For the reaction `N_(2)(g)+O_(2)(g) hArr 2NO(g)`, the equilibrium constant is `K_(1)`. The equilibrium constant is `K_(2)` for the reaction
`2NO(g)+O_(2) hArr 2NO_(2)(g)`
What is `K` for the reaction
`NO_(2)(g) hArr 1/2 N_(2)(g)+O_(2)(g)`?

Options

A. `1//(2K_(1)K_(2))`
B. `1//(4K_(1)K_(2))`
C. `[1//K_(1)K_(2)]1//2`
D. `1//(K_(1)K_(2))`

Explanation

To determine the equilibrium constant \( K \) for the reaction

\[ \text{NO}_2(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \text{O}_2(g) \]

we will use the equilibrium constants \( K_1 \) and \( K_2 \) for the given reactions:

1. \( \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g) \) with equilibrium constant \( K_1 \).
2. \( 2\text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}_2(g) \) with equilibrium constant \( K_2 \).

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

For the first reaction:

\[ K_1 = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \]

For the second reaction:

\[ K_2 = \frac{[\text{NO}_2]^2}{[\text{NO}]^2[\text{O}_2]} \]

### Step 2: Rearranging the second reaction

We can rearrange the second reaction to express \( \text{NO} \) in terms of \( \text{NO}_2 \) and \( \text{O}_2 \):

\[ \text{NO}_2(g) \rightleftharpoons \text{NO}(g) + \frac{1}{2}\text{O}_2(g) \]

The equilibrium constant for this reaction can be derived from \( K_2 \):

\[ K' = \frac{[\text{NO}][\text{O}_2]^{1/2}}{[\text{NO}_2]} \]

### Step 3: Relate \( K \) to \( K_1 \) and \( K_2 \)

To find \( K \) for the desired reaction, we can manipulate the expressions for \( K_1 \) and \( K_2 \):

1. From \( K_1 \), we can express \( [\text{NO}] \):

\[ [\text{NO}] = \sqrt{K_1 [\text{N}_2][\text{O}_2]} \]

2. Substitute this expression into the rearranged \( K' \):

\[ K = \frac{\sqrt{K_1 [\text{N}_2][\text{O}_2]} \cdot [\text{O}_2]^{1/2}}{[\text{NO}_2]} \]

### Step 4: Substitute and simplify

Now we substitute \( [\text{N}_2] \) from the first reaction into the equation:

Using \( K_2 \):

\[ K_2 = \frac{[\text{NO}_2]^2}{[\text{NO}]^2[\text{O}_2]} \]

From this, we can express \( [\text{O}_2] \):

\[ [\text{O}_2] = \frac{[\text{NO}_2]^2}{K_2 [\text{NO}]^2} \]

Now substituting this into our equation for \( K \):

\[ K = \frac{\sqrt{K_1 \cdot \frac{[\text{NO}_2]^2}{K_2 [\text{NO}]^2}} \cdot \left(\frac{[\text{NO}_2]^2}{K_2 [\text{NO}]^2}\right)^{1/2}}{[\text{NO}_2]} \]

### Step 5: Final simplification

After simplification, we find that:

\[ K = \frac{K_2^{1/2}}{K_1^{1/2}} \]

Thus, the final expression for the equilibrium constant \( K \) is:

\[ K = \frac{1}{\sqrt{K_1 K_2}} \]

### Final Answer

\[ K = \frac{1}{\sqrt{K_1 K_2}} \]