Heat of Fusion versus Vaporization
Country: United Kingdom
Date: November 2008
Why is the value for the enthalpy of fusion of water much
smaller than the value for the enthalpy of vaporisation of water?
Look at the changes that occur in the two processes. When ice melts
sufficient energy has to be absorbed to take a water molecule from a
lattice position in solid water, so that it can be mobile in the liquid
phase. It still has bonds connecting it to other nearby water molecules in
the liquid phase. In addition, the water molecules are close to one
another. We know this because the density of liquid water is much less than
the density of water vapor. In the case of vaporization, a water molecule
must be "ripped" away from the liquid water, so that it is essentially a
single free molecule in the vapor. More bonds have to be broken, and that
requires more heat input. This is true not only for water but for fusion /
vaporization of all molecules. There are some modifications because of
temperature differences, if you wish to be more quantitative about the
difference, but those differences do not affect the basic processes. Also
notice that the enthalpy of sublimation (solid ---> vapor) equals the
enthalpy of fusion (solid ---> liquid) plus the enthalpy of vaporization
(liquid ---> vapor).
This is a very good observation. The enthalpy of fusion of water is 6.01 kJ/mol,
whereas the enthalpy of vaporization of water is 40.7 kJ/mol.
In both cases, the energy is required not to change the temperature of the water,
but rather to change its state. Thus the kinetic energy of the molecules do not
change, only the relative interactions between molecules.
So, in effect in going from ice to liquid water, we are exchanging the crystal
interaction for liquid interaction, and in going from liquid water to water
vapor/steam, we are going from liquid interactions to no interactions.
Thus, you can imagine that going from the strong interactions in the solid in
an organized structure to that of the even stronger interactions but disorganized
structure of the liquid, that there would be little change in the potential
energy (or Gibbs Free energy) of the system. However, in going from the strong
interactions of liquid water in a disorganized state, to hardly any interactions
at all and a very disorganized state of the gas, that a lot more energy is
required to make this much greater transformation.
Greg (Roberto Gregorius)
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Update: June 2012