`A(g) + 3B(g) rarr 4C(g)`Initially concentration of A is equal to that of B. The equilibrium concentrations of A and C are equal. Kc is :

Options

A. 0.08
B. 8
C. 80
D. `1//8`

Explanation

To solve the problem step by step, we will analyze the given reaction and the information provided.

### Step 1: Write the balanced chemical equation.
The reaction is given as:
\[ A(g) + 3B(g) \rightarrow 4C(g) \]

### Step 2: Set up initial concentrations.
Let the initial concentration of A and B be equal to \( A_0 \). Thus, we can denote:
- \([A]_0 = A_0\)
- \([B]_0 = A_0\)
- \([C]_0 = 0\)

### Step 3: Define changes in concentration at equilibrium.
Let \( x \) be the change in concentration of A that reacts at equilibrium. According to the stoichiometry of the reaction:
- For A: \([A] = A_0 - x\)
- For B: \([B] = A_0 - 3x\)
- For C: \([C] = 4x\)

### Step 4: Use the information that equilibrium concentrations of A and C are equal.
We know that at equilibrium, the concentration of A is equal to the concentration of C:
\[ A_0 - x = 4x \]

### Step 5: Solve for \( x \).
Rearranging the equation gives:
\[ A_0 = 5x \]
Thus, we can express \( x \) as:
\[ x = \frac{A_0}{5} \]

### Step 6: Substitute \( x \) back into the equilibrium concentrations.
Now substitute \( x \) back into the expressions for the equilibrium concentrations:
- \([A] = A_0 - x = A_0 - \frac{A_0}{5} = \frac{4A_0}{5}\)
- \([B] = A_0 - 3x = A_0 - 3\left(\frac{A_0}{5}\right) = \frac{2A_0}{5}\)
- \([C] = 4x = 4\left(\frac{A_0}{5}\right) = \frac{4A_0}{5}\)

### Step 7: Write the expression for \( K_c \).
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[C]^4}{[A][B]^3} \]

### Step 8: Substitute the equilibrium concentrations into the \( K_c \) expression.
Substituting the values we found:
\[ K_c = \frac{\left(\frac{4A_0}{5}\right)^4}{\left(\frac{4A_0}{5}\right)\left(\frac{2A_0}{5}\right)^3} \]

### Step 9: Simplify the expression.
Calculating the numerator:
\[ \left(\frac{4A_0}{5}\right)^4 = \frac{256A_0^4}{625} \]

Calculating the denominator:
\[ \left(\frac{4A_0}{5}\right)\left(\frac{2A_0}{5}\right)^3 = \left(\frac{4A_0}{5}\right)\left(\frac{8A_0^3}{125}\right) = \frac{32A_0^4}{625} \]

### Step 10: Final calculation for \( K_c \).
Now substituting back into the \( K_c \) expression:
\[ K_c = \frac{\frac{256A_0^4}{625}}{\frac{32A_0^4}{625}} = \frac{256}{32} = 8 \]

### Conclusion:
Thus, the equilibrium constant \( K_c \) is:
\[ K_c = 8 \]