`A + B hArrC + D`. If finally the concentrations of A and B are both equal but at equilibrium concentration of D will be twice of that of A then what will be the equilibrium constant of reaction.

Options

A. (a) `(4)/(9)`
B. (b) `(9)/(4)`
C. (c) `(1)/(9)`
D. (d) `4`

Answer

To find the equilibrium constant \( K_c \) for the reaction \( A + B \rightleftharpoons C + D \) given the conditions of the problem, we can follow these steps:

### Step 1: Define the Initial Concentrations
Let's assume the initial concentrations of \( A \) and \( B \) are both \( A \) (in moles).

### Step 2: Define Changes at Equilibrium
Let \( X \) be the amount of \( A \) and \( B \) that reacts at equilibrium. Thus, at equilibrium:
- Concentration of \( A \) = \( A - X \)
- Concentration of \( B \) = \( A - X \)
- Concentration of \( C \) = \( X \)
- Concentration of \( D \) = \( X \)

### Step 3: Use Given Conditions
According to the problem, the concentration of \( D \) is twice that of \( A \):
\[ D = 2A \]
Since \( D = X \), we have:
\[ X = 2(A - X) \]

### Step 4: Solve for \( X \)
Rearranging the equation:
\[ X = 2A - 2X \]
Adding \( 2X \) to both sides:
\[ 3X = 2A \]
Thus,
\[ X = \frac{2}{3}A \]

### Step 5: Find Equilibrium Concentrations
Now substituting \( X \) back into the equilibrium concentrations:
- Concentration of \( A \) = \( A - X = A - \frac{2}{3}A = \frac{1}{3}A \)
- Concentration of \( B \) = \( A - X = \frac{1}{3}A \)
- Concentration of \( C \) = \( X = \frac{2}{3}A \)
- Concentration of \( D \) = \( X = \frac{2}{3}A \)

### Step 6: Write the Expression for the Equilibrium Constant
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C][D]}{[A][B]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{2}{3}A\right)\left(\frac{2}{3}A\right)}{\left(\frac{1}{3}A\right)\left(\frac{1}{3}A\right)} \]

### Step 7: Simplify the Expression
Calculating \( K_c \):
\[ K_c = \frac{\left(\frac{4}{9}A^2\right)}{\left(\frac{1}{9}A^2\right)} = \frac{4}{1} = 4 \]

### Conclusion
Thus, the equilibrium constant \( K_c \) for the reaction is:
\[ \boxed{4} \]