March 28, 2024, 12:26:04 PM
Forum Rules: Read This Before Posting


Topic: Understanding Le Châtelier's Principle  (Read 1975 times)

0 Members and 1 Guest are viewing this topic.

Offline mikasaur

  • Full Member
  • ****
  • Posts: 235
  • Mole Snacks: +27/-1
  • Gender: Male
  • Chemist in training
Understanding Le Châtelier's Principle
« on: November 23, 2015, 01:14:22 PM »
I'm learning about Le Châtelier's Principle and it makes sense to me, generally. But I'm having a hard time understanding the pressure aspect of it. Specifically, why it's not necessarily a change in pressure of a system that drives that system to a new equilibrium, but a change in pressure due to a change in volume.

From a mathematical standpoint it's extremely straightforward. For the reaction:

N2O4(g)  ::equil::  2NO2(g)

The value of Q is:

[itex]Q = \frac{(P_{\ce{NO2}}/P^{\circ})^2}{P_{\ce{N2O4}}/P^{\circ}}[/itex]

So if I compress the system and increase the partial pressures I'll drive the equilibrium towards more N2O4. BUT if I introduce a bunch of Ne gas to the system and maintain the volume (and increase the total pressure of my container) I will do nothing to the partial pressures and I won't drive the equilibrium in any direction.

What I'm having a hard time understanding is why. I've done some reading here, here, and here, but in each case they explain that this phenomenon occurs but not really why.

In my mental model I imagine the container being compressed and these molecules bumping around each other more and being "unhappy" which causes them to counteract that change by reducing the number of molecules and reducing the pressure. And so something similar should happen when Ne gas is introduced.

Since that doesn't happen, obviously my mental model is wrong/too simplistic. Why am I wrong and what is a better way to think about it?
Or you could, you know, Google it.

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3471
  • Mole Snacks: +526/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Understanding Le Châtelier's Principle
« Reply #1 on: November 23, 2015, 02:11:43 PM »
Fundamentally it's an issue of rates. What does changing the conditions of a reaction do the relative probabilities of a forward or backward reaction at any time? Partial pressures are basically a surrogate for concentrations, and concentrations are (among other things) a way to represent relative reaction probabilities. This is because a concentration (and a partial pressure) is the amount of potentially reacting stuff per unit volume. For a reaction to occur, the two reactants have to collide, and this becomes more probable if there is more of the reacting stuff in a smaller space.

In the case of your reaction A :lequil: 2B, the equilibrium position is determined by the point where the forward reaction equals the reverse reaction. The reverse reaction requires that two molecules of B collide with each other. If you change any condition that increases the likelihood of this happening, you shift the equilibrium to the left (all other things being equal). This means if you reduce the volume or add more B, you shift the equilibrium to the left. Adding a spectator gas will increase the total pressure, but it does nothing to change the likelihood of two B's coming together and creating an A. On the other hand, reducing the volume also increases the total pressure - but, more to the point, it increases the probability of two B's coming together and creating an A. So, adding a spectator gas doesn't shift the equilibrium, but changing the volume does. At least, to a first approximation.

(NB: For the most part, changing volume doesn't really change the rate of the forward reaction because this reaction doesn't depend on a collision. The forward rate does depend on the amount of stuff present, because the probability of a spontaneous decomposition does depend on how much is there (kind of like the probability of drawing the Jack of Spades depends on how many cards you get to draw) but not on the stuff per unit volume.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline mikasaur

  • Full Member
  • ****
  • Posts: 235
  • Mole Snacks: +27/-1
  • Gender: Male
  • Chemist in training
Re: Understanding Le Châtelier's Principle
« Reply #2 on: November 23, 2015, 03:17:24 PM »
I think my need to create a mental model in my head for scientific concepts is usually a good one but sometimes it gets me in trouble. It was good for kinetics and even E&M, not so good for relativity and quantum...

Your explanation makes sense though. The spectator gas would increase the total pressure but would not increase the likelihood of two NO2 colliding because the spectator would just get in the way. My collision-theory-based mental model holds.

Thanks for clarifying about the rate of the decomposition, too. It makes sense that it would be independent of pressure because it doesn't depend on a collision. So really the equilibrium state is determined by the rate of the reverse reaction, as far as pressure is concerned.
Or you could, you know, Google it.

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3471
  • Mole Snacks: +526/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Understanding Le Châtelier's Principle
« Reply #3 on: November 23, 2015, 03:28:14 PM »
Do be aware that this was a generalization. There are plenty of gas-phase reactions the rate of which depends on a spectator gas.  E.g., https://en.wikipedia.org/wiki/Lindemann_mechanism

I don't know the specific mechanism of the reaction you mentioned, but the decomposition of N2O5 apparently involves such a collision-mediated process. So, you have to be careful about what kind of conclusions you make just by inspecting the overall reaction equation.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Sponsored Links