For a reaction, `H_2+I_2hArr 2HI` at 721 K , the value of equilibrium constant is 50. If 0.5 moles each of `H_2` and `I_2` is added to the system the value of equilibrium constant will be :

Options

A. `0.02`
B. `0.2`
C. 50
D. 25

Answer

To solve the problem, we need to understand the concept of the equilibrium constant (K) for the given reaction:

**Reaction:**
\[ H_2 + I_2 \rightleftharpoons 2HI \]

**Given:**
- The equilibrium constant (K) at 721 K is 50.
- 0.5 moles each of \( H_2 \) and \( I_2 \) is added to the system.

**Step-by-step Solution:**

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

2. **Effect of Adding Reactants:**
When we add 0.5 moles of \( H_2 \) and \( I_2 \) to the system, we are changing the concentrations of the reactants. However, the equilibrium constant itself is dependent on the temperature and not on the concentrations of the reactants or products.

3. **Equilibrium Constant is Temperature Dependent:**
The value of the equilibrium constant (K) is constant at a specific temperature. Therefore, even though we are changing the concentrations by adding more \( H_2 \) and \( I_2 \), the equilibrium constant at 721 K remains unchanged.

4. **Conclusion:**
Since the equilibrium constant is constant at a given temperature, the value of the equilibrium constant after adding the moles of \( H_2 \) and \( I_2 \) will still be:
\[ K = 50 \]

Thus, the final answer is that the value of the equilibrium constant will remain **50**.

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