Chapter 14, Problem 14A.2ST

Interpretation Introduction

Interpretation:

The polarizability volume and the dipole moment of chlorobenzene 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.2ST

The dipole moment of chlorobenzene is $\underset{\_}{1.16D}$.

The polarizability volume of chlorobenzene is $\underset{\_}{1.349\times 10^{-29}{m}^3}$.

Explanation of Solution

The molar mass of chlorobenzene is $112.56\text{g/mol}$.

The constant density, $\rho$ of chlorobenzene is $1.11{\text{g/cm}}^{\text{3}}$.

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 relative permittivities of chlorobenzene at different temperatures are shown below.

(1) $5.71$ at $\text{20°C}$ or $\text{293}\text{K}$.

(2) $5.62$ at $\text{25°C}$ or $298\text{K}$.

According to the Debye equation, the polarization $\left(P_m\right)$ of chlorobenzene at $\text{293}\text{K}$ and $298\text{K}$ is calculated by the expression 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.
• $M$ is the molar mass of chlorobenzene.
• $\rho$ is the density of chlorobenzene.

Substitute the values of ${\epsilon }_r$ as $5.71$ and $\rho$ in equation (1) to calculate the polarization $\left(P_m\right)$ at $\text{293}\text{K}$.

    $\begin{array}{c}{P}_{\text{m(293}\text{K)}}=\left(\frac{5.71-1}{5.71+2}\right)\times \frac{112.56\text{g/mol}}{1.11{\text{g/cm}}^{\text{3}}}\\ =\left(\frac{4.71}{7.71}\right)\times 101.4054\\ =\frac{61.948{\text{cm}}^{\text{3}}\text{/mol}\times {10}^{-6}{\text{m}}^{\text{3}}}{1{\text{cm}}^{\text{3}}}\\ =61.948\times {10}^{-6}{\text{m}}^{\text{3}}\text{/mol}\end{array}$

Substitute the values of ${\epsilon }_r$ as $5.62$ in equation (1) to calculate the polarization $\left(P_m\right)$ at $\text{298}\text{K}$.

    $\begin{array}{c}{P}_{\text{m(298}\text{K)}}=\left(\frac{5.62-1}{5.62+2}\right)\times \frac{112.56\text{g/mol}}{1.11{\text{g/cm}}^{\text{3}}}\\ =\left(\frac{4.62}{7.62}\right)\times 101.4054\\ =\frac{61.482{\text{cm}}^{\text{3}}\text{/mol}\times {10}^{-6}{\text{m}}^{\text{3}}}{1{\text{cm}}^{\text{3}}}\\ =61.482\times {10}^{-6}{\text{m}}^{\text{3}}\text{/mol}\end{array}$

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

    $\begin{array}{c}\alpha +\frac{{\mu }^2}{3kT}=\frac{3{\epsilon }_{\text{ο}}{P}_m}{N_A}\\ \alpha =\frac{3{\epsilon }_{\text{ο}}{P}_m}{N_A}-\frac{{\mu }^2}{3kT}\end{array}$        (2)

In the given case, the polarizability remains constant and the polarization varies with respect to temperature. Therefore, the dipole moment is calculated by the formula shown below.

    $\begin{array}{c}\frac{{\mu }^2}{3k}\times \left(\frac{1}{T}-\frac{1}{T^{\text{'}}}\right)=\left(\frac{3{\epsilon }_{\text{ο}}}{N_A}\right)\times \left(P_{\text{m(298}\text{K)}}-P_{\text{m(293}\text{K)}}\right)\\ {\mu }^2=\frac{\left(\frac{9{\epsilon }_{\text{ο}}k}{N_A}\right)\times \left(P_{\text{m(298}\text{K)}}-P_{\text{m(293}\text{K)}}\right)}{\left(\frac{1}{T}-\frac{1}{T^{\text{'}}}\right)}\\ {\mu }^2=\frac{\left(9{\epsilon }_{\text{ο}}k\right)\times \left(P_{\text{m(298}\text{K)}}-P_{\text{m(293}\text{K)}}\right)}{N_A\left(\frac{1}{T}-\frac{1}{T^{\text{'}}}\right)}\end{array}$        (3)

