Chapter 17, Problem 122P

(a)

To determine

The energy that must be supplied to break a single hydrogen bond.

Answer to Problem 122P

The energy that must be supplied to break a single hydrogen bond is $3\times {10}^{-20}\text{ J}$ .

Explanation of Solution

The energy needed to break the bond will be the negative of the electrostatic energy of the configuration.

Write the equation for the electrostatic energy due to two point charges.

  $U_{q_1{q}_2}=\frac{k{q}_1{q}_2}{r}$        (I)

Here, $U_{q_1{q}_2}$ is the electrostatic energy due to the point charges $q_1$, $q_2$ , $k$  is the coulomb constant and $r$ is the distance between the charges.

Write the equation for the electrostatic energy due to two OH charges using equation (I).

  $U_{\text{OHOH}}=\frac{k{q}_{\text{OH}}{q}_{\text{OH}}}{r_{\text{OHOH}}}$        (II)

Here, $U_{\text{OHOH}}$ is the electrostatic energy due to the two OH charges, $q_{\text{OH}}$ is the charge of the OH and $r_{\text{OHOH}}$ is the distance between the two OH charges.

Write the equation for the electrostatic energy due to distant OH charge and the H charge using equation (I).

  $U_{\text{OHH}}=\frac{k{q}_{\text{OH}}{q}_{\text{H}}}{r_{\text{OHH}}}$        (III)

Here, $U_{\text{OHH}}$ is the electrostatic energy due to the distant OH charge and the H charge, $q_{\text{H}}$ is the charge of the H and $r_{\text{OHH}}$ is the distance between the distant OH charge and the H charge.

Write the equation for the electrostatic energy due to nearer OH charge and the H charge using equation (I).

  $U_{\text{HOH}}=\frac{k{q}_{\text{H}}{q}_{\text{OH}}}{r_{\text{HOH}}}$        (IV)

Here, $U_{\text{HOH}}$ is the electrostatic energy due to the nearer OH charge and the H charge and $r_{\text{HOH}}$ is the distance between the nearer OH charge and the H charge.

Write the equation for the electrostatic energy due to two H charges using equation (I).

  $U_{\text{HH}}=\frac{k{q}_{\text{H}}{q}_{\text{H}}}{r_{\text{HH}}}$        (V)

Here, $U_{\text{HH}}$ is the electrostatic energy due to two H charges and $r_{\text{HH}}$ is the distance between the two H charges.

Write the equation for the total energy of the configuration.

  $U_E=U_{\text{OHOH}}+U_{\text{OHH}}+U_{\text{HOH}}+U_{\text{HH}}$

Here, $U_E$ is the energy of the configuration.

Put equations (II), (III), (IV) and (V) in the above equation.

  $U_E=\frac{k{q}_{\text{OH}}{q}_{\text{OH}}}{r_{\text{OHOH}}}+\frac{k{q}_{\text{OH}}{q}_{\text{H}}}{r_{\text{OHH}}}+\frac{k{q}_{\text{H}}{q}_{\text{OH}}}{r_{\text{HOH}}}+\frac{k{q}_{\text{H}}{q}_{\text{H}}}{r_{\text{HH}}}$

Substitute $\left(-0.35e\right)$ for $q_{\text{OH}}$ and $\left(0.35e\right)$ for $q_{\text{H}}$ in the above equation.

  $\begin{array}{c}{U}_E=\frac{k\left(-0.35e\right)\left(-0.35e\right)}{r_{\text{OHOH}}}+\frac{k\left(-0.35e\right)\left(0.35e\right)}{r_{\text{OHH}}}+\frac{k\left(0.35e\right)\left(-0.35e\right)}{r_{\text{HOH}}}+\frac{k\left(0.35e\right)\left(0.35e\right)}{r_{\text{HH}}}\\ =k{\left(0.35e\right)}^2\left[\frac{1}{r_{\text{OHOH}}}-\frac{1}{r_{\text{OHH}}}-\frac{1}{r_{\text{HOH}}}+\frac{1}{r_{\text{HH}}}\right]\end{array}$        (VI)

Here, $e$ is the magnitude of the electronic charge.

