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

Question

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

Answer 1

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