Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What's behind Le Chatelier?

  1. May 23, 2012 #1
    What's the underlying principle which is behind Le Chatelier's principle that equilibria shift to oppose external changes? Is it the Second Law of Thermodynamics? If so, why? Or is it some other principle? Or just an observed fact we can't explain!? I've quoted this Law for years without realising I dont understand it!
  2. jcsd
  3. May 24, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It is inevitable in any circumstance of stable equilibrium. A system in stable equilibrium, such as a ball in a dish, recovers from a small perturbation. Normally one thinks of, say, pushing the ball to one side and releasing. But what if you keep your finger there instead of releasing the ball? The forces that would restore equilibrium when the finger is removed must already be acting.
    It follows that the rule can break down in some circumstances:
    - unstable equilibrium
    - metastable equilibrium
    - a perturbation that exceeds the range of stable equilibrium
    It may be that such circumstances are rarer in Chemistry then in some other disciplines.
  4. May 24, 2012 #3


    User Avatar
    Science Advisor

    Last edited by a moderator: May 6, 2017
  5. May 24, 2012 #4
    I think it can be described as minimization of free energy.

    When a reaction is exothermic, and the system is heated up, Le Chatelier says that the reaction equilibrium state will shift in the direction of the reactants. This makes sense because more heat will reduce the stability of the products relative to the reactants if it is exothermic.

    In other cases of equilibrium, I think it is the same, such as with pressure and gases.
  6. May 25, 2012 #5
    One other way to see it, is thanks to Harmonic Oscilators.
    Well it's not an HO, but!
    At every equilibrium point r0, you can expand your potential (which causes the force) in Taylor series. For small changes, you have:
    V(r+r0)= V(r0) + (r-r0) dV/dr|r=r0+(r-r0)2 d2V/dr2|r=r0+O(r3)

    the 2nd term on the right side because of equilibrium is ZERO.
    I can set V(r0)=0
    So every region around the equilibrium is like an HO, V(r)=A r2

    So if you disturb your system from its equilibrium a little bit, there will be a force appearing that will tend to bring it back to its initial state.
    Of course that is for very small perturbations , small enough that I can forget the terms of r3, but works fine.

    In that way, you can understand that any kind of system that is in an equilibrium state, if you drag it out of it, in "first orders" will try to return in a way. For what I used above I didn't use any kind of "determining what forces there are" only that my system was at an equilibrium and then something dragged it out of it.

    Does it work in everything?
    Well I guess yes. The only thing that is important, is to have a stable equilibrium states, and not an unstable-saddle ones. Otherwise my expansion and so peturbation would have no meaning.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook