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

For the given reaction

$$N_2(g)+O_2(g)\rightarrow 2\,NO(g)$$

the equilibrium constant is

$$K_{c1}=4\times 10^{-4}$$

Step 1 : Write the required reaction in terms of the given one.

The desired reaction is

$$NO(g)\rightarrow \tfrac12\,N_2(g)+\tfrac12\,O_2(g)$$

Notice two operations connect the two reactions:

1. Halving the stoichiometric coefficients.
2. Reversing the direction.

Step 2 : Effect of halving a reaction.

If every coefficient of a reaction is divided by 2, the new equilibrium constant is the square root of the original:

$$K_{c\,(\text{halved})}=K_{c1}^{1/2}$$

Thus

$$K_{c\,(\text{halved})}=(4\times10^{-4})^{1/2}=2\times10^{-2}=0.02$$

Step 3 : Effect of reversing a reaction.

Reversing a reaction inverts its equilibrium constant:

$$K_{c\,(\text{reversed})}=\frac1{K_{c\,(\text{halved})}}$$

Therefore

$$K_c=\frac1{0.02}=50.0$$

Step 4 : State the result.

The equilibrium constant for $$NO(g)\rightarrow \tfrac12\,N_2(g)+\tfrac12\,O_2(g)$$ at the same temperature is $$50.0$$.

Option D which is: $$50.0$$