For the reaction, `N_2(g) +3H_2(g)hArr 2NH_3(g), DeltaH^@=-ve` the number of moles of `H_2` at equilibrium will increases when

Options

A. volume of vessel is increased
B. volume of vessel is decreased
C. Ne gas is added at constant volume
D. `NH_3` is removed

Answer

To solve the problem, we need to analyze the given chemical reaction and how changes in conditions affect the equilibrium position according to Le Chatelier's principle.

**Given Reaction:**
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]
**ΔH° = -ve (exothermic reaction)**

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- We have a reaction involving nitrogen gas (N2), hydrogen gas (H2), and ammonia (NH3).
- The reaction is exothermic, meaning it releases heat.

2. **Understanding the Equilibrium Shift:**
- According to Le Chatelier's principle, if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium will shift to counteract that change.

3. **Analyze the Options:**
- **Option 1: Volume of the vessel is increased.**
- When the volume is increased, the pressure decreases. The equilibrium will shift towards the side with more moles of gas to increase pressure.
- On the left side, there are 4 moles (1 N2 + 3 H2) and on the right side, there are 2 moles (2 NH3).
- Therefore, the equilibrium will shift to the left, increasing the concentration of H2.

- **Option 2: Volume of the vessel is decreased.**
- This is the opposite of option 1. Decreasing the volume increases pressure, and the equilibrium will shift towards the side with fewer moles of gas.
- This means it would shift to the right (towards NH3), decreasing the concentration of H2. Thus, this option is incorrect.

- **Option 3: Neon gas is added at constant volume.**
- Adding an inert gas at constant volume does not change the concentration of the reactants or products. Therefore, this option does not affect the equilibrium position and is incorrect.

- **Option 4: NH3 is removed.**
- Removing NH3 decreases its concentration. According to Le Chatelier's principle, the equilibrium will shift to the right to produce more NH3, which means the concentration of H2 will decrease. Thus, this option is also incorrect.

4. **Conclusion:**
- The only option that increases the number of moles of H2 at equilibrium is **Option 1: Volume of the vessel is increased**.

### Final Answer:
The number of moles of H2 at equilibrium will increase when the volume of the vessel is increased.

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