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))`

Explanation

To solve the problem step-by-step, we need to analyze the given equilibrium reaction and the concentrations of the species involved.

### Step 1: Write the equilibrium reaction
The equilibrium reaction provided is:
\[ \text{Au}^{3+} (aq) + 3\text{Fe}^{2+} (aq) \rightleftharpoons 3\text{Fe}^{3+} (aq) + \text{Au} (s) \]

### Step 2: Determine the initial concentrations
We start with:
- 250 mL of \(\text{Au}^{3+}\) solution
- 250 mL of \(\text{Fe}^{2+}\) solution

The total volume of the solution after mixing is:
\[ V = 250 \, \text{mL} + 250 \, \text{mL} = 500 \, \text{mL} = 0.5 \, \text{L} \]

### Step 3: Define the initial concentrations
Assuming the initial concentrations of \(\text{Au}^{3+}\) and \(\text{Fe}^{2+}\) are \(C_{Au}\) and \(C_{Fe}\) respectively, the initial moles can be expressed as:
- Moles of \(\text{Au}^{3+}\) = \(C_{Au} \times 0.25\) (since 250 mL = 0.25 L)
- Moles of \(\text{Fe}^{2+}\) = \(C_{Fe} \times 0.25\)

### Step 4: Define changes at equilibrium
Let \(x\) be the change in moles of \(\text{Au}^{3+}\) that reacts at equilibrium. According to the stoichiometry of the reaction:
- \(\text{Au}^{3+}\) decreases by \(x\)
- \(\text{Fe}^{2+}\) decreases by \(3x\)
- \(\text{Fe}^{3+}\) increases by \(3x\)
- \(\text{Au}\) forms \(x\)

At equilibrium, we have:
- \(\text{Au}^{3+}\) = \(C_{Au} \times 0.25 - x\)
- \(\text{Fe}^{2+}\) = \(C_{Fe} \times 0.25 - 3x\)
- \(\text{Fe}^{3+}\) = \(3x\)
- \(\text{Au}\) = \(x\)

### Step 5: Express concentrations at equilibrium
The concentrations at equilibrium are:
- \([\text{Au}^{3+}] = \frac{C_{Au} \times 0.25 - x}{0.5}\)
- \([\text{Fe}^{2+}] = \frac{C_{Fe} \times 0.25 - 3x}{0.5}\)
- \([\text{Fe}^{3+}] = \frac{3x}{0.5}\)

### Step 6: Write the expression for the equilibrium constant \(K_c\)
The equilibrium constant \(K_c\) for the reaction is given by:
\[ K_c = \frac{[\text{Fe}^{3+}]^3 \times [\text{Au}]}{[\text{Au}^{3+}] \times [\text{Fe}^{2+}]^3} \]

Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{3x}{0.5}\right)^3 \times x}{\left(\frac{C_{Au} \times 0.25 - x}{0.5}\right) \times \left(\frac{C_{Fe} \times 0.25 - 3x}{0.5}\right)^3} \]

### Step 7: Simplify the expression
This simplifies to:
\[ K_c = \frac{27x^4}{(C_{Au} \times 0.25 - x)(C_{Fe} \times 0.25 - 3x)^3} \cdot 8 \]
\[ K_c = \frac{216x^4}{(C_{Au} \times 0.25 - x)(C_{Fe} \times 0.25 - 3x)^3} \]

### Conclusion
The final expression for the equilibrium constant \(K_c\) is derived, and we can analyze it based on the values of \(x\), \(C_{Au}\), and \(C_{Fe}\).