By adding inert gas at a constant volume, which of the following equilibrium will not be affected?

Options

A. `H_(2)(g)+I_(2)(g)hArr 2HI(g)`
B. `3H_(2)(g)+N_(2)(G)hArr 2NH_(3)(g)`
C. `PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)`
D. All of above

Answer

To solve the question regarding the effect of adding an inert gas at constant volume on chemical equilibria, we will analyze each given reaction to determine the change in the number of moles (Δn) of gases involved.

### Step-by-Step Solution:

1. **Understanding the Concept**:
- When an inert gas is added at constant volume, it does not change the partial pressures of the reactants and products involved in the equilibrium. The effect on equilibrium depends on the change in the number of moles of gaseous reactants and products.

2. **Identify the Reactions**:
- We are given three reactions to analyze.

3. **First Reaction**:
- \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \)
- Calculate Δn:
- Moles of products = 2 (from 2HI)
- Moles of reactants = 2 (from H2 and I2)
- Δn = Moles of products - Moles of reactants = 2 - 2 = 0

4. **Second Reaction**:
- \( 3H_2(g) + N_2(g) \rightleftharpoons 2NH_3(g) \)
- Calculate Δn:
- Moles of products = 2 (from 2NH3)
- Moles of reactants = 3 + 1 = 4 (from 3H2 and N2)
- Δn = 2 - 4 = -2

5. **Third Reaction**:
- \( PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \)
- Calculate Δn:
- Moles of products = 1 + 1 = 2 (from PCl3 and Cl2)
- Moles of reactants = 1 (from PCl5)
- Δn = 2 - 1 = 1

6. **Conclusion**:
- The first reaction has Δn = 0, which means that adding an inert gas at constant volume will not affect the equilibrium. The second and third reactions have changes in the number of moles (Δn = -2 and Δn = 1 respectively), indicating that their equilibria will be affected.

### Final Answer:
- The equilibrium that will not be affected by adding an inert gas at constant volume is the first reaction: \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \).