Chapter 17, Problem 17B.4AE

(i)

Interpretation Introduction

Interpretation:

The second order rate constant for the reaction ${\text{CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2(aq)}}^{\text{-}}{\text{+ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{OH}}_{\text{(aq)}}$ is $\text{0}{\text{.11dm}}^{\text{3}}{\text{mol}}^{\text{-1}}$.  The concentration of ester after $\text{20s}$ has to be found.

Explanation of Solution

  $\text{A + B }\stackrel{}{\to }\text{ Products}$

Integrated rate law for this reaction is:

    $\begin{array}{l}\text{ln}\left(\frac{{\text{[A]}}_{\text{o}}\text{[B]}}{{\text{[B]}}_{\text{o}}\text{[A]}}\right)\text{ = }\left({\text{[B]}}_{\text{o}}{\text{-[A]}}_{\text{o}}\right)\text{kt}\\ \text{k=rate constant}\\ {\text{[A]}}_{\text{o}}{\text{,[B]}}_{\text{o}}\text{-initial concentration of the reactants}\\ \text{[A],[B]- concentrations of A,B}\end{array}$

Given equation is:

    $\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2(aq)}}^{\text{-}}{\text{+ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{OH}}_{\text{(aq)}}\\ \text{ln}\left(\frac{{{\text{[CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\text{]}}_{\text{o}}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[OH]}}_{\text{o}}{\text{[CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\text{]}}\right)\text{=}\left({{\text{[OH}}^{\text{-}}\text{]}}_{\text{o}}{\text{-[CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}{\text{]}}_{\text{o}}\right)\text{kt}\\ {\text{Replace[OH}}^{\text{-}}{\text{] and [CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}{\text{] with [OH}}^{\text{-}}{\text{]}}_{\text{o}}{\text{-x and [CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}{\text{]}}_{\text{o}}\text{-x}\\ \text{ln}\left(\frac{{{\text{[CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\text{]}}_{\text{o}}{\text{[OH}}^{\text{-}}\text{]-x}}{{\text{[OH]}}_{\text{o}}{\text{[CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\text{]-x}}\right)\text{=}\left({{\text{[OH}}^{\text{-}}\text{]}}_{\text{o}}{\text{-[CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}{\text{]}}_{\text{o}}\right)\text{kt}\end{array}$

Then the concentration of ester after $\text{20s}$ when ethyl ethanoate is added to aqueous sodium hydroxide will be:

$\begin{array}{l}\text{ln}\left(\frac{\text{(0}{\text{.110moldm}}^{\text{-3}}\text{)(0}{\text{.060moldm}}^{\text{-3}}\text{-x)}}{\text{(0}{\text{.060moldm}}^{\text{-3}}\text{)(0}{\text{.110moldm}}^{\text{-3}}\text{-x)}}\right)\text{ = (0}{\text{.060moldm}}^{\text{-3}}\text{-0}{\text{.110moldm}}^{\text{-3}}\text{) (0}{\text{.110moldm}}^{\text{-3}}\text{) 20 s }\\ \text{ln}\left(\left(\frac{\text{0}\text{.110}}{\text{0}\text{.060}}\right)\frac{\text{(0}{\text{.060moldm}}^{\text{-3}}\text{-x)}}{\text{(0}{\text{.110moldm}}^{\text{-3}}\text{-x)}}\right)\text{ = -0}\text{.11}\\ \frac{\text{(0}{\text{.060moldm}}^{\text{-3}}\text{-x)}}{\text{(0}{\text{.110moldm}}^{\text{-3}}\text{-x)}}{\text{ = e}}^{\text{-0}\text{.11}}\left(\frac{\text{0}\text{.110}}{\text{0}\text{.060}}\right)\\ \text{0}{\text{.060moldm}}^{\text{-3}}\text{-x = 0}\text{.4886(0}{\text{.110moldm}}^{\text{-3}}\text{-x)}\\ \text{0}{\text{.060moldm}}^{\text{-3}}\text{-x = 0}{\text{.0538moldm}}^{\text{-3}}\text{- 0}\text{.4886x}\\ \text{x = 0}{\text{.0121moldm}}^{\text{-3}}\\ \text{So concentration of ester left after 20 seconds:}\\ \text{0}{\text{.110moldm}}^{\text{-3}}\text{-0}{\text{.0121moldm}}^{\text{-3}}\text{= 0}{\text{.098moldm}}^{\text{-3}}\end{array}$

(ii)

Interpretation Introduction

Interpretation:

The second order rate constant for the reaction ${\text{CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2(aq)}}^{\text{-}}{\text{+ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{OH}}_{\text{(aq)}}$ is $\text{0}{\text{.11dm}}^{\text{3}}{\text{mol}}^{\text{-1}}$.  The concentration of ester after $\text{15 minutes}$ has to be found.

