Starting with `250 ml` of `Au^(3+)` solution `250 ml` of `Fe^(2+)` solution, the following equilibrium is established `Au^(3+) (aq) +3Fe^(2+)(aq) hArr 3Fe^(3+) (aq) +Au(s)`
At equilibrium the equivalents of `Au^(3+), Fe^(2+), Fe^(3+)` and `Au` are x,y,z and w respectively. Then:

Options

A. `K_(c) = (Z^(3))/(6xy^(3))`
B. `K_(c) = (3Z^(3))/(2xy^(3))`
C. `K_(c) = (4Z^(3))/(9xy^(3))`
D. `K_(c) = (8Z^(3))/(9 xy^(3))`

Answer

B

Explanation

n factor of `Au^(+3), Fe^(+3), Fe^(+3) & Au` are 3,1,1,& 3 respectively.
Total volume `500 ml = 0.5 lit`
`K_(c) =((Z)^(3)(0.5))/(((x)/(3))((y)/(1))^(3))`
`K_(c) = (3Z^(3))/(2xy^(3))`