General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
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Chapter 17, Problem 17.22P
Interpretation Introduction

Interpretation:

The rate constant for the first order reaction of the below described equation at 500°C is 5.5×10-4s-1.

General Chemistry, Chapter 17, Problem 17.22P

The half-life of cyclopropane at 500°C has to be calculated.  Also, the concentration of cyclopropane that remains after 2.0hours has to be calculated, if the initial cyclopropane concentration is 1.00×10-3M at 500°C.

Concept Introduction:

The half-life of first order reaction is calculated by the following equation,

  t1/2=0.693kt1/2=half-lifeofthereactionk=Rateconstantofthereaction

The concentration of A that remains after time, t in a first order reaction can be calculated by the following equation:

  ln[A][A]0=-kt[A]=concentrationofAattime,t[A]0=initialconcentrationofAk=Rateconstantt=time

Expert Solution & Answer
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Explanation of Solution

Given,

  k=5.5×10-4s-1[cyclopropane]0=1.00×10-3Mt=2.0hours (1hour=3600s2.0hours=7200s)

The half-life of cyclopropane can be calculated as

  t1/2=0.693k=0.6935.5×10-4s-1=1260s

The concentration of cyclopropane that remains after 2.0hours can be calculated as

  ln[cyclopropane][cyclopropane]0=-ktln[cyclopropane][1.00×10-3M]=-(5.5×10-4s-1)(7200s)ln[cyclopropane][1.00×10-3M]=-3.96[cyclopropane][1.00×10-3M]=e-3.96[cyclopropane][1.00×10-3M]=0.019[cyclopropane]=0.019×(1.00×10-3M) =1.9×10-5M

The concentration of cyclopropane that remains after 2.0hours is 1.9×10-5M.

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