Calculate the Keq of the reversible reactions Cr2O7 + 2OH -> CrO4 + H2O and 2CrO42->Cr2O72 , given the following data:

HCrO4 -> H + CrO42
K = 3.2 x 10^-7
2HCrO4 -> Cr2O72 + H2O
K = 34
H2O -> H+ + OH-
K = 1 x 10^-14

HCL and NaOH = 1M
KCrO4 and KCr2O7 = 0.1M

The "reversible" reactions you write are not balanced. You should place a 2 for CrO4^2- in the first one.

For the bottom row of three equations, I will call them eqn 1, eqn 2 and eqn 3.
Reverse eqn 2.
Add to 2x the reverse of eqn 3
Add to 2x eqn 1.

When you reverse an equation, take the reciprocal of k. When you multiply a rxn, the coefficient becomes the exponent for k. For example, if K is 10 then the reverse is 1/10 and if
A + B = C K = 10 then
C = A + B K is 1/10
and 2A + 2B = 2C, K is 10^2
and 2C = 2A + 2B, K is 1/10^2
When you add equations you multiply k values.
A + B = C k1 = 10
D+ E = F k2 = 20
Then A + B + D + E = C + F K is k1k2

To calculate the Keq of the reversible reactions, we need to use the given data and apply the principles of equilibrium.

1. First, let's write the balanced equations for the reactions:
a) Cr2O7 + 2OH- -> CrO4 + H2O
b) 2CrO4^2- -> Cr2O7^2-

2. Next, we'll determine the expressions for each equilibrium constant (Keq) based on the balanced equations:
a) Keq1 = [CrO4][H2O] / [Cr2O7][OH-]^2
b) Keq2 = [Cr2O7^2-] / [CrO4^2-]^2

3. Now, let's find the concentrations of the species involved in each reaction:

From the given information:
- HCrO4 dissociates into H+ and CrO4^2- with a Keq = 3.2 x 10^-7
- 2HCrO4 dissociates into Cr2O7^2- and H2O with a Keq = 34
- H2O dissociates into H+ and OH- with a Keq = 1 x 10^-14

Let's assume that x mol/L of HCrO4 dissociates:
- HCrO4 -> H+ + CrO4^2- (Keq = 3.2 x 10^-7)
[HCrO4] = 0.1 - x
[H+] = x mol/L
[CrO4^2-] = x mol/L

So, [HCrO4] = 0.1 - x

- 2HCrO4 -> Cr2O7^2- + H2O (Keq = 34)
[HCrO4] = 0.1 - x
[Cr2O7^2-] = x mol/L
[H2O] = x mol/L

So, [HCrO4] = (0.1 - x)/2
[H2O] = x/2

- H2O -> H+ + OH- (Keq = 1 x 10^-14)
[H2O] = x mol/L
[H+] = x mol/L
[OH-] = x mol/L

4. Substituting the concentrations into the Keq expressions:

a) Keq1 = [CrO4][H2O] / [Cr2O7][OH-]^2
Keq1 = (x)(x/2) / (x) / (x)^2
Keq1 = (x^2 / 2) / (x) / (x^2)
Keq1 = 1 / 2x

b) Keq2 = [Cr2O7^2-] / [CrO4^2-]^2
Keq2 = (x) / (x)^2
Keq2 = 1 / x

5. Finally, let's calculate the values of Keq1 and Keq2:

Keq1 = 1 / 2(0.1 - x)
Keq2 = 1 / x

As we do not have a specific value for x, we cannot calculate the exact values of Keq1 and Keq2. However, the equations above provide the general expressions for Keq1 and Keq2 based on the given data and assumptions.

To calculate the equilibrium constant (Keq) for the given reversible reactions, we need to use the information provided and apply the principles of equilibrium.

Here are the steps to calculate the Keq for each reaction:

1. Write the balanced equations for the reactions:
a) Cr2O7 + 2OH- -> CrO4 + H2O
b) 2CrO4^2- -> Cr2O7^2-

2. Determine the Keq expression for each reaction based on the balanced equations.
a) For the reaction Cr2O7 + 2OH- -> CrO4 + H2O:
Keq = [CrO4][H2O] / [Cr2O7][OH-]^2

b) For the reaction 2CrO4^2- -> Cr2O7^2-:
Keq = [Cr2O7]^2 / [CrO4]^2

3. Identify the concentrations required to calculate Keq.

For the reaction Cr2O7 + 2OH- -> CrO4 + H2O:
- We need the concentrations of Cr2O7, OH-, CrO4, and H2O.

For the reaction 2CrO4^2- -> Cr2O7^2-:
- We need the concentrations of CrO4 and Cr2O7.

4. Using the given data, find the concentrations of the species needed for each reaction.

- Given: HCrO4 -> H+ + CrO4^2-
- K = 3.2 x 10^-7
- Let x be the concentration of H+
- At equilibrium, the concentration of CrO4^2- is also x.
- Therefore, [HCrO4] = [H+] + [CrO4^2-] = x + x = 2x
- So, [HCrO4] = 2x

- Given: 2HCrO4 -> Cr2O7^2- + H2O
- K = 34
- Let y be the concentration of Cr2O7^2- and H2O.
- At equilibrium, the concentrations of HCrO4 and H2O are also y.
- Therefore, [2HCrO4] = [Cr2O7^2-] + [H2O] = y + y = 2y
- So, [2HCrO4] = 2y

- Given: H2O -> H+ + OH-
- K = 1 x 10^-14
- Let z be the concentration of H2O, H+, and OH-.
- At equilibrium, [H+] = [OH-] = z
- Therefore, [H2O] = z

- Given: HCl and NaOH = 1M
- The concentrations of H+ and OH- are determined by the strong acid-strong base reaction.
- Since HCl and NaOH fully dissociate, [H+] = [OH-] = 1M

- Given: KCrO4 and KCr2O7 = 0.1M
- The initial concentrations of CrO4 and Cr2O7 are given as 0.1M.

5. Substitute the concentrations into the Keq expressions and calculate Keq for each reaction.

- For the reaction Cr2O7 + 2OH- -> CrO4 + H2O:
Keq = ([CrO4][H2O]) / ([Cr2O7][OH-]^2)

Substituting the concentrations:
Keq = ([0.1M][z]) / ([0.1M][1M]^2)
Keq = z / 10

- For the reaction 2CrO4^2- -> Cr2O7^2-:
Keq = ([Cr2O7]^2) / ([CrO4]^2)

Substituting the concentrations:
Keq = ([y]^2) / ([0.1M]^2)
Keq = y^2 / 0.01

Therefore, the Keq for the given reactions are:
- Keq(Cr2O7 + 2OH- -> CrO4 + H2O) = z / 10
- Keq(2CrO4^2- -> Cr2O7^2-) = y^2 / 0.01

Please note that the final values for Keq can be obtained by substituting the appropriate concentration values for z and y, which can be calculated depending on the actual concentrations of H2O, H+, OH-, Cr2O7, CrO4, and other factors.