Consider the reaction,
`NO_(2) rarr 1/2N_(2) + O_(2),K_(1) , N_(2)O_(4) rarr 2NO_(2) , K_(2)` Give the equilibrium constant for the formation of `N_(2)O_(4) "from" N_(2) "and" O_(2)`.

Options

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

Explanation

To find the equilibrium constant for the formation of \( N_2O_4 \) from \( N_2 \) and \( O_2 \), we will manipulate the given reactions and their equilibrium constants.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Equilibrium Constants**:
- The first reaction is:
\[ NO_2 \rightleftharpoons \frac{1}{2} N_2 + O_2 \quad (K_1) \]
- The second reaction is:
\[ N_2O_4 \rightleftharpoons 2 NO_2 \quad (K_2) \]

2. **Write the Desired Reaction**:
- We want to find the equilibrium constant for the reaction:
\[ N_2 + 2 O_2 \rightleftharpoons N_2O_4 \]

3. **Manipulate the First Reaction**:
- Multiply the first reaction by 2:
\[ 2 NO_2 \rightleftharpoons N_2 + 2 O_2 \]
- The equilibrium constant for this modified reaction becomes:
\[ K = K_1^2 \]

4. **Reverse the Modified First Reaction**:
- Now, reverse this reaction:
\[ N_2 + 2 O_2 \rightleftharpoons 2 NO_2 \]
- The equilibrium constant for this reversed reaction is:
\[ K = \frac{1}{K_1^2} \]

5. **Reverse the Second Reaction**:
- Reverse the second reaction:
\[ 2 NO_2 \rightleftharpoons N_2O_4 \]
- The equilibrium constant for this reversed reaction is:
\[ K = \frac{1}{K_2} \]

6. **Combine the Reactions**:
- Now, we can add the two modified reactions:
\[ N_2 + 2 O_2 \rightleftharpoons 2 NO_2 \quad \left(\frac{1}{K_1^2}\right) \]
\[ 2 NO_2 \rightleftharpoons N_2O_4 \quad \left(\frac{1}{K_2}\right) \]
- When we add these reactions, the \( 2 NO_2 \) cancels out, giving us:
\[ N_2 + 2 O_2 \rightleftharpoons N_2O_4 \]

7. **Calculate the Overall Equilibrium Constant**:
- The overall equilibrium constant \( K \) for the formation of \( N_2O_4 \) from \( N_2 \) and \( O_2 \) is the product of the equilibrium constants of the individual reactions:
\[ K = \left(\frac{1}{K_1^2}\right) \times \left(\frac{1}{K_2}\right) = \frac{1}{K_1^2 K_2} \]

### Final Answer:
The equilibrium constant for the formation of \( N_2O_4 \) from \( N_2 \) and \( O_2 \) is:
\[ K = \frac{1}{K_1^2 K_2} \]