`K_(1)` and `K_(2)` are equilibrium constants for reaction (i) and (ii)
`N_(2)(g)+O_(2)(g) hArr 2NO(g)` …(i)
`NO(g) hArr 1//2 N_(2)(g)+1//2O_(2)(g)` …(ii)
then,

Options

A. `K_(1)=(1//K_(2))^(2)`
B. `K_(1)=K_(2)^(2)`
C. `K_(1)=1//K_(2)`
D. `K_(1)=(K_(2))^(@)`

Explanation

To find the relationship between the equilibrium constants \( K_1 \) and \( K_2 \) for the given reactions, we will analyze each reaction step by step.

### Step 1: Write the Equations and Their Equilibrium Constants
1. The first reaction is:
\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \]
The equilibrium constant for this reaction is denoted as \( K_1 \).

2. The second reaction is:
\[ NO(g) \rightleftharpoons \frac{1}{2} N_2(g) + \frac{1}{2} O_2(g) \]
The equilibrium constant for this reaction is denoted as \( K_2 \).

### Step 2: Relate the Two Reactions
The second reaction can be seen as the reverse of the first reaction, with stoichiometric coefficients halved. To relate \( K_1 \) and \( K_2 \), we need to consider the following:

- When a reaction is reversed, the equilibrium constant for the reverse reaction is the reciprocal of the original equilibrium constant. Therefore, if we reverse the first reaction, we have:
\[ 2NO(g) \rightleftharpoons N_2(g) + O_2(g) \]
The equilibrium constant for this reversed reaction is:
\[ K' = \frac{1}{K_1} \]

### Step 3: Adjust for Stoichiometry
Since the second reaction is half of the reversed first reaction, we need to take the square root of the equilibrium constant:
- When the coefficients of a balanced equation are multiplied by a factor, the equilibrium constant is raised to the power of that factor. In this case, since we are halving the coefficients, we take the square root:
\[ K_2 = \sqrt{K'} \]
Substituting \( K' \):
\[ K_2 = \sqrt{\frac{1}{K_1}} = \frac{1}{\sqrt{K_1}} \]

### Step 4: Final Relationship
Thus, we can express the relationship between \( K_1 \) and \( K_2 \) as:
\[ K_2 = \frac{1}{\sqrt{K_1}} \]

### Step 5: Rewriting the Equation
To express \( K_1 \) in terms of \( K_2 \):
\[ K_1 = \frac{1}{K_2^2} \]

### Conclusion
The final relationship between the equilibrium constants is:
\[ K_1 = \frac{1}{K_2^2} \]