13
$\begingroup$

The formation of ice out of liquid water can be written down like this:

$$\ce{H2O (l) <=> H2O(s)}$$

We can calculate the change in standard Gibbs free energy (per mol substance) in the following way:

$$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$

If we do this at room temperature $(\pu{298 K}),$ we get: $\Delta G^\circ = \pu{546 J mol^-1}.$

So we can see that this process is very non-spontaneous, which is what you'd expect, you never see ice forming at room temperature. However if we calculate the equilibrium constant $K_\mathrm{eq}$ for this reaction we get:

$$K_\mathrm{eq} = \exp\left(-\frac{\Delta G^\circ}{RT}\right) \approx \mathrm e^{-0.22} \approx 0.80$$

It's obvious that this is wrong, because at room temperature we never see liquid water in equilibrium with a decent amount of ice. I just don't know what I did wrong, I have checked my units and calculations but can't seem to find what I did wrong.

$\endgroup$
4
  • 3
    $\begingroup$ You did it right. That's the thing with pure liquids and solids: even if the constant is not very far from 1, the reaction goes all the way to the end. $\endgroup$ May 15 at 10:22
  • 1
    $\begingroup$ How did you get the entropy change from liquid water to ice at 298 K and 1 bar? $\endgroup$ May 15 at 10:54
  • 1
    $\begingroup$ First of all, you should drive away this misconception of yours that every reaction needs to be in equilibrium, secondly, even if we have an equilibrium that doesn't mean the concentration of product is any appreciable value. $\endgroup$ May 15 at 10:59
  • 2
    $\begingroup$ It's best to avoid the phrase "very non-spontaneous". Reactions are spontaneous or not or at equilibrium. There is no degree of spontaneity. $\endgroup$
    – Andrew
    May 15 at 12:18
15
$\begingroup$

Now that you've got $K_{eq}$, you need to take a look at the expression for the equilibrium constant, which is $$K_{eq} = \frac{a_{\text{ice}}}{a_{\text{water}}},$$ where $a$ is the activity of each species.

For a pure liquid and a pure solid, the activity is defined as 1, so the expression on the right hand side of that equation (usually referred to as the reaction quotient $Q$) can only take on three values:

1 if ice and water are both present

0 if water is present but not ice

undefined if water is absent, whether or not there is ice.

So if a mixture of ice and water is put in a room temperature environment, the ice melts because the reaction quotient of 1 is greater than the equilibrium constant of 0.8. Only when all of the ice melts and it makes the step change to $Q=0$ does it get below the equilibrium constant. Then we have a situation where anytime any infinitesimally small amount of ice crystallizes, then $Q$ exceeds $K_{eq}$ and the crystal returns to liquid form.

$\endgroup$
4
  • $\begingroup$ But following the same train of thought, when the ice has melted and Q=0 the equilibrium constant is still 0.8. Doesn't that drive it to freeze again? After all, the system aims to be at $K_{eq}$? $\endgroup$
    – Buck Thorn
    May 15 at 15:47
  • 8
    $\begingroup$ @BuckThorn - Yes indeed. But as soon as the amount of ice is non-zero, the driving force immediately switches in favor of melting (b/c Q=1) so they never get beyond an amount that is functionally 0. If you observed every molecule in the liquid over time, you'd see frequent assemblies of a few molecules into a solid-like structure, but they'd immediately get hit by other molecules and broken apart. Only when the T gets down to 273 K is the average KE low enough that those assemblies can continue to grow without being broken up. $\endgroup$
    – Andrew
    May 15 at 15:53
  • 1
    $\begingroup$ @BuckThorn or you consider a third state, nuclei. They are dissolved in water, have a concentration, and will be in equilibrium with liquid water. $\endgroup$ May 16 at 3:06
  • $\begingroup$ @Andrew would equilibrium constant for same reaction at 273 K greater than or less than or equal to 1? $\endgroup$
    – Lllt
    Jun 13 at 14:49
0
$\begingroup$

There is a fallacy in the discussion that in this context the notion of an equilibrium constant retains its usual meaning (as rightfully questioned in a comment). It doesn't. An equilibrium constant arises when you assume that various chemical components (in the same or different phases) can coexist. The condition for coexistence of two phases is that the chemical potential of each component in all phases is the same. For liquid water in equilibrium with ice, one would write

$$\mu_\ce{H2O(\textrm{l})} = \mu_\ce{H2O(\textrm{s})} $$

But such an equation holds only at the melting point. At other points the chemical potentials are not equal, so that while a hypothetical equilibrium constant can be written, it is in practice without meaning as a proper "equilibrium constant". Such a parameter might be used to interpret the direction in which the system will change, as suggested in another answer, but it no longer represents a ratio of actual activities in the system at equilibrium. The interpretation is much simpler than that:

$$\begin{align} K<1 &\rightarrow \text{reaction proceeds completely to the right}\\ K>1 &\rightarrow \text{reaction proceeds completely to the left} \end{align}$$

$\endgroup$
3
  • $\begingroup$ I don't understand why this is not a "proper equilibrium constant". If Keq =0.8 and the system is allowed to equilibrate and then held at that temp, the time average of the ratio of the activities will converge to 0.8, just as with any other system in equilibrium. Your answer applies more to systems where Keq is so large that the reverse reaction is functionally impossible because it is so improbable. $\endgroup$
    – Andrew
    May 16 at 11:36
  • $\begingroup$ @Andrew it's not an equilibrium constant in the strict sense because you can't have equilibrium between the phases (except at the boiling point, when trivially $K_{eq}=1$ ). I am being literal. The equilibrium constant follows from an initial assumption that the chemical potentials of the species in question are equal (in equilibrium), but in this case they can't be. $\endgroup$
    – Buck Thorn
    May 16 at 11:44
  • $\begingroup$ From a practical standpoint, the phases are not in equilibrium because there's never enough solid to matter, but from a literal thermodynamic standpoint, we certainly do have two phases in equilibrium at any temperature, with the caveat that we must consider the time average, which is a caveat that applies to almost any system with a finite number of molecules. Because we can only have integer numbers of reactant and product molecules, Q at a specific time instant for any reaction "at equilibrium" is rarely exactly equal to K, but the limit of the time average of Q is K. $\endgroup$
    – Andrew
    May 16 at 12:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.