its rate constant. The slope of graph in exercise 45 is 10 . Hence, the rate constant is 10 .

Formula

The integral rate law equation of second order reaction is,

1[A]=kt+1[A]0kt=1[A]1[A]0

Where,

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

(b)

Interpretation Introduction

Interpretation: The graph of exercise 45 is to be considered to answer the question. The concentration of A is to be calculated for the given time. The first three half-lives is to be calculated for this experiment.

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. The differential rate law provides the rate of a reaction at specific reaction concentrations.

To determine: The first three half lives of the given experiment.

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