What is the value of `K_p` for the following reaction
`N_2(g)+2O_2(g) hArr N_2O_4(g)`
`{:("Reaction",K_p),(1/2N_2O_4(g)hArrNO_2(g),x),(1/2N_2(g)+O_2(g)hArrNO_2(g),y):}`

Options

A. `y/x`
B. `y^2/x^2`
C. `x^2/y^2`
D. `xy^2`

Answer

To find the value of \( K_p \) for the reaction

\[ N_2(g) + 2O_2(g) \rightleftharpoons N_2O_4(g) \]

we will relate it to the two given reactions with their respective \( K_p \) values, \( x \) and \( y \).

### Step 1: Identify the given reactions and their \( K_p \) values
- **Reaction 2**: \( \frac{1}{2} N_2O_4(g) \rightleftharpoons NO_2(g) \) with \( K_p = x \)
- **Reaction 3**: \( \frac{1}{2} N_2(g) + O_2(g) \rightleftharpoons NO_2(g) \) with \( K_p = y \)

### Step 2: Manipulate the reactions to derive the desired reaction
To derive the reaction \( N_2(g) + 2O_2(g) \rightleftharpoons N_2O_4(g) \), we can manipulate the given reactions as follows:

1. **Multiply Reaction 3 by 2**:
\[ N_2(g) + 2O_2(g) \rightleftharpoons 2NO_2(g) \quad (K_p = y^2) \]

2. **Reverse Reaction 2**:
\[ NO_2(g) \rightleftharpoons \frac{1}{2} N_2O_4(g) \quad (K_p = \frac{1}{x}) \]

### Step 3: Combine the manipulated reactions
Now, we will add the modified Reaction 3 and the reversed Reaction 2:

- From the modified Reaction 3, we have:
\[ N_2(g) + 2O_2(g) \rightleftharpoons 2NO_2(g) \]

- From the reversed Reaction 2, we have:
\[ 2NO_2(g) \rightleftharpoons N_2O_4(g) \]

Adding these two reactions gives:
\[ N_2(g) + 2O_2(g) \rightleftharpoons N_2O_4(g) \]

### Step 4: Write the expression for \( K_p \)
According to the rules of combining equilibrium constants:
\[ K_p = K_p(\text{Reaction 3}) \times K_p(\text{Reversed Reaction 2}) \]
Thus,
\[ K_p = y^2 \times \frac{1}{x} = \frac{y^2}{x} \]

### Final Answer
The value of \( K_p \) for the reaction \( N_2(g) + 2O_2(g) \rightleftharpoons N_2O_4(g) \) is:
\[ K_p = \frac{y^2}{x} \]

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