`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 step by step, let's break down the information given and apply the concepts of chemical equilibrium.

### Step 1: Define Initial Conditions
Let the initial moles of A and B be \( A \) (since they are equal, we can denote them both as \( A \)).

### Step 2: Define Change at Equilibrium
Let \( x \) be the moles of A that react at equilibrium. Therefore, the change in moles can be expressed as:
- Moles of A at equilibrium = \( A - x \)
- Moles of B at equilibrium = \( A - x \)
- Moles of C at equilibrium = \( x \)
- Moles of D at equilibrium = \( x \)

### Step 3: Use Given Information
According to the problem, the moles of C at equilibrium are three times the moles of A at equilibrium. Therefore:
\[ x = 3(A - x) \]

### Step 4: Solve for \( x \)
Now, we can solve the equation:
1. Expand the equation:
\[ x = 3A - 3x \]
2. Rearranging gives:
\[ x + 3x = 3A \]
\[ 4x = 3A \]
\[ x = \frac{3A}{4} \]

### Step 5: Calculate Moles at Equilibrium
Now that we have \( x \):
- Moles of A at equilibrium = \( A - x = A - \frac{3A}{4} = \frac{A}{4} \)
- Moles of B at equilibrium = \( A - x = \frac{A}{4} \)
- Moles of C at equilibrium = \( x = \frac{3A}{4} \)
- Moles of D at equilibrium = \( x = \frac{3A}{4} \)

### 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:
\[ K_c = \frac{[C][D]}{[A][B]} \]

### Step 7: Substitute the Equilibrium Values
Substituting the moles at equilibrium into the expression:
\[ K_c = \frac{\left(\frac{3A}{4}\right)\left(\frac{3A}{4}\right)}{\left(\frac{A}{4}\right)\left(\frac{A}{4}\right)} \]

### Step 8: Simplify the Expression
1. Calculate the numerator:
\[ \frac{3A}{4} \times \frac{3A}{4} = \frac{9A^2}{16} \]
2. Calculate the denominator:
\[ \frac{A}{4} \times \frac{A}{4} = \frac{A^2}{16} \]
3. Therefore, substituting back into the equation:
\[ K_c = \frac{\frac{9A^2}{16}}{\frac{A^2}{16}} = 9 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction is:
\[ K_c = 9 \]

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