`15` mol of `H_(2)` and `5.2` moles of `I_(2)` are mixed and allowed to attain eqilibrium at `500^(@)C` At equilibrium, the concentration of HI is founf to be `10` mol. The equilibrium constant for the formation of HI is.

Options

A. `50`
B. `25`
C. `200`
D. `15`

Explanation

To solve the problem of finding the equilibrium constant \( K_c \) for the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \) given the initial moles of \( H_2 \) and \( I_2 \), and the equilibrium moles of \( HI \), we can follow these steps:

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]

2. **Identify initial moles:**
- Initial moles of \( H_2 = 15 \, \text{mol} \)
- Initial moles of \( I_2 = 5.2 \, \text{mol} \)
- Initial moles of \( HI = 0 \, \text{mol} \)

3. **Set up the change in moles at equilibrium:**
Let \( x \) be the amount of \( I_2 \) that reacts. Since the stoichiometry of the reaction shows that 1 mole of \( I_2 \) reacts with 1 mole of \( H_2 \) to produce 2 moles of \( HI \), we can express the changes as:
- Moles of \( H_2 \) at equilibrium: \( 15 - x \)
- Moles of \( I_2 \) at equilibrium: \( 5.2 - x \)
- Moles of \( HI \) at equilibrium: \( 2x \)

4. **Use the information given at equilibrium:**
We know that at equilibrium, the moles of \( HI = 10 \, \text{mol} \). Therefore:
\[ 2x = 10 \implies x = 5 \]

5. **Calculate the moles of \( H_2 \) and \( I_2 \) at equilibrium:**
- Moles of \( H_2 \) at equilibrium:
\[ 15 - x = 15 - 5 = 10 \, \text{mol} \]
- Moles of \( I_2 \) at equilibrium:
\[ 5.2 - x = 5.2 - 5 = 0.2 \, \text{mol} \]

6. **Write the expression for the equilibrium constant \( K_c \):**
The equilibrium constant for the reaction is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
where \( [HI] \), \( [H_2] \), and \( [I_2] \) are the molar concentrations of the respective gases.

7. **Substitute the equilibrium values into the \( K_c \) expression:**
Assuming the volume of the container is \( V \):
- Concentration of \( HI = \frac{10}{V} \)
- Concentration of \( H_2 = \frac{10}{V} \)
- Concentration of \( I_2 = \frac{0.2}{V} \)

Thus, substituting these into the expression for \( K_c \):
\[ K_c = \frac{\left(\frac{10}{V}\right)^2}{\left(\frac{10}{V}\right)\left(\frac{0.2}{V}\right)} = \frac{100}{2} = 50 \]

### Final Answer:
The equilibrium constant \( K_c \) for the formation of \( HI \) is approximately \( 50 \).