Explanation of Solution

  $\text{A + B }\stackrel{}{\to }\text{ Products}$

Integrated rate law for this reaction is:

    $\begin{array}{l}\text{ln}\left(\frac{{\text{[A]}}_{\text{o}}\text{[B]}}{{\text{[B]}}_{\text{o}}\text{[A]}}\right)\text{ = }\left({\text{[B]}}_{\text{o}}{\text{-[A]}}_{\text{o}}\right)\text{kt}\\ \text{k=rate constant}\\ {\text{[A]}}_{\text{o}}{\text{,[B]}}_{\text{o}}\text{-initial concentration of the reactants}\\ \text{[A],[B]- concentrations of A,B}\end{array}$

Given equation is:

    $\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2(aq)}}^{\text{-}}{\text{+ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{OH}}_{\text{(aq)}}\\ \text{ln}\left(\frac{{{\text{[CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\text{]}}_{\text{o}}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[OH]}}_{\text{o}}{\text{[CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\text{]}}\right)\text{=}\left({{\text{[OH}}^{\text{-}}\text{]}}_{\text{o}}{\text{-[CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}{\text{]}}_{\text{o}}\right)\text{kt}\\ {\text{Replace[OH}}^{\text{-}}{\text{] and [CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}{\text{] with [OH}}^{\text{-}}{\text{]}}_{\text{o}}{\text{-x and [CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}{\text{]}}_{\text{o}}\text{-x}\\ \text{ln}\left(\frac{{{\text{[CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\text{]}}_{\text{o}}{\text{[OH}}^{\text{-}}\text{]-x}}{{\text{[OH]}}_{\text{o}}{\text{[CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\text{]-x}}\right)\text{=}\left({{\text{[OH}}^{\text{-}}\text{]}}_{\text{o}}{\text{-[CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}{\text{]}}_{\text{o}}\right)\text{kt}\end{array}$

Then the concentration of ester after $\text{15 minutes}$ when ethyl ethanoate is added to aqueous sodium hydroxide will be:

$\text{15min}\left(\frac{\text{60s}}{\text{1min}}\right)\text{ = 900s}$

$\begin{array}{l}\text{ln}\left(\frac{\text{(0}{\text{.110moldm}}^{\text{-3}}\text{)(0}{\text{.060moldm}}^{\text{-3}}\text{-x)}}{\text{(0}{\text{.060moldm}}^{\text{-3}}\text{)(0}{\text{.110moldm}}^{\text{-3}}\text{-x)}}\right)\text{ = (0}{\text{.060moldm}}^{\text{-3}}\text{-0}{\text{.110moldm}}^{\text{-3}}\text{) (0}{\text{.110moldm}}^{\text{-3}}\text{) 900 s }\\ \text{ln}\left(\left(\frac{\text{0}\text{.110}}{\text{0}\text{.060}}\right)\frac{\text{(0}{\text{.060moldm}}^{\text{-3}}\text{-x)}}{\text{(0}{\text{.110moldm}}^{\text{-3}}\text{-x)}}\right)\text{ = -4}\text{.95}\\ \frac{\text{(0}{\text{.060moldm}}^{\text{-3}}\text{-x)}}{\text{(0}{\text{.110moldm}}^{\text{-3}}\text{-x)}}{\text{ = e}}^{\text{-0}\text{.495}}\left(\frac{\text{0}\text{.060}}{\text{0}\text{.110}}\right)\\ \text{0}{\text{.060moldm}}^{\text{-3}}\text{-x = 7}\text{.083}\times {\text{10}}^{-3}\text{(0}{\text{.110moldm}}^{\text{-3}}\text{-x)}\\ \text{0}{\text{.060moldm}}^{\text{-3}}\text{-x = 0}{\text{.0538moldm}}^{\text{-3}}\text{- 0}\text{.4886x}\\ \text{x = 0}{\text{.596moldm}}^{\text{-3}}\\ \text{So concentration of ester left after 20 seconds:}\\ \text{0}{\text{.110moldm}}^{\text{-3}}\text{-0}{\text{.596moldm}}^{\text{-3}}\text{= 0}{\text{.050moldm}}^{\text{-3}}\end{array}$