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

    $\begin{array}{c}{\mu }^2=\frac{\begin{array}{l}\left(9\times 8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}\times 1.38\times {10}^{-23}\text{J}/\text{K}\right)\\ \times \left(61.948-61.482\right){10}^{-6}{\text{m}}^3{\text{mol}}^{-1}\end{array}}{6.022\times {10}^{23}{\text{mol}}^{-1}\left(\frac{1}{293}-\frac{1}{298}\right)}\\ =\frac{1.099\times {10}^{-33}\times 0.466\times {10}^{-6}}{6.022\times {10}^{23}\times 5.7\times {10}^{-5}}{\text{C}}^2{\text{m}}^2\\ =\frac{0.51213\times {10}^{-39}}{34.3254\times {10}^{18}}{\text{C}}^2{\text{m}}^2\\ =0.01492\times {10}^{-57}{\text{C}}^2{\text{m}}^2\end{array}$

Simplify the above equation.

    $\begin{array}{c}{\mu }^2=0.01492\times {10}^{-57}{\text{C}}^2{\text{m}}^2\\ \mu =\sqrt{0.01492\times {10}^{-57}{\text{C}}^2{\text{m}}^2}\\ =3.{\text{8611×10}}^{-30}\text{C}\text{m}\end{array}$

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.

    $\begin{array}{c}3.{\text{8611×10}}^{-30}\text{C}\text{m}=3.{\text{8611×10}}^{-30}\text{C}\text{m}\times 2.997\times {10}^{29}\text{D}\\ =\underset{\_}{1.16D}\end{array}$

Therefore, the dipole moment of chlorobenzene is $\underset{\_}{1.16D}$.

Substitute the value of ${\mu }^2=0.01492\times {10}^{-57}{\text{C}}^2{\text{m}}^2$ in equation (2) to calculate the polarizability.

    $\begin{array}{c}\alpha =\frac{3\times 8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}\times 61.948\times {10}^{-6}{\text{m}}^{\text{3}}\text{/mol}}{6.022\times {10}^{23}{\text{mol}}^{-1}}\\ -\frac{0.01492\times {10}^{-57}{\text{C}}^2{\text{m}}^2}{3\times 1.38\times {10}^{-23}\text{J}/\text{K}\times 293\text{K}}\\ =\frac{1.645\times {10}^{-15}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^2}{6.022\times {10}^{23}}-\frac{0.01492\times {10}^{-57}{\text{C}}^2{\text{m}}^2}{1.213\times {10}^{-20}}\\ =\left(2.732\times {10}^{-39}-1.23\times {10}^{-39}\right){\text{J}}^{-1}{\text{C}}^2{\text{m}}^2\\ =1.5\times {10}^{-39}{\text{J}}^{-1}{\text{C}}^{\text{2}}{\text{m}}^{\text{2}}\end{array}$

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 $1.5\times {10}^{-39}{\text{J}}^{-1}{\text{C}}^{\text{2}}{\text{m}}^{\text{2}}$ and ${\epsilon }_o$ in the above expression.

    $\begin{array}{c}\alpha \text{'}=\frac{1.5\times {10}^{-39}{\text{J}}^{-1}{\text{C}}^{\text{2}}{\text{m}}^{\text{2}}}{4\times 3.14\times 8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}}\\ =\frac{1.5\times {10}^{-27}{\text{m}}^{\text{3}}}{111.206}\\ =\underset{\_}{1.349\times 10^{-29}{m}^3}\end{array}$

Therefore, the polarizability volume of chlorobenzene is $\underset{\_}{1.349\times 10^{-29}{m}^3}$.