At constant pressure, the addition of argon to `N_(2(g)) +3H_(2(g)) to 2NH_(3(g))`

Options

A. Reduces the formation of ammonia from nitrogen and hydrogen
B. Increases the formation of ammonia from nitrogen and hydrogen
C. Does not effect the equilibrium of the reaction in which ammonia is formed from nitrogen and dasan
D. Reuces the dissociation of ammonia

Answer

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

### Step-by-Step Solution:

1. **Identify the Reaction and Equilibrium**:
The equilibrium reaction is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]
At equilibrium, the concentrations of the reactants and products remain constant.

2. **Understand the Effect of Adding Argon**:
Argon is an inert gas. When added to a system at constant pressure, it does not react with the components of the equilibrium system. However, it increases the total volume of the system.

3. **Analyze the Change in Volume**:
When the volume of the system increases (due to the addition of argon), the partial pressures of the gases involved in the reaction decrease. According to Le Chatelier's principle, the system will respond to counteract this change.

4. **Determine the Direction of the Shift**:
In the given reaction:
- Reactants: \( N_2 + 3H_2 \) (Total = 4 moles)
- Products: \( 2NH_3 \) (Total = 2 moles)

Since there are more moles of gas on the reactant side (4 moles) than on the product side (2 moles), the equilibrium will shift towards the side with more moles of gas to increase the pressure.

5. **Conclusion**:
Therefore, the addition of argon at constant pressure will shift the equilibrium to the left (towards the reactants), which means the formation of ammonia (\( NH_3 \)) from nitrogen and hydrogen will be reduced.

### Final Answer:
The addition of argon at constant pressure **reduces the formation of ammonia from nitrogen and hydrogen**.

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