A 1-litre vessel contains 2 moles each of gases A, B, C and D at equilibrium. If 1 mole each of A and B are removed, K. for A+B C +D will be

Options

A. 4
B. 1
C. `1//4`
D. 2

Answer

To solve the problem, we need to determine the equilibrium constant \( K_c \) for the reaction \( A + B \rightleftharpoons C + D \) after removing 1 mole each of gases A and B from a 1-litre vessel that initially contains 2 moles of each gas at equilibrium.

### Step-by-Step Solution:

1. **Initial Conditions**:
- Before any gas is removed, we have:
- \( [A] = 2 \, \text{mol/L} \)
- \( [B] = 2 \, \text{mol/L} \)
- \( [C] = 2 \, \text{mol/L} \)
- \( [D] = 2 \, \text{mol/L} \)

2. **Write the Expression for \( K_c \)**:
- The equilibrium constant \( K_c \) for the reaction \( A + B \rightleftharpoons C + D \) is given by:
\[ K_c = \frac{[C][D]}{[A][B]} \]

3. **Calculate Initial \( K_c \)**:
- Substituting the initial concentrations into the expression:
\[ K_c = \frac{(2)(2)}{(2)(2)} = \frac{4}{4} = 1 \]

4. **Removing Moles of A and B**:
- After removing 1 mole each of A and B, the new concentrations will be:
- \( [A] = 2 - 1 = 1 \, \text{mol/L} \)
- \( [B] = 2 - 1 = 1 \, \text{mol/L} \)
- \( [C] = 2 \, \text{mol/L} \) (unchanged)
- \( [D] = 2 \, \text{mol/L} \) (unchanged)

5. **Recalculate \( K_c \)**:
- Now, substituting the new concentrations into the \( K_c \) expression:
\[ K_c = \frac{(2)(2)}{(1)(1)} = \frac{4}{1} = 4 \]

6. **Conclusion**:
- The equilibrium constant \( K_c \) after removing 1 mole each of A and B is 4.

### Final Answer:
The value of \( K_c \) for the reaction \( A + B \rightleftharpoons C + D \) after removing 1 mole each of A and B is **4**.