The equilibrium constant for the reaction $$N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$$ at temperature $$T$$ is $$4 \times 10^{-4}$$. The value of $$K_c$$ for the reaction $$NO(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g)$$ at the same temperature is

When a reaction is reversed, its equilibrium constant becomes the reciprocal of the original constant ($$K' = \frac{1}{K_1}$$):

$$2NO(g) \rightleftharpoons N_2(g) + O_2(g) \quad \implies \quad K' = \frac{1}{4 \times 10^{-4}} = \frac{10^4}{4} = 2500$$

When a reaction is multiplied by a factor $$n$$, the new equilibrium constant is raised to the power of $$n$$:

$$NO(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) \quad \implies \quad K_c = (K')^{1/2} = \sqrt{2500}$$    
$$K_c = 50$$