`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)^0`

Explanation

To solve the problem regarding the equilibrium constants \( K_1 \) and \( K_2 \) for the given reactions, we need to analyze the reactions step by step.

### Step 1: Write down the reactions and their equilibrium constants
The two reactions given are:
1. \( N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \) with equilibrium constant \( K_1 \)
2. \( NO(g) \rightleftharpoons \frac{1}{2} N_2(g) + \frac{1}{2} O_2(g) \) with equilibrium constant \( K_2 \)

### Step 2: Relate the equilibrium constants
To relate \( K_1 \) and \( K_2 \), we need to manipulate the second reaction. The second reaction can be rewritten in a way that helps us find a relationship with the first reaction.

The second reaction can be multiplied by 2:
\[ 2NO(g) \rightleftharpoons N_2(g) + O_2(g) \]
When we multiply a reaction by a coefficient, the equilibrium constant is raised to the power of that coefficient. Therefore, the equilibrium constant for this modified reaction is:
\[ K' = K_2^2 \]

### Step 3: Establish the relationship
Now, we can see that the modified second reaction is the reverse of the first reaction. The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward reaction:
\[ K' = \frac{1}{K_1} \]

### Step 4: Combine the equations
From the previous steps, we have:
\[ K_2^2 = \frac{1}{K_1} \]
Taking the square root of both sides gives us:
\[ K_2 = \frac{1}{\sqrt{K_1}} \]

### Final Result
Thus, the relationship between the equilibrium constants \( K_1 \) and \( K_2 \) is:
\[ K_2 = \frac{1}{\sqrt{K_1}} \]