   Chapter 12, Problem 97CWP

Chapter
Section
Textbook Problem

# A reaction of the form aA → Products gives a plot of ln[A] versus time (in seconds), which is a straight line with a slope of −7.35 X 10−3. Assuming [A]0 = 0.0100 M, calculate the time (in seconds) required for the reaction to reach 22.9% completion.

Interpretation Introduction

Interpretation: For the given reaction the time required for the 22.9% completion of the reaction is 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.

To determine: The amount of time required for the reaction to reach 22.9% completion.

Explanation

Explanation

Given

The reaction is given as,

aAproducts

The plot of ln[A] Vs time (sec) is a straight line with a slop of 7.35×103 .

The initial concentration of the reactant (A) is 0.0100M .

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

For the first order reaction the integrated rate law is given by the equation,

ln[A]=kt+ln[A0]

Where,

• ln[A] is the concentration of the reactant.
• ln[A0] is the initial concentration of the reactant.
• k is the slop and is known as rate constant.
• t is the time.
• The negative sign shows the negative slop of the straight line.

The above exactly a form of equation of straight line that is given as,

y=mx+c

Where,

• y is the vertical axis.
• x is the horizontal axis.
• m is the slope of the straight line.
• c is the intercept on y -axis.

The first order reaction clearly indicates the straight line plot of ln[A] Vs time. The negative value of slope is 7.35×103 . Therefore, the value of k is 7.35×103

The half life of the first order reaction is given by the formula,

kt1/2=0.693

Where,

• t1/2 is the half life.

Substitute the value of rate constant in the above equation.

kt1/2=0.693(7.35×103)t1/2=0.693t1/2=0.6937.35×103=94

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