Conclusion:

The value of $k$ is $8.988\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2$ and the value of $e$ is $1.602\times {10}^{-19}\text{ C}$ .

Substitute $8.988\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2$ for $k$ , $1.602\times {10}^{-19}\text{ C}$ for $e$ , $0.27\text{ nm}$ for $r_{\text{OHOH}}$ , $0.37\text{ nm}$ for $r_{\text{OHH}}$ , $0.17\text{ nm}$ for $r_{\text{HOH}}$ and $0.27\text{ nm}$ for $r_{\text{HH}}$ in equation (VI) to find $U_E$ .

  $\begin{array}{c}{U}_E=\left(8.988\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2\right){\left(0.35\left(1.602\times {10}^{-19}\text{ C}\right)\right)}^2\left[\frac{1}{0.27\text{ nm}\frac{1\text{ m}}{{10}^9\text{ nm}}}-\frac{1}{0.37\text{ nm}\frac{1\text{ m}}{{10}^9\text{ nm}}}-\frac{1}{0.17\text{ nm}\frac{1\text{ m}}{{10}^9\text{ nm}}}+\frac{1}{0.27\text{ nm}\frac{1\text{ m}}{{10}^9\text{ nm}}}\right]\\ =-3.3\times {10}^{-20}\text{ J}\\ \approx -3\times {10}^{-20}\text{ J}\end{array}$

The energy needed to break the hydrogen bond will be the negative of the determined electrostatic energy.

Therefore, the energy that must be supplied to break a single hydrogen bond is $3\times {10}^{-20}\text{ J}$ .

(b)

To determine

The energy that must be supplied to break the hydrogen bonds in $1\text{ kg}$ of liquid water and to compare it with the heat of vaporization of water.

Answer to Problem 122P

The energy that must be supplied to break the hydrogen bonds in $1\text{ kg}$ of liquid water is $1\text{ MJ/kg}$ and it is equal in order of magnitude of the heat of vaporization of water and it is due the fact that the hydrogen bonds in the liquid phase must be broken to form a gas.

Explanation of Solution

Write the equation for the energy that must be supplied to break the hydrogen bonds in $1\text{ kg}$ of liquid water.

  $E=N{U}_H\frac{N_A}{M}$        (VII)

Here, $E$ is the energy that must be supplied to break the hydrogen bonds in $1\text{ kg}$ of liquid water, $N$ is the number of hydrogen bonds per a water molecule, $U_H$ is the energy needed to break a single hydrogen bond, $N_A$ is the Avogadro number and $M$ is the molar mass of water.

Conclusion:

Each oxygen in a water molecule forms one hydrogen bond with hydrogen in another water molecule so that there will be one hydrogen bond per molecule. The value of $N_A$ is $6\times {10}^{23}\text{ molecules/mol}$ and the molar mass of water is $0.018\text{ kg/mol}$ .

Substitute $1\text{ bond/molecule}$ for $N$ , $3.3\times {10}^{-20}\text{ J}$ for $U_H$ , $6\times {10}^{23}\text{ molecules/mol}$ for $N_A$ and $0.018\text{ kg/mol}$ for $M$ in equation (VII) to find $E$ .

  $\begin{array}{c}E=\left(1\text{ bond/molecule}\right)\left(3.3\times {10}^{-20}\text{ J}\right)\frac{6\times {10}^{23}\text{ molecules/mol}}{0.018\text{ kg/mol}}\\ =1\times {10}^6\text{ J/kg}\frac{1\text{ MJ}}{{10}^6\text{ J}}\\ =1\text{ MJ/kg}\end{array}$

The heat of vaporization of water is $2.256\text{ MJ/kg}$ . From the comparison of the two values, it can be seen that energy that must be supplied to break the hydrogen bonds in $1\text{ kg}$ of liquid water is equal in order of magnitude of heat of vaporization of water. This is not a coincidence since the hydrogen bonds in the liquid phase must be broken to form a gas.

Therefore, the energy that must be supplied to break the hydrogen bonds in $1\text{ kg}$ of liquid water is $1\text{ MJ/kg}$ and it is equal in order of magnitude of the heat of vaporization of water and it is due the fact that the hydrogen bonds in the liquid phase must be broken to form a gas.