`A(G)+B(g)hArrC(g)+D(g)`
Above equilibrium is established by taking A& B in a closed container. Initial concentration of A is twice of the initial concentration of B. At equilibrium concentraons of B and C are equal. Then find the equilibrium constant for the reaction, `C(g)+D(g)hArrA(g)+B(g)`.

Answer

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

### Step 1: Write the reaction and initial conditions
The reaction is given as:
\[ A(g) + B(g) \rightleftharpoons C(g) + D(g) \]

We are told that the initial concentration of A is twice that of B. Let's denote the initial concentration of B as \( x \). Therefore, the initial concentration of A will be \( 2x \).

### Step 2: Set up the initial concentrations
At the start (initially), we have:
- \([A] = 2x\)
- \([B] = x\)
- \([C] = 0\)
- \([D] = 0\)

### Step 3: Define the change in concentration at equilibrium
Let \( y \) be the change in concentration of B that reacts to form C and D. Therefore, at equilibrium, the concentrations will be:
- \([A] = 2x - y\)
- \([B] = x - y\)
- \([C] = y\)
- \([D] = y\)

### Step 4: Use the condition given in the problem
According to the problem, at equilibrium, the concentrations of B and C are equal:
\[ [B] = [C] \]
This gives us the equation:
\[ x - y = y \]

### Step 5: Solve for \( y \)
From the equation \( x - y = y \), we can rearrange it:
\[ x = 2y \]

### Step 6: Substitute \( y \) back to find concentrations
Now we can substitute \( y \) back into the expressions for the equilibrium concentrations:
- For A:
\[ [A] = 2x - y = 2x - 2y = 2(2y) - 2y = 4y - 2y = 2y \]
- For B:
\[ [B] = x - y = 2y - y = y \]
- For C:
\[ [C] = y \]
- For D:
\[ [D] = y \]

Thus, we have:
- \([A] = 2y\)
- \([B] = y\)
- \([C] = y\)
- \([D] = y\)

### Step 7: Write the equilibrium constant expression for the first reaction
The equilibrium constant \( K_c \) for the reaction \( A + B \rightleftharpoons C + D \) is given by:
\[ K_c = \frac{[C][D]}{[A][B]} \]

Substituting the equilibrium concentrations:
\[ K_c = \frac{(y)(y)}{(2y)(y)} = \frac{y^2}{2y^2} = \frac{1}{2} \]

### Step 8: Write the equilibrium constant expression for the reverse reaction
Now, we need to find the equilibrium constant for the reverse reaction:
\[ C(g) + D(g) \rightleftharpoons A(g) + B(g) \]

The equilibrium constant \( K' \) for this reaction is the reciprocal of \( K_c \):
\[ K' = \frac{1}{K_c} = \frac{1}{\frac{1}{2}} = 2 \]

### Final Answer
Thus, the equilibrium constant for the reaction \( C(g) + D(g) \rightleftharpoons A(g) + B(g) \) is:
\[ K' = 2 \]