# melting and boiling …really equilibrium? [closed]

Why are melting and boiling considered equilibrium processes even though the amount (concentration) of both phases keep changing i.e from solid to liquid and so on?

• There can only be an equilibrium if the process is adiabatic. – MaxW Oct 25 '19 at 18:23
• If the amount of phases is changing, this is not an equilibrium. – Ivan Neretin Oct 25 '19 at 18:28
• By equilibrium I suppose you mean reversible. – Buck Thorn Oct 25 '19 at 18:40

I think what you are asking is this: Equilibria for chemical reactions typically* (see note at end) require specific ratios of products to reactants (as expressed by the equilibrium constant). By contrast, equilibria for phase transitions don't require specific ratios of products to reactants. [For instance, at the phase transition between ice and water, ice and water can be in equilibrium with any relative amounts of ice and water.] Why is this?

Consider a gas-phase chemical reaction, say

$$\ce{A(g) + B(g) <=> C(g)}.$$

At equilibrium, the chemical potentials of all species are the same. Associated with this is an equilibrium constant,

$$K_p(T) = \frac{\displaystyle\frac{p_\ce{C}}{p^\circ}}{\displaystyle\frac{p_\ce{A}}{p^\circ}\cdot\frac{p_\ce{B}}{p^\circ}},$$

which specifies a strict relationship between the partial pressures of the reactants and products.

Alternately, consider a solid-liquid phase transition,

$$\ce{A(s) -> A(l)}.$$

At the melting temperature, the chemical potentials of the two phases are the same. However, by contrast with the example gas-phase reaction, this is true independent of the relative amounts of $$\ce{A(s)}$$ and $$\ce{A(l)}.$$

The question then arises: why the difference between the two? The answer is that the particular mathematical form of the equilibrium constant—and the attendant constraint it imposes on the relative amounts of reactants and products—arises from the entropy of mixing, and there is no entropy of mixing in a phase transition! At equilibrium, in the example gas-phase reaction, the entropy of mixing just balances the relative free energies of the reactants and products. That's why we have an equilibrium. If it weren't for the entropy of mixing, reactions would always go completely to the side whose pure components had the lower free energy (which is exactly what happens in a phase transition).

By contrast, there is no entropy of mixing term in a phase change, because the difference in phases keeps the two components separated. And without an entropy of mixing term, the relative amounts of the two phases become irrelevant to their relative chemical potential, and thus irrelevant to the equilibrium state (by which I mean any relative amount of the two phases is allowed at the phase transition temperature).

This also explains why, when we are (say) somewhat above the melting temperature, we don't have a small amount of solid in equilibrium with a large amount of liquid (as would be the case if this behaved like a chemical reaction). The reason for this is there's no entropy of mixing to favor this mixed state over the liquid alone, i.e., over the pure substance with the lower chemical potential (which, above the melting temperature, is the liquid state). [Or more accurately, while there is an entropy of mixing, because the two phases aren't mixed at a molecular level—phase is a bulk property (i.e., a collective property of many molecules), not a molecular one—the entropy of mixing term (which is essentially an exchange entropy) is negligible in this case.]

N.B.: Just for clarity, let me emphasize that, as a substance is melting or boiling, it is not at equilibrium. It's not at equilibrium during a phase transition until there is no macroscopic change in the system, i.e., until it settles into a fixed ratio of one phase to another (and that it can be at equilibrium at any ratio).

*Incidentally, this is also why, when we have a mixed-phase chemical reaction where one phase can't mix with the others, that phase isn't included in the equilibrium expression. For instance, for

$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)},$$ we find that

$$K_p(T) = \frac{p_\ce{CO2}}{p^\circ}$$

• How does this address boiling? – Buck Thorn Oct 26 '19 at 14:06
• @BuckThorn It's the same thing. Imagine the liquid and gaseous forms of water together in an adiabatic piston at T= 100C and p = 1 atm. Under these conditions, the chemical potentials of both phases are the same, and we have phase coexistence. Now we temporarily make it non-adiabatic and allow some heat to flow in, converting some of the liquid water to gas (but keeping the temp at 100 C). We are still at equilibrium between the liquid and the gas, even though the relative amounts have changed. I.e., equilibrium, for phase coexistence, requires no specific ratio of liquid to gas. – theorist Oct 26 '19 at 14:46
• Why does it have to be adibatic at all? At the BP T and p, no net heat transfer should occur even if it is not adiabatic. – Buck Thorn Oct 26 '19 at 14:56
• @BuckThorn It may not need to be adiabatic. The point is that, to keep a certain ratio of liquid to gas, there needs to be no heat flow. So I specified adiabatic to give a clear example. But, as you say, if the conditions are isothermal, classically there should likewise be no heat flow. Except I'm wondering about this: If we consider fluctuations, it seems possible that, under isothermal conditions, the system could drift from its initial conditions, attaining a new equilibrium state at that same T and p (i.e., here there would be heat flow, as a result of this "thermal random walk"). – theorist Oct 26 '19 at 15:46

[OP] Why are melting and boiling considered equilibrium processes [...]

