Chapter 14, Problem 14A.8P

Interpretation Introduction

Interpretation:

The molecular informations such as polarizability volume and the dipole moment of methanol have to be calculated.

Concept introduction:

The dipole forces are attractive forces between two permanent dipoles.  The interaction is between the positive end of one polar molecule and the negative end of another polar molecule.  The induced dipole interaction occurs when an ion or a dipole induces a temporary dipole in an atom or a molecule with no dipole.  The induced dipole forces are the attraction forces between molecules having temporary dipoles.

Answer to Problem 14A.8P

The dipole moment of methanol is $\underset{\_}{0.256D}$.

The polarizability volume of methanol is $\underset{\_}{1.43\times 10^{-23}\text{}c{m}^3}$.

Explanation of Solution

The molar mass of methanol is $32.04\text{g/mol}$.

The value of ${\epsilon }_o$ is $8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}$.

The value of Boltzmann constant is $1.38\times {10}^{-23}\text{J}/\text{K}$.

The value of Avogadro’s number is $6.022\times {10}^{23}{\text{mol}}^{-1}$.

The constant density of methanol is $0.791\text{g}{\text{cm}}^{-3}$.

The relative permittivities of methanol at different temperatures are shown below.

$\theta /°\text{C}$	$-185$	$-170$	$-150$	$-140$	$-110$	$-80$	$-50$	$-20$	$0$	$20$
$\begin{array}{l}T/\text{K}\\ \left(273.15+°\text{C}\right)\end{array}$	$88.15$	$103.15$	$123.15$	$133.15$	$163.15$	$193.15$	$223.15$	$253.15$	$273.15$	$293.15$
${\epsilon }_r$	$3.2$	$3.6$	$4.0$	$5.1$	$67$	$57$	$49$	$43$	$38$	$34$

According to the Debye equation, for methanol is given below.

    $\begin{array}{c}\frac{{\epsilon }_r-1}{{\epsilon }_r+2}=\left(\frac{\rho P_{\text{m}}}{M}\right)\\ P_{\text{m}}=\left(\frac{{\epsilon }_r-1}{{\epsilon }_r+2}\right)\times \frac{M}{\rho }\end{array}$        (1)

Where,

• ${\epsilon }_r$ is the relative permittivity.
• $\left(P_m\right)$ is polarization of methanol.
• $M$ is the molar mass of methanol.
• $\rho$ is the density of methanol.

Substitute the values of ${\epsilon }_r$ as $3.2$ and $\rho$ as $0.791\text{g}{\text{cm}}^{-3}$ in equation (1) to calculate the polarization $\left(P_m\right)$ at $88.15\text{K}$.

    $\begin{array}{c}{P}_{\text{m(298}\text{K)}}=\left(\frac{3.2-1}{3.2+2}\right)\times \frac{32.04\text{g/mol}}{0.791\text{g}{\text{cm}}^{-3}}\\ =\left(\frac{2.2}{5.2}\right)\times 40.51\\ =17.14{\text{cm}}^{\text{3}}\text{/mol}\end{array}$

The calculated values of polarization $\left(P_m\right)$ for methanol at different temperatures are shown below.

$\theta /°\text{C}$	$-185$	$-170$	$-150$	$-140$	$-110$	$-80$	$-50$	$-20$	$0$	$20$
$\begin{array}{l}T/\text{K}\\ \left(273.15+°\text{C}\right)\end{array}$	$88.15$	$103.15$	$123.15$	$133.15$	$163.15$	$193.15$	$223.15$	$253.15$	$273.15$	$293.15$
$\begin{array}{l}{P}_{\text{m}}=\left(\frac{{\epsilon }_r-1}{{\epsilon }_r+2}\right)\times \frac{M}{\rho }\\ \left({\text{cm}}^{\text{3}}\text{/mol}\right)\end{array}$	$17.14$	$18.81$	$20.25$	$23.39$	$38.74$	$38.45$	$38.12$	$37.8$	$37.47$	$37.13$

The polarization $\left(P_m\right)$, polarizability $\left(\alpha \right)$ and dipole moment $\left(\mu \right)$ are related to each other by the expression shown below.

    $P_m=\left(\frac{N_A}{3{\epsilon }_{\text{ο}}}\right)\times \left(\alpha +\frac{{\mu }^2}{3kT}\right)$        (1)

The data for $1/T$ and $P_m$ is given in the table below.

$\theta /°\text{C}$	$-185$	$-170$	$-150$	$-140$	$-110$	$-80$	$-50$	$-20$	$0$	$20$
$\begin{array}{l}T/\text{K}\\ \left(273.15+°\text{C}\right)\end{array}$	$88.15$	$103.15$	$123.15$	$133.15$	$163.15$	$193.15$	$223.15$	$253.15$	$273.15$	$293.15$
$\begin{array}{l}1/T\\ \left({\text{K}}^{-1}\times {10}^{-3}\right)\end{array}$	$11.34$	$9.69$	$8.12$	$7.51$	$6.13$	$5.18$	$4.48$	$3.95$	$3.66$	$3.41$
$\begin{array}{l}{P}_{\text{m}}=\left(\frac{{\epsilon }_r-1}{{\epsilon }_r+2}\right)\times \frac{M}{\rho }\\ \left({\text{cm}}^{\text{3}}\text{/mol}\right)\end{array}$	$17.14$	$18.81$	$20.25$	$23.39$	$38.74$	$38.45$	$38.12$	$37.8$	$37.47$	$37.13$

A graph is plotted between $1/T$ and $P_m$ as shown below.

