   Chapter 12, Problem 56E

Chapter
Section
Textbook Problem

# Theophylline is a pharmaceutical drug that is sometimes used to help with lung function. You observe a ease where the initial lab results indicate that the concentration of theophyllinein a patient’s body decreased from 2.0 × 10−3 M to 1.0 × 10−3M in 24 hours. In another 12 hours the drug concentration was found to be 5.0 × 10−4 M. What is the value of the rate constant for the metabolism of this drug in the body?

Interpretation Introduction

Interpretation: It is given that value of concentration of theophylline is decreasing in 24hours . The value of concentration of theophylline is further reduced in another 12hours . The rate constant of this drug is to be calculated.

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 rate constant of metabolism of theophylline

Explanation

The initial concentration of theophyllineis 2.0×103M .

The concentration of theophylline after 24hours is 1.0×103M .

The concentration of theophylline after 12hours is 5.0×103M .

It is clear from the given value that,half life is decreasing as the concentration of theophylline is decreasing. But this trend of decrease in value of half life is found in zero order reaction. Hence, the given reaction is zero order reaction.

Formula

The integral rate law equation of zero order reaction is,

[A]=kt+[A]0

Where,

• k is rate constant.
• t is time (24hours) .
• [A]0 is initial concentration.
• [A] is concentration at time t (1.0×103M) .

Substitute the values of [A]0,[A] and t in the above equation

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