To the system `H_(2)(g)+I_(2)(g) hArr 2HI(g)`
in equilibrium, some `N_(2)` gas was added at constant volume. Then,

Options

A. `K_(p)` will remain constant and `K_(c)` will change
B. `K_(c)` will remain constant and `K_(p)` will change
C. Both `K_(p)` and `K_(c)` will remain constant
D. Both `K_(p)` and `K_(c)` will change

Answer

To solve the problem regarding the equilibrium system \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \) when \( N_2 \) gas is added at constant volume, we need to analyze how the addition of an inert gas affects the equilibrium constants \( K_p \) and \( K_c \).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The given reaction is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]
This is a gaseous equilibrium reaction.

2. **Effect of Adding \( N_2 \)**:
When \( N_2 \) gas is added to the system at constant volume, it does not participate in the reaction. However, it increases the total pressure of the system because the total number of moles of gas in the container increases.

3. **Equilibrium Constants**:
The equilibrium constants \( K_p \) and \( K_c \) are defined as follows:
- \( K_p = \frac{(P_{HI})^2}{(P_{H_2})(P_{I_2})} \)
- \( K_c = \frac{[HI]^2}{[H_2][I_2]} \)
Both \( K_p \) and \( K_c \) depend only on the temperature of the system.

4. **Temperature Consideration**:
Since the problem states that the volume is constant and we are adding an inert gas (\( N_2 \)), the temperature of the system does not change. The addition of \( N_2 \) does not affect the concentrations of \( H_2 \), \( I_2 \), or \( HI \) because it is not involved in the reaction.

5. **Conclusion**:
Since both \( K_p \) and \( K_c \) depend solely on temperature and the temperature remains constant, we can conclude that:
- \( K_p \) will remain constant.
- \( K_c \) will also remain constant.

Thus, the correct answer is:
**Both \( K_p \) and \( K_c \) will remain constant.**

### Final Answer:
Both \( K_p \) and \( K_c \) will remain constant.

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