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 follow these steps:

### Step 1: Write the Given Reactions
We have two reactions with their equilibrium constants:
1. \( NO_2 \rightleftharpoons \frac{1}{2} N_2 + O_2 \) with equilibrium constant \( K_1 \)
2. \( N_2O_4 \rightleftharpoons 2 NO_2 \) with equilibrium constant \( K_2 \)

### Step 2: Reverse the First Reaction
To find the equilibrium constant for the formation of \( N_2O_4 \) from \( N_2 \) and \( O_2 \), we need to reverse the first reaction:
\[ \frac{1}{2} N_2 + O_2 \rightleftharpoons NO_2 \]
When we reverse a reaction, the equilibrium constant becomes the reciprocal:
\[ K' = \frac{1}{K_1} \]

### Step 3: Multiply the Reversed Reaction
Next, we need to multiply the reversed reaction by 2 to match the stoichiometry for \( N_2O_4 \):
\[ N_2 + 2 O_2 \rightleftharpoons 2 NO_2 \]
The equilibrium constant for this reaction becomes:
\[ K'' = (K')^2 = \left(\frac{1}{K_1}\right)^2 = \frac{1}{K_1^2} \]

### Step 4: Write the Second Reaction
Now, we will consider the second reaction:
\[ 2 NO_2 \rightleftharpoons N_2O_4 \]
The equilibrium constant for this reaction is \( K_2 \).

### Step 5: Combine the Reactions
Now, we can add the two reactions:
1. \( N_2 + 2 O_2 \rightleftharpoons 2 NO_2 \) with \( K = \frac{1}{K_1^2} \)
2. \( 2 NO_2 \rightleftharpoons N_2O_4 \) with \( K = K_2 \)

When we add these reactions, the \( 2 NO_2 \) cancels out, giving us:
\[ N_2 + 2 O_2 \rightleftharpoons N_2O_4 \]

### Step 6: Calculate the Overall Equilibrium Constant
The overall equilibrium constant for the formation of \( N_2O_4 \) from \( N_2 \) and \( O_2 \) is the product of the individual equilibrium constants:
\[ K_c = K'' \cdot K_2 = \left(\frac{1}{K_1^2}\right) \cdot K_2 = \frac{K_2}{K_1^2} \]

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

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