1 mole of 'A' 1.5 mole of 'B' and 2 mole of 'C' are taken in a vessel of volume one litre. At equilibrium concentration of C is 0.5 mole /L .Equilibrium constant for the reaction , `A_((g))+B_((g) hArr C_((g))` is

Options

A. 0.66
B. 0.066
C. 66
D. 6.6

Answer

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

\[ A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} \]

### Step-by-Step Solution:

1. **Identify Initial Moles and Concentrations**:
- Initial moles of \( A = 1 \) mole
- Initial moles of \( B = 1.5 \) moles
- Initial moles of \( C = 2 \) moles
- Volume of the vessel = 1 L
- Therefore, initial concentrations:
\[ [A]_{initial} = 1 \, \text{mol/L}, \quad [B]_{initial} = 1.5 \, \text{mol/L}, \quad [C]_{initial} = 2 \, \text{mol/L} \]

2. **Determine the Change in Concentration**:
- At equilibrium, the concentration of \( C \) is given as \( 0.5 \, \text{mol/L} \).
- The initial concentration of \( C \) was \( 2 \, \text{mol/L} \).
- Change in concentration of \( C \):
\[ \Delta[C] = 2 - 0.5 = 1.5 \, \text{mol/L} \]

3. **Establish Changes for \( A \) and \( B \)**:
- Since the reaction produces \( C \) from \( A \) and \( B \), the change in concentrations for \( A \) and \( B \) will be equal to the change in \( C \):
\[ \Delta[A] = -1.5 \, \text{mol/L}, \quad \Delta[B] = -1.5 \, \text{mol/L} \]

4. **Calculate Equilibrium Concentrations**:
- Equilibrium concentration of \( A \):
\[ [A]_{equilibrium} = [A]_{initial} + \Delta[A] = 1 - 1.5 = -0.5 \, \text{mol/L} \, (\text{not possible, hence check the approach}) \]
- Since \( C \) decreased by \( 1.5 \, \text{mol/L} \), we need to consider that \( C \) is actually increasing from the reverse reaction:
\[ [C]_{equilibrium} = 2 + x \quad \text{where } x = 1.5 \text{ (the amount that reacted)} \]
- Thus, \( [C]_{equilibrium} = 0.5 \) is consistent with the decrease in \( C \) from \( 2 \).

5. **Recalculate \( x \)**:
- Let \( x \) be the amount of \( C \) that dissociated:
\[ [C]_{equilibrium} = 2 - x = 0.5 \implies x = 1.5 \]
- Therefore, at equilibrium:
\[ [A]_{equilibrium} = 1 - x = 1 - 1.5 = -0.5 \, \text{(not valid)} \]
- This indicates that the reaction has shifted to the left, and we need to adjust our calculations.

6. **Final Equilibrium Concentrations**:
- \( [A]_{equilibrium} = 1 + 1.5 = 2.5 \, \text{mol/L} \)
- \( [B]_{equilibrium} = 1.5 + 1.5 = 3 \, \text{mol/L} \)
- \( [C]_{equilibrium} = 0.5 \, \text{mol/L} \)

7. **Calculate the Equilibrium Constant \( K_c \)**:
- The expression for \( K_c \) is:
\[ K_c = \frac{[C]_{equilibrium}}{[A]_{equilibrium} \cdot [B]_{equilibrium}} = \frac{0.5}{(2.5)(3)} \]
- Calculate \( K_c \):
\[ K_c = \frac{0.5}{7.5} = \frac{1}{15} \approx 0.0667 \]

### Final Answer:
The equilibrium constant \( K_c \) for the reaction is approximately \( 0.0667 \).