Question

Using the heats of fusion and vaporization for water, calculate the change in enthalpy for the sublimation of water:

${\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Using the ΔH value given in Exercise 24 and the number of hydrogen bonds formed to each water molecule, estimate what portion of the intermolecular forces in ice can be accounted for by hydrogen bonding.

Short answer

The change in enthalpy for sublimation of water will be 46.7 kJ/mol

90.% of the portion in ice can be accounted for hydrogen- bonding

Step-by-step solution

Introduction

Conversion of solid to liquid is considered as fusion whereas conversion of liquid to gas is known as vaporization. Very few fractions of hydrogen bonds are broken while converting from solid to liquid phase. As a result, most of the hydrogen bonds remains in the liquid phase which must be broken during the transition of liquid to gas phase. Direct conversion of solid to gas is known as sublimation

Determine the change in enthalpy

$H_{2}O\left(s\right)\stackrel{6.0kJ}{\to }{H}_{2}O\left(l\right)\stackrel{40.7kJ}{\to }{H}_{2}O\left(g\right)$

From the above data we can write

$$\begin{gathered} ∆H_{sub}=∆H_{fus}+∆H_{vap} \\ ∆H_{sub}=\left(6+40.7\right)kJ \\ ∆H_{sub}=46.7kJ \end{gathered}$$

Determine portion of intermolecular force in ice

It requires 46.7 kJ/mol of energy to convert $H_{2}O\left(s\right)\to H_{2}O\left(g\right)$

A single hydrogen bond has the strength of 21 kJ/mol. Therefore, for 2 hydrogen bonds take 42 kJ/mol of energy to break all the hydrogen bonds in water.

The amount of energy to disrupt all of the intermolecular forces in ice is calculated below (42 ÷46.7) × 100 = 90% of the portion in ice can be accounted for hydrogen-bonding