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

Explanation

To solve the problem, we need to determine the amount of C at equilibrium for the reaction:

\[ A + B \rightleftharpoons C + D \]

Given that \( K_c = 9 \) and A and B are taken in equal amounts, we can follow these steps:

### Step 1: Set up the initial conditions
Assume we start with 1 mole of A and 1 mole of B. Therefore, the initial concentrations are:
- \([A] = 1 \, \text{mol}\)
- \([B] = 1 \, \text{mol}\)
- \([C] = 0 \, \text{mol}\)
- \([D] = 0 \, \text{mol}\)

### Step 2: Define the change in concentrations at equilibrium
Let \( x \) be the amount of A and B that react to form C and D. At equilibrium, the concentrations will be:
- \([A] = 1 - x\)
- \([B] = 1 - x\)
- \([C] = x\)
- \([D] = x\)

### Step 3: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by the formula:

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

Substituting the equilibrium concentrations into the expression, we get:

\[ K_c = \frac{x \cdot x}{(1 - x)(1 - x)} = \frac{x^2}{(1 - x)^2} \]

### Step 4: Substitute the value of \( K_c \)
We know \( K_c = 9 \), so we can set up the equation:

\[ 9 = \frac{x^2}{(1 - x)^2} \]

### Step 5: Solve for \( x \)
To solve for \( x \), we can cross-multiply:

\[ 9(1 - x)^2 = x^2 \]

Expanding this gives:

\[ 9(1 - 2x + x^2) = x^2 \]

This simplifies to:

\[ 9 - 18x + 9x^2 = x^2 \]

Rearranging the equation results in:

\[ 8x^2 - 18x + 9 = 0 \]

### Step 6: Use the quadratic formula
Now we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 8, b = -18, c = 9 \):

\[ x = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 8 \cdot 9}}{2 \cdot 8} \]
\[ x = \frac{18 \pm \sqrt{324 - 288}}{16} \]
\[ x = \frac{18 \pm \sqrt{36}}{16} \]
\[ x = \frac{18 \pm 6}{16} \]

Calculating the two possible values for \( x \):

1. \( x = \frac{24}{16} = 1.5 \) (not possible since we started with 1 mole)
2. \( x = \frac{12}{16} = 0.75 \)

### Step 7: Conclusion
The amount of C at equilibrium is equal to \( x \), which we found to be:

\[ [C] = x = 0.75 \, \text{mol} \]

Thus, the amount of C in equilibrium is **0.75 moles**.