In which of following reactions, increase in the volume at constant temperature does not affect the number of moles of at equilibrium?

Options

A. `2NH_(3)hArrN_(2)+3H_(2)`
B. `C(g)+(1//2)O_(2)(g)hArrCO(g)`
C. `H_(2)(g)+O_(2)(g)hArrH_(2)O_(2)(g)`
D. None of these

Explanation

To determine in which of the given reactions an increase in volume at constant temperature does not affect the number of moles at equilibrium, we need to analyze the change in the number of moles of gas (Δn_g) for each reaction. The equilibrium will not be affected by a change in volume if Δn_g = 0.

### Step-by-Step Solution:

1. **Understanding Δn_g**:
- Δn_g is defined as the difference between the number of moles of gaseous products and the number of moles of gaseous reactants.
- Formula: Δn_g = moles of products - moles of reactants.

2. **Analyzing Each Reaction**:
- **Reaction 1**: 2NH₃(g) ⇌ N₂(g) + 3H₂(g)
- Moles of products = 1 (N₂) + 3 (H₂) = 4 moles
- Moles of reactants = 2 (NH₃)
- Δn_g = 4 - 2 = 2 (not equal to 0)
- Conclusion: Equilibrium will shift towards the left (reactants) when volume increases.

- **Reaction 2**: CO(g) + 1.5O₂(g) ⇌ 2CO₂(g)
- Moles of products = 2 (CO₂)
- Moles of reactants = 1 (CO) + 1.5 (O₂) = 2.5 moles
- Δn_g = 2 - 2.5 = -0.5 (not equal to 0)
- Conclusion: Equilibrium will shift towards the right (products) when volume increases.

- **Reaction 3**: H₂(g) + O₂(g) ⇌ H₂O₂(g)
- Moles of products = 1 (H₂O₂)
- Moles of reactants = 1 (H₂) + 1 (O₂) = 2 moles
- Δn_g = 1 - 2 = -1 (not equal to 0)
- Conclusion: Equilibrium will shift towards the right (products) when volume increases.

3. **Final Conclusion**:
- For all three reactions analyzed, Δn_g is not equal to 0. Therefore, an increase in volume at constant temperature will affect the number of moles at equilibrium in all cases.
- Thus, the correct answer is that none of the reactions listed will have their equilibrium unaffected by an increase in volume.