Chapter 14, Problem 14B.8P

Interpretation Introduction

Interpretation:

The dependence of the molar potential energy of interaction on the angle $\theta$ has to be calculated.

Concept introduction:

Hydrogen bonding occurs when a hydrogen atom is covalently bonded to oxygen, nitrogen or fluorine and it develops some partial charge on both the atom.  Hydrogen bonds can also be formed between the lone pair of an electronegative atom and hydrogen atom attached with an electronegative atom.

Answer to Problem 14B.8P

The dependence of the molar potential energy of interaction on the angle $\theta$ is shown below.

    $V_{\text{m}}=1.386\times {10}^{-4}\text{J}{\text{mol}}^{-1}\left(-1.17\times {10}^{-29}+\frac{-9.562\times {10}^{-39}}{{\left(4.916\times {10}^{-20}-3.828\cos \theta \right)}^{1/2}}\right)$

Explanation of Solution

The arrangement of a system containing an $\text{O}-\text{H}$ group and $\text{O}$ group is shown below.

Figure 1

The expression for the molar potential energy ($V_{\text{m}}$) of an electrostatic model of hydrogen bonding is shown below.

    $V_{\text{m}}=\frac{N_{\text{A}}{e}^2}{4\text{π}{\epsilon }_0}\left\{\frac{{\delta }_{\text{O}}{\delta }_{\text{H}}}{r_{\text{O}-\text{H}}}+\frac{{\delta }_{\text{O}}{\delta }_{\text{H}}}{r_{\text{O}\cdot \cdot \cdot \text{H}}}+\frac{{\delta }_{\text{O}}^2}{r_{\text{O}-\text{O}}}\right\}$        (1)

Where,

• $N_{\text{A}}$ is the Avogadro's number. ($6.022\times {10}^{23}{\text{mol}}^{-1}$)
• ${\epsilon }_0$ is the vacuum permittivity. ($8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}$)
• $\delta$ is the charge.
• $r$ is the distance.
• $e$ is the charge on an electron. ($1.6\times {10}^{-19}\text{C}$)

Equation (1) is solved further as shown below.

    $V_{\text{m}}=\frac{N_{\text{A}}{e}^2}{4\text{π}{\epsilon }_0}\left\{{\delta }_{\text{O}}{\delta }_{\text{H}}\left(\frac{1}{r_{\text{O}-\text{H}}}+\frac{1}{r_{\text{O}\cdot \cdot \cdot \text{H}}}\right)+\frac{{\delta }_{\text{O}}^2}{r_{\text{O}-\text{O}}}\right\}$        (2)

The value of $r_{\text{O}\cdot \cdot \cdot \text{H}}$ is calculated by the formula given below.

    $r_{\text{O}\cdot \cdot \cdot \text{H}}={\left(r_{\text{O}-\text{H}}^2+r_{\text{O}-\text{O}}^2-2r_{\text{O}-\text{H}}{r}_{\text{O}-\text{O}}\cos \theta \right)}^{1/2}$        (3)

The value of $r_{\text{O}-\text{H}}$ is $\text{95}\text{.7}\text{pm}$.

The conversion of $\text{pm}$ to $\text{m}$ is shown below.

    $1\text{pm}={10}^{-12}\text{m}$

Therefore, the conversion of $\text{95}\text{.7}\text{pm}$ to $\text{m}$ is shown below.

    $\text{95}\text{.7}\text{pm}=\text{95}\text{.7}\times {10}^{-12}\text{m}$

The value of $r_{\text{O}-\text{O}}$ is $\text{200}\text{pm}$.

The conversion of $\text{pm}$ to $\text{m}$ is shown below.

    $1\text{pm}={10}^{-12}\text{m}$

Therefore, the conversion of $\text{200}\text{pm}$ to $\text{m}$ is shown below.

    $\text{200}\text{pm}=\text{200}\times {10}^{-12}\text{m}$

Substitute the value of $r_{\text{O}-\text{O}}$ and $r_{\text{O}-\text{H}}$ in equation (3).

