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 equilibrium cosntant `K_(C)` for the reaction is

Options

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

Answer

To find the equilibrium constant \( K_c \) for the reaction \( A(g) + 2B(g) \rightleftharpoons C(g) \), we can follow these steps:

### Step 1: Write the expression for \( K_c \)

The equilibrium constant \( K_c \) for the reaction is given by the formula:

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

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

### Step 2: Substitute the 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 \( K_c \) expression:

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

### Step 3: Calculate \( K_c \)

First, calculate \((0.12)^2\):

\[ (0.12)^2 = 0.0144 \]

Now, substitute this back into the equation:

\[ K_c = \frac{0.216}{(0.06)(0.0144)} \]

Next, calculate the denominator:

\[ 0.06 \times 0.0144 = 0.000864 \]

Now, substitute this value into the equation for \( K_c \):

\[ K_c = \frac{0.216}{0.000864} \]

Finally, perform the division:

\[ K_c = 250 \]

### Conclusion

The equilibrium constant \( K_c \) for the reaction is \( 250 \).

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