The equilibrium constant for the reaction N2( g)+O2( g)⇌2NO(g) at temperature T is 4×10-4. The value of Kc for the reaction: NO(g)⇌12 N2( g)+12O2( g) at the same temperature is

The equilibrium constant for the reaction $N_{2} (g) + O_{2} (g) ⇌ 2 NO (g)$ at temperature T is $4 \times 10^{- 4}$ . The value of $K_{c}$ for the reaction: $NO (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g)$ at the same temperature is

A
$2.5 \times 10^{2}$
B
50
C
$4 \times 10^{- 4}$
D
0.02

Solution:

We have

$N_{2} (g) + O_{2} (g) ⇌ 2 NO (g) K_{C} = 4 \times 10^{- 4}$

For the reaction $NO (g) ⇌ (1 / 2) N_{2} (g) + (1 / 2) O_{2} (g)$ . $K_{c}^{'} = \frac{1}{\sqrt{K_{c}}} = \frac{1}{\sqrt{4 \times 10^{- 4}}} = 50$

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