The equilibrium constant for the reaction
`N_(2)(g)+O_(2)(g) hArr 2NO(g)`
at temperature T is `4xx10^(-4)`.
The value of `K_(c)` for the reaction
`NO(g) hArr 1/2 N_(2)(g)+1/2 O_(2)(g)`
at the same temperature is

Options

A. 0.02
B. `2.5 xx 10^(2)`
C. `4 xx 10^(-4)`
D. 50.0

Answer

To solve the problem, we need to find the equilibrium constant \( K_c \) for the reaction:

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

given that the equilibrium constant for the reaction

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

is \( K_c = 4 \times 10^{-4} \).

### Step 1: Write the equilibrium constant expression for the first reaction.

For the reaction:

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

The equilibrium constant \( K_{c1} \) is given by:

\[ K_{c1} = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \]

Given that \( K_{c1} = 4 \times 10^{-4} \).

### Step 2: Write the equilibrium constant expression for the second reaction.

For the reaction:

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

The equilibrium constant \( K_{c2} \) is given by:

\[ K_{c2} = \frac{[\frac{1}{2} \text{N}_2][\frac{1}{2} \text{O}_2]}{[\text{NO}]} \]

### Step 3: Relate \( K_{c2} \) to \( K_{c1} \).

To find \( K_{c2} \), we first note that the second reaction is the reverse of the first reaction divided by 2.

1. **Reverse the first reaction**: When we reverse the reaction, the equilibrium constant becomes the reciprocal:

\[ K_{c1}^{-1} = \frac{[\text{N}_2][\text{O}_2]}{[\text{NO}]^2} \]

2. **Divide by 2**: When we divide the entire reaction by 2, the new equilibrium constant is the old equilibrium constant raised to the power of \( \frac{1}{2} \):

\[ K_{c2} = (K_{c1}^{-1})^{\frac{1}{2}} = \left(\frac{[\text{N}_2][\text{O}_2]}{[\text{NO}]^2}\right)^{\frac{1}{2}} = \frac{[\text{N}_2]^{\frac{1}{2}}[\text{O}_2]^{\frac{1}{2}}}{[\text{NO}]} \]

### Step 4: Calculate \( K_{c2} \).

Now we can calculate \( K_{c2} \):

\[ K_{c2} = (K_{c1})^{-1/2} = \left(4 \times 10^{-4}\right)^{-1/2} \]

Calculating this gives:

\[ K_{c2} = \frac{1}{\sqrt{4 \times 10^{-4}}} = \frac{1}{2 \times 10^{-2}} = \frac{1}{0.02} = 50 \]

### Final Answer:

Thus, the value of \( K_c \) for the reaction \( \text{NO}(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \) is:

\[ \boxed{50} \]