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Enthalpies of melting and boiling for pure elements versus temperatures of transition, demonstrating Trouton's rule.

The enthalpy of fusion of a substance, also known as (latent) heat of fusion, is the change in its enthalpy resulting from providing energy, typically heat, to a specific quantity of the substance to change its state from a solid to a liquid, at constant pressure. For example, when melting 1 kg of ice (at 0°C under a wide range of pressures), 333.55 kJ of energy is absorbed with no temperature change. The heat of solidification (when a substance changes from liquid to solid) is equal and opposite.

This energy includes the contribution required to make room for any associated change in volume by displacing its environment against ambient pressure. The temperature at which the phase transition occurs is the melting point or the freezing point, according to context. By convention, the pressure is assumed to be 1 atm (101.325 kPa) unless otherwise specified.

Overview

The 'enthalpy' of fusion is a latent heat, because during melting the heat energy needed to change the substance from solid to liquid at atmospheric pressure is latent heat of fusion, as the temperature remains constant during the process. The latent heat of fusion is the enthalpy change of any amount of substance when it melts. When the heat of fusion is referenced to a unit of mass, it is usually called the specific heat of fusion, while the molar heat of fusion refers to the enthalpy change per amount of substance in moles.

The liquid phase has a higher internal energy than the solid phase. This means energy must be supplied to a solid in order to melt it and energy is released from a liquid when it freezes, because the molecules in the liquid experience weaker intermolecular forces and so have a higher potential energy (a kind of bond-dissociation energy for intermolecular forces).

When liquid water is cooled, its temperature falls steadily until it drops just below the line of freezing point at 0 °C. The temperature then remains constant at the freezing point while the water crystallizes. Once the water is completely frozen, its temperature continues to fall.

The enthalpy of fusion is almost always a positive quantity; helium is the only known exception.^[1]Helium-3 has a negative enthalpy of fusion at temperatures below 0.3 K. Helium-4 also has a very slightly negative enthalpy of fusion below 0.77 K (−272.380 °C). This means that, at appropriate constant pressures, these substances freeze with the addition of heat.^[2] In the case of ⁴He, this pressure range is between 24.992 and 25.00 atm (2,533 kPa).^[3]

Standard enthalpy change of fusion of period three.

Standard enthalpy change of fusion of period two of the periodic table of elements.

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

Examples

A) To heat 1 kg (1.00 liter) of water from 283.15 K to 303.15 K (10 °C to 30 °C) requires 83.6 kJ. However, to melt ice also requires energy. We can treat these two processes independently; thus, to heat 1 kg of ice from 273.15 K to water at 293.15 K (0 °C to 20 °C) requires:

(1) 333.55 J/g (heat of fusion of ice) = 333.55 kJ/kg = 333.55 kJ for 1 kg of ice to melt

PLUS

(2) 4.18 J/(g·K) × 20K = 4.18 kJ/(kg·K) × 20K = 83.6 kJ for 1 kg of water to increase in temperature by 20 K

= 417.15 kJ

From these figures it can be seen that one part ice at 0 °C will cool almost exactly 4 parts water from 20 °C to 0 °C.

B) Silicon has a heat of fusion of 50.21 kJ/mol. 50 kW of power can supply the energy required to melt about 100 kg of silicon in one hour, after it is brought to the melting point temperature:

50 kW = 50kJ/s = 180000kJ/h

180000kJ/h * (1 mol Si)/50.21kJ * 28gSi/(mol Si) * 1kgSi/1000gSi = 100.4kg/h

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)displaystyle (x_2) of solute at saturation is a function of the heat of fusion, the melting point of the solid (Tfus)displaystyle (T_mathit fus) and the temperature (T) of the solution:

ln⁡x2=−ΔHfus∘R(1T−1Tfus)displaystyle ln x_2=-frac Delta H_mathit fus^circ Rleft(frac 1T-frac 1T_mathit fusright)

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

x2=exp⁡(−28100 J mol−18.314 J K−1 mol−1(1298−1442))=0.0248displaystyle x_2=exp left(-frac 28100mbox J mol^-18.314mbox J K^-1mbox mol^-1left(frac 1298-frac 1442right)right)=0.0248

This equals to a solubility in grams per liter of:

0.0248∗1000 g18.053 mol−11−0.0248∗151.17 mol−1=213.4displaystyle frac 0.0248*frac 1000mbox g18.053mbox mol^-11-0.0248*151.17mbox 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.^[5]

Proof

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

μsolid∘=μsolution∘displaystyle mu _solid^circ =mu _solution^circ ,

or

μsolid∘=μliquid∘+RTln⁡X2displaystyle mu _solid^circ =mu _liquid^circ +RTln X_2,

with Rdisplaystyle R, the gas constant and Tdisplaystyle T, the temperature.

Rearranging gives:

RTln⁡X2=−(μliquid∘−μsolid∘)displaystyle RTln X_2=-(mu _liquid^circ -mu _solid^circ ),

and since

ΔGfus∘=μliquid∘−μsolid∘displaystyle Delta G_mathit fus^circ =mu _liquid^circ -mu _solid^circ ,

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

RTln⁡X2=−(ΔGfus∘)displaystyle RTln X_2=-(Delta G_mathit fus^circ ),

Application of the Gibbs–Helmholtz equation:

(∂(ΔGfus∘T)∂T)p=−ΔHfus∘T2displaystyle left(frac partial (frac Delta G_mathit fus^circ T)partial Tright)_p,=-frac Delta H_mathit fus^circ T^2

ultimately gives:

(∂(ln⁡X2)∂T)=ΔHfus∘RT2displaystyle left(frac partial (ln X_2)partial Tright)=frac Delta H_mathit fus^circ RT^2

or:

∂ln⁡X2=ΔHfus∘RT2∗δTdisplaystyle partial ln X_2=frac Delta H_mathit fus^circ RT^2*delta T

and with integration:

∫X2=1X2=x2δln⁡X2=ln⁡x2=∫TfusTΔHfus∘RT2∗ΔTdisplaystyle int _X_2=1^X_2=x_2delta ln X_2=ln x_2=int _T_mathit fus^Tfrac Delta H_mathit fus^circ RT^2*Delta T

the end result is obtained:

ln⁡x2=−ΔHfus∘R(1T−1Tfus)displaystyle ln x_2=-frac Delta H_mathit fus^circ Rleft(frac 1T-frac 1T_mathit fusright)

See also

• Heat of vaporization


• Heat capacity


• Thermodynamic databases for pure substances


• 
  Joback method (Estimation of the heat of fusion from molecular structure)


• Latent heat

Notes

References

• 
  Atkins, Peter; Jones, Loretta (2008), Chemical Principles: The Quest for Insight (4th ed.), W. H. Freeman and Company, p. 236, ISBN 0-7167-7355-4
  


• 
  Ott, BJ. Bevan; Boerio-Goates, Juliana (2000), Chemical Thermodynamics: Advanced Applications, Academic Press, ISBN 0-12-530985-6