   # The conversion of cyclopropane to propene (Example 14.5) occurs with a first-order rate constant of 2.42 ×10 −2 h −1 . How long will it take for the concentration of cyclopropane to decrease from an initial concentration of 0.080 mol/L to 0.020 mol/L? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 14, Problem 18PS
Textbook Problem
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## The conversion of cyclopropane to propene (Example 14.5) occurs with a first-order rate constant of 2.42 ×10−2 h−1. How long will it take for the concentration of cyclopropane to decrease from an initial concentration of 0.080 mol/L to 0.020 mol/L?

Interpretation Introduction

Interpretation: The time required for the reaction has to be found.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

aA + bBxXRate of reaction = k [A]m[B]n

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

### Explanation of Solution

The time required for the reaction is calculated as,

Reactionrate = k [cyclopropane].Given:The reaction is first order reaction,rate law is , ln[cyclopropane]t=-kt+ln[cyclopropane]0Therefore,ln[cyclopropane]t=-kt+ ln[cyclopropane]0ln (0.020mol/L)     = (2.42×10-2h-1)(t)+ ln(0

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