Chapter 14, Problem 30PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Radioactive gold-198 is used in the diagnosis of liver problems. The half-life of this isotope is 2.7 days. If you begin with a 5.6-mg sample of the isotope, how much of this sample remains after 1.0 day?

Interpretation Introduction

Interpretation: The amount of sample remains after 1 day 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.

Explanation

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

â€‚Â Reactionâ€‰rateÂ =Â kÂ [198Au]1.Given:t1/2â€‰=â€‰2.7Â days;t1/2=0.693kk=0.6932.7Â days=â€‰0.2567â€‰d-1[198Au]0â€‰â€‰=â€‰â€‰5.6â€‰mg[198Au]t=?Therefore,FirstÂ orderÂ rateÂ law,ln[198Au]tâ€‰=â€‰-â€‰kâ€‰tÂ +Â ln[198Au]0ln[198Au]tâ€‰=â€‰-â€‰(0

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