`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, let's break it down step by step.

### Step 1: Understand the Reaction and Equilibrium Constant
The given reaction is:
\[ 2 \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \]

The equilibrium constant \( K_c \) for this reaction at 300 K is given as 6.4. The equilibrium constant expression for this reaction is:
\[ K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \]

### Step 2: Analyze the Addition of Reactants
We are adding 0.25 moles of \( \text{H}_2 \) and \( \text{I}_2 \) to the system. This means that the concentrations of both \( \text{H}_2 \) and \( \text{I}_2 \) will increase.

### Step 3: Apply Le Chatelier's Principle
According to Le Chatelier's Principle, if a system at equilibrium is disturbed, the system will shift in a direction that counteracts the disturbance. In this case, since we are increasing the concentrations of \( \text{H}_2 \) and \( \text{I}_2 \), the equilibrium will shift to the left (towards the reactants) to reduce the concentrations of these products.

### Step 4: Determine the Effect on the Equilibrium Constant
It is important to note that while the equilibrium position may shift due to the addition of products, the value of the equilibrium constant \( K_c \) itself remains unchanged at a given temperature. Therefore, even after the addition of \( \text{H}_2 \) and \( \text{I}_2 \), the equilibrium constant will still be:
\[ K_c = 6.4 \]

### Conclusion
The equilibrium constant after adding 0.25 moles of \( \text{H}_2 \) and \( \text{I}_2 \) will remain:
\[ \text{Equilibrium Constant} = 6.4 \]

### Final Answer
The equilibrium constant will be **6.4**.

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