    $\begin{array}{c}{r}_{\text{O}\cdot \cdot \cdot \text{H}}={\left({\left(\text{95}\text{.7}\times {10}^{-12}\text{m}\right)}^2+{\left(\text{200}\times {10}^{-12}\text{m}\right)}^2-\left(2\times \text{95}\text{.7}\times {10}^{-12}\text{m}\times \text{200}\times {10}^{-12}\text{m}\times \cos \theta \right)\right)}^{1/2}\\ ={\left(9.16\times {10}^{-21}{\text{m}}^2+4\times {10}^{-20}{\text{m}}^2-\left(3.828\times {10}^{-20}\cos \theta {\text{m}}^2\right)\right)}^{1/2}\\ ={\left(4.916\times {10}^{-20}-\left(3.828\times {10}^{-20}\cos \theta \right)\right)}^{1/2}\text{m}\end{array}$

The value of ${\delta }_{\text{O}}$ is $-0.83e$.

The value of ${\delta }_{\text{H}}$ is $+0.45e$.

Substitute the value of $N_{\text{A}}$, ${\epsilon }_0$, ${\delta }_{\text{O}}$, ${\delta }_{\text{H}}$, $r_{\text{O}-\text{O}}$, $r_{\text{O}-\text{H}}$, and $r_{\text{O}\cdot \cdot \cdot \text{H}}$ in equation (2).

    $V_{\text{m}}=\left(\begin{array}{c}\frac{6.022\times {10}^{23}{\text{mol}}^{-1}\times e^2}{4\times 31.4\times 8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}}\\ \times \left\{\begin{array}{l}\left(-0.83e\right)\left(+0.45e\right)\left(\begin{array}{l}\frac{1}{\text{95}\text{.7}\times {10}^{-12}\text{m}}\\ +\frac{1}{{\left(4.916\times {10}^{-20}-3.828\times {10}^{-20}\cos \theta \right)}^{1/2}\text{m}}\end{array}\right)\\ +\frac{{\left(-0.83e\right)}^2}{\text{200}\times {10}^{-12}\text{m}}\end{array}\right\}\end{array}\right)$

Substitute the value of $e$ in the above equation.

    $\begin{array}{c}{V}_{\text{m}}=\left(\begin{array}{c}\frac{6.022\times {10}^{23}{\text{mol}}^{-1}\times {\left(1.6\times {10}^{-19}\text{C}\right)}^2}{4\times 3.14\times 8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}}\\ \times \left\{\begin{array}{l}\left(-0.83\times 1.6\times {10}^{-19}\text{C}\right)\left(+0.45\times 1.6\times {10}^{-19}\text{C}\right)\left(\begin{array}{l}\frac{1}{\text{95}\text{.7}\times {10}^{-12}\text{m}}\\ +\frac{1}{{\left(4.916\times {10}^{-20}-3.828\cos \theta \right)}^{1/2}\text{m}}\end{array}\right)\\ +\frac{{\left(-0.83\times 1.6\times {10}^{-19}\text{C}\right)}^2}{\text{200}\times {10}^{-12}\text{m}}\end{array}\right\}\end{array}\right)\\ =1.386\times {10}^{-4}\text{J}{\text{mol}}^{-1}\text{m}\left(\begin{array}{l}-9.562\times {10}^{-39}{\text{m}}^{-1}\left(\begin{array}{l}1.045\times {10}^{10}\\ +\frac{1}{{\left(4.916\times {10}^{-20}-3.828\cos \theta \right)}^{1/2}}\end{array}\right)\\ +8.81792\times {10}^{-29}{\text{m}}^{-1}\end{array}\right)\\ =1.386\times {10}^{-4}\text{J}{\text{mol}}^{-1}\left(\begin{array}{l}-9.99\times {10}^{-29}+\frac{-9.562\times {10}^{-39}}{{\left(4.916\times {10}^{-20}-3.828\cos \theta \right)}^{1/2}}\\ +8.81792\times {10}^{-29}\end{array}\right)\\ =1.386\times {10}^{-4}\text{J}{\text{mol}}^{-1}\left(-1.17\times {10}^{-29}+\frac{-9.562\times {10}^{-39}}{{\left(4.916\times {10}^{-20}-3.828\cos \theta \right)}^{1/2}}\right)\end{array}$

Therefore, the dependence of the molar potential energy of interaction on the angle is shown below.

    $V_{\text{m}}=1.386\times {10}^{-4}\text{J}{\text{mol}}^{-1}\left(-1.17\times {10}^{-29}+\frac{-9.562\times {10}^{-39}}{{\left(4.916\times {10}^{-20}-3.828\cos \theta \right)}^{1/2}}\right)$