They should not be considered equilibrium processes. If melting is defined as the process where there is a net change from solid to liquid phase, this is not an equilibrium. If boiling is defined as the process where liquid turns into vapor (rolling boil with bubbles forming below the surface), this is not an equilibrium either.

Wikipedia, in the article on melting point retrieved 10/25/2019, makes a statement that is perhaps too general:

[Wikipedia] At the melting point the solid and liquid phase exist in equilibrium.

If you slowly transfer heat to the system, its temperature will remain at the melting point but solid will turn into liquid, which shows that the system is not at equilibrium.

On the other hand, it is correct to say "when the solid and liquid phase exist in equilibrium, we call the temperature of the system the melting point of that substance". Similarly, when the liquid and vapor phase (both pure) exist in equilibrium, the temperature is called the boiling point for the prevailing pressure.

When you actually melt a substance, you are transferring heat into the system and the system is not a thermal equilibrium. Typically, the overall temperature is higher than the melting point. Also, it is not uniform because the melting process is endothermic, influencing the local temperature when the substance gradually melts. Similarly, we typically boil liquids by heating them from the bottom, with a temperature gradient inside the sample (even if you use a microwave, the temperature does not rise uniformly). Complicating matters, there are supercooled or superheated liquids that have a kinetic barrier to starting the process of freezing or boiling, respectively, so freezing or boiling might not even occur as you reach the freezing and boiling point.

[OP] Why are melting and boiling considered equilibrium processes even though the amount (concentration) of both phases keep changing i.e from solid to liquid and so on?

Let's look at the second part of the statement. If both phases are pure, the concentrations (or activities) do not change at a given temperature, so that part of the statement is not accurate (the amounts do change while solid is melting or liquid is freezing/crystallizing). So at the melting temperature, you could have more liquid present or more solid, and in both cases the system would be at equilibrium if not heat is transferred in or out. Both systems would have the same reaction quotient $$Q = 1$$. Another way of defining the melting point is to say it is the temperature where the equilibrium constant for the phase transition is equal to one, $$K_\mathrm{eq} = 1$$.

In an adiabatic system, if the temperature is a bit higher than the melting point, some solid will melt until you reach the melting temperature. At that point, the system will be at equilibrium, and no net changes occur. If the temperature is a bit lower than the melting point, some liquid will freeze until you reach the melting temperature. At that point, the system will be at equilibrium, and no net changes occur.

Take-home message

You can have a liquid/solid system at equilibrium at the melting temperature, but in this case no net melting occurs. You can have a liquid/vapor system at equilibrium at the boiling temperature, but in this case no net evaporation or boiling occurs.

Two different phases of a substance in contact with each other in a closed system at some uniform temperature and pressure (thermal and mechanical equilibrium) will be in equilibrium if the chemical potential of the substance is the same in both phases. It turns out that at its boiling point, a liquid has the same chemical potential as its vapor at that pressure and temperature. Similarly, the liquid and solid phases share the same chemical potential at the melting T and p. Provided surface effects are ignored and the quantities are macroscopic, any proportion of the two phases will be an equilibrium state at that T and p. Another way of saying the same thing is that, for a total amount of substance, at phase equilibrium the free energy of the substance will be independent of the amount of each phase. Since there is no change in free energy when performing a transition from one mole ratio to another, the process is reversible, which also implies equilibrium.

As an illustration, consider that at a given temperature, a condensed phase (liquid or solid) will be in equilibrium with vapor at a particular pressure (this is in accordance with Gibbs' phase rule). You can change the relative volumes (or fractions) of the two phases and they will remain in equilibrium at that p and T. In addition, because any of those different fractional compositions represents an equilibrium point, you can perform a reversible transformation by progressively changing the relative amount of each phase at constant T and p, all the while remaining "in equilibrium". You would not say, however, that the two systems of different composition are in equilibrium with each other at that point, even though you might envision some way of coupling them somehow in which case you could say that.

Clarification: As rightly pointed out in other answers, a process is not an equilibrium state. Therefore, "boiling" and "melting" processes are not equilibrium states, they are processes. The game is given away by the use of the gerund. However, melting point and boiling point are terms used to refer to states (not processes) at which liquid and solid or gas and liquid are in equilibrium with each other. The best analogy I can think of is moving an object along a perfectly flat surface: sitting still is an equilibrium state, while moving the object over the surface is a process that potentially follows a series of equilibrium states (if performed reversibly it proceeds along a series of equilibrium states).