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

Options

A. `H_2(g)+ 1/2 O_2(g) hArr H_2O(g)`
B. `H_2(g)+ I _2(g) hArr 2HI(g)`
C. `C(g) +1/2 O_2 hArr CO(g)`
D. `2NH_3(g) hArr N_2(g) + 3H_2(g)`

Answer

To determine in which of the given reactions an increase in pressure at constant temperature does not affect the moles at equilibrium, we can apply Le Chatelier's principle. This principle states that if an external change is applied to a system at equilibrium, the system will adjust to counteract that change.

### Step-by-Step Solution:

1. **Understand Le Chatelier's Principle**:
- When pressure is increased, the equilibrium will shift towards the side with fewer moles of gas. This is because the system tries to reduce the pressure by favoring the side with fewer gaseous molecules.

2. **Identify the Reactions**:
- We need to analyze each of the four reactions provided to see how many moles of gas are present on both sides of the equation.

3. **Analyze Each Reaction**:
- **Reaction 1**:
- Reactants: 1 mole of H₂ (g) + 0.5 moles of O₂ (g) = 1.5 moles of gas
- Products: 1 mole of H₂O (g) = 1 mole of gas
- **Conclusion**: The equilibrium will shift to the right (forward direction) due to fewer moles of gas on the product side.

- **Reaction 2**:
- Reactants: 1 mole of H₂ (g) + 1 mole of I₂ (g) = 2 moles of gas
- Products: 2 moles of HI (g) = 2 moles of gas
- **Conclusion**: The number of moles of gas is the same on both sides (2 moles), so increasing pressure does not affect the equilibrium.

- **Reaction 3**:
- Reactants: 1 mole of C (g) + 0.5 moles of O₂ (g) = 1.5 moles of gas
- Products: 1 mole of CO (g) = 1 mole of gas
- **Conclusion**: The equilibrium will shift to the right (forward direction) due to fewer moles of gas on the product side.

- **Reaction 4**:
- Reactants: 2 moles of NH₃ (g) = 2 moles of gas
- Products: 1 mole of N₂ (g) + 3 moles of H₂ (g) = 4 moles of gas
- **Conclusion**: The equilibrium will shift to the left (backward direction) due to more moles of gas on the product side.

4. **Final Conclusion**:
- The only reaction where an increase in pressure does not affect the moles at equilibrium is **Reaction 2**, where the number of gaseous moles is equal on both sides.

### Answer:
**The answer is Reaction 2.**

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