In the two gaseous reactions (i) and (ii) at `250^@C`
(i) `2NO(g)+(1)/(2)O_2(g)hArrNO_2(g)K_1`
(ii) `2NO_2(g) hArr2NO(g)+O_2(g),K_2` the equilibrium constants `K_1` and `K_2` are releated as

Options

A. `K_2=(1)/(K_1)`
B. `K_2=K_1^(1//2)`
C. `K_2=(K_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 will analyze each reaction step by step.

### Step 1: Write the expressions for \( K_1 \) and \( K_2 \)

For the first reaction:
\[ 2NO(g) + \frac{1}{2}O_2(g) \rightleftharpoons NO_2(g) \]
The equilibrium constant \( K_1 \) is given by:
\[ K_1 = \frac{[NO_2]}{[NO]^2 \cdot [O_2]^{1/2}} \]

For the second reaction:
\[ 2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g) \]
The equilibrium constant \( K_2 \) is given by:
\[ K_2 = \frac{[NO]^2 \cdot [O_2]}{[NO_2]^2} \]

### Step 2: Relate \( K_1 \) and \( K_2 \)

To find the relationship between \( K_1 \) and \( K_2 \), we can manipulate the expression for \( K_2 \).

From the expression for \( K_1 \), we can rearrange it:
\[ [NO]^2 \cdot [O_2]^{1/2} = \frac{[NO_2]}{K_1} \]

Now, we can substitute this into the expression for \( K_2 \):
\[ K_2 = \frac{[NO]^2 \cdot [O_2]}{[NO_2]^2} \]

### Step 3: Substitute \( [NO]^2 \cdot [O_2] \)

We can express \( [O_2] \) in terms of \( K_1 \):
\[ [O_2] = \frac{[NO_2]}{K_1 \cdot [NO]^2} \]

Substituting this back into the \( K_2 \) expression gives:
\[ K_2 = \frac{[NO]^2 \cdot \left(\frac{[NO_2]}{K_1 \cdot [NO]^2}\right)}{[NO_2]^2} \]

### Step 4: Simplify the expression

This simplifies to:
\[ K_2 = \frac{[NO_2]}{K_1 \cdot [NO_2]} = \frac{1}{K_1} \]

### Conclusion

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