1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – Karsten
    May 15, 2019 at 14:40
  • $\begingroup$ 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! $\endgroup$
    – Kdbmvp
    May 15, 2019 at 15:47
  • $\begingroup$ 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) $\endgroup$
    – Karsten
    May 15, 2019 at 15:51

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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