Consider the reactions `1/2N_2+O_2hArrNO_2``K_1`
`2NO_2hArrN_2O_4``K_2`
Using above equations , write down expression for K of the following reaction `N_2O_4hArrN_2+2O_2 K`

Options

A. `K_1K_2`
B. `(K_2^2)/K_1`
C. `1/(K_1K_2^(2))`
D. `1/(K_1^2K_2)`

Answer

To derive the expression for the equilibrium constant \( K \) for the reaction \( N_2O_4 \rightleftharpoons N_2 + 2O_2 \), we will use the two given reactions and manipulate them accordingly.

### Step-by-Step Solution:

1. **Identify the Given Reactions:**
- Reaction 1:
\[ \frac{1}{2} N_2 + O_2 \rightleftharpoons NO_2 \quad (K_1) \]
- Reaction 2:
\[ 2 NO_2 \rightleftharpoons N_2O_4 \quad (K_2) \]

2. **Manipulate Reaction 1:**
- Multiply the entire Reaction 1 by 2:
\[ N_2 + 2O_2 \rightleftharpoons 2NO_2 \]
- The equilibrium constant for this modified reaction becomes:
\[ K' = K_1^2 \]

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

4. **Combine the Reactions:**
- Now, add the modified Reaction 1 and the reversed Reaction 2:
\[ (N_2 + 2O_2 \rightleftharpoons 2NO_2) + (N_2O_4 \rightleftharpoons 2NO_2) \]
- This results in:
\[ N_2O_4 \rightleftharpoons N_2 + 2O_2 \]

5. **Determine the Overall Equilibrium Constant \( K \):**
- Since we added the two reactions, the overall equilibrium constant \( K \) is the product of the individual equilibrium constants:
\[ K = K' \times K'' = K_1^2 \times \frac{1}{K_2} \]

6. **Final Expression for \( K \):**
- Therefore, the expression for the equilibrium constant \( K \) for the reaction \( N_2O_4 \rightleftharpoons N_2 + 2O_2 \) is:
\[ K = \frac{K_1^2}{K_2} \]