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For the reaction: $$\ce{P4(s) + 5O2(g) <=> P4O10(s)}$$

Let's say the reaction is at equilibrium, and SOME $\ce{P4O10(s)}$ is removed from the reaction. Will that affect equilibrium? My assumption is no, since $\ce{P4O10(s)}$ is not included in the equilibrium expression.

However, for me, it gets confusing when let's say $\ce{P4O10(s)}$ is COMPLETELY removed. In this case, I would assume that the reaction shifts to the products (since no products are present). I'm not sure if this is right.

Help is appreciated. Thank you!

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  • $\begingroup$ There isn't much of an equilibrium here; the reaction is shifted pretty much all the way to the right anyway. That being said, your reasoning is correct. $\endgroup$ – Ivan Neretin Feb 21 at 0:25
  • $\begingroup$ @IvanNeretin That makes sense. Thank you. I'm just confused because I thought solids/pure liquids do not effect the equilibrium? $\endgroup$ – user105067 Feb 21 at 0:32
  • $\begingroup$ If there's no equilibrium then there's no effect on equilibrium. Find a valid example. $\endgroup$ – Mithoron Feb 21 at 0:45
  • $\begingroup$ Hmm but in my example, there was an equilibrium. It's just that I'm saying what happens in P4O10 is completely removed? $\endgroup$ – user105067 Feb 21 at 1:19
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Nothing would happen. The reaction would not shift to produce more product.

Let's suppose you allow this reaction to equilibrate in a sealed container. Equilibrate means you've reached the equilibrium value for $p_{\text{O}_2}$ at that temperature. Now further suppose you were able to remove all the product without changing the free volume (say you slightly decreased the volume in the container by an amount equal to the volume of the solid product you removed). The system is still at equilibrium ($p_{\text{O}_2}$ is unchanged). Thus nothing will happen—no new product would be formed.

If you're not convinced, imagine that you had some fluctuation in the equilibrium such that a bit of product was produced. In that case, $p_{\text{O}_2}$ would increase above its equilibrium value, and thus the product would need to disappear (the reaction would need to shift back to the left) for the system to go back to equilibrium.

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    $\begingroup$ I agree with "nothing would happen" but think it is important to say that something was happening before (dynamic equilibrium), and now the underlying molecular processes (forward and reverse reaction) are no longer in place. $\endgroup$ – Karsten Theis Feb 21 at 16:43
  • $\begingroup$ @KarstenTheis That's a v. good point—that while nothing changes macroscopically, there are changes microscopically. But while you wouldn't have fwd and rev rxns as they would normally occur in an equil sys, I wouldn't go so far as to say that the fwd and rev rxns no longer take place, since there would still be fwd and rev rxns due to fluctuations about equil (as describred in my last paragraph). So this raises a very interesting ques: How would the nature and magnitude of the fluctuations about equil differ between the original equil sys (where product was present) and this "one-sided" sys? $\endgroup$ – theorist Feb 21 at 19:20
  • $\begingroup$ I would agree with your last paragraph for a homogeneous equilibrium. If the reaction involves forming a solid (as in a precipitation reaction) and the solution is not saturated, no solid is formed. On the microscopic level, there might be some ion pairs formed, or even some transient nuclei but I don't think they rise to the concept of solid. $\endgroup$ – Karsten Theis Mar 3 at 14:10
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I am using a saturated sodium chloride solution as an example here because it is more common and easier to reproduce at home. This system is at equilibrium, with salt dissolving and crystallizing at the same rate.

If you take away the solid NaCl from a saturated solution, both the forward and the reverse reactions stop. Obviously, salt crystals are no longer dissolving because there aren't any. Less obviously, there is no crystallization process anymore (there are no crystals that solvated ions could attach to). So this is not a dynamic equilibrium, i.e. the reaction or process no longer is observed at all.

This is different from a system out of equilibrium where one direction of the process is faster than the other direction. In that case, the system will approach equilibrium, and the process (in either direction) will continue to happen when equilibrium is reached (which is why it is called a dynamic equilibrium).

You might ask how we can ever get solid sodium chloride again. If the solution is sufficiently supersaturated, nuclei will form (in solution, or on the container or on some impurity), and then crystal growth will be observed again.

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    $\begingroup$ What do you mean by "it is not longer an equilibrium?" If it isn't, wouldn't a change follow? $\endgroup$ – Buck Thorn Feb 21 at 10:40
  • $\begingroup$ @BuckThorn Sorry, sloppy language. Either we have a reaction or process where both forward and reverse reactions are happening. This can either be at equilibrium or approaching equilibrium. Or, like in this case, neither forward nor reverse reactions are happening. $\endgroup$ – Karsten Theis Feb 21 at 12:19
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    $\begingroup$ Ok, I see your point. @theorist made the additional good point that you have to keep things constant, e.g. pressure, to speak of equilibtrium when you remove the salt. $\endgroup$ – Buck Thorn Feb 21 at 14:38
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It might help to think of the system as containing three phases: $\ce{P4(s)}$, $\ce{O2(g)}$ and $\ce{P4O10(s)}$. As long as the pressure, temperature and compositions of each phase remain constant (and P,T and chemical potentials across phases are in equilibrium i.e. equal), we can alter their extents arbitrarily without altering the condition for equilibrium to be sustained.

An example is freezing of water. When water is cooled to $\pu{0^\circ C}$ at 1 atm pressure it can remain at equilibrium without forming ice (it will not freeze), provided no energy is removed (as heat) from the liquid phase. Otherwise ice will begin to form. If you remove an ice cube from a glass of water at $\pu{0^\circ C}$ more ice does not spontaneously form, provided T and p are not altered (say by providing a means for the water in the glass to lose heat to the surroundings, which would of course describe a nonequilibrium situation).

This is admittedly somewhat strange and not the end of the story because we have considered the system to be an abstract body in which the interfaces between phases are not important. In fact they can be, and we might describe a system in which substituting the interface between two phases with another type of interface might cause a disequilibrium.

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