Equilibrium of H, OH, CrO4 and Cr2O7

  • Thread starter Thread starter ND3G
  • Start date Start date
  • Tags Tags
    Equilibrium
AI Thread Summary
The discussion focuses on the equilibrium reactions involving H, OH, CrO4, and Cr2O7. The balanced equations for the reactions are confirmed, with the correct equilibrium expressions derived for each reaction. It is clarified that water (H2O) should not be included in the equilibrium constant expressions. A follow-up question addresses the effect of adding Ca(OH)2 to solutions of K2CrO4 and K2Cr2O7, with the explanation that the solubility of Ca(OH)2 and its interaction with water limits its impact on the equilibrium. The change in color of Cr2O7 to pale is noted, indicating a shift in equilibrium, but it is emphasized that the overall equilibrium remains largely unaffected.
ND3G
Messages
79
Reaction score
0
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:
Physics news on Phys.org
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.
 
Last edited:
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top