   Chapter 14, Problem 29PS

Chapter
Section
Textbook Problem

The radioactive isotope 64Cu is used in the form of copper(II) acetate to study Wilson’s disease. The isotope has a half-life of 12.70 hours. What fraction of radioactive copper(II) acetate remains after 64 hours?

Interpretation Introduction

Interpretation: The fraction of radioactive copper isotope remains after 64 hours has to be given.

Concept Introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time.  The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of mol/(L.s).

Integrated rate law for first order reaction:

Consider A as substance, that gives the product based on the equation,

aAproducts

Where a= stoichiometric co-efficient of reactant A.

Consider the reaction has first-order rate law,

Rate=-Δ[A]Δt=k[A]

The integrated rate law equation can be given as,

ln[A]t[A]o=-kt

The above expression is called integrated rate law for first order reaction.

Half-life for first order reactions:

The half-life for the first order reaction is constant and it is independent of the reactant concentration.

Half-life period of first order reaction can be calculated using the equation,

t1/2=0.693k

Explanation

The fraction of radioactive copper isotope remains after 64 hours is calculated as,

Reactionrate = k [Xe(CF3)2]1.Given:t1/2=12.70 hours;t1/2=0.693kk=0.69312.70 hours=0.0546h-1[64Cu]0=7.50mg[64Cu]t=0.25mgTherefore,First order rate law,ln[64Cu]t=-kt + ln[64Cu]

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