If Ar is added to the equilibrium `N_2(g) +3H_2(g) hArr 2NH_3` at constant volume , the equilibrium will

Options

A. Shift in forward direction
B. Shift in reverse direction
C. Not shift in any direction
D. All are incorrect

Answer

To solve the problem of what happens to the equilibrium when Argon (Ar) is added to the reaction \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) at constant volume, we can follow these steps:

### Step 1: Understand the Reaction
The given equilibrium reaction is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]
This reaction involves nitrogen gas (N2), hydrogen gas (H2), and ammonia (NH3).

### Step 2: Identify the Effect of Adding Argon
When Argon (Ar), an inert gas, is added to the system at constant volume, it does not react with any of the components of the equilibrium. The addition of Argon increases the total pressure in the container because the total number of gas moles increases.

### Step 3: Analyze Partial Pressures
The partial pressure of each gas in a mixture is given by the formula:
\[ P_i = \frac{n_i}{V}RT \]
where \( P_i \) is the partial pressure, \( n_i \) is the number of moles of the gas, \( V \) is the volume, \( R \) is the gas constant, and \( T \) is the temperature.

Since the volume is constant and the number of moles of the reacting gases (N2, H2, NH3) does not change when Argon is added, the partial pressures of N2, H2, and NH3 remain unchanged.

### Step 4: Conclusion on Equilibrium
Because the partial pressures of the reactants and products remain unchanged, the position of equilibrium does not shift. The equilibrium constant \( K_c \) for the reaction is dependent on the concentrations (or partial pressures) of the reactants and products, which have not changed.

### Final Answer
Therefore, the equilibrium will remain unchanged when Argon is added to the system at constant volume.

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