aqueous dilution LCP collision theory explanation (1 Viewer)

[indeed]

hi all,
i'm a bit confused on the collision theory reasoning behind dilution effect in regards to Le Chatelier's principle:
let's say for this e.g. (4:1 aq molar ratio)

initially all species conc decreases (becaused diluted), then the side with more aqueous moles is favoured. I don't get why the side with more aq moles is favoured (if someone could please explain using collision theory)

 

[synthesisFR]

I’ve never fully understood eqm in relation to water completely myself tbh

Is it just like here water is an actual product, and reactant of the reverse reaction, so increasing water will increase the reverse rate due to more product collisions…

But yeah I think there was also something like this in esterification where u include water in the eqm expression and stuff it’s so annoying

 

[indeed]

I’ve never fully understood eqm in relation to water completely myself tbh

Is it just like here water is an actual product, and reactant of the reverse reaction, so increasing water will increase the reverse rate due to more product collisions…

But yeah I think there was also something like this in esterification where u include water in the eqm expression and stuff it’s so annoying

Oh ok yeah fair enough, thanks anyways

 

[indeed]

I had a breakthrough and worked it out (kinda):

 

[carrotsss]

You can explain this with Keq. Keq must remain constant throughout this ordeal, since the temperature is the same. However, if you halve all of the concentrations in this equation, you’ll find that Q is 4 times greater than Keq, and hence the reaction will shift towards the reactants in order to restore equilibrium.

 

[wizzkids]

It's not about the water! The water is a spectator.
The concentration of the water is practically constant during this equilibrium, so it can't be caused by any change in the water.
It is caused by the fact that equilibrium occurs when the rates of the forward and reverse reactions are equal.
There are twice as many moles of chromate CrO₄^2- as dichromate Cr₂O₇^2- ions on the left and right hand sides.
When you change the dilution, let's say you suddenly dump more water into the system so that all the species suddenly are halved in concentration, according to collision theory the rates of the forward and reverse reactions will both decrease (because the frequency of collisions goes down) but not by the same amount. The reaction direction towards CrO₄^2- will not decrease as much.
If this was a reaction between equal numbers of moles on the left and right hand sides, you would not see the same effect.