Equilibrium of H, OH, CrO4 and Cr2O7

In summary, the equilibrium between K2CrO4 and K2CR2O7 is not greatly affected by the addition of Ca(OH)2. The Cr2O7 becomes pale due to less available to react with.
  • #1
ND3G
79
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:
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  • #2
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.
 
  • #3
Sorry, do you mean right, H2O should not be included.

Also, I corrected (I think) all the coefficients.
 
  • #4
yes, H2O should not be present. corrected the coefficients? so you got hte correct expression?
 
  • #5
I believe so

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

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

c) H + 2CrO4 --> OH + Cr2O7
 
  • #6
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:

1. What is the equilibrium constant for the reaction of H2O with OH-?

The equilibrium constant (Kw) for the reaction of water with hydroxide ions is equal to 1.0 x 10-14 at 25°C. This value is a constant and remains the same regardless of the concentration of H2O and OH-.

2. How does the presence of CrO4 and Cr2O7 affect the equilibrium of H2O and OH-?

The presence of CrO4 and Cr2O7 ions in a solution can shift the equilibrium of H2O and OH- to the left, decreasing the concentration of hydroxide ions and increasing the concentration of water molecules. This is because these ions can react with H2O and consume hydroxide ions, reducing their concentration and shifting the equilibrium to the left.

3. How does temperature affect the equilibrium constant of H2O and OH-?

The equilibrium constant (Kw) for the reaction of water with hydroxide ions is temperature-dependent. As the temperature increases, the value of Kw also increases. This is because the reaction is endothermic, meaning it requires energy to proceed. As the temperature rises, more energy is available, leading to an increase in the concentration of products (H2O and OH-) and a higher value of Kw.

4. How does the presence of a catalyst affect the equilibrium of H2O and OH-?

A catalyst does not affect the equilibrium of H2O and OH- as it only speeds up the rate of the reaction in both the forward and reverse directions. The concentrations of H2O and OH- at equilibrium will remain the same with or without the presence of a catalyst.

5. How can Le Chatelier's principle be applied to the equilibrium of H2O and OH-?

According to Le Chatelier's principle, a system at equilibrium will respond to any changes imposed on it in a way that tends to minimize the effect of the change. In the case of the equilibrium of H2O and OH-, if the concentration of H2O or OH- is increased, the system will shift to the opposite side to consume the excess ions and maintain equilibrium. Similarly, if the temperature is increased, the system will shift to the side with fewer molecules to consume the excess energy and maintain equilibrium.

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