If the volume of the racion flask is reduced to half of its initial value and temperature is kept constant the in which of the following cases the position of equilibrium will not shift?

Options

A. `CO(g)+H_(2)O(g)hArrCO(2)(g)+H_(2)(g)`
B. `I_(2)(g)hArr2I(g)`
C. `NH_(4)HS_(s)hArrNH_(3)(g)+H_(2)S(g)`
D. `2NOCI(g)hArr2NO(g)+CI_(2)(g)`

Answer

To solve the question, we need to analyze how the change in volume affects the position of equilibrium for each of the given reactions. The key principle to remember is that if the volume of the reaction flask is reduced, the concentrations of the gases will increase, which can shift the equilibrium position according to Le Chatelier's principle.

### Step-by-Step Solution:

1. **Understanding the Effect of Volume Change**:
- When the volume of the reaction flask is halved, the concentration of all gaseous reactants and products will double. This is because concentration (C) is defined as the number of moles (N) divided by volume (V): \( C = \frac{N}{V} \). If V is halved, C will double.

2. **Analyzing Each Reaction**:
- We will analyze each reaction to determine whether the equilibrium position will shift or remain unchanged.

3. **Option 1: CO + H2O ⇌ CO2 + H2**:
- **Initial Concentration**: \( K_{C1} = \frac{[CO2][H2]}{[CO][H2O]} \)
- **After Volume Reduction**: The concentrations double:
\[ K_{C2} = \frac{[2CO2][2H2]}{[2CO][2H2O]} = \frac{4[CO2][H2]}{4[CO][H2O]} = K_{C1} \]
- **Conclusion**: The equilibrium constant remains the same, so the position of equilibrium does not shift.

4. **Option 2: I2 ⇌ 2I**:
- **Initial Concentration**: \( K_{C1} = \frac{[I]^2}{[I2]} \)
- **After Volume Reduction**: The concentrations double:
\[ K_{C2} = \frac{[2I]^2}{[2I2]} = \frac{4[I]^2}{2[I2]} = 2K_{C1} \]
- **Conclusion**: The equilibrium constant increases, so the position of equilibrium shifts to the left (towards reactants).

5. **Option 3: NH4HS(s) ⇌ NH3(g) + H2S(g)**:
- **Initial Concentration**: \( K_{C1} = \frac{[NH3][H2S]}{1} \) (solids do not appear in the expression)
- **After Volume Reduction**: The concentrations double:
\[ K_{C2} = \frac{[2NH3][2H2S]}{1} = 4K_{C1} \]
- **Conclusion**: The equilibrium constant increases, so the position of equilibrium shifts to the right (towards products).

6. **Option 4: 2NOCl ⇌ 2NO + Cl2**:
- **Initial Concentration**: \( K_{C1} = \frac{[NO]^2[Cl2]}{[NOCl]^2} \)
- **After Volume Reduction**: The concentrations double:
\[ K_{C2} = \frac{[2NO]^2[2Cl2]}{[2NOCl]^2} = \frac{4[NO]^2 \cdot 2[Cl2]}{4[NOCl]^2} = 2K_{C1} \]
- **Conclusion**: The equilibrium constant increases, so the position of equilibrium shifts to the right (towards products).

### Final Conclusion:
- The only reaction where the position of equilibrium does not shift when the volume is halved is **Option 1: CO + H2O ⇌ CO2 + H2**.