For the system `A(g)+2B(g) hArr C(g)` the equilibrium concentration is
`A=0.06 mol L^(-1), B=0.12 mol L^(-1)`
`C=0.216 mol L^(-1)` The `K_(eq)` for the reaction is

Options

A. `250`
B. `416`
C. `4xx10^(-3)`
D. `125`

Explanation

To find the equilibrium constant \( K_{eq} \) for the reaction \( A(g) + 2B(g) \rightleftharpoons C(g) \), we can use the equilibrium concentrations provided.

### Step-by-Step Solution:

1. **Write the expression for the equilibrium constant \( K_{eq} \)**:
The equilibrium constant expression for the reaction is given by:
\[ K_{eq} = \frac{[C]}{[A][B]^2} \]
where \([C]\), \([A]\), and \([B]\) are the equilibrium concentrations of the respective species.

2. **Substitute the given equilibrium concentrations**:
From the problem, we have:
- \([A] = 0.06 \, \text{mol L}^{-1}\)
- \([B] = 0.12 \, \text{mol L}^{-1}\)
- \([C] = 0.216 \, \text{mol L}^{-1}\)

Plugging these values into the \( K_{eq} \) expression:
\[ K_{eq} = \frac{0.216}{(0.06)(0.12)^2} \]

3. **Calculate \((0.12)^2\)**:
First, calculate \((0.12)^2\):
\[ (0.12)^2 = 0.0144 \]

4. **Calculate the denominator**:
Now calculate the denominator:
\[ [A][B]^2 = 0.06 \times 0.0144 = 0.000864 \]

5. **Calculate \( K_{eq} \)**:
Now substitute the values into the \( K_{eq} \) expression:
\[ K_{eq} = \frac{0.216}{0.000864} \]

Performing the division:
\[ K_{eq} = 250 \]

### Final Answer:
Thus, the equilibrium constant \( K_{eq} \) for the reaction is:
\[ K_{eq} = 250 \]