The reaction `A(g) + B(g)hArr2C (g)` is occurred by mixing of 3 moles of A and 1 mole of B in one litre container. If a of B is `(1)/(3)` , then `K_(c)` for this reaction is:

Options

A. `0.12`
B. `0.25`
C. `0.50`
D. `0.75`

Explanation

To solve the problem, we need to find the equilibrium constant \( K_c \) for the reaction:

\[ A(g) + B(g) \rightleftharpoons 2C(g) \]

**Step 1: Determine Initial Concentrations**
- We start with 3 moles of \( A \) and 1 mole of \( B \) in a 1-liter container.
- Therefore, the initial concentrations are:
\[ [A] = 3 \, \text{mol/L}, \quad [B] = 1 \, \text{mol/L}, \quad [C] = 0 \, \text{mol/L} \]

**Step 2: Define Degree of Dissociation**
- The degree of dissociation of \( B \) is given as \( \alpha = \frac{1}{3} \).
- This means that \( \frac{1}{3} \) of the initial amount of \( B \) will react.

**Step 3: Calculate Moles at Equilibrium**
- The amount of \( B \) that reacts is:
\[ \text{Moles of } B \text{ that reacted} = \alpha \times \text{initial moles of } B = \frac{1}{3} \times 1 = \frac{1}{3} \]
- The remaining moles of \( B \) at equilibrium:
\[ [B]_{eq} = 1 - \frac{1}{3} = \frac{2}{3} \]
- For every mole of \( B \) that reacts, 1 mole of \( A \) reacts and 2 moles of \( C \) are produced. Thus:
- Moles of \( A \) that reacted = \( \frac{1}{3} \)
- Remaining moles of \( A \):
\[ [A]_{eq} = 3 - \frac{1}{3} = \frac{8}{3} \]
- Moles of \( C \) produced:
\[ [C]_{eq} = 2 \times \frac{1}{3} = \frac{2}{3} \]

**Step 4: Write the Expression for \( K_c \)**
- The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C]^2}{[A][B]} \]

**Step 5: Substitute Equilibrium Concentrations**
- Substitute the equilibrium concentrations into the expression:
\[ K_c = \frac{\left(\frac{2}{3}\right)^2}{\left(\frac{8}{3}\right)\left(\frac{2}{3}\right)} \]

**Step 6: Simplify the Expression**
- Calculate the numerator:
\[ \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]
- Calculate the denominator:
\[ \left(\frac{8}{3}\right)\left(\frac{2}{3}\right) = \frac{16}{9} \]
- Therefore:
\[ K_c = \frac{\frac{4}{9}}{\frac{16}{9}} = \frac{4}{16} = \frac{1}{4} = 0.25 \]

**Final Answer:**
\[ K_c = 0.25 \]

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