The following equilibrium exists in a closed vessel in `1L` capacity `A(g)+3B(g)hArr4C(g)`
initial cocentration of `A(g)` is equal to that `B(g)`. The equilibrium concentration of `A(g)` and `C(g)` are equal. `K_(c)` for the reaction is

Options

A. `0.08`
B. `0.8`
C. 8
D. 80

Answer

To solve the problem, we need to analyze the equilibrium reaction given:

\[ A(g) + 3B(g) \rightleftharpoons 4C(g) \]

### Step 1: Define Initial Concentrations
Let the initial concentration of both \( A \) and \( B \) be \( 1 \, \text{mol/L} \). Therefore, we have:
- \([A]_0 = 1 \, \text{mol/L}\)
- \([B]_0 = 1 \, \text{mol/L}\)
- \([C]_0 = 0 \, \text{mol/L}\)

### Step 2: Set Up the Change in Concentration
Let \( x \) be the amount of \( A \) that reacts at equilibrium. According to the stoichiometry of the reaction:
- \( A \) decreases by \( x \)
- \( B \) decreases by \( 3x \)
- \( C \) increases by \( 4x \)

### Step 3: Write Equilibrium Concentrations
At equilibrium, the concentrations will be:
- \([A]_{eq} = 1 - x\)
- \([B]_{eq} = 1 - 3x\)
- \([C]_{eq} = 4x\)

### Step 4: Use Given Information
It is given that the equilibrium concentrations of \( A \) and \( C \) are equal:
\[ [A]_{eq} = [C]_{eq} \]
Substituting the equilibrium concentrations:
\[ 1 - x = 4x \]

### Step 5: Solve for \( x \)
Rearranging the equation:
\[ 1 = 4x + x \]
\[ 1 = 5x \]
\[ x = \frac{1}{5} \]

### Step 6: Calculate Equilibrium Concentrations
Now substitute \( x \) back to find the equilibrium concentrations:
- \([A]_{eq} = 1 - \frac{1}{5} = \frac{4}{5} \, \text{mol/L}\)
- \([B]_{eq} = 1 - 3 \times \frac{1}{5} = 1 - \frac{3}{5} = \frac{2}{5} \, \text{mol/L}\)
- \([C]_{eq} = 4 \times \frac{1}{5} = \frac{4}{5} \, \text{mol/L}\)

### Step 7: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C]_{eq}^4}{[A]_{eq}^1 [B]_{eq}^3} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{4}{5}\right)^4}{\left(\frac{4}{5}\right)^1 \left(\frac{2}{5}\right)^3} \]

### Step 8: Simplify the Expression
Calculating the numerator and denominator:
- Numerator: \(\left(\frac{4}{5}\right)^4 = \frac{256}{625}\)
- Denominator: \(\left(\frac{4}{5}\right) \left(\frac{2}{5}\right)^3 = \left(\frac{4}{5}\right) \left(\frac{8}{125}\right) = \frac{32}{625}\)

Now substituting back:
\[ K_c = \frac{\frac{256}{625}}{\frac{32}{625}} = \frac{256}{32} = 8 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is:
\[ \boxed{8} \]