Le Chatelier dilution- shift to which side?

[keikoblue2]

Does diluting a solution cause the reaction to shift to the side with more moles or less moles?

Dilution = decreasing the concentration, so the reaction counteracts the stress by increasing the concentration. Initially, I thought this meant that the reaction would shift to favor the side with more moles.


But TBR says that it shifts to the side with less moles. Why? Is this an error? The only possible explanation that I see is that when you're diluting the solution, both the reactants and products get diluted. But the side with more moles gets diluted less (its concentration is bigger) than the side with less moles. Thus, to increase the concentration, the reaction favors the side with less moles.

Am I correct in justifying it this way or did TBR make an extremely rare mistake?

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[onedirection]

Diluting something shouldn't be influencing anything unless the reaction isn't at equilibrium yet because adding water shouldn't do anything {What were the Keq and Q values as well as the reaction?}

 

[keikoblue2]

Diluting something shouldn't be influencing anything unless the reaction isn't at equilibrium yet because adding water shouldn't do anything {What were the Keq and Q values as well as the reaction?}

Ah sorry, I think they were talking about an aqueous solvent, not pure liquid water. And it was a hypothetical situation (like what happens if you increase pressure? Rxn shifts to the side with less moles).

Here's what TBR wrote:
"...the shifts are similar for a solution-phase equilibrium, except that concentrations are considered, rather than volume changes of the container. Changes in the concentration can result from changes in the volume (quantity ) of solvent. The only effect is that situations #3 (increase external pressure, shift to less moles) and #4 (increase volume, shift to more moles) are now dilution and evaporation of solvent. The system still reacts by asserting the inverse of the stress done upon it. when diluted, it reacts to increase its concentration. When solvent is removed, increasing the concentration, it reacts to reduce the concentration." Thank you!

 

[Romz]

I wouldv'e gotten this wrong aswell.

I would figure that both reactants and products were being diluted at equal amounts.

I guess this counts as a change in concentration which I know is affected by Le Chatelier's principle

 

[onedirection]

Couldn you post the whole question

 

[The_Sunny_Doc]

I just read this, so you made me curious. Looked back at the page and it says, verbatim, "When diluted [equilibrium] shifts to increase its concentration."

When concentrations are too low, equilibrium shifts to the side producing more moles, like it says for "Situation #4 Increase the volume" on pg 181.

And here:

"If the stress is a pressure increase, the reaction will shift in the direction that decreases the number of moles of reactants and products. If the stress is a dilution, then equilibrium shifts to increase the number of moles in the reaction mixture.

These adjustments do not change the numerical value of Keq. If the temperature is raised, however, the reaction will shift in the direction in which heat is absorbed, and will change."

I'm glad you asked about it this. Finally got that concept down!