At constant pressure , addition of helium to the reaction system : `N_2(g)+3H_3(g)hArr2NH_3(g)`

Options

A. Favorus the formation of ammonia
B. Reduces the formation of ammonia
C. Does not affect the position of equilibrium
D. Reduces the dissociation of ammonia

Answer

To solve the question regarding the effect of adding helium gas to the reaction system \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) at constant pressure, we can follow these steps:

### Step 1: Understand the Reaction
The reaction given is the formation of ammonia from nitrogen and hydrogen gases. The balanced equation is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 2: Analyze the Moles of Gas
In the reaction:
- On the reactant side, we have 1 mole of \( N_2 \) and 3 moles of \( H_2 \), totaling 4 moles of gas.
- On the product side, we have 2 moles of \( NH_3 \).

### Step 3: Determine the Effect of Adding Helium
When helium gas is added to the system at constant pressure, it does not participate in the reaction. However, it increases the total number of moles of gas in the system.

### Step 4: Apply Le Chatelier's Principle
According to Le Chatelier's Principle, if a system at equilibrium is disturbed, the system will shift in a direction that counteracts the disturbance. In this case, adding an inert gas (helium) at constant pressure will increase the total pressure but will not change the partial pressures of the reacting gases.

Since the reaction produces fewer moles of gas (4 moles of reactants to 2 moles of products), the addition of helium will shift the equilibrium towards the side with more moles of gas, which is the reactants' side.

### Step 5: Conclusion
Thus, the addition of helium gas will reduce the formation of ammonia \( (NH_3) \) because the equilibrium will shift to the left (towards the reactants).

### Final Answer
The addition of helium gas at constant pressure reduces the formation of ammonia.

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