How Does Adding Water Affect the Equilibrium in Le Chatelier's Principle?

[i_love_science]

Homework Statement

[Co(H2O)6]2+(aq) + 4Cl−(aq) <-> CoCl42−(aq) + 6H2O(l)

Relevant Equations

equilibrium

I think that adding water here shifts the reaction to the left in favour of reactants (since the amount of product is increased). This will also decrease the concentration of [Co(H2O)6]2+, Cl-, and CoCl42− too (dilution), but I'm not sure how much that would affect the direction that the reaction shifts to. Thanks.

 

[docnet]

Homework Statement:: [Co(H2O)6]2+(aq) + 4Cl−(aq) <-> CoCl42−(aq) + 6H2O(l)
Relevant Equations:: equilibrium

I think that adding water here shifts the reaction to the left in favour of reactants (since the amount of product is increased). This will also decrease the concentration of [Co(H2O)6]2+, Cl-, and CoCl42− too (dilution), but I'm not sure how much that would affect the direction that the reaction shifts to. Thanks.

increasing the product concentration and decreasing the reactant concentration have the same effect of shifting the equilibrium to the left, so you are finished.

 

[i_love_science]

increasing the product concentration and decreasing the reactant concentration have the same effect of shifting the equilibrium to the left, so you are finished.

But the concentration of one of the products (CoCl₄^2-) also decreases. So does this not change the combined effect of decrease in all reactant concentration and increase in amount of one product?

 

[docnet]

After looking at the reaction more carefully, I actually think you should not look at H₂O on the right side as a product. In this reaction, H₂O is the solvent and we assume there is plenty of it to start with. This is a "precipitation" reaction where it makes sense for the equilibrium to shift to the right when adding more water (solvent) as you just explained.

https://chemed.chem.purdue.edu/genchem/demosheets/12.10.html

 

[Borek]

LeChatelier's principle is quite limited and its predictions are sometimes problematic. In this case I would use reaction quotient to see how it changes with the dilution.

This is actually quite simple. Assume the reaction was at equilibrium, so the Q=K, add water to double the volume - concentrations of all substances but water decrease twice. What is the new Q? If it is lower than K, reaction will proceed to the right, if it is higher than K, reaction will proceed to the left.

 

[mjc123]

OP, try to use the subscript and superscript functions when writing formulae (in the 3 dots menu above the reply box). There is no such thing as CoCl₄₂^- or CoCl^42-, but it sure looks as if you meant one of them!

 

[i_love_science]

Thanks.
I want to know if this is a fixed rule. When water is a product, does it always shift the reaction to left, or is this variable (or should I just compare Q and K every time)?

 

[docnet]

Thanks.
I want to know if this is a fixed rule. When water is a product, does it always shift the reaction to left, or is this variable (or should I just compare Q and K every time)?

I would just compare $Q$ and $K$ every time, in that process you are required to be more exact and mathematical, and in the process you have to decide whether water is a product or not.

That's an interesting question though, because water tends to be a very stable molecule and a very good solvent. As such, water is usually the solvent in many chemical reactions, and not the reactant or product. Where water is a product of a reaction, it is often part of irreversible and highly exothermic gaseous reactions. For example, a reaction of hydrogen and oxygen
$$O_{2(g)}+2H_{2(g)}=2H_{2}O_{(l)}$$
is one of many many reactions that highly favor the product side. All this reaction requires is a spark, and a radical reaction propagates until essentially all of one or both of the gases are depleted. The reverse reaction does not occur without a significant source of energy to break the $H-O$ bonds. i tried to find counterexamples.

 

[Borek]

In this case water is both product and solvent so changes in its amount change both equilibrium and concentration of substances involved (actually the only thing that doesn't change is the water concentration ).