Equilibrium constant question?

Question

If K=47.6 for 4AB -> 2A2 + 4B, then for A2 + 2B -> 2AB, K=
a. 47.6
b. -0.145
c. 6.90
d. 2.63.
e. 0.145
f. 3.24

Now, the second equation is half of the first one and reversed. So I have taken half of 47.6 and obtained it's inverse which=.0420. I've done several other methods and I don't get any answer from above. Can anyone offer me some insight? Thanks.

Answer 1

Let us write an equilbrium constant for the first reaction.

4AB -> 2A2 + 4B

K1 = ([A2]^2)* [B^4]/([AB]^4)

Let us write an equilbrium constant for the second reaction

A2 + 2B -> 2AB

K2 = [AB^2]/[ A2]*[B^2]

we can inverse K2 to yield

1/K2 = [ A2]*[B^2]/[AB^2]

We can square it

1/(K2^2) = [ A2]^2*[B^4]/[AB^4]

The RHS is equal to K1

K1 = 1/(K2^2)

Therefore

K2^2 = 1/K1

K2 = sqrt(1/K1)

K2 = sqrt(1/47.6) = 0.145

The answer is e.

Answer 2

As mentioned, the equilibrium constant for the second equation would be the square root of the inverse of the first equation.

Answer 3

it is K"= squareroot(1/47.6) =0.145

K= [A2]^2*[B]^4 / [AB]^4 = ([A2][B]^2 / [AB]^2 )^2 = (1/K")^2 =>
K" =squareroot (1/K)

It helps if you write down the K expressions for the two reactions:

K= [A2]^2*[B]^4 / [AB]^4 for the first and
K"= [AB]^2 / [A2][B]^2