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, we need to analyze the effects of adding nitrogen gas (N₂) to the equilibrium system of the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \).

### Step-by-Step Solution:

1. **Understanding the Reaction**: The equilibrium reaction is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]
Here, we have 1 mole of \( H_2 \) and 1 mole of \( I_2 \) producing 2 moles of \( HI \).

2. **Identifying the Equilibrium Constants**: The equilibrium constants are defined as:
- \( K_p \) (in terms of partial pressures)
- \( K_c \) (in terms of molar concentrations)

The relationship between \( K_p \) and \( K_c \) is given by:
\[ K_p = K_c (RT)^{\Delta n} \]
where \( \Delta n \) is the change in the number of moles of gas, which is calculated as:
\[ \Delta n = n_{products} - n_{reactants} \]
In this case, \( n_{products} = 2 \) (from \( 2HI \)) and \( n_{reactants} = 1 + 1 = 2 \) (from \( H_2 + I_2 \)), thus:
\[ \Delta n = 2 - 2 = 0 \]

3. **Effect of Adding N₂**: When nitrogen gas is added at constant volume, it does not participate in the reaction. Therefore, the concentrations of \( H_2 \), \( I_2 \), and \( HI \) remain unchanged because the volume is constant and only the total pressure increases due to the addition of \( N_2 \).

4. **Impact on K_p and K_c**: Since both \( K_p \) and \( K_c \) depend only on temperature and the concentrations of the reactants and products do not change, we conclude that:
- \( K_p \) remains constant.
- \( K_c \) remains constant.

5. **Final Conclusion**: Therefore, both \( K_p \) and \( K_c \) will remain constant after the addition of \( N_2 \).

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