The value of the equilibrium constant for the reaction :
H_(2) (g) +I_(2) (g) hArr 2HI (g)`
at 720 K is 48. What is the value of the equilibrium constant for the reaction :
`1//2 H_(2)(g) + 1//2I_(2)(g) hArr HI (g)`

Answer

To find the equilibrium constant for the reaction \( \frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \rightleftharpoons HI(g) \) given that the equilibrium constant \( K_1 \) for the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \) is 48 at 720 K, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Equilibrium Constants:**
- Let \( K_1 \) be the equilibrium constant for the reaction:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]
Given \( K_1 = 48 \).
- Let \( K_2 \) be the equilibrium constant for the reaction:
\[ \frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \rightleftharpoons HI(g) \]

2. **Relate \( K_1 \) and \( K_2 \):**
- The second reaction is essentially half of the first reaction. When the coefficients of a balanced equation are halved, the equilibrium constant for the new reaction is the square root of the original equilibrium constant.
- Therefore, we have:
\[ K_2 = \sqrt{K_1} \]

3. **Calculate \( K_2 \):**
- Substitute the value of \( K_1 \):
\[ K_2 = \sqrt{48} \]

4. **Simplify \( \sqrt{48} \):**
- We can break down \( 48 \) as follows:
\[ 48 = 16 \times 3 \]
- Thus,
\[ K_2 = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3} \]

5. **Final Answer:**
- The value of the equilibrium constant \( K_2 \) for the reaction \( \frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \rightleftharpoons HI(g) \) is:
\[ K_2 = 4\sqrt{3} \]