For a reaction `H_(2(g))+I_(2(g))rArr2HI_((g))at 721^(@)C` the value of equilibrium constant is 50 . If 0.5 moles each of `H_(2)` & `I(2)` is added to the system the value of equilibrium constant will be .

Options

A. 100
B. 25
C. 50
D. 200

Answer

To solve the problem, we need to understand the concept of the equilibrium constant (K) and how it behaves under different conditions.

### Step-by-Step Solution:

1. **Identify the Reaction and Given Information**:
- The reaction is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]
- The equilibrium constant (K) at 721°C is given as 50.

2. **Understanding the Equilibrium Constant**:
- The equilibrium constant (K) for the reaction is defined as:
\[ K = \frac{[HI]^2}{[H_2][I_2]} \]
- This expression shows that K is dependent on the concentrations of the reactants and products at equilibrium.

3. **Effect of Adding Reactants**:
- The problem states that 0.5 moles of \(H_2\) and \(I_2\) are added to the system.
- However, it is important to note that the equilibrium constant is only affected by temperature and not by the concentrations of the reactants or products.

4. **Conclusion**:
- Since the temperature remains constant at 721°C, the value of the equilibrium constant (K) will also remain constant at 50, regardless of the changes in concentration due to the addition of 0.5 moles of \(H_2\) and \(I_2\).

5. **Final Answer**:
- Therefore, the value of the equilibrium constant after adding the moles will still be:
\[ K = 50 \]