# Why does increasing the pressure cause ice to melt more?

This question was given under chemical equilibrium.

Why does increasing the pressure on this system cause ice to melt more?
$$\ce{H2O(s) <=> H2O(l)}$$

I'm pretty sure this has to be solved using Le Chatelier's principle; the hint given is that The molar volume of water is more that of ice.

My understanding of Le Chatelier's principle is that a system tries to oppose any change.

So the equilibrium must shift towards the side where the pressure decreases.

Given this hint, I fail to see the correlation between molar volumes and pressure.

## 2 Answers

I'm pretty sure this has to be solved using Le Chatelier's principle the hint given is that The molar volume of water is more that of ice

The hint is exactly backwards.

Per here, at $\pu{0 ^\circ C}$:

\begin{align} v_\ce{H2O(\ell)} &= \pu{18.0182 cm3/mol} \\ v_\ce{H2O\!(s)} &= \pu{19.66 cm3/mol} \end{align}

In this situation, the pressure change is the Le Chatelier stimulus, and a reduction in total system volume is the response, in a fashion loosely analogous to the compression of, say, a block of rubber when pressure is applied. Since liquid water has a smaller molar volume than ice, one of the 'options' the system has for decreasing its total volume in response to applied pressure is to undergo the phase change to liquid water.

This is analogous to the situation where a change in the number of moles between reactants and products leads to a Le Chatelier response to a pressure change. Consider the following reaction:

$$\ce{N2 + 3H2 <=> 2 NH3}$$

Here, increasing the pressure will also drive the equilibrium to the right. In this case, the main 'option' the system has for responding to the increase in pressure is to react four moles of reactants to form two moles of products. As above, the Le Chatelier response is in the direction that works to mitigate the applied pressure increase.

• But to counter the increase in pressure i.e. the decrease in volume, isn't the equilibrium supposed to shift in a direction with a larger molar volume? – Gaurang Tandon Mar 27 '18 at 14:59
• The liquid and solid, to a reasonable approximation, can be considered incompressible (i.e., dV=0) – Chet Miller Mar 27 '18 at 15:47
• @GaurangTandon Let's continue in chat – hBy2Py Mar 27 '18 at 15:48
• @GaurangTandon No. The decrease in the total volume of the system is the Le Chatelier response to the increased pressure, and the melting phase change is one of the mechanisms available to the system to decrease its volume. – hBy2Py Mar 27 '18 at 16:20
• Le Chataliers principle can get you into trouble when reactions/products are 50/50 in this reaction... – B. Kelly Feb 22 '19 at 5:29

At equilibrium, the change in Gibbs free energy is zero. So, to maintain equilibrium, if the pressure increases, the temperature must change in tandem in such a way that the change in free energy of the solid is the same as that of the liquid. We have, for either phase, $$\mathrm{d}G=-S\mathrm{d}T+V\mathrm{d}P$$So, $$\mathrm{d}(\Delta G)=-\Delta S \mathrm{d}T+\Delta V\mathrm{d}P=0$$where the deltas refer to the changes between liquid and solid. We also have that, at equilibrium, $$\Delta S=\frac{\Delta H}{T}$$So, combining these equations gives $$\mathrm{d}T=\frac{T(V_\mathrm{l}-V_\mathrm{s})}{(H_\mathrm{l}-H_\mathrm{s})}\mathrm{d}P$$Since $V_\mathrm{l}<V_\mathrm{s}$, this equation indicates that the equilibrium temperature for the solid-liquid transition decreases as the pressure increases.

This shows that the melting temperature decreases. So if the pressure is increased, and the actual temperature is not changed, the temperature of the ice will be higher than the melting temperature, and the ice will melt.

• Interesting. But I was taught that $\mathrm{d}G=\mathrm{d}H-T\mathrm{d}S$. From your first equation, it seems that you've taken $\mathrm{d}H=V\mathrm{d}P$. How did you derive that? I thought that this relation only holds in adiabatic conditions. But aren't we heating ice to change it into solid? (so it isn't an adiabatic process?) Also, shouldn't it be $T\mathrm{d}S$ instead of $S\mathrm{d}T$? – Gaurang Tandon Mar 27 '18 at 14:28
• Actually, the correct equation is $dG=dH-d(TS)=dH-TdS-SdT$. When combined with $dH=TdS+VdP$, this becomes $dG=-SdT+VdP$. This equation is in every thermo book. – Chet Miller Mar 27 '18 at 15:14