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 will follow these steps:

### Step 1: Write the expression for the equilibrium constant
The equilibrium constant \( K_{eq} \) for the reaction is given by the formula:

\[ K_{eq} = \frac{[C]}{[A][B]^2} \]

where:
- \([C]\) is the concentration of the product \( C \),
- \([A]\) is the concentration of reactant \( A \),
- \([B]\) is the concentration of reactant \( B \).

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

Now, substituting these values into the \( K_{eq} \) expression:

\[ K_{eq} = \frac{0.216}{(0.06)(0.12)^2} \]

### Step 3: Calculate the denominator
First, calculate \((0.12)^2\):

\[ (0.12)^2 = 0.0144 \]

Now, calculate the denominator:

\[ (0.06)(0.0144) = 0.000864 \]

### Step 4: Calculate \( K_{eq} \)
Now, substituting back into the \( K_{eq} \) expression:

\[ K_{eq} = \frac{0.216}{0.000864} \]

Now, perform the division:

\[ K_{eq} = 250 \]

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

\[ K_{eq} = 250 \]