1. Given reaction and its equilibrium constant

The reaction is:

$N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$

with

$K_c = 4 \times 10^{-4}$

2. Required reaction

We need $K_c$ for:

$NO(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g)$

3. Relate the required reaction to the given reaction

First, reverse the given reaction:

$2NO(g) \rightleftharpoons N_2(g) + O_2(g)$

When a reaction is reversed, the equilibrium constant becomes its reciprocal:

$K_c' = \frac{1}{4 \times 10^{-4}} = 2.5 \times 10^3$

4. Now divide the entire reaction by 2

Dividing

$2NO(g) \rightleftharpoons N_2(g) + O_2(g)$

by 2 gives:

$NO(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g)$

When all stoichiometric coefficients are divided by 2, the new equilibrium constant becomes the square root:

$K_c'' = \left(2.5 \times 10^3\right)^{1/2}$

5. Calculate

$K_c'' = \sqrt{2500} = 50$

6. Check options

• A: $2.5 \times 10^2 = 250$ ❌
• B: $4 \times 10^{-4}$ ❌
• C: $50$ ✅
• D: $0.02$ ❌

Therefore, the correct answer is:

$\boxed{50}$