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I was reading the section on Le Chatelier's principle (section 6.8) from Chemical Principles by Steven S. Zumdahl. On page 191 where I'm currently at, I do not understand why, to consider pressure changes after a system has reached equilibrium, instead I have to consider volume changes under constant pressure? I mean the pressure is changing, right?

When the volume of the container is changed, the concentrations (and thus the partial pressures) of both reactants and products are changed. We could calculate Q and predict the direction of the shift. However, for systems involving gaseous components, there is an easier way: We focus on the volume. The central idea is that when the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume. This is done by decreasing the total number of gaseous molecules in the system.

To see how this works, we can rearrange the ideal gas law to give $V = (R T/P) n$ or at constant T and P $$V \propto n$$

That is, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas present.

Suppose we have a mixture of gaseous nitrogen, hydrogen, and ammonia at equilibrium (a Fig. 6.9). If we suddenly reduce the volume, what will happen to the equilibrium position? The reaction system can reduce its volume by reducing the number of molecules present. Consequently, the reaction

$$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$$ will shift to the right, since in this direction four molecules (one of nitrogen [...]

Since the pressure goes down after a rise (the system responds thus), the net effect is as if the pressure didn't change much at all, instead a reduction in volume caused the number of moles to reduce. $V \propto n$ under almost constant $P$. Sounds pretty wacky, but perhaps this is what the author is trying to tell?

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  • $\begingroup$ There are tons of resources about the principle. Try to search this site, Wikipedia, chem.libretexts.org or general internet. $\endgroup$
    – Poutnik
    Oct 21, 2021 at 19:12

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However, for systems involving gaseous components, there is an easier way: We focus on the volume.

This does not seem easier. First, it is about changing the pressure. If your system behaves like an ideal gas, it is clear that in order to change the pressure, something else must change. So the volume. But then the volume change is associated with a change in amount of substance $n$. That is not correct, the change in volume was at constant $n$, resulting in a change in pressure.

The example illustrates this. However, the example is wacky, too. It says when you reduce the volume, the reaction system can reduce its volume, too. There is only one volume; if you reduce it, that's the new volume.

It would be more helpful to say that the pressure goes up. If the reaction goes in the direction of fewer gas particles, that will lower the pressure again.

On page 191 where I'm currently at, I do not understand why, to consider pressure changes after a system has reached equilibrium, instead I have to consider volume changes under constant pressure? I mean the pressure is changing, right?

I don't understand either, and applying Le Chatelier's principle to situations where more than one thing changes (in this case, all the concentrations) is tricky, see e.g. here: https://doi.org/10.1021/ed086p514

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  • $\begingroup$ Since the pressure goes down after a rise (the system responds thus), the net effect is as if the pressure didn't change much at all, instead a reduction in volume caused the number of moles to reduce. $V \propto n$ under almost constant $P$. Sounds pretty wacky, but perhaps this is what the author was trying to tell? What do you think? $\endgroup$
    – HERO
    Oct 22, 2021 at 4:23
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    $\begingroup$ The pressure will change. The reaction counteracts the initial change in conditions, but only partially. $\endgroup$ Oct 22, 2021 at 17:25
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It is apparently a problem of chemical equilibrium. Suppose you are studying a mixture of $\ce{N2, H2}$ and $\ce{NH3}$ which is at equilibrium in a big syringe. It means that the concentrations of these three gases are not changing in time. Now you are to change the volume by pressing on the piston, or compressing this gaseous mixture. Both volume and pressure will change. Pressure will increase, and volume will decrease. As a consequence, the number of gaseous molecules in a given volume will increase.

But how will the equilibrium react ? There are two possibilities : the reaction may go to the right, or to the left-hand-side. Which one is chosen by nature ?

Well ! Nature chooses a way that produces an effect opposed to the cause of the disruption. Here the cause of the disruption is an increase of gaseous molecules par volume unit due to compression. The effect will be something that reduces this number of gaseous molecules per volume unit. Well ! How to do this chemically ? In the reaction, there are more molecules ($4$} on the left-hand-side ($\ce{N2 + 3 H2}$, than on the right-hand side ($\ce{2 NH3}$). So Mother nature will favorize this side and destroy the cumbersome $\ce{N2 + H2}$ molecules in order to create more $\ce{NH3}$. This will help decrease the number of gaseous molecules per volume unit.

This is an application of the so-called Le Chatelier principle, which can be simply stated so : If a chemical equilibrium is disturbed by a disruption, Nature reacts by creating an effect which is opposed to the cause.

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