`K_(c) = 9` for the reaction, `A + B hArrC + D`. If A and B are taken in equal amounts, then amount of C in equilibrium is:

Options

A. 1
B. `0.25`
C. `0.75`
D. None of these

Explanation

To solve the problem step by step, we will analyze the reaction and use the equilibrium constant to find the amount of C at equilibrium.

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

### Step 2: Set up the initial amounts
We are told that A and B are taken in equal amounts. Let's assume we start with 1 mole of A and 1 mole of B:
- Initial moles of A = 1
- Initial moles of B = 1
- Initial moles of C = 0
- Initial moles of D = 0

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the amount of A and B that reacts at equilibrium. Therefore, the changes in moles will be:
- Moles of A at equilibrium = \( 1 - x \)
- Moles of B at equilibrium = \( 1 - x \)
- Moles of C at equilibrium = \( x \)
- Moles of D at equilibrium = \( x \)

### Step 4: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is defined as:
\[ K_c = \frac{[C][D]}{[A][B]} \]
Substituting the equilibrium concentrations, we have:
\[ K_c = \frac{x \cdot x}{(1 - x)(1 - x)} = \frac{x^2}{(1 - x)^2} \]

### Step 5: Substitute the value of \( K_c \)
We are given that \( K_c = 9 \). Therefore, we can set up the equation:
\[ \frac{x^2}{(1 - x)^2} = 9 \]

### Step 6: Solve the equation
Taking the square root of both sides gives:
\[ \frac{x}{1 - x} = 3 \quad \text{or} \quad \frac{x}{1 - x} = -3 \]
Since the amount of substance cannot be negative, we only consider the first equation:
\[ x = 3(1 - x) \]
Expanding this gives:
\[ x = 3 - 3x \]
Combining like terms:
\[ 4x = 3 \]
Thus,
\[ x = \frac{3}{4} = 0.75 \]

### Step 7: Conclusion
The amount of C at equilibrium is:
\[ \text{Amount of C} = x = 0.75 \text{ moles} \]