Given the equilibrium constant valuesI N2g-1-2O2g- N2Og- KC-2-7- 10-18II N2O4g- 2NOg- KC-4-6- 10-3III 1-2N2g-O2g- NO2g KC4-1- 10-9Thus for the reaction-2N2Og-3O2g- 2N2O4g-KC is

(I) N_2(g)+1/2O_2(g)⇌ N_2O(g); K_C=2.7× 10^ 18 (II) N_2O_4(g)⇌ 2NO(g); K_C=4.6× 10^ 3 (III) 1/2N_2(g)+O_2(g)⇌ NO_2(g) K_C= 4.1× 10^ 9 Required equilibriu ...

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Standard XII

Chemistry

Effective Equilibrium Constant

Given the equ...

Question

Given the equilibrium constant values
$(I) N_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ N_{2} O (g); K_{C} = 2.7 \times 10^{- 18}$
$(I I) N_{2} O_{4} (g) ⇌ 2 N O (g); K_{C} = 4.6 \times 10^{- 3}$
$(I I I) \frac{1}{2} N_{2} (g) + O_{2} (g) ⇌ N O_{2} (g) K_{C} 4.1 \times 10^{- 9}$
Thus for the reaction,
$2 N_{2} O (g) + 3 O_{2} (g) ⇌ 2 N_{2} O_{4} (g), K_{C}$ is

5.46 \times 10^{7}

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5.46 \times 10^{- 7}

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1.832 \times 10^{- 6}

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1.832 \times 10^{6}

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Solution

The correct option is D $1.832 \times 10^{6}$
$(I) N_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ N_{2} O (g); K_{C} = 2.7 \times 10^{- 18}$
$(I I) N_{2} O_{4} (g) ⇌ 2 N O (g); K_{C} = 4.6 \times 10^{- 3}$
$(I I I) \frac{1}{2} N_{2} (g) + O_{2} (g) ⇌ N O_{2} (g) K_{C} = 4.1 \times 10^{- 9}$

Required equilibrium is obtained if we operate.
$2 N_{2} O (g) + 3 O_{2} (g) ⇌ 2 N_{2} O_{4} (g)$
$E q . (I I I) \times 4 - E q . (I) \times 2 - E q . (I I I) \times 2$

$K_{C} = \frac{[N_{2} O_{4}]^{2}}{[N_{2} O]^{2} [O_{2}]^{3}}$
$K_{C} = \frac{(4.1 \times 10^{- 9})^{4}}{(2.7 \times 10^{- 18})^{2} (4.6 \times 10^{- 3})^{2}}$
$K_{C} == 1.832 \times 10^{6}$

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Effective Equilibrium Constant

Standard XII Chemistry

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Given the equilibrium constant values (I) N2(g)+12O2(g)⇌N2O(g); KC=2.7×10−18 (II) N2O4(g)⇌2NO(g); KC=4.6×10−3 (III) 12N2(g)+O2(g)⇌NO2(g) KC4.1×10−9 Thus for the reaction, 2N2O(g)+3O2(g)⇌2N2O4(g),KC is