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Why does ice melt slower at higher altitudes and how does Le Chatelier's principle play a role in this?

This question appeared right after this chapter in my text book and I've also read a very rough explanation from another source which cited this principle.

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closed as off-topic by Mithoron, airhuff, andselisk, Pritt says Reinstate Monica, M.A.R. Nov 5 '17 at 10:30

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    $\begingroup$ What a poor question. Dumping that statement with its tremendous lack of context and forcing the link to be found leaves it wide open to misinterpretation and speculation, and is likely to either cause confusion or establish wrong connections. $\endgroup$ – Nicolau Saker Neto May 20 '17 at 4:11
  • $\begingroup$ I'm sorry but I've found this question in my book Just as I've wrote and as concerning to context, I'm only asking is there a link to La Chateliers Principle to the phenomenon. $\endgroup$ – Saadman Yasar May 20 '17 at 5:15
  • $\begingroup$ Sorry, I didn't mean to imply you are at fault, it's the book's author who should done a better job. You're fine! $\endgroup$ – Nicolau Saker Neto May 20 '17 at 5:18
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A simplified attempt.

For a simplified answer that attempts at a solution using Le Chatelier's principle, you'll need a constraint: Ice and water are at an equilibrium at $\pu{0^\circ C}$.

$$\ce{Ice <=> Water}$$

Simple enough, now pull up the values of densities of ice and water, $\rho_{ice}= 0.9167~\pu{g/cm3}, \rho_{water}= 0.9998 ~\pu{g/cm3}$. Mass being the same, you can conclude that volume of ice is greater than that of water, so in accordance with Le Chatelier's principle, when you move at higher altitudes, the atmospheric pressure falls, thus the equilibrium is favored towards a side where the species occupy a higher volume, i.e. towards the formation of ice. This works antagonistically for $\ce{Ice->Water}$.


Alternatively, have you heard of phase diagrams? This might help you to understand things a bit more clearly after you go through its definition from other sources: https://chemistry.stackexchange.com/a/72732/43942

The phase diagrams for water are different though.

https://d2gne97vdumgn3.cloudfront.net/api/file/m1GFJVbQSBmwkDXRvglV

enter image description here

This, too indicates that for water, at lower pressures, the formation of solid (i.e. ice) is more favored.


A realistic insight.

To be more realistic, as MaxW pointed in the comments, the pressure difference is rather small. The summit pressure at Mt. Everest is $\pu{33.7 kPa}$, while at the sea level it is $101.3$, which gives us a difference of $\pu{67.6 kPa}$ which converts to a mere $\pu{0.676}$ Bar.

Although this is quite a significant pressure (see Jon's comment) for our perception, it doesn't affect freezing that much. For a comparison, this answer https://chemistry.stackexchange.com/a/8883/43942 states that the pressure required to solidify $\ce{CO2}$ to make it into dry-ice at room temperature, you'll need a pressure of about $\pu{56 bar}$.

Freezing at higher latitudes, requires us to also ponder on the ambient temperature. You probably wouldn't want to talk about latitudes as high as the Everest, because then the temperatures will be extremely low (averaging about $\pu{-19^\circ C}$ in summers!) and you'll have to apply an external pressure to adjust the melting point. Latitudes where the ambient temperature is (say) $\pu{3^\circ C}$ have a less temperature gradient when compared to a place (probably at the sea level) with an ambient temperature of (say) $\pu{25^\circ C}$. A higher magnitude of a gradient means a high rate of heat transfer, which hints at a slower melting of ice at higher latitudes.

Also, airhuff suggested that thermal conductivity of thin air at high latitudes may have a role to play in this. I searched this up, you might want to look at this paper: http://nist.gov/data/PDFfiles/jpcrd269.pdf

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    $\begingroup$ Even on Mt. Everest the pressure difference will be small. The main effect on ice melting is temperature not pressure. $\endgroup$ – MaxW May 18 '17 at 19:50
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    $\begingroup$ The pressure at the summit of Everest is roughly 1/3 of an atmosphere. You would notice it... $\endgroup$ – Jon Custer May 18 '17 at 20:40
  • $\begingroup$ @MaxW This has nothing to do with temperature. We're talking about the equilibrium between air and water at 0°C (which I've stated in the first line of my answer). Also, if you're so keen about the temperature of the surroundings, Everest's (summit) temperature averages -19°C in summer and -36°C in winter. Apply Fourier's law (law of heat conduction); it tells you that heat will now flow from ice to the surroundings, thus further lowering the temperature of ice. $\endgroup$ – Berry Holmes May 19 '17 at 9:44
  • $\begingroup$ @BerryHolmes - I think you constrained the problem beyond how it was stated. The original problem makes no mention of constraining the consideration to 0°C. $\endgroup$ – MaxW May 19 '17 at 16:48
  • $\begingroup$ @MaxW Talking about the ice-water equilibrium (probably you're thinking of an elevated melting point) at yet another temperature asks for a changed (lower in your case) pressure in itself (see the phase diagram), which contradicts your first comment. $\endgroup$ – Berry Holmes May 19 '17 at 16:57
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Easy. As you get higher, there is less air, making the area colder. So, the ice is surrounded by colder and colder tempatures.

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