For the reaction `3AhArr C+2D`, initially A was taken. At equilibrium the concentration of D is twice that of A. The value of `K_(c)` is

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The given reaction is:
\[ 3A \rightleftharpoons C + 2D \]

### Step 2: Set up the initial concentrations
Let the initial concentration of A be \( [A]_0 \). At the start, the concentrations of C and D are both 0:
- \( [A] = [A]_0 \)
- \( [C] = 0 \)
- \( [D] = 0 \)

### Step 3: Define the change in concentrations at equilibrium
Let \( x \) be the amount of A that reacts at equilibrium. Then, the changes in concentrations will be:
- \( [A] = [A]_0 - 3x \)
- \( [C] = x \)
- \( [D] = 2x \)

### Step 4: Use the information given in the problem
According to the problem, at equilibrium, the concentration of D is twice that of A:
\[ [D] = 2[A] \]

Substituting the expressions for D and A:
\[ 2x = 2([A]_0 - 3x) \]

### Step 5: Solve for \( x \)
Expanding the equation:
\[ 2x = 2[A]_0 - 6x \]

Rearranging gives:
\[ 2x + 6x = 2[A]_0 \]
\[ 8x = 2[A]_0 \]
\[ x = \frac{[A]_0}{4} \]

### Step 6: Find the equilibrium concentrations
Now substituting \( x \) back into the expressions for the concentrations:
- \( [A] = [A]_0 - 3x = [A]_0 - 3\left(\frac{[A]_0}{4}\right) = [A]_0 - \frac{3[A]_0}{4} = \frac{[A]_0}{4} \)
- \( [C] = x = \frac{[A]_0}{4} \)
- \( [D] = 2x = 2\left(\frac{[A]_0}{4}\right) = \frac{[A]_0}{2} \)

### Step 7: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C][D]^2}{[A]^3} \]

Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{[A]_0}{4}\right)\left(\frac{[A]_0}{2}\right)^2}{\left(\frac{[A]_0}{4}\right)^3} \]

### Step 8: Simplify the expression
Calculating the numerator:
\[ [C][D]^2 = \left(\frac{[A]_0}{4}\right)\left(\frac{[A]_0^2}{4}\right) = \frac{[A]_0^3}{16} \]

Calculating the denominator:
\[ [A]^3 = \left(\frac{[A]_0}{4}\right)^3 = \frac{[A]_0^3}{64} \]

Now substituting back into the \( K_c \) expression:
\[ K_c = \frac{\frac{[A]_0^3}{16}}{\frac{[A]_0^3}{64}} \]

### Step 9: Final calculation
This simplifies to:
\[ K_c = \frac{64}{16} = 4 \]

### Conclusion
The value of \( K_c \) is:
\[ \boxed{4} \]