Apparent paradox in the formation of ice at room temperature The formation of ice out of liquid water can be written down like this:
$$\ce{H2O (l) &lt;=&gt; H2O(s)}$$
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.
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.