1
$\begingroup$

I am well aware of Le Chatelier's principle, but what causes molecules to behave in such a way, my guess is that it all has to do with rates of reactions and frequency of collisions.

When at a constant volume, the addition of an inert gas causes no change in the equilibrium, since the concentration of each species will remain constant, I understand that there will be the same amount of molecules per unit volume as before the addition, however wouldn't the rate or probability of collision increase since now each unit of volume has less space in which the molecules can move about due to the addition of the inert gas in that unit of volume, hence if a reaction had more moles in the forward then the reverse the probability of the forward occurring would be greater, causing a forward shift?

$\endgroup$
  • $\begingroup$ The inert gas does not affect the equilibrium if the volume is help constant because it is present on both the reactant and product sides of the equation and cancels out. As the partial pressures are important increasing volume will decrease these and more reaction towards side with greater number of moles. $\endgroup$ – porphyrin Jun 5 '17 at 8:37
2
$\begingroup$

The molecules don't know, they aren't alive in any way for that matter!

Here's an easy way to see it (note that I'm avoiding a mathematical approach to give an intuitive reason):

Consider a hypothetical reaction where $\ce{A}$ and $\ce{B}$ combine and gives $\ce{C}$:

$$\ce{A + B <=> C}$$

Let's say initially, there's an equilibrium. The rate of combination of $\ce{A}$ with $\ce{B}$ is the same as the rate of dissociation of $\ce{C}$. Let's say we added an excess of $\ce{B}$. If there's lots of $\ce{B}$ around, molecules of $\ce{A}$ have more chances of bumping into molecules of $\ce{B}$. This pushes the equilibrium forward, until there's enough $\ce{C}$ to counteract the amount of $\ce{A}$ and $\ce{B}$ present. A similar argument can be given if $\ce{A}$ was in excess than in $\ce{B}$.

You could now think about what happens if we removed some amount of $\ce{C}$. In such a case, the rate dissociation of the limited amount of $\ce{C}$ in the vessel would not balance with the rate of combination of $\ce{A}$ and $\ce{B}$, and thus equilibrium would again flow towards the products.

Now, lets see what happens if you add an inert gas $\ce{D}$ to that mixture. In such a case, $\ce{C}$ will continue to dissociate in it's usual rate, but $\ce{A}$ and $\ce{B}$ cannot combine with each other as fast because they may bump into $\ce{D}$ instead of each other. Thus equilibrium shifts towards the reactants.

I hope this makes sense. Leave a comment if you're not understanding any part of this.

$\endgroup$
  • $\begingroup$ Your answer makes alot of sense ...however I still have trouble understanding equillibrium shifts concerning addition of inert gases $\endgroup$ – LM26 Jun 4 '17 at 19:41
  • $\begingroup$ @LM26 Edited the answer $\endgroup$ – Pritt Balagopal Jun 5 '17 at 3:03
  • $\begingroup$ Well imagine that it is already in a gaseous environment. When you add an inert gas, all you're doing is replacing that 'air' environment with an inert gas. $\endgroup$ – Equinox Jun 5 '17 at 4:51
  • $\begingroup$ In that case, your not removing air are you? Then the $\ce{A}$ and $\ce{B}$ would bump into air and inert gas as well. $\endgroup$ – Pritt Balagopal Jun 5 '17 at 4:54
  • $\begingroup$ Note that adding inert gas does not change equilibrium unless the volume is also changed. $\endgroup$ – porphyrin Jun 5 '17 at 8:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.