Chapter 6, Problem 6C.10E

(a)

Interpretation Introduction

Interpretation:

The ester concentration left after 15 seconds of reaction between ethyl acetate and sodium hydroxide with given initial concentration has to be calculated.

Concept Introduction:

Second order reaction: If here are two reactant molecules are involving in a reaction, then the order of the reaction is two. So, such a reaction is known as a second order reaction.

General representation of a second order reaction is as follows:

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

Rate law for second order reaction is given below:

$\begin{array}{l}\text{Rate}=\text{K}\left[\text{A}\right]\left[\text{B}\right]\\ \text{K}=\text{is}\text{the}\text{proportionality}\text{constant}.\end{array}$

Integrated rate law for the second order reaction is given below:

$\left[\text{P}\right]\text{=}\frac{{\left[\text{A}\right]}_{\text{o}}{\left[\text{B}\right]}_{\text{o}}\left({\text{1- e}}^{\left({\left[\text{B}\right]}_{\text{o}}\text{-}{\left[\text{A}\right]}_{\text{o}}\right)}{}^{k_{r}t}\right)}{{\left[\text{A}\right]}_{\text{o}}-{\left[\text{B}\right]}_{\text{o}}{\text{e}}^{\left({\left[\text{B}\right]}_{\text{o}}\text{-}{\left[\text{A}\right]}_{\text{o}}\right)}{}^{k_{r}t}}$

$\begin{array}{l}{\left[\text{A}\right]}_{\text{t}}=\text{concentration}\text{of}\text{the}\text{reactant}\text{after}\text{certain}\text{time}\text{'}\text{t'}.\\ {\left[\text{A}\right]}_{\text{o}}=\text{initial}\text{concentration}\text{of}\text{the}\text{reactant}\text{.}\\ \text{'}\text{t'}=\text{time}\text{taken}\text{by}\text{the}\text{reactant}\text{to}\text{attain}\text{its}\text{final}\text{concentration}\text{from}\text{its}\text{initial}\text{concentration}.\end{array}$

Explanation of Solution

The ester concentration left after 15 seconds for given reaction is given as follows,

$\begin{array}{c}\left[\text{P}\right]\text{=}\frac{{\left[\text{A}\right]}_{\text{o}}{\left[\text{B}\right]}_{\text{o}}\left({\text{1- e}}^{\left({\left[\text{B}\right]}_{\text{o}}\text{-}{\left[\text{A}\right]}_{\text{o}}\right)}{}^{{\text{k}}_{\text{r}}\text{t}}\right)}{{\left[\text{A}\right]}_{\text{o}}\text{-}{\left[\text{B}\right]}_{\text{o}}{\text{e}}^{\left({\left[\text{B}\right]}_{\text{o}}\text{-}{\left[\text{A}\right]}_{\text{o}}\right)}{}^{{\text{k}}_{\text{r}}\text{t}}}\\ \\ \left[\text{P}\right]\text{=}\frac{\left(\text{0}\text{.150}\times \text{0}\text{.055}\right)\left({\text{1- e}}^{\left(\text{0}\text{.055 - 0}\text{.150}\right)}{}^{0.11\times 15}\right)}{\text{0}\text{.150}\text{-}\text{0}\text{.055}{\text{e}}^{\left(\text{0}\text{.055-0}\text{.150}\right)}{}^{0.11\times 15}}\\ \\ =1.16\times {10}^{-2}\text{mol}{\text{dm}}^{\text{-3}}\\ \\ \left[{\text{CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\right]=0.150\text{mol}{\text{dm}}^{\text{-3}}-1.16\times {10}^{-2}\text{mol}{\text{dm}}^{\text{-3}}\\ \\ =0.1384\text{mol}{\text{dm}}^{\text{-3}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The ester concentration left after 15 minutes of reaction between ethyl acetate and sodium hydroxide with given initial concentration has to be calculated.

Concept Introduction:

Second order reaction: If here are two reactant molecules are involving in a reaction, then the order of the reaction is two. So, such a reaction is known as a second order reaction.

General representation of a second order reaction is as follows:

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

Rate law for second order reaction is given below:

$\begin{array}{l}\text{Rate}=\text{K}\left[\text{A}\right]\left[\text{B}\right]\\ \text{K}=\text{is}\text{the}\text{proportionality}\text{constant}.\end{array}$

Integrated rate law for the second order reaction is given below:

$\left[\text{P}\right]\text{=}\frac{{\left[\text{A}\right]}_{\text{o}}{\left[\text{B}\right]}_{\text{o}}\left({\text{1- e}}^{\left({\left[\text{B}\right]}_{\text{o}}\text{-}{\left[\text{A}\right]}_{\text{o}}\right)}{}^{k_{r}t}\right)}{{\left[\text{A}\right]}_{\text{o}}-{\left[\text{B}\right]}_{\text{o}}{\text{e}}^{\left({\left[\text{B}\right]}_{\text{o}}\text{-}{\left[\text{A}\right]}_{\text{o}}\right)}{}^{k_{r}t}}$

$\begin{array}{l}{\left[\text{A}\right]}_{\text{t}}=\text{concentration}\text{of}\text{the}\text{reactant}\text{after}\text{certain}\text{time}\text{'}\text{t'}.\\ {\left[\text{A}\right]}_{\text{o}}=\text{initial}\text{concentration}\text{of}\text{the}\text{reactant}\text{.}\\ \text{'}\text{t'}=\text{time}\text{taken}\text{by}\text{the}\text{reactant}\text{to}\text{attain}\text{its}\text{final}\text{concentration}\text{from}\text{its}\text{initial}\text{concentration}.\end{array}$

Explanation of Solution

The ester concentration left after 15 seconds for given reaction is given as follows,

$\begin{array}{c}\left[\text{P}\right]\text{=}\frac{{\left[\text{A}\right]}_{\text{o}}{\left[\text{B}\right]}_{\text{o}}\left({\text{1- e}}^{\left({\left[\text{B}\right]}_{\text{o}}\text{-}{\left[\text{A}\right]}_{\text{o}}\right)}{}^{{\text{k}}_{\text{r}}\text{t}}\right)}{{\left[\text{A}\right]}_{\text{o}}\text{-}{\left[\text{B}\right]}_{\text{o}}{\text{e}}^{\left({\left[\text{B}\right]}_{\text{o}}\text{-}{\left[\text{A}\right]}_{\text{o}}\right)}{}^{{\text{k}}_{\text{r}}\text{t}}}\\ \\ \left[\text{P}\right]\text{=}\frac{\left(\text{0}\text{.150}\times \text{0}\text{.055}\right)\left({\text{1- e}}^{\left(\text{0}\text{.055 - 0}\text{.150}\right)}{}^{0.11\times 900}\right)}{\text{0}\text{.150}\text{-}\text{0}\text{.055}{\text{e}}^{\left(\text{0}\text{.055-0}\text{.150}\right)}{}^{0.11\times 900}}here,15\min =900s\\ \\ =5.5\times {10}^{-2}\text{mol}{\text{dm}}^{\text{-3}}\\ \\ \left[{\text{CH}}_{\text{3}}{\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\right]=0.150\text{mol}{\text{dm}}^{\text{-3}}-5.5\times {10}^{-2}\text{mol}{\text{dm}}^{\text{-3}}\\ \\ =0.095\text{mol}{\text{dm}}^{\text{-3}}\end{array}$