Melting Point

Written by Clayton Maike

The melting point of a material is an intensive physical property that indicates the temperature at which the substance transforms from a solid to a liquid or vice versa.

Properties of Matter

On the most basic level, there are two types of properties of matter: chemical and physical properties. Chemical properties are classified as those that change when the substance or material undergoes a chemical reaction involving a fundamental change in the identity of the material. On the other hand, a physical property involves only a change in appearance of the material.

For example, consider the boiling a pot of water to cook a bowl of spaghetti. After sometime, the water begins to boil and form a vapor (i.e. steam). The water that boils off as steam has undergone a physical change in appearance, thus we can consider the boiling point of a substance as a physical property.

Now consider, a loaf of bread accidentally left out after a trip to the grocery store. Over the course of a week or two, mold will form on the bread indicating a chemical change in the identity of this bread molecules. The specific property that lead to this reaction would then be considered a chemical property.

Physical Properties

Physical properties are distinguished into two subgroups: intensive and extensive physical properties. Intensive properties are independent of the quantity of the material present where as extensive properties are not. For example, the density of a material is an intensive property while the mass of a material is an extensive property.

Melting Point

As mentioned previously, the melting point of a substance indicates the temperature at which the phase transition from solid to liquid or liquid to solid occurs. The melting point of a substance will be the same regardless of the amount of the material present thus making the melting point an intensive property. Most individuals are familiar with standard melting points of various substances such as water, ethanol, or nitrogen. This standard melting point refers to the temperature at which melting occurs at atmospheric pressure. This knowledge of standard conditions is often very useful as most transformations and reactions will occur in a open environment (atmospheric pressure), although it is important to note that melting point is not independent of pressure. That is, water at a pressure of 10 atmospheres will melt at a very different temperature than water at standard pressure.

Clausius-Clapeyron Equation

The melting and boiling point of a substance is dependent upon the pressure. This relationship is modeled by the Clausius-Clapeyron equation.

It is written as:

[math]\displaystyle{ \frac{dP}{dT} = \frac{PL}{T^2R} }[/math]

Using separation of variables and integrating from [math]\displaystyle{ P_1 }[/math] to [math]\displaystyle{ P_2 }[/math] and [math]\displaystyle{ T_1 }[/math] to [math]\displaystyle{ T_2 }[/math] this equation becomes:

[math]\displaystyle{ \ln\frac{P_1}{P_2} = -\frac{H_F}{R}(\frac{1}{T_1}-\frac{1}{T_2}) }[/math]

where [math]\displaystyle{ H_F }[/math] is equal the enthalpy of fusion, which is equal to the amount of energy that must be taken out or put into the system per mole of material for the phase transformation to occur. R is the gas constant.

Note: This equation should be used only in idealized situations as it does not include the temperature dependence of the heat of fusion. For this reason, there is some inaccuracies involved in this calculation.

Melting Point Depression

It is likely that you or someone you know has, at one point, spread salt on the his or her driveway before the onset of a cold front. Somehow, this greatly reduces the amount of ice that forms on the covered surfaces. This helpful winter trick occurs due to a phenomenon know as melting point depression. While a much more in-depth explanation could be given, here is a easy way to think about. When a liquid freezes, the molecules are attempting to orient and pack themselves in a way so as to form a solid. If foreign particles are in the liquid, they will partially block the liquid particles from forming into a solid thereby lowering the freezing point of the liquid.

See [Freezing-point Depression] for a closer look on how this process occurs and the math behind calculating the change in melting point!

Examples

Consider the hypothetical element Greconium, which is a liquid at room temperature and has a standard melting point of 273.15K. If the pressure is increased to 5 atmospheres, what will be the melting point of Greconium at this elevated pressure? Assume the heat of fusion is constant and equal to [math]\displaystyle{ 45000\frac{J}{mol} }[/math] for this problem.

Using the simplified version of the Clausius-Clapeyron equation derived above, we must solve for [math]\displaystyle{ T_2 }[/math]:

[math]\displaystyle{ \ln\frac{P_1}{P_2} = -\frac{H_F}{R}(\frac{1}{T_1}-\frac{1}{T_2}) }[/math]

Rearranging and solving for [math]\displaystyle{ T_2 }[/math]:

[math]\displaystyle{ T_2 = \frac{1}{\frac{1}{T_1}+\ln(\frac{P_1}{P_2})\frac{R}{H_F}} }[/math]

Plugging in for known variables:

[math]\displaystyle{ T_2 = \frac{1}{\frac{1}{273.15}+\ln(\frac{1}{5})\frac{8.314}{45000}} }[/math]

We find:

[math]\displaystyle{ T_2 = 297.3 K }[/math]

In this case, the increase in pressure causes the melting point of Greconium to rise to 297.3K.

Note: This problem was meant only to demonstrate how to use the Clausius-Clapeyron equation.

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