   Chapter 12, Problem 98CWP

Chapter
Section
Textbook Problem

# A certain reaction has the form aA → Products At a particular temperature, concentration versus time data were collected. A plot of 1/ [A] versus Lime (in seconds) gave a straight line with a slope of 6.90 × 10−2. What is the differential rate law for this reaction? What is the integrated rate law for this reaction? What is the value of the rate constant for this reaction? If [A]0 for this reaction is 0.100 M, what is the first half-life (in seconds)? If the original concentration (at t = 0) is 0.100 M, what is the second half-life (in seconds)?

Interpretation Introduction

Interpretation: For the given reaction the differential and integrated rate law is to be stated. The rate constant is to be calculated. The first and second half-life values for the given concentration are to be calculated.

Concept introduction: Rate constant is a proportionality coefficient that relates the rate of any chemical reaction at a specific temperature to the concentration of the reactant or the concentration of the product.

Order of reaction is a part of reaction which present as a power of concentration of reactants.

Half-life is the time in which the amount of the species reduces to just half.

To determine: The differential and integral rate law; the rate constant; the first and second half-life for the given reaction.

Explanation

Explanation

Given

The reaction is given as,

aAproducts

The plot of l/[A] Vs time (sec) is a straight line with a slop of 6.90×102 .

The given straight line plot is a compulsory condition of second order reaction. Therefore, the given reaction is the second order reaction.

For the second order reaction the differential rate law is expressed as,

aAProductsRate=k[A]2dxdt=k[A]2

Where,

• dxdt is the differential rate representation.
• k is the rate constant.
• [A] is the concentration.
• 2 is the order of reaction.

The integrated form of rate law is obtained by expression,

aAProductstime=0a0time=t(ax)x

The above representation shows that the change in concentration from initial time to a certain time t .

Therefore the differential rate law is also written as,

dxdt=k[A]2=k(ax)2

Where,

• (ax) is the change in concentration.

Integrate the above expression within the limit of 0t with respect to time and 0x for concentration as,

dxdt=k(ax)20x1(ax)2dx=k0tdt

Simplify the above expression

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