Calculating melting temperature of water with thermodynamical data

Question

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?

Answer 1

I offer an alternative solution. We will apply the Clapeyron equation at the point of interest $(P_m,T_m)$ lying in the P-T diagram on the solid-liquid equilibrium \begin{align} \dfrac{dP^\pu{sat}}{dT} &= \dfrac{\Delta_\pu{fus} H}{T_m \Delta V} \tag{1} \\ \end{align} Here we make our bold assumption: we consider that the solid-liquid equilibrium curve can be approximated by a straight line, in the range of the triple point $(P_t,T_t)$ to the melting point $(P_m,T_m)$; obviously $P_m = \pu{101325 Pa}$. Thus, we can approximate the derivative in the left-hand side of Eq. (1) by \begin{align} \dfrac{P_m - P_t}{T_m - T_t} &= \dfrac{\Delta_\pu{fus} H}{T_m (V_\pu{l} - V_\pu{s})} \\ T_m &= \dfrac{T_t}{1 - \dfrac{(V_\pu{l} - V_\pu{s})(P_m - P_t)}{\Delta_\pu{fus} H}} \tag{2} \end{align}

Useful data from Wikipedia and knowing that the molar mass is $\pu{18.015 g/mol}$:

1. $\rho_\pu{s} = 916.7 \; \pu{kg/m3}$ or $V_\pu{s} = \pu{1.9652E-5 mol/m3}$.
2. $\rho_\pu{l} = 999.8 \; \pu{kg/m3}$ or $V_\pu{l} = \pu{1.8019E-5 mol/m3}$.
3. $\Delta_\pu{fus} H = 6008 \; \pu{J/mol}$.
4. The triple point has the coordinates $P_t = \pu{611.657 Pa}$ and $T_t = \pu{273.16 K}$.

Replacing in Eq. (2) yields $$ \boxed{T_m = 273.1525 \; \pu{K} = 0.002521 \; ^\circ\pu{C}} \tag{3} $$ Pretty near to the exact value.