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

Explanation

To solve the problem, we need to understand how the equilibrium constant changes when we alter the coefficients of the reactants and products in a balanced chemical equation.

### Step-by-Step Solution:

1. **Identify the Given Reaction and Equilibrium Constant**:
The original 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 reaction can be expressed as:
\[ K = \frac{[NO]^2}{[N_2][O_2]} \]

3. **Consider the New Reaction**:
The new reaction we need to analyze is:
\[ \frac{1}{2} N_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons NO(g) \]

4. **Relate the New Reaction to the Original Reaction**:
Notice that the new reaction is derived from the original reaction by dividing all coefficients by 2.

5. **Apply the Property of Equilibrium Constants**:
When the coefficients of a balanced equation are multiplied by a factor \( n \), the equilibrium constant is raised to the power of \( n \). In this case, since we are dividing by 2, we can express the new equilibrium constant \( K' \) as:
\[ K' = K^{\frac{1}{2}} \]

6. **Final Expression for the New Equilibrium Constant**:
Therefore, the equilibrium constant for the new reaction is:
\[ K' = \sqrt{K} \]

### Conclusion:
The equilibrium constant for the reaction \( \frac{1}{2} N_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons NO(g) \) will be \( \sqrt{K} \).