# Equilibrium equations

1. May 24, 2007

### ND3G

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: May 24, 2007
2. May 24, 2007

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. May 24, 2007

### ND3G

Sorry, do you mean right, H2O should not be included.

Also, I corrected (I think) all the coefficients.

4. May 24, 2007

yes, H2O should not be present. corrected the coefficients? so you got hte correct expression?

5. May 24, 2007

### ND3G

I believe so

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

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

c) H + 2CrO4 --> OH + Cr2O7

6. May 24, 2007

### ND3G

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