`A + B hArrC + D`. If finally the concentrations of A an d 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. `(4)/(9)`
B. `(9)/(4)`
C. `(1)/(9)`
D. 4

Answer

To solve the problem, we need to determine the equilibrium constant for the reaction given the conditions provided. Let's break it down step by step.

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

### Step 2: Define the Equilibrium Concentrations
We are told that at equilibrium:
- The concentrations of A and B are equal.
- The concentration of D is twice that of A.

Let’s denote the equilibrium concentration of A as \( [A] \). Since the concentrations of A and B are equal, we can write:
\[ [A] = [B] \]

Let’s denote this common concentration as \( x \):
\[ [A] = [B] = x \]

According to the problem, the concentration of D is twice that of A:
\[ [D] = 2[A] = 2x \]

### Step 3: Determine the Concentration of C
From the stoichiometry of the reaction, we know that the amount of C produced will be equal to the amount of D produced since one mole of A and one mole of B produce one mole of C and one mole of D. Therefore:
\[ [C] = [D] = 2x \]

### Step 4: Write the Equilibrium Expression
The equilibrium constant \( K_c \) for the reaction is defined as:
\[ K_c = \frac{[C][D]}{[A][B]} \]

Substituting the values we have:
\[ K_c = \frac{[C][D]}{[A][B]} = \frac{(2x)(2x)}{(x)(x)} \]

### Step 5: Simplify the Expression
Now, simplifying the expression:
\[ K_c = \frac{4x^2}{x^2} = 4 \]

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

### Final Answer
The equilibrium constant of the reaction is **4**.

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