   Chapter 14, Problem 62GQ

Chapter
Section
Textbook Problem

NOx, mixture of NO and NO2, plays an essential role in the production of pollutants found in photochemical smog. The NOx in the atmosphere is slowly broken down to N2 and O2 in a first-order reaction. The average half-life of NOx in the smoke-stack emissions in a large city during daylight is 3.9 hours. (a) Starting with 1.50 mg in an experiment, what quantity of NOx remains after 5.25 hours? (b) How many hours of daylight must have elapsed to decrease 1.50 mg of NOx to 2.50 × 10−6 mg?

(a)

Interpretation Introduction

Interpretation: Amount of NOx present after 5.25 hours should be determined.

Concept introduction:

Order of a reaction:  The order of each reactant is represented by the exponential term of the respective reactant present in the rate law and the overall order of the reaction is the sum of all the exponents of all reactants present in the chemical reaction. The order of the reaction is directly proportional to the concentration of the reactants.

The integrated rate law for the first order is as follows:

ln[A]=ln[A0]ktwhere,[A0]= initialconcentrationk=rate constantt=time[A]=concentrationaftertimet

Half–life period: It is the time required for the reactant (substrate) concentration to reduce to the one-half of its initial concentration.

The half–life period for first order reaction is as follows:

t1/2=0.693kWhere,k = Rate constantt1/2= half-life

Explanation

Given,

Half-time,t1/2=3.9hTime,t=5.25hInitialconcentration,No=1.50mg

It is given that the given reaction is a first order reaction.

Half–life is determined by the expression as follows:

t1/2=0.693kWhere,k = Rate constantt1/2= half-life

The rate constant of the reaction can be determined as follows:

Substitute the values as given below:

k=0.693t1/2k=0.6933.9h=0.1777hr1

The rate constant of the reaction is 0.1777hr1

The integrated rate law for the first sorder is as follows:

ln[A]=ln[A0]ktwhere,[A0]= initialconcentrationk=rate constantt=time[A]=concentrationaftertimet

Rearranging the vales as given below:

k=2

(b)

Interpretation Introduction

Interpretation: Time required to decrease 1.50mg of NOx to 2.50×106mg should be determined.

Concept introduction:

Order of a reaction:  The order of each reactant is represented by the exponential term of the respective reactant present in the rate law and the overall order of the reaction is the sum of all the exponents of all reactants present in the chemical reaction. The order of the reaction is directly proportional to the concentration of the reactants.

The integrated rate law for the first order is as follows:

ln[A]=ln[A0]ktwhere,[A0]= initialconcentrationk=rate constantt=time[A]=concentrationaftertimet

Half–life period: It is the time required for the reactant (substrate) concentration to reduce to the one-half of its initial concentration.

The half–life period for first order reaction is as follows:

t1/2=0.693kWhere,k = Rate constantt1/2= half-life

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 