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 find the equilibrium constant for the reaction \( A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} \), we will follow these steps:

### Step 1: Write the initial concentrations
We are given:
- 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

Thus, the initial concentrations are:
- \([A]_{initial} = \frac{1 \text{ mole}}{1 \text{ L}} = 1 \text{ M}\)
- \([B]_{initial} = \frac{1.5 \text{ moles}}{1 \text{ L}} = 1.5 \text{ M}\)
- \([C]_{initial} = \frac{2 \text{ moles}}{1 \text{ L}} = 2 \text{ M}\)

### Step 2: Set up the change in concentrations at equilibrium
Let \( x \) be the amount of A and B that react to form C at equilibrium. Thus, at equilibrium:
- \([A] = 1 - x\)
- \([B] = 1.5 - x\)
- \([C] = 2 + x\)

### Step 3: Use the equilibrium concentration of C
We are given that the equilibrium concentration of C is 0.5 M. Therefore:
\[ 2 + x = 0.5 \]
Solving for \( x \):
\[ x = 0.5 - 2 = -1.5 \]

### Step 4: Calculate the equilibrium concentrations of A and B
Now substituting \( x = -1.5 \) into the expressions for A and B:
- \([A]_{equilibrium} = 1 - (-1.5) = 1 + 1.5 = 2.5 \text{ M}\)
- \([B]_{equilibrium} = 1.5 - (-1.5) = 1.5 + 1.5 = 3 \text{ M}\)

### Step 5: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[C]_{equilibrium}}{[A]_{equilibrium} \cdot [B]_{equilibrium}} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{0.5}{(2.5)(3)} \]

### Step 6: Calculate \( K_c \)
Calculating the denominator:
\[ (2.5)(3) = 7.5 \]
Now substituting back into the equation for \( K_c \):
\[ K_c = \frac{0.5}{7.5} = \frac{1}{15} \approx 0.0667 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is approximately:
\[ K_c \approx 0.0667 \]