The equilibrium constant Kc for the reaction N2g-O2g- 2 NO g at temperature T is 4- 10-4- The value of Kc for the reaction- NOg-12N2g- 12O2g at the same temperature is-

The correct option is D 50.0 For N_2(g)+O_2(g)→ 2NO(g) , K=4× 10^ 4 #160;NO(g) →1/2 N_2(g)+ 1/2 O_2(g) , K'=√(1/K)=√(1/ 4× 10^ 4)=50 ...

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Standard XII

Chemistry

Order of Reaction

The equilibri...

Question

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

0.02

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2.5 \times 10^{2}

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4 \times 10^{- 4}

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50.0

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Solution

The correct option is D $50.0$
For $N_{2} (g) + O_{2} (g) \to 2 N O (g)$ , $K = 4 \times 10^{- 4}$

$N O (g) \to \frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g)$ , $K^{'} = \sqrt{\frac{1}{K}} = \sqrt{\frac{1}{4 \times 10^{- 4}}} = 50$

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

The correct option is D 50.0For N2(g)+O2(g)→2NO(g), K=4×10−4NO(g)→12N2(g)+12O2(g), K′=√1K=√14×10−4=50

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

The correct option is D $50.0$
For $N_{2} (g) + O_{2} (g) \to 2 N O (g)$ , $K = 4 \times 10^{- 4}$

$N O (g) \to \frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g)$ , $K^{'} = \sqrt{\frac{1}{K}} = \sqrt{\frac{1}{4 \times 10^{- 4}}} = 50$