I understand Le Chatelier's principle and how every change to an equilibrium system causes an opposing reaction from the system. I also understand how, when pressure is increased, the equilibrium shifts to the side with the lowest number of moles of gas and vice versa.

So, why does this happen? How is it more energetically favourable - or is it something else?

Also, are these explanations of why the ration changes for the other factors right:

  • Temperature - equilibrium moves to the endothermic side of the reaction as temperature increases because the activation enthalpy of the exothermic reaction increases as the overall energy of the system is less
  • Concentration - ratio of products/reactants changes immediately after one part of the reaction's concentration increases, it then approaches the same ratio (assuming one whole side of the reaction is added)
  • Catalysts do nothing more than decrease the time it takes to reach equilibrium as they catalyse both sides of a reversible reaction and reduce the activation enthalpy

Le Chatelier's principle is taught to every schoolchild (well, every schoolchild that studies chemistry) but you won't find it used by any research chemist and using it is unlikely to get you many marks in university exams. That's because Le Chatelier's principle is really just an observation and isn't based on any fundamental principles.

If you have some reaction:

$$ A + B + etc \rightarrow H + I + J + etc $$

then there is a Gibbs free energy change, $\Delta$G associated with the reaction. There is also an equilibrium constant, K, that tells you how far the reaction goes, and the two are related by:

$$ K = e^{-\Delta G/RT} $$

When you change the conditions, e.g. the temperature and pressure, you change the Gibbs free energy of both the reagents and the products, but they are unlikely to change by the same amount. That means changing the experimental conditions changes $\Delta$G for the reaction and therefore it changes the equilibrium constant. The only reliable way to find out what effect changing the conditions has is to calculate the changes in free energy and then calculate the equilibrium constant.

  • $\begingroup$ But is there a clear reason for the ΔG to necessarily change the ratio of products:reactants to favour the side with the lowest number of moles of gas? $\endgroup$
    – Kian
    Feb 11 '13 at 18:44
  • 1
    $\begingroup$ This previous question: physics.stackexchange.com/q/51894/4952 should help $\endgroup$
    – Sankaran
    Feb 11 '13 at 19:09
  • $\begingroup$ Thanks, Sankaran @John, you gave the equation e^-ΔG/RT - what do R and T refer to? $\endgroup$
    – Kian
    Feb 11 '13 at 19:30
  • $\begingroup$ T is temperature and R is the gas constant - en.wikipedia.org/wiki/Gas_constant $\endgroup$ Feb 11 '13 at 19:35
  • $\begingroup$ Thanks. Also - are my assumptions right about the other factors? $\endgroup$
    – Kian
    Feb 11 '13 at 20:20

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