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The volume of certain saturated solution is greater than the sum of the volumes of the water and salt from which it is made. On increasing pressure solubility of this salt

(a) Increases

(b) Decreases

(c) Remains unaffected

(d) Can't be predicted

The solution says

Density of product < Density of reactant [mass of reactant = mass of product] So on increasing pressure equilibrium shift toward higher density solid or liquid.

According to Le Chatelier's Principle, shouldn't equilibrium be shifted towards the side of low density when pressure is increased?

Is my interpretation of the Principle wrong or is the solution of the answer wrong?

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  • $\begingroup$ You're wrong, maybe you mixed density and volume? $\endgroup$ – Mithoron Mar 27 '15 at 14:35
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The answer is "(b) Decreases." The reason is that the saturated solution occupies more volume than the starting reagents. At higher pressures, creating this extra volume requires more $P dV$ work be done by the dissolving reagents.

According to Le Chatelier's Principle, shouldn't equilibrium be shifted towards the side of low density when pressure is increased?

I find it easier to think in terms of volumes, not densities. High pressure should favor low volumes. Imagine a balloon filled with saturated solution. If we squeeze the balloon by raising the pressure on the outside of it, the balloon will "want to" shrink. In the example given, the balloon can shrink only by saturated solution converting back into solid salt and liquid water, so that is what would be thermodynamically favored.

Update: Based on discussion in the comments, I wanted to add a point about compressibility and thermodynamic stability. The degree to which a system opposes a change in pressure is called the compressibility and is related to dV/dP for a given system, which you can find from the equation of state. In general, systems in thermodynamic equilibrium cannot have a positive dV/dP value. That is, the volume must shrink as pressure rises for mechanically stable thermodynamic systems.

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  • $\begingroup$ "High pressure should favor low volumes." According to me, high pressure creates low volumes and the system should oppose such a change and try to regain its earlier higher volume. What's wrong with this interpretation? $\endgroup$ – Binary Geek Mar 27 '15 at 15:31
  • $\begingroup$ The degree to which the system opposes such a change is callled the compressibility and is related to dV/dP for a given system, which you can find from the equation of state. In general, systems in thermodynamic equilibrium cannot have a positive dV/dP value. That is, the volume must shrink as pressure rises for mechanically stable thermodynamic system. Your interpretation implies the opposite. $\endgroup$ – Curt F. Mar 27 '15 at 16:15
  • $\begingroup$ Please edit your post to include this point. I feel this make the answer more complete. $\endgroup$ – Binary Geek Mar 27 '15 at 16:43
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It shifts towards high density and not low.

You are right to suggest that when applying Le Chatelier's Principle, it's often useful to write and equation. We are told: graphite <--> diamond We are also given that diamond has a greater density meaning that it occupies less volume than the same mass of graphite, so the balanced equation could be written as: graphite <--> diamond + volume Now think of it this way: if you were to add volume, you would shift to the left according to Le Chatelier's Principle. Therefore, an increase in pressure (= decrease in volume) must shift to the right.

Source: (http://www.mcat-prep.com/forum/gs-1-physical-sciences-f8/topic2040.html)

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There was another answer here which got removed because it just contained a link. The link was also useful to me in addition to Curt's answer. I'll quote the relevant portion here.

The topic in discussion was over the conversion of graphite to diamond but the concept applied was the same as in this question.

You are right to suggest that when applying Le Chatelier's Principle, it's often useful to write and equation. We are told:

$\ce{graphite <=> diamond}$

We are also given that diamond has a greater density meaning that it occupies less volume than the same mass of graphite, so the balanced equation could be written as:

$\ce{graphite <=> diamond + volume}$

Now think of it this way: if you were to add volume, you would shift to the left according to Le Chatelier's Principle. Therefore, an increase in pressure (= decrease in volume) must shift to the right.

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