`2HI(g) rarr H_(2)(g) + I_(2)(g)`
The equilibrium constant of the above reaction is 6.4 at 300 K. If 0.25 mole each of `H_(2) "and" I_(2)` are added to the system, the equilibrium constant will be

Options

A. 6.4
B. 0.8
C. 3.2
D. 1.6

Explanation

To solve the problem, we need to understand the relationship between the equilibrium constant and the conditions of the reaction. The given reaction is:

\[ 2HI(g) \rightleftharpoons H_2(g) + I_2(g) \]

The equilibrium constant \( K_c \) for this reaction at 300 K is given as 6.4.

### Step-by-Step Solution:

1. **Understanding the Equilibrium Constant**:
The equilibrium constant \( K_c \) is defined as:
\[ K_c = \frac{[H_2][I_2]}{[HI]^2} \]
where \([H_2]\), \([I_2]\), and \([HI]\) are the molar concentrations of the respective species at equilibrium.

2. **Effect of Adding Products**:
It is important to note that the equilibrium constant \( K_c \) is only dependent on the temperature for a given reaction. It does not change with the addition of reactants or products, as long as the temperature remains constant.

3. **Adding 0.25 Moles of \( H_2 \) and \( I_2 \)**:
When 0.25 moles of \( H_2 \) and \( I_2 \) are added to the system, it will shift the equilibrium position according to Le Chatelier's principle, but it will not change the value of the equilibrium constant \( K_c \).

4. **Conclusion**:
Since the temperature is not changed and the equilibrium constant depends solely on temperature, the equilibrium constant after adding the moles of \( H_2 \) and \( I_2 \) will still be:
\[ K_c = 6.4 \]

### Final Answer:
The equilibrium constant will remain **6.4**.

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