Quick question about equilibrium of a reaction

  • Thread starter mjolnir80
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  • #1
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Homework Statement


we have a solution of
Cocl[tex]^{}2-_{}4[/tex](alcohol) + 6H[tex]_{}2[/tex]O(l) <---> Co(H[tex]_{}2+[/tex]O)[tex]_{}6^{}2+[/tex](aq) + 4Cl-(aq)
at equilibrium
what would happen if we were to add more H2O?






The Attempt at a Solution


we cant really change the concentration of the water so nothing would happen?

edit: sorry abot the bad reaction formula
 

Answers and Replies

  • #2
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this is an equilibrium, what happens when one side of the equilibrium is disturbed? what chemical principle is this related to?
 
  • #3
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its related to le chatelires principle but im not sure whether the water has any effect at all because we cant really change the concentration of the water
 
  • #4
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any reactant or solvent in a chemical reaction has a concentration, even water. So what will the equilibrium do if more water is added?
 
Last edited:
  • #5
Borek
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It is hard to tell from your reaction, but isn't it done in alcohol, in which case water concentration can be easily controlled?
 
  • #6
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likely the CoCl2 4- was prepared with an alcohol content in its solution to maintain this species and not revert to the water product.


Homework Statement


we have a solution of
Cocl[tex]^{}2-_{}4[/tex](alcohol) + 6H[tex]_{}2[/tex]O(l) <---> Co(H[tex]_{}2+[/tex]O)[tex]_{}6^{}2+[/tex](aq) + 4Cl-(aq)
at equilibrium
what would happen if we were to add more H2O?
water is a reactant here.
 
  • #7
chemisttree
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Homework Statement


we have a solution of
CoCl2-4(alcohol) + 6H2O(l) <---> Co(H2O)6+2(aq) + 4Cl-(aq)
at equilibrium
what would happen if we were to add more H2O?
Here is a cleaned up version of your equation. Hit 'quote' to see how I did it.

I think it should be:

CoCl4-2(alcoholic) + 6H2O(l) <---> Co(H2O)6+2(aq) + 4Cl-(alcoholic)
 
Last edited:

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