Equilibrium of H, OH, CrO4 and Cr2O7

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Discussion Overview

The discussion focuses on the equilibrium reactions involving H, OH, CrO4, and Cr2O7, including the derivation of equilibrium expressions and the effects of adding Ca(OH)2 to solutions of K2CrO4 and K2Cr2O7. Participants explore the correct formulation of balanced equations and the inclusion of substances in equilibrium expressions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Experimental/applied

Main Points Raised

  • Participants discuss the equilibrium expressions for the reactions between H and CrO4 to produce Cr2O7 and H2O, and between OH and Cr2O7 to reproduce CrO4 and H2O.
  • There is a question regarding whether to include H2O in the equilibrium expressions, with some participants asserting that it should not be included.
  • One participant expresses uncertainty about the correctness of their balanced equations and equilibrium expressions, prompting others to suggest double-checking coefficients.
  • A follow-up question is raised about the effect of adding Ca(OH)2 to the solutions, with a participant proposing that the solubility of Ca(OH)2 and its reaction with water affects the equilibrium.
  • Participants discuss the observation that CrO4 remains yellow while Cr2O7 becomes very pale, with one participant seeking clarification on the implications of this observation.

Areas of Agreement / Disagreement

Participants generally agree that H2O should not be included in the equilibrium expressions. However, there is uncertainty regarding the balanced equations and the implications of the color change in Cr2O7, indicating that the discussion remains unresolved in these areas.

Contextual Notes

There are limitations in the discussion regarding the accuracy of the balanced equations and the assumptions made about the effects of Ca(OH)2 on the equilibrium, which remain unresolved.

ND3G
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Write the equations for the equilibrium, showing

a) the reaction between H and CrO4 to produce Cr2O7 and H2O

b) the reaction between OH and Cr2O7 to reproduce CrO4 and H2O

c) a summation equation showing both H and OH in the equilibrium reaction between CrO4 and Cr2O7


First I found the balanced equations and then wrote:

a) k = [Cr2O7] / [H]^2[CrO4]^2

*I rememeber reading somewhere that you do not include either solids or liquids in the equation. Is that correct or should I add [H2O]*

b) k = [CrO4]^2 / [OH]^2[Cr2O7]

c) k = [Cr2O7][OH] / [H][CrO4]^2

Am I on the right track here or playing a different sport?
 
Last edited:
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ND3G said:

*I rememeber reading somewhere that you do not include either solids or liquids in the equation. Is that correct or should I add [H2O] to the denominator?*

right.

in the reaction cC + dD <=> aA + bB :

k = ([A]^a * ^b )/ ([C]^c * [D]^d)

but i think you should double check your balanced equations to make sure that all your coefficients are correct. if they're flawed, your equilibrium expression will be flawed as well.
 
Sorry, do you mean right, H2O should not be included.

Also, I corrected (I think) all the coefficients.
 
yes, H2O should not be present. corrected the coefficients? so you got hte correct expression?
 
I believe so

a) 2H + 2CrO4 --> Cr2O7 +H2O

b) 2OH + Cr2O7 --> 2CrO4 + H2O

c) H + 2CrO4 --> OH + Cr2O7
 
I have a follow up question:

Pour a solution of K2CrO4 and K2CR2O7 into separate test tubes. Add Ca(OH)2 to the solution.

Results: CrO4 stays yellow, Cr2O7 becomes very pale

Explain why Ca(OH)2 has little effect on the equilibrium


I answered:

Ca(OH)2 is moderately soluble in water. Heat is liberated in this dissolving process and the systems tendency towards minimum energy favours the solution of one in the other. As the Ca(OH)2 is reacting with the water present less is available to react with the CrO4 or Cr2O7.

Does that sound ok. Also, I am unsure why the Cr2O7 becomes pale and what that means. Apparently, it has little to do with Cr2O7 as the equilibrium is not greatly affected.
 
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