The equilibrium constant $(\mathrm{K}_{\mathrm{C}})$ for the reaction $\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})$ at temperature T is $4 \times 10^{-4}$. The value of $\mathrm{K}_{\mathrm{C}}$ for the reaction, $\mathrm{NO}(\mathrm{g}) \rightarrow \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})$ at the same temperature is :

JEE Main · 2012 · 29 Apr 2012 · Offline · Equilibrium

The equilibrium constant

(K_{C})

for the reaction

N_{2} (g) + O_{2} (g) \to 2 NO (g)

at temperature T is

4 \times 1 0^{- 4}

. The value of

K_{C}

for the reaction,

NO (g) \to \frac{1}{2} N_{2} (g) + 1/2 O_{2} (g)

at the same temperature is :

Answer: (D)

The given reaction

N_{2} + O_{2} \to 2NO

has

K_{C} = 4 \times 1 0^{- 4}

. Reversing it inverts the constant to

\frac{1}{4 \times 1 0 ^{- 4}} = 2500

, and writing it for half a mole of

NO

— halving all the coefficients — takes the square root. So the constant for

NO \to \frac{1}{2} N_{2} + \frac{1}{2} O_{2}

2500 = 50

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option