The equilibrium constant for `N_(2)(g)+O_(2)(g)hArr2NO "is" K_(1)`
and that for `NO(g)hArr(1)/(2)N_(2)(g)+(1)/(2)O_(2)(g)` is `K_(2). K_(1) "and" K_(2)`
will be related as

Options

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

Explanation

To find the relationship between the equilibrium constants \( K_1 \) and \( K_2 \) for the given reactions, we can follow these steps:

### Step 1: Write the reactions and their equilibrium constants

1. The first reaction is:
\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \quad (K_1) \]

2. The second reaction is:
\[ 2NO(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) \quad (K_2) \]

### Step 2: Reverse the second reaction

When we reverse the first reaction, we get:
\[ 2NO(g) \rightleftharpoons N_2(g) + O_2(g) \]
The equilibrium constant for the reversed reaction, denoted as \( K' \), is given by:
\[ K' = \frac{1}{K_1} \]

### Step 3: Relate \( K_2 \) to \( K' \)

The second reaction is half of the reversed first reaction. When we take half of a reaction, the equilibrium constant is related to the original constant by taking the square root:
\[ K_2 = \sqrt{K'} \]

### Step 4: Substitute \( K' \) in the equation

Substituting \( K' \) from Step 2 into the equation from Step 3:
\[ K_2 = \sqrt{\frac{1}{K_1}} = \frac{1}{\sqrt{K_1}} \]

### Step 5: Rearranging the equation

To express \( K_1 \) in terms of \( K_2 \), we can square both sides:
\[ K_2^2 = \frac{1}{K_1} \]
Rearranging gives:
\[ K_1 = \frac{1}{K_2^2} \]

### Conclusion

Thus, the relationship between \( K_1 \) and \( K_2 \) is:
\[ K_1 = \frac{1}{K_2^2} \]