حرارة الإنصهار

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.

حرارة الانصهار أو طاقة الذوبان هي كمية الطاقة الحرارية اللازمة لتحويل مول واحد من مادة ما من الحالة الصلبة إلى الحالة السائلة أو بالعكس، وتدعى درجة الحرارة التي تحدث عندها العملية بنقطة الانصهار.

 القيم المرجعية للمواد الشائعة

Standard enthalpy change of fusion of period three.

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

هذه القيم مأخوذة من CRC Handbook of Chemistry and Physics, 62nd edition. The conversion between cal/g and kJ/kg 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..

 الاستخدامات

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

 توقع قابلية الذوبان

The heat of fusion can also be used to predict solubility for solids in liquids. Provided an ideal solution is obtained the mole fraction ${\displaystyle{\displaystyle(x_{2})}}$ of solute at saturation is a function of the heat of fusion, the melting point of the solid ${\displaystyle{\displaystyle(T_{fus})}}$ and the temperature (T) of the solution:

${\displaystyle{\displaystyle\ln x_{2}=-{\frac{\Delta H_{fus}^{\circ}}{R}}\left({\frac{1}{T}}-{\frac{1}{T_{fus}}}\right)}}$

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

${\displaystyle{\displaystyle\ln x_{2}=-{\frac{28100{\mbox{ J mol}}^{-1}}{8.314{\mbox{ J K}}^{-1}{\mbox{ mol}}^{-1}}}\left({\frac{1}{298}}-{\frac{1}{442}}\right)=0.0248}}$

This equals to a solubility in grams per liter of:

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

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

 البرهان

عند الاتزان، يتساوى الكمون الكيميائي للمذيب النقي وذلك للصلب النقي:

${\displaystyle{\displaystyle\mu_{solid}^{\circ}=\mu_{solution}^{\circ}\,}}$

أو

${\displaystyle{\displaystyle\mu_{solid}^{\circ}=\mu_{liquid}^{\circ}+RT\ln X_{2}\,}}$

with ${\displaystyle{\displaystyle R\,}}$ the gas constant and ${\displaystyle{\displaystyle T\,}}$ the temperature.

إعادة تنظيم الحدود يعطينا:

${\displaystyle{\displaystyle RT\ln X_{2}=-(\mu_{liquid}^{\circ}-\mu_{solid}^{\circ})\,}}$

and since

${\displaystyle{\displaystyle\Delta G_{fus}^{\circ}=-(\mu_{liquid}^{\circ}-\mu_{solid}^{\circ})\,}}$

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

${\displaystyle{\displaystyle RT\ln X_{2}=-(\Delta G_{fus}^{\circ})\,}}$

Application of the Gibbs-Helmholtz equation:

${\displaystyle{\displaystyle\left({\frac{\partial({\frac{\Delta G_{fus}^{\circ}}{T}})}{\partial T}}\right)_{p\,}={\frac{\Delta H_{fus}^{\circ}}{T^{2}}}}}$

ultimately gives:

${\displaystyle{\displaystyle\left({\frac{\partial(\ln X_{2})}{\partial T}}\right)={\frac{\Delta H_{fus}^{\circ}}{RT^{2}}}}}$

or:

${\displaystyle{\displaystyle\partial\ln X_{2}={\frac{\Delta H_{fus}^{\circ}}{RT^{2}}}*\delta T}}$

and with integration:

${\displaystyle{\displaystyle\int_{x_{2}=1}^{x_{2}=x_{2}}\delta\ln X_{2}=\ln x_{2}=\int_{T_{f}us}^{T}{\frac{\Delta H_{fus}^{\circ}}{RT^{2}}}*\Delta T}}$

the end result is obtained:

${\displaystyle{\displaystyle\ln x_{2}=-{\frac{\Delta H_{fus}^{\circ}}{R}}\left({\frac{1}{T}}-{\frac{1}{T_{fus}}}\right)}}$

 طالع أيضاً

• حرارة التبخر
• السعة الحرارية
• السعة الحرارية النوعية
• Thermodynamic databases for pure substances
• Joback method (Estimation of the heat of fusion from molecular structure)