Chapter 17, Problem 17.22P

Interpretation Introduction

Interpretation:

The rate constant for the first order reaction of the below described equation at ${\text{500}}^{\text{°}}\text{C}$ is $\text{5}\text{.5}\text{×}{\text{10}}^{\text{-4}}{\text{s}}^{\text{-1}}$.

The half-life of cyclopropane at ${\text{500}}^{\text{°}}\text{C}$ has to be calculated.  Also, the concentration of cyclopropane that remains after $\text{2}\text{.0}\text{hours}$ has to be calculated, if the initial cyclopropane concentration is $\text{1}\text{.00}\text{×}{\text{10}}^{\text{-3}}\text{M}$ at ${\text{500}}^{\text{°}}\text{C}$.

Concept Introduction:

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

  $\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\\ \\ {\text{t}}_{\text{1/2}}\text{=}\text{half-life}\text{of}\text{the}\text{reaction}\\ \\ \text{k}\text{=}\text{Rate}\text{constant}\text{of}\text{the}\text{reaction}\end{array}$

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

  $\begin{array}{l}\text{ln}\frac{\left[\text{A}\right]}{{\left[\text{A}\right]}_{\text{0}}}\text{=}\text{-kt}\\ \\ \left[\text{A}\right]\text{=}\text{concentration}\text{of}\text{A}\text{at}\text{time,t}\\ \\ {\left[\text{A}\right]}_{\text{0}}\text{=}\text{initial}\text{concentration}\text{of}\text{A}\\ \\ \text{k}\text{=}\text{Rate}\text{constant}\\ \\ \text{t}\text{=}\text{time}\end{array}$

Explanation of Solution

Given,

  $\begin{array}{l}\text{k}\text{=}\text{5}\text{.5}\text{×}{\text{10}}^{\text{-4}}{\text{s}}^{\text{-1}}\\ \\ {\left[\text{cyclopropane}\right]}_{\text{0}}\text{=}\text{1}\text{.00}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ \\ \text{t}\text{=}\text{2}\text{.0}\text{hours}\left(\begin{array}{l}\text{1}\text{hour}\text{=}\text{3600}\text{s}\\ \text{2}\text{.0}\text{hours}\text{=}\text{7200}\text{s}\end{array}\right)\end{array}$

The half-life of cyclopropane can be calculated as

  $\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\\ \text{=}\frac{\text{0}\text{.693}}{\text{5}\text{.5}\text{×}{\text{10}}^{\text{-4}}{\text{s}}^{\text{-1}}}\\ \text{=}\text{1260}\text{s}\end{array}$

The concentration of cyclopropane that remains after $\text{2}\text{.0}\text{hours}$ can be calculated as

  $\begin{array}{l}\text{ln}\frac{\left[\text{cyclopropane}\right]}{{\left[\text{cyclopropane}\right]}_{\text{0}}}\text{=}\text{-kt}\\ \text{ln}\frac{\left[\text{cyclopropane}\right]}{\left[\text{1}\text{.00}\text{×}{\text{10}}^{\text{-3}}\text{M}\right]}\text{=}\text{-}\left(\text{5}\text{.5}\text{×}{\text{10}}^{\text{-4}}{\text{s}}^{\text{-1}}\right)\left(\text{7200}\text{s}\right)\\ \text{ln}\frac{\left[\text{cyclopropane}\right]}{\left[\text{1}\text{.00}\text{×}{\text{10}}^{\text{-3}}\text{M}\right]}\text{=}\text{-}\text{3}\text{.96}\\ \\ \frac{\left[\text{cyclopropane}\right]}{\left[\text{1}\text{.00}\text{×}{\text{10}}^{\text{-3}}\text{M}\right]}\text{=}{\text{e}}^{\text{-3}\text{.96}}\\ \frac{\left[\text{cyclopropane}\right]}{\left[\text{1}\text{.00}\text{×}{\text{10}}^{\text{-3}}\text{M}\right]}\text{=}\text{0}\text{.019}\\ \\ \left[\text{cyclopropane}\right]\text{=}\text{0}\text{.019}\text{×}\left(\text{1}\text{.00}\text{×}{\text{10}}^{\text{-3}}\text{M}\right)\\ \text{=}\text{1}\text{.9}\text{×}{\text{10}}^{\text{-5}}\text{M}\end{array}$

The concentration of cyclopropane that remains after $\text{2}\text{.0}\text{hours}$ is $\text{1}\text{.9}\text{×}{\text{10}}^{\text{-5}}\text{M}$.