My textbook said Le Chatelier's principle normally applies to gases but there is no explanation.

I did some Googling and it said if you increase the pressure, it will shift to the side with less moles of gas.

What does this really mean? Could someone clarify Le Chatelier's principle for me?

  • $\begingroup$ In the balanced equation for the reaction, the equilibrium will shift to the side with fewer moles of gas when pressure is increased. What exactly are you confused about? $\endgroup$
    – bon
    Jul 30, 2015 at 7:39

2 Answers 2


Le Chatalier's principle states that if a stress is applied to a system at equilibrium, then the system, will shift its equilibrium position in order to bring the system back to its equilibrium.

So the stress applied to the system could be temperature, pressure and the concentration of one of the components in the system.

I will choose to explain about pressure because apparently you have the problem understanding that.

Pressure. If you add pressure to a system then the stress will be to decrease the pressure by shifting the equilibrium position towards the side with less number of particles. If you remove pressure from a system, then the stress will be to increase the pressure of the system by shifting the equilibrium position towards the side with the higher number of particles.

So it is true that when you increase the increase the pressure in a system, then the system is stressed to reduce the stress. In order to do that, the system's equilibrium will shift to the side with less number of particles.

For example:

2Fe$_2$O$_3$ + 3C -----> 4Fe + 3CO$_2$

So if you add pressure to this above system, then the stress is to reduce the pressure. On your reactants side the sum of the coefficients is 5 and on your products side, the sum is 7. Since you need to bring it down then the shift will be to the side with the least sum which is to the left-hand side. Therefore, the shift will be to the left-hand side.

Hope this helped you with the problem.


Just as if it was a chemical reaction of aqueous solutions, a gaseous chemical reaction has its reagents and products and these exist in an equilibrium kind of state that can be shifted either way depending on the conditons the reaction is in and the amount of reagents and products you have (in moles). Adding pressure to the reaction is like compressing the gasses into a smaller volume and this causes more molecule collisions. Since the space between the molecules involved in the chemical reaction is decreased, you have more molecule collitions among them, this makes the molecules to react with each other more rapidly and speeds up the reaction. The reaction will go the way the equilibrium constant tells it to go, but with this addition of pressure it will do it faster. Molecule collitions also creates and destroys molecule's bonds, thus this can create or absorb heat depending on the reaction that is taking place.

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    $\begingroup$ I'm not really sure this answers the question about Le Chatelier's principle. $\endgroup$
    – bon
    Jul 30, 2015 at 17:36
  • $\begingroup$ What is the question exactly then? Is it what is the Le Chatelier's principle? or is it what happens when you add pressure to a chemical reaction taking into consideration the Le Chatelier's principle? The question is not even stated correctly, how can you say that my answer didn't answer the question. $\endgroup$ Jul 30, 2015 at 19:30
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    $\begingroup$ @EduardoZurita - The increased pressure will increase collisions, which will increase rate. However the definition of equilibrium is the state at which the forward and reverse reactions are both occurring with equal rates. Thus, any discussion of reaction rates changing in an equilibrium situation is not helpful. The system will shift under stimuli to achieve a state of equal rates. The OP is probably stuck at the part you brush off: The reaction will go the way the equilibrium constant tells it to go. How do you know which way that is? $\endgroup$
    – Ben Norris
    Jul 30, 2015 at 20:16
  • $\begingroup$ If you are unsure what the question is you should ask for clarification from the OP as I have done. I also agree with @BenNorris. $\endgroup$
    – bon
    Jul 30, 2015 at 20:18
  • $\begingroup$ Delete my answer if that makes you all happy. I got on this website becuase I needed help with an issue I had at work. I received a message from this website saying that I am supposed to help out other ppl answer their questions. I did my best trying to answer a question and I only got critized for not answer it correctly. I don't have a college education like most of you all do, man, I dont even speak English well. So could any of you who are just picking on me help me out answer the question I posted instead of just being all ruth to me. Please! $\endgroup$ Jul 31, 2015 at 18:59

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