For the reaction
`N_(2)(G)+3H_(2)(g)hArr2NH_(3)(g),Delta=-93.6KJmol^(-1)`
The number of moles of `H_(2)` at equilibrium will increase If :

Options

A. (a) volume is increased
B. (b) volume is decreased
C. (c) argon gas is added at constant volume
D. (d) `NH_(3)` Is removed

Answer

To determine how the number of moles of \( H_2 \) at equilibrium will increase, we need to analyze the given reaction and the factors that can affect it. The reaction is:

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

with an enthalpy change \( \Delta H = -93.6 \, \text{kJ/mol} \), indicating that the reaction is exothermic.

### Step 1: Understand the Reaction
The reaction involves 1 mole of nitrogen gas and 3 moles of hydrogen gas producing 2 moles of ammonia gas. This means that on the left side (reactants), there are a total of 4 moles of gas (1 mole of \( N_2 \) + 3 moles of \( H_2 \)), while on the right side (products), there are only 2 moles of gas (2 moles of \( NH_3 \)).

### Step 2: Analyze the Effect of Volume Change
According to Le Chatelier's principle, if we change the conditions of a system at equilibrium, the system will adjust to counteract that change.

- **Increasing Volume**: When the volume of the system is increased, the equilibrium will shift towards the side with more moles of gas to increase the pressure. In this case, the left side has 4 moles (1 \( N_2 \) + 3 \( H_2 \)) compared to the right side with 2 moles (2 \( NH_3 \)). Therefore, increasing the volume will shift the equilibrium to the left, increasing the concentration of \( H_2 \).

### Step 3: Consider Other Factors
1. **Adding Inert Gas (Argon)**: If an inert gas is added at constant volume, it does not affect the equilibrium position because it does not change the partial pressures of the reactants or products. Thus, this will not increase the moles of \( H_2 \).

2. **Removing \( NH_3 \)**: If \( NH_3 \) is removed, the equilibrium will shift to the right to produce more \( NH_3 \), which would decrease the moles of \( H_2 \) rather than increase them.

### Conclusion
The only factor that would lead to an increase in the number of moles of \( H_2 \) at equilibrium is the increase in volume. Therefore, the correct answer is:

**The number of moles of \( H_2 \) at equilibrium will increase if the volume is increased.**