If the equilibrium constant for
`N_(2) (g) + O_(2)(g) hArr 2NO(g)` is K , the equilibrium
`" constant for " 1/2 N_(2) (g) +1/2 O_(2) (g) hArr NO (g)` will be

Options

A. `1/2 K`
B. `K`
C. `K^(2)`
D. `K^(1//2)`

Explanation

To solve the problem, we need to understand how equilibrium constants change when the stoichiometry of the reaction is altered.

### Step-by-Step Solution:

1. **Identify the Given Reaction and its Equilibrium Constant**:
The first reaction is:
\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \]
The equilibrium constant for this reaction is given as \( K \).

2. **Write the Expression for the Equilibrium Constant**:
The equilibrium constant \( K \) for the above reaction can be expressed as:
\[ K = \frac{[NO]^2}{[N_2][O_2]} \]

3. **Consider the Modified Reaction**:
The modified reaction we need to analyze is:
\[ \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons NO(g) \]
We denote the equilibrium constant for this reaction as \( K' \).

4. **Relate the Modified Reaction to the Original Reaction**:
The modified reaction is essentially half of the original reaction. When the coefficients in a balanced equation are multiplied by a factor, the equilibrium constant is raised to the power of that factor.

5. **Determine the New Equilibrium Constant**:
Since the coefficients in the modified reaction are all halved (i.e., multiplied by \( \frac{1}{2} \)), the new equilibrium constant \( K' \) can be expressed in terms of \( K \):
\[ K' = K^{\frac{1}{2}} = \sqrt{K} \]

### Final Answer:
Thus, the equilibrium constant for the reaction \( \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons NO(g) \) is:
\[ K' = \sqrt{K} \]