Enthalpy of fusion

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Molar heat content of zinc above 298.15 K and at 1 atm pressure, showing discontinuities at the melting and boiling points. The enthalpy of melting (ΔH°m) of zinc is 7323 J/mol, and the enthalpy of vaporization (ΔH°v) is 115 330 J/mol.
Molar heat content of zinc above 298.15 K and at 1 atm pressure, showing discontinuities at the melting and boiling points. The enthalpy of melting (ΔH°m) of zinc is 7323 J/mol, and the enthalpy of vaporization (ΔH°v) is 115 330 J/mol.

The standard enthalpy of fusion (symbol: ΔHfus), also known as the heat of fusion or specific melting heat, is the amount of thermal energy which must be absorbed or evolved for 1 mole of a substance to change states from a solid to a liquid or vice versa. It is also called the latent heat of fusion or the enthalpy change of fusion, and the temperature at which it occurs is called the melting point.

When you withdraw thermal energy from a liquid or solid, the temperature falls. When you add heat energy the temperature rises. However, at the transition point between solid and liquid (the melting point), extra energy is required (the heat of fusion). To go from liquid to solid, the molecules of a substance must become more ordered. For them to maintain the order of a solid, extra heat must be withdrawn. In the other direction, to create the disorder from the solid crystal to liquid, extra heat must be added.

The heat of fusion can be observed if you measure the temperature of water as it freezes. If you plunge a closed container of room temperature water into a very cold environment (say −20 °C), you will see the temperature fall steadily until it drops just below the freezing point (0 °C). The temperature then rebounds and holds steady while the water crystallizes. Once completely frozen, the temperature will fall steadily again.

The temperature stops falling at (or just below) the freezing point due to the heat of fusion. The energy of the heat of fusion must be withdrawn (the liquid must turn to solid) before the temperature can continue to fall.

The units of heat of fusion are usually expressed as:

  1. kiloJoules per mole (the SI units)
  2. calories per gram (old metric units now little used, except for a different, larger calorie used in nutritional contexts)
  3. British thermal units per pound or Btu per pound-mole
  • Note: These are not the calories found in food. The calories found in food are more properly known as kilocalories—equal to 1000 calories. 1000 calories = 1 kilocalorie = 1 food calorie. Food calories are sometimes abbreviated as kcal as if small calories were being used, while calories are abbreviated as cal. Another distinguishing method, though often confusing, uses capitalization. A Calorie is a food calorie, or 1000 calories. So 1 Cal = 1000 cal or 1 kcal

Contents

Reference values of common substances

Standard enthalpy change of fusion of period three.
Standard enthalpy change of fusion of period three.
Standard enthalpy change of fusion of period two of the periodic table of elements.
Standard enthalpy change of fusion of period two of the periodic table of elements.
Substance Heat of fusion
(cal/g)		 Heat of fusion
(J/g)
water 79.72 333.55
methane 13.96 58.41
ethane 22.73 95.10
propane 19.11 79.96
methanol 23.70 99.16
ethanol 26.05 108.99
glycerol 47.95 200.62
formic acid 66.05 276.35
acetic acid 45.91 192.09
acetone 23.42 97.99
benzene 30.45 127.40
myristic acid 47.49 198.70
palmitic acid 39.18 163.93
stearic acid 47.54 198.91

These values are from the CRC Handbook of Chemistry and Physics, 62nd edition. The conversion between cal/g and kJ/kg in the above table uses the thermochemical calorie (calth) = 4.184 joules rather than the International Steam Table calorie (calINT) = 4.1868 joules..

Applications

To heat one kilogram (about 1 litre) of water from 10 °C to 30 °C requires 20 kcal.
However, to melt ice and raise the resulting water temperature 20 °C requires extra energy. To heat ice from 0 °C to water at 20 °C requires:

(1) 80 cal/g (heat of fusion of ice) = 80 kcal for 1 kg
PLUS
(2) 1 cal/(g·°C) = 20 kcal for 1 kg to go up 20 °C
= 100 kcal

Solubility prediction

The heat of fusion can also be used to predict solubility for solids in liquids. Provided an ideal solution is obtained the mole fraction (x2) of solute at saturation is a function of the heat of fusion, the melting point of the solid (Tfus) and the temperature (T) of the solution:

\ln x_2 = - \frac {\Delta H^\circ_{fus}}{R} \left(\frac{1}{T}- \frac{1}{T_{fus}}\right)

Here, R is the gas constant. For example the solubility of paracetamol in water at 298 K is predicted to be:

\ln x_2 = - \frac {28100 \mbox{ J mol}^{-1}} {8.314 \mbox{ J K}^{-1} \mbox{ mol}^{-1}}\left(\frac{1}{298}- \frac{1}{442}\right) = 0.0248

This equals to a solubility in grams per liter of:

\frac{0.0248*\frac{1000 \mbox{ g}}{18.053 \mbox{ mol}^{-1}}}{1-0.0248}*151.17 \mbox{ mol}^{-1} = 213.4

which is a deviation from the real solubility (240 g/L) of 11%. This error can be reduced when an additional heat capacity parameter is taken into account [1]

Proof

At equilibrium the chemical potentials for the pure solvent and pure solid are identical:

\mu^\circ_{solid} = \mu^\circ_{solution}\,

or

\mu^\circ_{solid} = \mu^\circ_{liquid} + RT\ln X_2\,


with R\, the gas constant and T\, the temperature.

Rearranging gives:

RT\ln X_2 = - (\mu^\circ_{liquid} - \mu^\circ_{solid})\,

and since

\Delta G^\circ_{fus} = - (\mu^\circ_{liquid} - \mu^\circ_{solid})\,

the heat of fusion being the difference in chemical potential between the pure liquid and the pure solid, it follows that

RT\ln X_2 = - ( \Delta G^\circ_{fus})\,

Application of the Gibbs-Helmholtz equation:

\left( \frac{\partial ( \frac{\Delta G^\circ_{fus} } {T} ) } {\partial T} \right)_{p\,} = \frac {\Delta H^\circ_{fus}} {T^2}

ultimately gives:

\left( \frac{\partial ( \ln X_2 ) } {\partial T} \right) = \frac {\Delta H^\circ_{fus}} {RT^2}

or:

\partial \ln X_2 = \frac {\Delta H^\circ_{fus}} {RT^2}*\delta T

and with integration:


\int^{x_2=x_2}_{x_2 = 1} \delta \ln X_2 = \ln x_2 = \int_{T_fus}^T \frac {\Delta H^\circ_{fus}} {RT^2}*\Delta T

the end result is obtained:

\ln x_2 = - \frac {\Delta H^\circ_{fus}} {R}\left(\frac{1}{T}- \frac{1}{T_{fus}}\right)

See also

References

  1. ^ Measurement and Prediction of Solubility of Paracetamol in Water-Isopropanol Solution. Part 2. Prediction H. Hojjati and S. Rohani Org. Process Res. Dev.; 2006; 10(6) pp 1110 - 1118; (Article) doi:10.1021/op060074g

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  • This page was last modified on 6 October 2008, at 13:29.

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