A one litre vessel contains 3 moles each of gases A, B, C and D at equilibrium. If one mole each of A and B are removed `K_(C)` for `A(g)+B(g)hArrC(g)+D(g)` will be

Options

A. 2.5
B. 4
C. 1
D. `1//4`

Explanation

To solve the problem, we need to analyze the equilibrium reaction and the effect of removing moles of reactants on the equilibrium constant \( K_c \).

### Step-by-Step Solution:

1. **Write the Equilibrium Reaction**:
The reaction is given as:
\[ A(g) + B(g) \rightleftharpoons C(g) + D(g) \]

2. **Initial Concentrations**:
Initially, we have 3 moles of each gas in a 1-liter vessel. Therefore, the initial concentrations are:
\[ [A] = 3 \, \text{mol/L}, \quad [B] = 3 \, \text{mol/L}, \quad [C] = 3 \, \text{mol/L}, \quad [D] = 3 \, \text{mol/L} \]

3. **Calculate the Initial Equilibrium Constant \( K_c \)**:
The equilibrium constant \( K_c \) is defined as:
\[ K_c = \frac{[C][D]}{[A][B]} \]
Substituting the initial concentrations:
\[ K_c = \frac{(3)(3)}{(3)(3)} = \frac{9}{9} = 1 \]

4. **Effect of Removing Moles of A and B**:
If we remove 1 mole of gas A and 1 mole of gas B, the new concentrations will be:
\[ [A] = 3 - 1 = 2 \, \text{mol/L}, \quad [B] = 3 - 1 = 2 \, \text{mol/L}, \quad [C] = 3 \, \text{mol/L}, \quad [D] = 3 \, \text{mol/L} \]

5. **Recalculate the Equilibrium Constant \( K_c \)**:
The equilibrium constant does not change with the change in concentration of reactants or products at equilibrium. Thus, even after the removal of A and B, the equilibrium constant remains:
\[ K_c = \frac{[C][D]}{[A][B]} = \frac{(3)(3)}{(2)(2)} = \frac{9}{4} \]
However, since the equilibrium constant is a property of the reaction at a given temperature, it remains constant regardless of the changes in concentrations.

6. **Final Conclusion**:
Therefore, the value of \( K_c \) remains the same as it was initially calculated, which is:
\[ K_c = 1 \]

### Final Answer:
The value of \( K_c \) after removing 1 mole each of A and B is **1**.