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. `2.5xx10^(2)`
B. 50
C. `4xx10^(-4)`
D. `0.02`

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

### Step-by-step Solution:

1. **Write the expression for the first reaction**:
The equilibrium constant \( K \) for the reaction \( \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{NO}(g) \) can be expressed as:
\[ K = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \]
Given \( K = 4 \times 10^{-4} \).

2. **Write the 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 \( K' \) can be expressed as:
\[ K' = \frac{[\text{N}_2]^{1/2}[\text{O}_2]^{1/2}}{[\text{NO}]} \]

3. **Relate the two equilibrium constants**:
The second reaction is the reverse of the first reaction and involves halving the stoichiometric coefficients. The relationship between the equilibrium constants for a reaction and its reverse is given by:
\[ K' = \frac{1}{K^{1/2}} \]

4. **Substitute the known value of \( K \)**:
Now substituting the value of \( K \):
\[ K' = \frac{1}{(4 \times 10^{-4})^{1/2}} = \frac{1}{\sqrt{4 \times 10^{-4}}} \]

5. **Calculate \( K' \)**:
\[ K' = \frac{1}{2 \times 10^{-2}} = \frac{1}{0.02} = 50 \]

### 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 \( 50 \).