Chapter 16, Problem 116IL

(a)

Interpretation Introduction

Interpretation:

Each prduct as a lewis acid or a lewis base is to be stated.

Concept Introduction:

A Lewis acid is a substance that contains an empty orbital which is capable of accepting an electron pair. A Lewis base is a substance that has a filled orbital containing an electron pair which is not involved in bonding but may form a dative bond with a lewis acid.

Answer to Problem 116IL

On the product side, $B{F}_3$ can act as a lewis acid and ${\left(C{H}_3\right)}_{2}O$ can act as a lewis base.

Explanation of Solution

A lewis acid can accept a pair of electrons from a lewis base. The boron in ${\text{BF}}_{\text{3}}$ is electron poor and has an empty orbital, so it can accept a pair of electrons, making it as a lewis acid.

Methyl ether $\left({\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\right)$ is a lewis base, it can donate it’s lone pair of electrons. Chemical reaction between $\left({\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\right)$ and ${\text{BF}}_{\text{3}}$, the lone pair from methyl ether will form a dative bond with the empty orbital of ${\text{BF}}_{\text{3}}$ to form an adduct.

    ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}+{\text{BF}}_3\rightleftharpoons \left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O:}\to {\text{BF}}_{\text{3}}\right]$

On the product side, $B{F}_3$ can act as a lewis acid and ${\left(C{H}_3\right)}_{2}O$ can act as a lewis base.

(b)

Interpretation Introduction

Interpretation:

Total pressure at equilibrium, partial pressure of lewis acid and lewis base has to be calculated.

Concept Introduction:

A Lewis acid is a substance that contains an empty orbital which is capable of accepting an electron pair. A Lewis base is a substance that has a filled orbital containing an electron pair which is not involved in bonding but may form a dative bond with a lewis acid.

Ideal gas equation,

$P_{\text{T}}V=nRT$ $\left(1\right)$

Here,

$P_{\text{T}}$ is total pressure of the gas.

$V$ is volume of the gas container.

$n$ is total number of moles of gas particles.

$R$ is gas constant.

$T$ is temperature.

Answer to Problem 116IL

The total pressure in the flask at Equilibrium is $0.404atm$ and the partial pressure of ${\text{BF}}_{\text{3}}\text{,}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}$ and ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O:}\to {\text{BF}}_{\text{3}}$ are $0.14atm,0.14atm$ and $0.12atm$ respectively.

Explanation of Solution

From ideal gas equation,

$P_{\text{T}}V=nRT$ $\left(1\right)$

Here,

$P_{\text{T}}$ is total pressure of the gas.

$V$ is volume of the gas container.

$n$ is total number of moles of gas particles.

$R$ is gas constant.

$T$ is temperature.

An equilibrium constant $\left(K_{\text{P}}\right)$ can be written in terms of partial pressure $\left(P\right)$ of gases.

    $\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O:}\to {\text{BF}}_{\text{3}}\right]\left(\text{g}\right)\rightleftharpoons {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\left(\text{g}\right)+{\text{BF}}_3\left(\text{g}\right)$ $\left(2\right)$

An equilibrium constant $\left(K_{\text{P}}\right)$ for the above reaction is,

$K_{\text{P}}=\frac{P_{{\text{BF}}_{\text{3}}}\times P_{{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}}}{P_{{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\to {\text{BF}}_{\text{3}}}}$ $\left(3\right)$

Where, partial pressure of gases is,

$\begin{array}{l}{P}_{{\text{BF}}_{\text{3}}}=x_{{\text{BF}}_{\text{3}}}\times P_{\text{T}}\\ P_{{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}}=x_{{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}}\times P_{\text{T}}\\ P_{{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\to {\text{BF}}_3}=x_{{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\to {\text{BF}}_3}\times P_{\text{T}}\end{array}$ $\left(4\right)$

Where $x$ is the mole fraction of the species which equals to the ratio of moles of the substance to the total number of moles,

$x_{{\text{BF}}_{\text{3}}}=\frac{n_{{\text{BF}}_{\text{3}}}}{n_{{\text{BF}}_{\text{3}}}+n_{{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}}+n_{{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\to {\text{BF}}_{\text{3}}}}$

Given:

The mass of boron ether complex is $1\text{g}$.

The volume of the flask is $0.565\text{L}$.

The temperature is $398\text{K}$.

The gas constant $R$ is $0.0821\text{atm}\cdot \text{L}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Molecular weight of complex is $141.98\text{g}\cdot {\text{mol}}^{-1}$.

Total number of moles is equal to the mass of the complex divided by the molecular weight of the complex,

$\begin{array}{c}n=\frac{1\text{g}}{141.98\text{g}\cdot {\text{mol}}^{-1}}\\ =0.007\text{mol}\end{array}$

Total pressure at equilibrium is calculated from equation $\left(1\right)$ as follows,

$\begin{array}{c}{P}_T=\frac{0.007\left(\text{mol}\right)\times 0.0821\left(\text{atm}\cdot \text{L}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\times 398\left(\text{K}\right)}{0.565\left(\text{L}\right)}\\ =0.404\text{atm}\end{array}$

Set up an ICE table for the chemical reaction given in equation $\left(2\right)$

$\begin{array}{cccccc}\text{Equilibrium}& \left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O:}\to {\text{BF}}_{\text{3}}\right]\left(g\right)& \rightleftharpoons & B{F}_3\left(g\right)& +& {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\left(g\right)\\ \text{Initial}\text{moles}& 0.007& & 0& & 0\\ \text{Change}& -y& & +y& & +y\\ \text{At}\text{Equilibrium}& 0.007-y& & y& & y\end{array}$

Total number moles at Equilibrium is equal to $0.007+y$.

Substitute these values in equation $\left(3\right)$ to calculate change in moles at Equilibrium,

$\begin{array}{c}{K}_{\text{P}}=\frac{{\left(\frac{y}{0.007+y}\right)}^2\times P_{\text{T}}}{\left(\frac{0.007-y}{0.007+y}\right)}\\ 0.17=\frac{y^2\times 0.404}{\left(4.9\times {10}^{-5}\right)-y^2}\\ y=3.81\times {10}^{-3}\end{array}$

Now substitute the value of $y$ in equation $\left(4\right)$ to calculate partial pressure of given gases,

$\begin{array}{c}{P}_{{\text{BF}}_{\text{3}}}=\frac{\left(3.81\times {10}^{-3}\right)}{\left(7\times {10}^{-3}\right)+\left(3.81\times {10}^{-3}\right)}\times \left(0.404\right)\\ =0.14\end{array}$

Similarly,

$\begin{array}{c}{P}_{{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}}=0.14\\ P_{{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\to {\text{BF}}_{\text{3}}}=0.12\end{array}$

The total pressure in the flask at Equilibrium is $0.404atm$ and the partial pressure of ${\text{BF}}_{\text{3}}\text{,}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}$ and ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O:}\to {\text{BF}}_{\text{3}}$ are $0.14atm,0.14atm$ and $0.12atm$ respectively.