Figure 1

The slope of the above graph is calculated as follows.

    $\begin{array}{c}\frac{y_2-y_1}{x_2-x_1}=\frac{36.4-36.0}{1.0-0}\times {10}^3{\text{cm}}^3\text{K}{\text{mol}}^{-1}\\ =\frac{0.4\times {10}^3{\text{cm}}^3\times {10}^{-6}{\text{m}}^3}{1{\text{cm}}^{\text{3}}}\text{K}{\text{mol}}^{-1}\\ =0.4\times {10}^{-3}{\text{m}}^3\text{K}{\text{mol}}^{-1}\end{array}$

After the extrapolation, the intercept of the graph is $36.0\text{}{\text{cm}}^{\text{3}}{\text{mol}}^{-1}$.

The value of polarizability is calculated by the formula shown below.

    $\begin{array}{c}\frac{N_A\alpha }{3{\epsilon }_{\text{ο}}}=36.0\text{}{\text{cm}}^{\text{3}}{\text{mol}}^{-1}\\ \alpha =\frac{36.0\text{}{\text{cm}}^{\text{3}}{\text{mol}}^{-1}\times 3{\epsilon }_{\text{ο}}}{N_A}\end{array}$        (2)

The dipole moment is calculated by the formula shown below.

    $\begin{array}{c}\frac{{\mu }^2}{3k}\left(\frac{3{\epsilon }_{\text{ο}}}{N_A}\right)=0.4\times {10}^{-3}{\text{m}}^3\text{K}{\text{mol}}^{-1}\\ {\mu }^2=\frac{9k{\epsilon }_{\text{ο}}}{N_A}\left(0.4\times {10}^{-3}{\text{m}}^3\text{K}{\text{mol}}^{-1}\right)\end{array}$        (3)

Substitute the values of ${\epsilon }_o$, $k$, $N_A$, molar polarizations and temperatures in equation (3) to calculate the dipole moment.

    $\begin{array}{c}{\mu }^2=\frac{9\times 8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}\times 1.38\times {10}^{-23}\text{J}/\text{K}}{6.022\times {10}^{23}{\text{mol}}^{-1}}\left(0.4\times {10}^{-3}{\text{m}}^3\text{K}{\text{mol}}^{-1}\right)\\ =\frac{1.099\times {10}^{-33}\times 0.4\times {10}^{-3}}{6.022\times {10}^{23}}{\text{C}}^2{\text{m}}^2\\ =7.304\times {10}^{-61}{\text{C}}^2{\text{m}}^2\end{array}$

Simplify the above equation.

    $\begin{array}{c}{\mu }^2=7.304\times {10}^{-61}{\text{C}}^2{\text{m}}^2\\ \mu =\sqrt{7.304\times {10}^{-61}{\text{C}}^2{\text{m}}^2}\\ =0.855\times {10}^{-30}\text{C}\text{m}\end{array}$

The conversion of C m to Debye is done as shown below.

    $1\text{C}\text{m}=2.997\times {10}^{29}\text{D}$

Therefore, the conversion of $\text{5}{\text{.531×10}}^{-30}\text{C}\text{m}$ to Debye is done as shown below.

    $\begin{array}{c}0.855\times {10}^{-30}\text{C}\text{m}=0.855\times {10}^{-30}\text{C}\text{m}\times 2.997\times {10}^{29}\text{D}\\ =\underset{\_}{0.256D}\end{array}$

Therefore, the dipole moment of methanol is $\underset{\_}{0.256D}$.

The formula to calculate the polarizability volume $\left(\alpha \text{'}\right)$ is given below.

    $\alpha \text{'}=\frac{\alpha }{4\pi {\epsilon }_0}$

Substitute the value of $\alpha$ as $\frac{36.0\text{}{\text{cm}}^{\text{3}}{\text{mol}}^{-1}\times 3{\epsilon }_{\text{ο}}}{N_A}$ in the above expression.

    $\alpha \text{'}=\frac{36.0\text{}{\text{cm}}^{\text{3}}{\text{mol}}^{-1}\times 3{\epsilon }_{\text{ο}}}{4\pi {\epsilon }_0{N}_A}$

Substitute the values of ${\epsilon }_o$, $k$, $N_A$, molar polarizations and temperatures in the above equation to calculate the polarizability volume.

    $\begin{array}{c}\alpha \text{'}=\frac{36.0\text{}{\text{cm}}^{\text{3}}{\text{mol}}^{-1}\times 3}{4\times 3.14\times 6.022\times {10}^{23}{\text{mol}}^{-1}}\\ =\frac{108}{75.6363}\times {10}^{-23}\text{}{\text{cm}}^{\text{3}}\\ =\underset{\_}{1.43\times 10^{-23}\text{}c{m}^3}\end{array}$

Therefore, the polarizability volume of methanol is $\underset{\_}{1.43\times 10^{-23}\text{}c{m}^3}$.