`A+B rarr C+D`
Initially moles of A and B are equal. At equilibrium, moles of C are three times of A. The equilibrium constant of the reaction will be

Options

A. 1
B. 2
C. 4
D. 9

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define Initial Conditions
Let the initial moles of A and B be \( X \). Since the moles of A and B are equal, we have:
- Moles of A = \( X \)
- Moles of B = \( X \)
- Moles of C = 0
- Moles of D = 0

### Step 2: Define Changes at Equilibrium
Let \( a \) be the moles of A that react at equilibrium. Since the stoichiometric coefficients for A and B are equal (1:1), the moles of B that react will also be \( a \). Thus, at equilibrium, we have:
- Moles of A = \( X - a \)
- Moles of B = \( X - a \)
- Moles of C = \( a \)
- Moles of D = \( a \)

### Step 3: Use Given Information
We are given that the moles of C are three times the moles of A at equilibrium:
\[ a = 3(X - a) \]
This equation states that the moles of C (which is \( a \)) are three times the moles of A left at equilibrium (which is \( X - a \)).

### Step 4: Solve for \( a \)
Rearranging the equation gives:
\[ a = 3X - 3a \]
\[ 4a = 3X \]
\[ a = \frac{3X}{4} \]

### Step 5: Calculate Moles at Equilibrium
Now substituting \( a \) back into the expressions for moles at equilibrium:
- Moles of A at equilibrium = \( X - a = X - \frac{3X}{4} = \frac{X}{4} \)
- Moles of B at equilibrium = \( X - a = \frac{X}{4} \)
- Moles of C at equilibrium = \( a = \frac{3X}{4} \)
- Moles of D at equilibrium = \( a = \frac{3X}{4} \)

### Step 6: Write the Expression for the Equilibrium Constant \( K_c \)
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 (assuming volume \( V \) cancels out):
\[ K_c = \frac{\left(\frac{3X}{4V}\right)\left(\frac{3X}{4V}\right)}{\left(\frac{X}{4V}\right)\left(\frac{X}{4V}\right)} \]
This simplifies to:
\[ K_c = \frac{\left(\frac{3X}{4}\right)^2}{\left(\frac{X}{4}\right)^2} \]
\[ K_c = \frac{9X^2/16}{X^2/16} \]
\[ K_c = 9 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction is \( 9 \).

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