Why do impurities lower the melting point of an isolated substance?

Question

It is known that impurities in a desired isolated product lower the melting point of the mixture, even if the impurities' melting point is much higher than the desired product. Why is that so?

Answer 1

It's a very general statement, but it's not always true. I'll explain why it's often true, and give a counter-example at the end.

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Your majority component B and the impurity (let's call it A) form a binary system. In most cases, such binary mixtures exhibit a solid–liquid phase diagram as follows:

(image taken from these lecture notes).

This binary phase diagram has pure A on the left, pure B on the right. A and B form, somewhere, a eutectic. It is the point here at concentration e and temperature y. Because the existence of a eutectic point is guaranteed for any A/B binary system, and because the eutectic corresponds to a lower temperature, your liquidus curve decreases with increasing impurity concentration, and the impurity thus lowers the melting point.

However, not all binary mixtures form a eutectic. In the words of Wikipedia:

Not all binary alloys have a eutectic point; for example, in the silver-gold system the melt temperature (liquidus) and freeze temperature (solidus) both increase monotonically as the mix changes from pure silver to pure gold.

The corresponding phase diagram is as follows:

Answer 2

Thermodynamically, you're considering the chemical potentials ($\mu$) of the liquid and solid(s), specifically the temperature where they're equal. In a mixture, the potential is lower as the disorder (entropy) has increased, so all things equal, it will favor the liquid over a purer solid where there can be more disorder. Chemical systems seek to lower their potential through spontaneous chemical changes (e.g. of phase), minimizing the free energy ($G$). Equilibrium occurs where $dG = 0$. In a constant temperature, constant pressure system,

$$dG =\sum_i\mu_idn_i$$

$\mu$ is the chemical potential of a given species (some compound in some phase) and $n$ is the amount of that compound.

Assuming our solid generated is pure (valid sometimes, not always), our reaction (freezing) is $$\require{mhchem} \ce{A_{(l)} -> A_{(s)}}$$

so the relationship between the two species is equal and opposite (generating one mole of solid consumes one mole of liquid). $$-dn_{(l)} = dn_{(s)}$$

therefore,

$$\begin{align} dG = 0 &= \mu{(l)}dn_{(l)} + \mu{(s)}dn_{(s)}\\ &= \mu{(l)}dn_{(l)} - \mu{(s)}dn_{(l)}\\ 0 &= \mu{(l)} - \mu{(s)}\\ \mu{(s)} &= \mu{(l)}\\ \end{align}$$

As the solid is pure (prior assumption, note: $\star$ denotes pure compound),

$$\mu{(s)} = \mu^\star{(s)}$$

In an ideal mixture (assumes interactions between all components are equal) with mole fraction $\chi_A$ and $T$ being the freezing point of the mixture,

$$\mu{(l)} = \mu^\star{(l)} + RT \ln \chi_A$$

(If non-ideal, a general $a$ term is used $\mu{(l)} = \mu^\star{(l)} + RT \ln a_A$). All together:

$$\begin{align} \mu^\star{(s)} &= \mu{(s)} = \mu{(l)} = \mu^\star{(l)} + RT \ln \chi_A\\ \mu^\star{(s)} - \mu^\star{(l)} &= RT \ln \chi_A\\ -\Delta G^\star_{m, fus} &= RT \ln \chi_A\\ -(\Delta H^\star_{m, fus} - T \Delta S^\star_{m, fus}) &= RT \ln \chi_A\\ \frac{\Delta S^\star_{m, fus}}{R} - \frac{\Delta H^\star_{m, fus}}{RT} &= \ln \chi_A\\ \end{align}$$

If the mixture is pure (freezing point at $T^\star$), $\chi_A = 1$, so $\ln \chi_A = 0$,

$$\begin{align} -\frac{\Delta H^\star_{m, fus}}{RT^\star} + \frac{\Delta S^\star_{m, fus}}{R} &= 0\\ \frac{\Delta S^\star_{m, fus}}{R} &= \frac{\Delta H^\star_{m, fus}}{RT^\star} \\ \end{align}$$

Combined...

$$\begin{align} \frac{\Delta H^\star_{m, fus}}{RT^\star} - \frac{\Delta H^\star_{m, fus}}{RT} &= \ln \chi_A\\ \frac{\Delta H^\star_{m, fus}}{R}\left(\frac{1}{T} - \frac{1}{T^\star}\right) &= \ln \chi_A\\ \frac{\Delta H^\star_{m, fus}}{R}\left(\frac{T^\star - T}{TT^\star}\right) &= \ln \chi_A\\ \end{align}$$

As $T \approx T^\star$ and if we define $\Delta T$ as the change in equilibrium temperature (melting point) from the pure substance,

$$\begin{align} \frac{\Delta H^\star_{m, fus}}{R}\left(\frac{\Delta T}{T^{\star2}}\right) &= \ln \chi_A\\ \end{align}$$

This can be taken further to derive the coefficient for the freezing point depression, but from here we can see that because $\Delta H$, all $T$'s, and $R$ are positive and $\ln \chi_A$ is guaranteed to be negative ($\chi_A$ must be less than zero), $\Delta T$ must be negative.

To recap the assumptions, this is an ideal mixture and the solid formed is pure. Addressing the first, if non-ideal, we can use the general $a_A$ instead of $\chi_A$. The mole fraction can never exceed 1, but I'm not sure about $a$; if it does exceed 1, then there would be a freezing point elevation.

As for the second, if the solid formed is a mixture (e.g. a metal alloy), the that throws another wrench in the works that I don't definitively know how to address. I believe it would then be based on the difference of interaction between the components in liquid phase versus solid phase as well as the relative concentration in each.

Answer 3

The two above answers are academic/scholastic. I'll give an intuitive one. When impurity is in a solid, it usually (not always, as said in Answer 1) weakens the connections/forces between molecules, and hence makes it more vulnerable to heat (read lower melting point). A solid is like an army in rank and file. When you put a civilian in the army, it doesn't matter whether that civilian is as strong as Arnold Schwarzenegger, the formation will be messed to some extent.
However, weaker intermolecular forces do not mean that solid will becomes softer. It may become harder and more brittle. An example is that copper is changed to a harder bronze if tin is added.