# Calculating melting temperature of water with thermodynamical data

So at zero degrees celsius($$\pu{273 K}$$), for clean water, we have the equilibrium eq.:

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

As it is at equilibrium, we know that $$\Delta G = 0$$. As we are dealing with clean water and clean ice we also know that $$\Delta G_\text{standard} = 0$$. (From the $$\Delta G_\text{standard} =-RT\ln K$$ relationship, where $$K = 1$$ results in $$\Delta G_\text{standard} = 0$$).

We also have the relationship:

$$\Delta G_\text{standard} = \Delta H_\text{standard}-T\,\Delta S_\text{standard}$$.

which we can re-arrange to:

$$\Delta H_\text{standard}= T\,\Delta S_\text{standard}$$.

And we can calculate $$T$$ which will be our melting temperature.

I found thermodynamical data for the reaction for $$\ce{H2O(s)}$$ and $$\ce{H2O(l)}$$, and got the equilibrium temperature to be $$\pu{241 K}$$! Which seems weird. What is wrong?

EDIT: I know that $$\Delta H$$ and $$\Delta S$$ aren't necessarily temperature independent, although I guess that is a prerequisite for the way I have done it?

• Could you show us the values for entropy and enthalpy of melting (fusion) and tell us where you found them? Also, how is the title of your question related to the body of your question? – Karsten Theis May 15 '19 at 14:40
• deltaH_fusion = 6 kJ/mole. deltaS = -29 J/ K mole. All data from SI Chemical Data, 6th edition. You are right about the title, I will change it! – Kdbmvp May 15 '19 at 15:47
• I googled +6.02 kJ/mol and +22.0 J/(K mol), both for 273.15 K, see en.wikipedia.org/wiki/Water_(data_page) – Karsten Theis May 15 '19 at 15:51