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

Options

A. `4//9`
B. `9//4`
C. `1//9`
D. `4`

Explanation

To solve the problem step by step, we will analyze the reaction and the given conditions systematically.

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

### Step 2: Define initial concentrations
We are told that the initial concentrations of \( A \) and \( B \) are equal. Let's assume:
\[ [A]_0 = [B]_0 = 1 \, \text{M} \]

### Step 3: Define the change in concentrations at equilibrium
Let \( \alpha \) be the degree of dissociation. At equilibrium, the concentrations will change as follows:
- For \( A \) and \( B \):
\[ [A] = 1 - \alpha \]
\[ [B] = 1 - \alpha \]
- For \( C \) and \( D \):
\[ [C] = \alpha \]
\[ [D] = \alpha \]

### Step 4: Use the information given about \( C \)
We know that the concentration of \( C \) at equilibrium is twice that of \( A \):
\[ [C] = 2[A] \]
Substituting the expressions we have:
\[ \alpha = 2(1 - \alpha) \]

### Step 5: Solve for \( \alpha \)
Now, we can solve the equation:
\[ \alpha = 2(1 - \alpha) \\ \alpha = 2 - 2\alpha \\ \alpha + 2\alpha = 2 \\ 3\alpha = 2 \\ \alpha = \frac{2}{3} \]

### Step 6: Calculate the equilibrium concentrations
Now we can find the equilibrium concentrations:
- For \( A \):
\[ [A] = 1 - \alpha = 1 - \frac{2}{3} = \frac{1}{3} \]
- For \( B \):
\[ [B] = 1 - \alpha = \frac{1}{3} \]
- For \( C \):
\[ [C] = \alpha = \frac{2}{3} \]
- For \( D \):
\[ [D] = \alpha = \frac{2}{3} \]

### Step 7: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C][D]}{[A][B]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)}{\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)} = \frac{\frac{4}{9}}{\frac{1}{9}} = 4 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is:
\[ K_c = 4 \]