To determine what happens to the equilibrium of the given reaction upon adding argon gas to the container, we need to analyze the effect of an inert gas on the equilibrium system. Let's walk through the process step-by-step:
### Step-by-Step Solution:
1. Identify the Reaction and Its Characteristics:
The given reaction at equilibrium is:
[tex]\[
N_2(g) + 3 H_2(g) \rightleftarrows 2 NH_3(g)
\][/tex]
This reaction involves nitrogen gas ([tex]\(N_2\)[/tex]), hydrogen gas ([tex]\(H_2\)[/tex]), and ammonia gas ([tex]\(NH_3\)[/tex]).
2. Determine the Nature of the Added Gas:
Argon ([tex]\(Ar\)[/tex]) is an inert gas. This means it does not react with any of the substances in the reaction mixture.
3. Consider the Effect of Adding Argon Gas:
- Since argon is an inert gas, it will not chemically interact with [tex]\(N_2\)[/tex], [tex]\(H_2\)[/tex], or [tex]\(NH_3\)[/tex].
- Adding argon gas increases the total pressure in the container since more gas molecules are present.
- However, because argon is inert and we assume that the volume of the container is kept constant, the partial pressures of [tex]\(N_2\)[/tex], [tex]\(H_2\)[/tex], and [tex]\(NH_3\)[/tex] do not change. The partial pressure of each gas in the mixture is critical in determining how equilibrium is affected.
4. Impact on Equilibrium:
According to Le Chatelier's Principle, equilibrium is disturbed if a change is applied to a system. The system will adjust in a way that counteracts the applied change.
- In our case, adding an inert gas under constant volume conditions does not change the partial pressures of [tex]\(N_2\)[/tex], [tex]\(H_2\)[/tex], and [tex]\(NH_3\)[/tex].
- As a result, the ratios of the concentrations or partial pressures of the reactants and products (which define the equilibrium constant, [tex]\(K\)[/tex]) remain unchanged.
5. Conclusion:
Since the concentrations of [tex]\(N_2\)[/tex], [tex]\(H_2\)[/tex], and [tex]\(NH_3\)[/tex] do not change following the addition of argon gas under constant volume conditions, the system will stay at equilibrium without any shift.
Therefore, the correct answer is:
A. The concentrations of [tex]\(N_2\)[/tex], [tex]\(H_2\)[/tex], and [tex]\(NH_3\)[/tex] would remain the same.
